This section covers:

 Introduction to Sequences and Series
 Summary of Formulas for Sequences and Series
 Sequences and Series Terms
 Explicit Formulas Versus Recursive Formulas
 Arithmetic Sequences
 Geometric Sequences
 Writing Formulas
 Arithmetic Series
 Summation Notation
 Geometric Series
 Applications of Sequences and Series
 More Practice
Introduction to Sequences and Series
Sequences and Series are basically just numbers or expressions in a row that make up some sort of a pattern; for example, January, February, March, …, December is a sequence that represents the months of a year. Each of these numbers or expressions are called terms or elements of the sequence.
Sequences are the list of these items, separated by commas, and series are the sum of the terms of a sequence (if that sum makes sense; it wouldn’t make sense for months of the year).
You may have heard the term inductive reasoning, which is reasoning based on patterns, say from a sequence (as opposed to deductive reasoning, which is reasoning from rules or definitions). For example, let’s try to find the next few terms in the following sequences:
Sequence  Next Two Terms 
1, 8, 27, __ , __ 
Every term is cubed. Next two terms are \({{4}^{3}},\,\,{{5}^{3}}\), or 64, 125. 
3, 7, 11, 15, __ , __ 
Every term is 4 more than the previous one. Just add 4 to get to the next term. Next two terms are \(15+4,\,\,15+4+4\), or 19, 23. 
1000, 100, 10, 1, ___ , ___ 
Every term is the previous term divided by 10, or .1 times the previous one. Just multiply by .1 to get to the next term. Next two terms are \(1\times .1,\,\,1\times .1\times .1\), or .1, .01. 
\(\displaystyle \frac{{{{x}^{3}}}}{{x+1}},\,\,\frac{{{{x}^{5}}}}{{{{{\left( {x+1} \right)}}^{2}}}},\,\frac{{{{x}^{7}}}}{{{{{\left( {x+1} \right)}}^{3}}}},\,\,\_\_\_\,,\,\_\_\_\)  Sometimes we need to look at both the top and bottom of a fraction, and different things are going on. On the top (numerator), the next term is the previous multiplied by \({{x}^{2}}\), and the bottom (denominator), the next term is the previous term multiplied by \(\left( {x+1} \right)\). The next two terms are \(\displaystyle \frac{{{{x}^{9}}}}{{{{{\left( {x+1} \right)}}^{4}}}},\,\,\frac{{{{x}^{{11}}}}}{{{{{\left( {x+1} \right)}}^{5}}}}\). 
\(\displaystyle \frac{1}{2},\,\frac{2}{4},\,\frac{3}{8},\,\,\_\_\,,\,\,\_\_\) 
On the top (numerator), the next term is 1 more than the previous one, and the bottom (denominator), the next term is the previous term multiplied by 2. The next two terms are \(\displaystyle \frac{4}{{16}},\,\,\frac{5}{{32}}\). 
We will do more like this in the Writing Formulas section below.
Summary of Formulas for Sequences and Series
Before we get started, here is a summary of the main formulas for Sequences and Series:
\(\begin{array}{l}{{a}_{1}}=\,\,\text{first term}\\{{a}_{n}}=\,\,n\text{th term}\end{array}\) 
Arithmetic \(d=\) common difference 
Geometric \(r=\) common ratio 
Sequences 
\({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\)
(Recursive: \({{a}_{n}}=a,\,\,\,\,\,{{a}_{n}}={{a}_{{n1}}}+d\))

\({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\)
(Recursive: \({{a}_{n}}=a,\,\,\,\,\,{{a}_{n}}=r{{a}_{{n1}}}\) ) 
Finite Series (Sum) 
\(\displaystyle {{S}_{n}}=\frac{n}{2}\left( {{{a}_{1}}+{{a}_{n}}} \right)\,\,\,=\,\,\,\frac{n}{2}\left( {2{{a}_{1}}+\left( {n1} \right)d} \right)\)

\(\displaystyle {{S}_{n}}=\frac{{{{a}_{1}}\left( {1{{r}^{n}}} \right)}}{{1r}}\) 
Infinite Series (Sum) 
Not Applicable 
\(\displaystyle {{S}_{\infty }}=\frac{{{{a}_{1}}}}{{1r}},\,\,\,\,\,\,\,\left r \right<1\)

Sequences and Series Terms
OK, so I have to admit that this is sort of a play on words since each element in a sequence is called a term, and we’ll talk about the terms (meaning words) that are used with sequences and series, and the notation.
Let’s first compare sequences to relations or functions from the Algebraic Functions section. Think of the \(x\) part of the relation (the independent variable) as the numbers in a row that represent each part of the sequence; these typically start at 1 (one). So the \(x\) part will typically be 1, 2, 3, and so on (natural numbers). This is usually represented by the variable \(n\).
Note that the \(n\) part will always be a positive integer.
Then think of the \(y\) part of the sequence as the actual term, or what the element is for that particular \(n\) value; this is usually represented by \({{a}_{n}}\) (but sometimes you see this as \(f\left( n \right)\) instead) .
For the sequence 3, 7, 11, 15, and so one, we have:
1^{st} term 
2^{nd} term 
3^{rd} term 
4^{th} term 
\(n\)^{th} term 

\(n\, =\)  1  2  3  4 
… \(n\) (like the “\(x\)”) 
\({{a}_{n}}\) or \(f\left( n \right)\,=\) 
3  7  11  15 
… \({{a}_{n}}\) (like the “\(y\)”) 
Think of the “points” or “coordinates” of a sequence as \(\left( {n,\,\,{{a}_{n}}} \right)\).
Note that sequences are said to be convergent if as the domain (the \(n\)’s) approach infinity, the range (the \({{a}_{n}}\)’s) approach some number, called a limit. (We saw this when studying the end behavior of a function in the Parent Function and Transformations, and Graphing Rational Functions, including Asymptotes sections). A sequence is divergent if it is not convergent. We will talk about Limits of Sequences here in the Limits and Continuity section.
Explicit Formulas versus Recursive Formulas
Explicit formulas are formulas that are computed for each term; in other words, you can look at the formula for a term and know exactly how to get that term (you don’t rely on the previous term). We’ll talk about the two main types of explicit formulas, Arithmetic and Geometric, later.
Recursive formulas are formulas that use the previous term to get to the next one, and we always have to indicate what the first term is (\({{a}_{{1}}}\)), so we can get started. Notice that to reference the previous term, we use \({{a}_{{n1}}}\) (sometimes \(f\left( {n1} \right)\)), which makes sense, since we need the term before this one.
Since the use of recursive formulas are very limited, we typically don’t deal with them very much.
Here are examples; for most of these, we’ll come up with some tools to make finding the formulas much easier. And don’t worry if you don’t see how to get these right away; these take practice!
Sequence 
Explicit Formula 
Recursive Formula 
\(\displaystyle \begin{array}{l}\,2,6,10,14,\,\,…\\\,\,\vee \,\,\,\vee \,\,\,\vee \,\\+4\,\,+4\,\,+4\,\end{array}\) 
\(\displaystyle {{a}_{n}}=4n2\). Notice that the difference between the terms is 4, and, it turns out that the explicit formula then will have a “\(4n\)” in it.
To get back to the first term, we have to subtract by 2, since \(4n\) or \(4(1)=4\), but the first term is actually 2.
This is actually an arithmetic formula, and we’ll see an easier way to find this formula later. 
\({{a}_{1}}=2;\,\,\,\,{{a}_{n}}={{a}_{{n1}}}+4,\,\,\text{for}\,\,\,n>1\). We first have to indicate that the first term (\({{a}_{1}}\)) is 2.
Then we use this term figure out how to get to the next term; we add 4 to each previous term, for \(n=2,3,…\). 
\(\displaystyle \begin{array}{l}\,\,1,\,\,2,\,\,4,\,\,8,\,\,…\\\,\,\,\vee \,\,\,\,\,\vee \,\,\,\,\,\vee \,\,\\\,\times 2\,\,\times 2\,\,\times 2\,\end{array}\)  \({{a}_{n}}={{2}^{{n1}}}\). Notice that every term is twice the previous one, so we are multiplying by 2 each time to get to the next term. Do you see how if we raise 2 to a power, we can get to the next term?
But since \(n\) starts with 1, and \({{2}^{1}}=2\), we have to start with \({{2}^{0}}\); it turns out that for each term, we raise 2 to the power of the \(n\) of the last term, or \({{2}^{{n1}}}\). Tricky!
This is actually a geometric sequence, and we’ll see an easier way to find the formula later. 
\({{a}_{1}}=1;\,\,\,\,{{a}_{n}}=2{{a}_{{n1}}},\,\,\text{for}\,\,\,n>1\). We first have to indicate that the first term (\({{a}_{1}}\)) is 1.
Then we use this term figure out how to get to the next term; we multiply by 2. In this case, using the recursive formula is easier. 
\(\displaystyle 2,\,\,9,\,\,28,\,\,65,\,\,\,…\)  \({{a}_{n}}={{n}^{3}}+1\). This is a tough one; to get to the next term, we can’t simply add or multiply the previous term. But we do notice something; each term is one more than the cube of 1, 2, 3, and so on (the \(n\) part).  I really haven’t seen these types of sequences represented recursively, but I imagine we could do something like this:
\(\begin{array}{l}{{a}_{1}}=2;\,\,\,\,{{a}_{n}}={{\left( {\sqrt[3]{{{{a}_{{n1}}}1}}+1} \right)}^{3}}+1,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{for}\,\,\,n>1\end{array}\) 
\(\displaystyle \begin{array}{l}\,\,x,\,\,\,{{x}^{2}},{{x}^{3}},\,\,{{x}^{4}},\,…\\\,\,\,\,\,\,\,\,\,\vee \,\,\,\,\,\,\,\,\vee \,\,\,\,\,\,\,\,\vee \\\times \left( {x} \right)\times \left( {x} \right)\times \left( {x} \right)\end{array}\)  \({{a}_{n}}={{\left( x \right)}^{n}}{{\left( {1} \right)}^{n}}\,\,\,\text{or}\,\,\,\,{{a}_{n}}={{\left( {x} \right)}^{n}}\). To get to the next term, we have to both multiply by \(x\), and also change the sign.
We commonly do this by multiply by \(\left( {1} \right)\) raised to the power of \(n\) or \(n1\), since this will make the sign go back and forth. In this case, we can also multiply by \(x\) to get to the next term. This is another geometric sequence. 
\({{a}_{1}}=x;\,\,\,\,{{a}_{n}}=\left( {x} \right){{a}_{{n1}}},\,\,\,\,\,\,\text{for}\,n>1\). We first have to indicate that the first term is \(x,\) and then just multiply by \(x\) to get to the next term. 
Another Recursive Example
Here’s an example of a more complicated recursive formula; let’s get the first 5 terms: \(\left\{ \begin{array}{l}{{a}_{1}}=3\\{{a}_{2}}=2\\{{a}_{n}}=3{{a}_{{n1}}}4{{a}_{{n2}}};\,\,n>2\end{array} \right\}\).
We have the first two terms, and we need to get the third by using those first two terms, given the \({{a}_{n}}\) rule. Note that \({{a}_{n1}}\) means one term back and \({{a}_{n2}}\) means two terms back from the present term.
Therefore, the sequence is: \(\begin{array}{l}\,\,\,3,\,\,\,2,\,\,3\left( 2 \right)4\left( {3} \right),\,\,…\\=3,\,\,2,\,\,18,\,\,3\left( {18} \right)4\left( 2 \right),\,…\\=3,\,\,2,\,\,18,\,\,46,\,\,3\left( {46} \right)4\left( {18} \right),…\\=3,\,\,2,\,\,18,\,\,46,\,\,66,…\end{array}\). Pretty complicated!
Arithmetic Sequences
Arithmetic sequences are those where the difference between the terms is always the same. They can be defined recursively as \({{a}_{1}}=\,\,a;\,\,\,\,{{a}_{n}}={{a}_{{n1}}}+\,\,d,\,\,\text{for}\,\,\,n>1\). Here, “\(a\)” is the first term, and “\(d\)” is the common difference.
Arithmetic sequences are somewhat like linear equations (with the difference between terms like a slope), except sequences in general are discreet (just points) instead of continuous (a whole line).
The way I like to tell if a sequence is arithmetic is to see if “second term – (minus) first term” equals “third term – second term” equals “fourth term – third term”, and so on. Then once we have this number that is always the same, we have the common difference, or “\(d\)”. Note that “\(d\)” can be negative, a fraction, or a decimal. (“\(n\)” has to always be a positive integer).
Formula for an Arithmetic Sequence
We saw above that the if difference between the terms is \(d\) for an arithmetic sequence, the explicit formula then will have a “\(dn\)” in it (such as “\(4n\)” if \(d=4\)). We won’t prove it here, but it turns out the explicit formula for an arithmetic sequence is the formula below, for each term \({{a}_{n}}\). This makes sense since we always start out with the first term, and then we are adding the common difference \(d\) “\(n1\)” more times!
This is called the formula for a general term for an arithmetic sequence, and you’ll want to memorize this formula.
Note that all arithmetic sequences diverge; they never get closer and closer to any one number.
Now let’s do some problems:
Arithmetic Sequence Problem 
Solution 
Find the general formula for the \(n\)th term, and then find the 18^{th} term (\({{a}_{{18}}}\)) of the sequence:
4, 7, 10, 13, … 
We can see that the second term – the first = the third term – the second = 3, so this is the common difference.
We also see that the first term is 4, so now we can plug these numbers into the general formula of the equation: \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\), or \({{a}_{n}}=4+\left( {n1} \right)3=4+3n3=3n+1\). The general formula is \({{a}_{n}}=3n+1\).
To get the 18^{th} term, we’ll plug in 18 for \(n\), so we have \({{a}_{{18}}}=3\left( {18} \right)+1=55\). 
Find the general formula for the \(n\)th term, and then find the 12^{th} term (\({{a}_{{12}}}\)) of the sequence:
8, –1, –10, –19, … 
We can see that the second term – the first = the third term – the second = –1 – 8 = –9 (watch the negative signs!), so this is the common difference.
We also see that the first term is 8, so now we can plug these numbers into the general formula of the equation: \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\), or \({{a}_{n}}=8+\left( {n1} \right)\left( {9} \right)=89n+9=9n+17\). The general formula is \({{a}_{n}}=9n+17\).
To get the 12^{th} term, we’ll plug in 12 for \(n\), so we have \({{a}_{{12}}}=9\left( {12} \right)+17=91\). 
Find the general formula for the \(n\)th term, and then find the 8^{th} term (\({{a}_{{8}}}\)) of the sequence:
\(r,\,\,rs,\,\,r2s,\,…\)

We can see that the second term – the first = the third term – the second = \(\left( {rs} \right)r=0s=s\) (watch the negative signs!), so this is the common difference.
We also see that the first term is \(r\), so now we can plug these numbers into the general formula of the equation: \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\), or \({{a}_{n}}=r+\left( {n1} \right)\left( {s} \right)\) (we can just leave it like this). To get the 8^{th} term, we’ll plug in 8 for \(n\), so we have \({{a}_{8}}=r7s\). 
It turns out that if we are given the value of two specific terms of a sequence (and what terms they are – the “\(n\)”), we can derive the equation of that sequence. What we can do is subtract the \(n\)’s and subtract the terms (\({{a}_{n}}\))’s and then divide these two numbers to get \(d\), the common difference. This makes sense since we are multiplying by \(d\) to get the successive terms.
Then to get the first term (\({{a}_{1}}\)), we can use the formula \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\), using one of the \({{a}_{1}}\) and \(n\) combinations, and the \(d\) we just got. Pretty cool!
Again, think of the common difference \(d\) as a slope of this “linear equation”; we are finding this slope using our familiar slope formula for two points of the form \(\left( {n,\,\,{{a}_{n}}} \right)\) as change of \(y\)’s over change of \(x\)’s: \(\displaystyle d=\,\frac{{{{a}_{n}}\,\text{(second)}{{a}_{n}}\text{(first)}}}{{n\text{(second)}n\text{(first)}}}\)
Let’s try some problems:
Arithmetic Sequence Problem 
Solution 
Find the general formula for the \(n\)th term:
5^{th} term is 14 11^{th} term is 32 
Let’s use \(\displaystyle d=\,\frac{{{{a}_{n}}\,\text{(second)}{{a}_{n}}\text{(first)}}}{{n\text{(second)}n\text{(first)}}}\). Subtract the term numbers (\(n\)’s) to get \(115=6\) to see how many terms we “jump up”. Then subtract the actual value of the terms (\({{a}_{n}}\))’s to get \(3214=18\). Divide this second number by the first to get 3. Now we have the common difference \(d\) (sort of like a slope)!
We still need to get the first term to get the general formula of the sequence, so we can plug \(d\) and one of the \(\left( {n,{{a}_{n}}} \right)\) points to get \({{a}_{1}}\); let’s use \(\left( {5,14} \right)\): \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d;\,\,\,\,\,\,\,14={{a}_{1}}+\left( {51} \right)3;\,\,\,\,\,\,\,\,{{a}_{1}}=1412=2\) The general formula then is \({{a}_{n}}=2+\left( {n1} \right)3\) or \({{a}_{n}}=3n1\), and the sequence looks like this: 2, 5, 8, 11, 14, …. Try it – it works! 
Find the general formula for the \(n\)th term:
6^{th} term is –15 9^{th} term is –27 
Let’s use \(\displaystyle d=\,\frac{{{{a}_{n}}\,\text{(second)}{{a}_{n}}\text{(first)}}}{{n\text{(second)}n\text{(first)}}}\). Subtract the term numbers (\(n\)’s) to get \(96=3\) to see how many terms we “jump up”. Then subtract the actual value of the terms (\({{a}_{n}}\))’s to get \(27(15)=12\) (watch negative signs!). Divide this second number by the first to get –4. Now we have the common difference \(d\).
Plug in point \(\left( {6,15} \right)\) to get \({{a}_{1}}\): \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d;\,\,\,\,\,\,15={{a}_{1}}+\left( {61} \right)\left( {4} \right);\,\,\,\,\,\,\,\,{{a}_{1}}=15\left( {20} \right)=5\) The general formula then is \({{a}_{n}}=5+\left( {n1} \right)\left( {4} \right)\) or \({{a}_{n}}=4n+9\), and the sequence looks like this: 5, 1, –3, –7, –11, …. Try it – it works! 
The 5^{th} term of an arithmetic sequence is –2 and the sum of the first 10 terms of the sequence is –35. Find the sequence.

Let’s write down what we know, using the formulas we know. Since the 5^{th} term of the arithmetic sequence is –2, we know \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\), or \(2={{a}_{1}}+\left( {51} \right)d\), so \(2={{a}_{1}}+4d\).
We have two unknown variables, so we’ll need another equation to solve. Since the sum of the first 10 terms is –35, we have \(\displaystyle {{S}_{n}}=\frac{n}{2}\left( {{{a}_{1}}+{{a}_{n}}} \right)\), or \(\displaystyle 35=\frac{{10}}{2}\left( {{{a}_{1}}+{{a}_{{10}}}} \right);\,\,\,\,\,35=\frac{{10}}{2}\left[ {{{a}_{1}}+{{a}_{1}}+\left( {101} \right)d} \right];\,\,\,\,35=10{{a}_{1}}+45d\) Now we have two variables and two unknowns, so we have a solvable system. We can solve with substitution: \(\begin{array}{c}2={{a}_{1}}+4d;\,\,\,\,\,\,\,\,{{a}_{1}}=24d\\35=10{{a}_{1}}+45d;\,\,\,\,\,\,\,35=10\left( {24d} \right)+45d;\,\,\,\,\,\,15=5d;\,\,\,\,\,\,d=3\\{{a}_{1}}=24\left( {3} \right);\,\,\,\,\,\,\,{{a}_{1}}=\,\,10\end{array}\) This sequence then is \({{a}_{n}}=10+\left( {n1} \right)\left( {3} \right)=3n+13\). Try it – it works! 
Find \(x\) such that the following are consecutive terms of an arithmetic sequence. Then find the common difference for this sequence:
\(2x,\,\,3x+4,\,\,5x2,\,\,…\) 
We know that for an arithmetic sequence, we must have “second term – (minus) first term” equals “third term – second term”, and so on. (This is the common difference). So we have:
\(\begin{align}\left( {3x+4} \right)\left( {2x} \right)&=\left( {5x2} \right)\left( {3x+4} \right)\\x+4&=2x6\\x&=10\end{align}\) Since \(x=10\), the sequence looks like 20, 34, 48, …. The common difference is “second term – (minus) first term” \(=14\).

Geometric Sequences
Whereas arithmetic sequences are those where the difference between the terms is the same, geometric sequences are those where the quotient of the terms is always the same. They can be defined recursively as \({{a}_{1}}=\,\,a;\,\,\,\,{{a}_{n}}=r{{a}_{{n1}}},\,\,\text{for}\,\,\,n>1\). Here, “\(a\)” is the first term, and “\(r\)” is the common ratio. Note that “\(r\)” can be negative, a fraction, or a decimal. (“\(n\)” has to always be a positive integer).
The way I like to tell if a sequence is geometric is to see if “second term /(divided by) first term” equals “third term/second term” equals “fourth term/third term”, and so on. Then once we have this number that is always the same, we have the common ratio!
Geometric sequences are somewhat like exponential functions (with the common ratio like an exponential base), except sequences in general are discreet (just points) instead of continuous (a whole line).
Formula for a Geometric Sequence
It turns out the explicit formula for a geometric sequence is the formula below, for each term. This makes sense since we always start out with the first term, and then we are multiplying the common ratio \(r\) “\(d\)” more times!
This is called the formula for a general term for a geometric sequence, and you’ll want to memorize this formula.
What this says is the \(n\)th term of an arithmetic sequence is the first time times \(r\) (common ratio) raised to the (\(d\)).
IMPORTANT NOTE:
The above explicit formulas assume that the sequence begins at \(n=1\). Some textbooks show sequences starting with \(n=0\), and thus the Geometric Explicit formula would be \({{a}_{n}}=a{{\left( r \right)}^{n}}\), where \(a\) is the term where \(n=0\). This can be confusing, but the formula above will work if you assume the first term is in the first place (not place 0).
Now let’s do some problems:
Geometric Sequence Problem  Solution 
Find the general formula for the \(n\)th term, and then find the 8^{th} term (\({{a}_{8}}\)) of the sequence:
\(\displaystyle \frac{1}{8},\,\,\frac{1}{2},2\,\,,\,\,8,\,\,…\) 
We can see that the second term/the first = the third term/the second \(=4\) (watch fractions), so this is the common ratio.
We also see that the first term is \(\displaystyle \frac{1}{8}\), so now we can plug these numbers into the general formula of the equation \({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\) to get \(\displaystyle {{a}_{n}}=\frac{1}{8}{{\left( 4 \right)}^{{n1}}}\).
To get the 8^{th} term, plug in 8 for n, so we have \(\displaystyle {{a}_{8}}=\frac{1}{8}{{\left( 4 \right)}^{{81}}}=2048\). 
Find the general formula for the \(n\)th term, and then find the 10^{th} term (\({{a}_{{10}}}\)) of the sequence:
8, –16, 32, –64, … 
We can see that the second term/the first = the third term/the second \(=2\) (watch negative sign), so this is the common ratio.
We also see that the first term is 8, so now we can plug these numbers into the general formula of the equation \({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\) to get \({{a}_{n}}=8{{\left( {2} \right)}^{{n1}}}\).
To get the 10^{th} term, plug in 10 for n, so we have \({{a}_{{10}}}=8{{\left( {2} \right)}^{{101}}}=4096\). See how fast (the absolute value of) these numbers get large with geometric sequences? 
Find the general formula for the \(n\)th term, and then find the 6^{th} term (\({{a}_{6}}\)) of the sequence:
.02, .002, .0002, … 
We can see that the second term/the first = the third term/the second \(=.1\) (watch decimals), so this is the common ratio.
We also can see that the first term is .02, so now we can plug these numbers into the general formula of the equation \({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\) to get \({{a}_{n}}=02{{\left( {.1} \right)}^{{n1}}}\).
To get the 6^{th} term, plug in 6 for n: \({{a}_{6}}=.02{{\left( {.1} \right)}^{{61}}}=2{{\left( {10} \right)}^{{7}}}=.0000002\). 
Find the general formula for the \(n\)th term, and then find the 5^{th} term (\({{a}_{5}}\)) of the sequence:
\(\displaystyle r,\,\,\frac{r}{s},\,\,\frac{r}{{{{s}^{2}}}},\,\,…\,\,\) 
We can see that the second term/the first \(=\) the third term/the second \(\displaystyle =\frac{1}{s}\) (watch fractions), so this is the common ratio.
We also see that the first term is r, so now we can plug these numbers into the general formula of the equation \({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\) to get \(\displaystyle {{a}_{n}}=r{{\left( {\frac{1}{s}} \right)}^{{n1}}}\). To get the 5^{th} term, plug in 5 for n, so we have \(\displaystyle {{a}_{5}}=r{{\left( {\frac{1}{s}} \right)}^{{51}}}=r\left( {\frac{1}{{{{s}^{4}}}}} \right)=\frac{r}{{{{s}^{4}}}}\). 
Just like for arithmetic sequences, it turns out that if we are given the value of two specific terms of a sequence (and what terms they are – the “\(n\)”), we can derive the equation of that sequence. What we can do is subtract the \(n\)’s (say get \(x\)) and divide the terms (\({{a}_{n}}\) )’s (say get \(y\)) and then take the \(x\)’th root of \(y\) to get \(r\), the common ratio (and we have to watch even roots, as shown in an example below). This makes sense since we are raising \(r\) to successive powers to get the successive terms.
Then to get the first term (\({{a}_{1}}\) ), we can use the formula \({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\), using one of the \({{a}_{1}}\) and \(n\) combinations, and the \(r\) we just got. Pretty cool!
Let’s try some problems:
Geometric Sequence Problem  Solution 
Find the general formula for the nth term:
4^{th} term is 27 9^{th} term is 6561 
We can subtract the term numbers (n’s) to get \(94=5\), and then divide the actual value of the terms \(({{a}_{n}})\)’s to get \(\displaystyle \frac{{6561}}{{27}}=243\).
Then we can take the 5^{th} root of 243 to get 3 (we can use the calculator and put in 243^(1/5)); this is the common ratio r. We still need to get the first term to get the general formula of the sequence, so we can plug r and one of the \(\left( {n,\,\,{{a}_{n}}} \right)\) points to get \({{a}_{1}}\) back; let’s use \(\left( {4,\,\,27} \right)\): \({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\), or \(27={{a}_{1}}{{\left( 3 \right)}^{{41}}};\,\,\,{{a}_{1}}=27\div 27=1\). The general formula is \({{a}_{n}}=\,\,1{{\left( 3 \right)}^{{n1}}}\) or \({{a}_{n}}=\,\,{{\left( 3 \right)}^{{n1}}}\), and the sequence looks like this: 1, 3, 9, 27, 81, …. Try it – it works! 
Find the general formula for the nth term:
3^{rd} term is \(\displaystyle \frac{1}{2}\) 9^{th} term is \(\displaystyle \frac{1}{128}\) 
We can subtract the term numbers (n’s) to get \(93=6\) and then divide the actual value of the terms \(({{a}_{n}})\)’s to get \(\displaystyle \frac{{\frac{1}{{128}}}}{{\frac{1}{2}}}=\frac{1}{{64}}\) (watch fractions; we can put (–1/128)/(–1/2) in calculator and then MathFracEnter).
Then take the 6^{th} root of \(\displaystyle \frac{1}{{64}}\) to get \(\displaystyle \frac{1}{2}\) (we can use the calculator and put in (1/64)^(1/6)). Now we have the common ratio r. The problem is that if we raised \(\displaystyle \frac{1}{2}\) to the 6^{th} root, we’d also get \(\displaystyle \frac{1}{{64}}\), and since some terms are negative, we will need to have a negative common ratio r, so our r is \(\displaystyle \frac{1}{2}\). Tricky! Now we still need to get the first term to get the general formula of the sequence, so we can plug r and one of the \(\left( {n,\,\,{{a}_{n}}} \right)\) points to get \({{a}_{1}}\) back; let’s use \(\displaystyle \left( {3,\,\,\frac{1}{2}} \right)\): \({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\), or \(\displaystyle \frac{1}{2}={{a}_{1}}{{\left( {\frac{1}{2}} \right)}^{{\left( {31} \right)}}};\,\,\,{{a}_{1}}=\frac{1}{2}\times 4=2\). The general formula is \(\displaystyle {{a}_{n}}=\,\,\left( {2} \right){{\left( {\frac{1}{2}} \right)}^{{n1}}}\), and the sequence looks like this: \(\displaystyle 2,\,\,1,\,\,\frac{1}{2},\,\,\,\frac{1}{4},\,…\).Try it – it works! 
Here’s one more type of problem you may see:
Geometric Sequence Problem  Solution 
Find x such that the following are consecutive terms of a geometric sequence. Then find the common ratio for this sequence:
\(x,\,\,x+4,\,\,x+6,\,\,…\) 
We know that for a geometric sequence, we must have “second term/first term” equals “third term/second term”, and so on. (This is the common ratio, or r).
We have the following, and notice that we had to cross multiply. Also notice that if we had gotten a value of 0 or –4 for x, we’d have to “throw these away” since we can’t divide by 0: \(\displaystyle \frac{{x+4}}{x}=\frac{{x+6}}{{x+4}}\) \(\require{cancel} \begin{array}{c}\left( {x+4} \right)\left( {x+4} \right)=x\left( {x+6} \right)\\\cancel{{{{x}^{2}}}}+8x+16=\cancel{{{{x}^{2}}}}+6x\\2x=16\\x=8\end{array}\)
\(x=8\), and the sequence looks like –8, –4, –2, …. The common ratio is “second term/ first term” = \(\displaystyle \frac{1}{2}\) . 
Writing Formulas
Now let’s try to distinguish between arithmetic and geometric sequences, and also write formulas for those sequences that are neither arithmetic nor geometric.
Remember again to look for second – first = third – second, and so on, for arithmetic sequences, and second / first = third / second, and so on for arithmetic sequences.
If neither of these work, look for squares or cubes in a row, added to or subtracted by a certain number.
Here’s another trick, if it appears the sequence is a quadratic. Let’s use this example: 3, 6, 10, 15, 21, …
Take the differences of the terms ( \({{a}_{n}}\)) in the sequence and take the differences again (see A to the left). If this second difference is constant (we get “1”), this is a quadratic.
Now, this is the tricky part (and, honestly, I’m not sure why this works): Divide the second differences (“1”) by 2 (always by 2) to get the coefficient of the \({{n}^{2}}\); we have \(\displaystyle \frac{1}{2}=.5\); so far, we have \(.5{{n}^{2}}\).
Now subtract each term (\({{a}_{n}}\)) in the sequence by \(.5{{n}^{2}}\) (for \(n=1,2,\,\text{and}\,3\)), take the differences of these numbers, and you will magically see the same number when you do this (see B): we get “1.5”. This is the coefficient of the \(n\); so far, we have \(.5{{n}^{2}}+1.5n\).
To get the last number (constant) of the quadratic, we have to plug in what we have so far, so see “how far off we are” (to know what to add or subtract); for example, for term \(n=1\), we need to have a 3 for \({{a}_{1}}\). For this term, we’ll have \(.5{{n}^{2}}+1.5n\), or \(.5{{\left( 1 \right)}^{2}}+1.5\left( 1 \right)=2\). But we need to get 3, so we’ll add 1 (\(2+1=3\)).
The sequence is a quadratic: is \({{a}_{n}}=.5{{n}^{2}}+1.5n+1\). Tricky! 
Of course, it might easier to just do a regression on your calculator with the sequence: use the \(n\) for \(x\) values and \({{a}_{n}}\) for \(y\) values. We did regressions like this here in the Scatterplots, Correlation, and Regression section.
Here are some problems; we’ll be working again with explicit formulas (as opposed to recursive):
Writing Formula Problem  Solution 
Find the general formula for the \(n\)th term of the sequence:
\(\displaystyle 1,\frac{1}{4},\,\,\frac{1}{9},\,\,\frac{1}{{16}}\,\,…\)

We can see that the sequence isn’t arithmetic or geometric by looking at the differences and quotients of the terms.
We do notice that numerators are all “1” and the denominators all contain a perfect square: 1, 4, 9, and so on. We can write the sequence like this: \(\displaystyle {{a}_{n}}=\frac{1}{{{{n}^{2}}}}\) 
Find the general formula for the \(n\)th term of the sequence:
\(x,\,\,x2,\,\,x4,\,\,…\)

We can see that “the second term – the first term” = “the third term – the second term = –2 (watch negative sign), so this is the common difference.
We also see that the first term is \(x\), so now we can plug these numbers into the general formula of an arithmetic sequence: \({{a}_{n}}=x+\left( {n1} \right)\left( {2} \right)=x2n+2\) (it will still have an \(x\) in it). We have: \(\displaystyle {{a}_{n}}=x2n+2\) 
Find the general formula for the \(n\)th term of the sequence:
\(\displaystyle \frac{2}{1},\frac{3}{2},\,\,\frac{4}{6},\,\,\frac{5}{{24}},\,\,\frac{6}{{120}}\,…\) 
We can see that the sequence isn’t arithmetic or geometric by looking at the differences and quotients of the terms.
We do notice that the numerators start at 2 and increase by 1 each time (\(n+1\), since \(n\) starts at 1), and the denominators consist of the factorial of \(n\) (tricky!). We can then write the sequence like this: \(\displaystyle {{a}_{n}}=\frac{{n+1}}{{n!}}\) 
Find the general formula for the \(n\)th term of the sequence:
\(10,\,\,22,\,\,44,\,\,76,\,\,…\) 
We can see that the sequence isn’t arithmetic or geometric by looking at the differences and quotients of the terms.
As in the method above for quadratics, if we take the difference of the terms we get 12, 22, 32. If we take the differences again, we get 10. Then we take half of this to get the coefficient of \({{n}^{2}}\), so we have \(5{{n}^{2}}\) so far (somehow, this works!). To get the coefficient of the \(n\), we can subtract \(5{{n}^{2}}\) from 10, 22, and 44, where \(n\) is 1, 2, and 3, respectively. We get 5 (\(105{{\left( 1 \right)}^{2}}\)), 2 (\(225{{\left( 2 \right)}^{2}}\)), and –1 (\(1445{{\left( 3 \right)}^{2}}\)). When we subtract these terms, we get a difference of –3, So now we have \(5{{n}^{2}}3n\). To get the last number (constant) of the quadratic, we have to plug in what we have so far, so see “how far off we are”; for example, for term \(n=1\), we need to have a 10 for \({{a}_{1}}\). But if we plug in 1 for \(n\), we get \(5{{n}^{2}}3n\), or \(5{{\left( 1 \right)}^{2}}3\left( 1 \right)=2\). We need to get 10, so we’ll add 8 (\(2+8=10\)). The sequence then is the quadratic: \({{a}_{n}}=5{{n}^{2}}3n+8\).

Arithmetic Series
When we add up some or all of the terms of a sequence, we have a series. So when we add up the terms of an Arithmetic Sequence, we have an Arithmetic Series.
For example, the formula for the arithmetic sequence with \({{a}_{1}}=1\) and \(d=2\) is \({{a}_{n}}=1+\left( {n1} \right)2=2n1\). For \(n=1\,\,\,\text{to }5\), this sequence is 1, 3, 5, 7, 9. We would write the arithmetic series as the sum of these elements, or 1 + 3 + 5 + 7 + 9. The partial sum (since we are only adding up 5 terms) of this series is 1 + 3 + 5 + 7 + 9 = 25.
Another way to write this is to use summation notation (also called the summation formula) \(\sum\limits_{{k=1}}^{5}{{\,1+\left( {k1} \right)2}}=\sum\limits_{{k=1}}^{5}{{\,2k1}}\) (we usually use the variable “\(k\)”). The sigma sign, or \(\sum{{}}\) symbol means take the equation after it, start with the lower bound, or what’s at the bottom of the sigma sign (1 in this case), plug in for \(n\), and add up everything until you get to the upper bound, or what’s at the top of the sigma sign (5 in this case).
Here’s an example of how we might use an arithmetic series. Let’s say we have a theater that we are building and we know every row has 2 more seats than the row before it and there are 20 seats in the first row. We also know that we want 100 rows of the theater, and we are about to order the total number of seats in the theater. Do you see how we would then need to get the sum of an arithmetic series of 100 terms (with common difference 2) to find out how many seats to order? This series would be \(\sum\limits_{{k=1}}^{{100}}{{\,20+\left( {k1} \right)}}2\).
Formula for an Arithmetic Series
We actually have a formula to get this sum; this is called a partial sum formula. Note that we can never get the total or infinite sum of an arithmetic series, since this sum would just go on forever (extremely large or extremely small). We’ll talk how this will be a limit later.
Here is the partial sum formula for an arithmetic sequence, and you’ll want to memorize this formula:
\(\displaystyle {{S}_{n}}\,\,=\,\,\frac{n}{2}\left( {{{a}_{1}}+{{a}_{n}}} \right),\,\,\,\,\text{where }{{a}_{n}}\,\,=\,\,{{a}_{1}}+\left( {n1} \right)d\) or \(\displaystyle {{S}_{n}}\,\,=\,\,\frac{n}{2}\left[ {2{{a}_{1}}+\left( {n1} \right)d} \right]\)
Notice how the we’re using the formula for the \(n\)^{th} term of an arithmetic sequence (\({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\)) to get the second formula. (Personally, I prefer not to use the second formula, but just memorize the first, since we already have memorized the \({{a}_{n}}\) formula.)
It’s pretty straightforward to use this formula to solve partial sum arithmetic series problems:
Arithmetic Series Problem 
Solution 
Find an equation both for the \(n\)th term of the arithmetic sequence and the \(n\)th partial sum \({{S}_{n}}\) with:
\({{a}_{1}}=2,\,\,\,\,d=4\)
Then find \({{S}_{{40}}}\). 
We know the formula for the \(n\)th term for an arithmetic sequence is \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\), so for this sequence we have \({{a}_{n}}=2+\left( {n1} \right)4;\,\,{{a}_{n}}=4n2\). The sequence looks like 2, 6, 10, …, and the series looks like 2 + 6 + 10 + …., which is also \(\displaystyle \sum\limits_{{k=1}}^{n}{{2+\left( {k1} \right)4}}=\sum\limits_{{k=1}}^{n}{{4k2}}\).
The equation then for the series’ \(n\)th partial sum is \(\displaystyle {{S}_{n}}\,\,=\,\,\frac{n}{2}\left[ {2+\left( {4n2} \right)} \right]=\frac{n}{2}\left( {4n} \right)=2{{n}^{2}}\). To get the sum of the first 40 terms, we have \({{S}_{{40}}}\,=\,2{{\left( {40} \right)}^{2}}=3200\). 
Find an equation for the \(n\)th partial sum \({{S}_{n}}\):
\(100+75+50+…\) 
We can see that this is an arithmetic series, since the second term – the first = the third term – the second = –25; this is \(d\), or the common difference.
The first term is 100, so \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d=100+\left( {n1} \right)\left( {25} \right)=25n+125\). The equation for this series’ partial sum is \(\displaystyle {{S}_{n}}\,\,=\,\,\frac{n}{2}\left[ {100+\left( {25n+125} \right)} \right]=\frac{n}{2}\left( {25n+225} \right)\). (We can also write this \(\displaystyle {{S}_{n}}\,\,=\,\frac{{25{{n}^{2}}+225n}}{2}\).) Try this for \(n=2\) (\({{S}_{n}}=175\)); it works! 
Find an equation for the \(n\)th partial sum \({{S}_{n}}\), and find \({{S}_{{20}}}\):
\(\left( {5+t} \right)+\,\,\left( {5+3t} \right)+\,\,\left( {5+5t} \right)+…\) 
We can see that this is an arithmetic sequence with \(d=2t\). The first term is “\(5+t\)”, so we have \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d=\left( {5+t} \right)+\left( {n1} \right)\left( {2t} \right)=5+2ntt\).
The equation then for this series’ partial sum is \(\displaystyle {{S}_{n}}\,\,=\,\,\frac{n}{2}\left[ {\left( {5+t} \right)+\left( {5+2ntt} \right)} \right]=\frac{n}{2}\left( {2nt+10} \right)\). So \(\displaystyle {{S}_{{20}}}\,\,=\,\frac{{20}}{2}\left( {2\left( {20} \right)t+10} \right)=400t+100\). 
Evaluate (find the sum):
\(33\frac{1}{2}4\,…\) to 50 terms 
We can see that this is an arithmetic series, since the second term – the first = the third term – the second = –3.5 –(–3) = –.5; this is the \(d\), or the common difference.
The first term is –3, so \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d=3+\left( {n1} \right)\left( {.5} \right)=.5n2.5\). The equation then for this series’ partial sum is \(\displaystyle {{S}_{n}}\,\,=\,\,\frac{n}{2}\left[ {3+\left( {.5n2.5} \right)} \right]=\frac{n}{2}\left( {.5n5.5} \right)\). So \(\displaystyle {{S}_{{50}}}\,\,=\,\frac{{50}}{2}\left( {.5\left( {50} \right)5.5} \right)=762.5\) . 
Evaluate: \(\sum\limits_{{k=1}}^{n}{{\left( {8k+2} \right)}}\)
Find the sum: \(\sum\limits_{{k=1}}^{{40}}{{\left( {8k+2} \right)}}\) 
This series looks like 10, 18, 26, …. The first term is 10 and \(d=8\), so \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d=10+\left( {n1} \right)\left( 8 \right)=8n+2\). (Actually, we knew that from looking at the summation!)
The equation then for this series’ partial sum is \(\displaystyle {{S}_{n}}\,\,=\,\,\frac{n}{2}\left[ {10+\left( {8n+2} \right)} \right]=\frac{n}{2}\left( {8n+12} \right)=4{{n}^{2}}+6n\). So \(\displaystyle {{S}_{{40}}}\,\,=\,\frac{{40}}{2}\left( {8\left( {40} \right)+12} \right)=6640\). 
Find the sum:
\(\sum\limits_{{n=6}}^{{54}}{{\left( {5n3} \right)}}\) 
This one’s a little tricky since we aren’t starting with 1 for the lower limit: the lower limit is 6 and the upper limit is 54.
I always like to start out with writing out the first several terms of the sequence/series. Plug in 6, 7, and 8 for \(n\), knowing that we have to go to \(n=54\): \(\begin{array}{l}n\,\,\,=\,\,\,6,\,\,\,\,7,\,\,\,\,8,\,\,…\,\,\,54\\{{a}_{n}}\,=27,\,\,32,\,\,37,\,\,…\,\,\,?\end{array}\) We see that \(d=5\), and we still need to figure out how many terms we have; you might think it’s \(546\). It’s actually \(546+1=49\), since we typically start with 1, and not 0. (For example, if we have a series going from 1 to 8, we actually have \(81+1=8\) terms). In our case, since \(n=49\), the \(n\)th term is \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d=27+\left( {491} \right)5=267\), and the partial sum is \(\displaystyle {{S}_{n}}=\frac{n}{2}\left( {{{a}_{1}}+{{a}_{n}}} \right)=\frac{{49}}{2}\left( {27+267} \right)=7203\). Tricky!

Summation Notation
Before moving on to Geometric Series, we should get a little more practice with Summation Notation, since you’ll probably be seeing this in more advanced math classes. We will practice with writing out sums, and also expressing sums in summation notation:
Summation Notation Problem  Solution 
Write out the sum of the following series:
\(\sum\limits_{{k=2}}^{n}{{{{{\left( {1} \right)}}^{k}}}}{{2}^{{k+1}}}\)

Since we’re starting with 2 (from the “\(k=2\)” at the bottom of the sigma sign), we first plug in 2 in the \({{\left( {1} \right)}^{k}}{{2}^{{k+1}}}\) part; the first term is \({{\left( {1} \right)}^{2}}{{2}^{{2+1}}}=1\cdot 8=8\). For the next term, plug in 3 to get \({{\left( {1} \right)}^{3}}{{2}^{{3+1}}}=1\cdot 16=16\).
Note that to get to the next term, the sign changes and the previous term is multiplied by 2 (or, it’s like multiplying the previous term by –2). This is a geometric series. We have \(816+3264+…\,\,{{\left( {1} \right)}^{n}}{{\left( 2 \right)}^{{n+1}}}\) (since we are going until \(k=n\)). Note that when we use \({{\left( {1} \right)}^{k}}\), every other term is negative. 
Express the sum in summation notation:
\(13.5+68.5+…18.5\)

Disregarding the negative signs, we can see that this is an arithmetic series, since the second term – the first = the third term – the second = 2.5; this is d, or the common difference.
The first term is 1, so \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d=1+\left( {n1} \right)\left( {2.5} \right)=2.5n1.5\). But we also have to have the sign go back and forth, so we’ll use \({{\left( {1} \right)}^{{n1}}}\) (we need to have the first term positive, so we need a nonodd number, or 0, as an exponent). The series goes up to –18.5, so we need to disregard the negative sign and set 18.5 to \(2.5n1.5\) to get n (upper bound). We see \(n=8\) when \({{a}_{n}}=18.5\). We can write this series as \(\sum\limits_{{k=1}}^{8}{{{{{\left( {1} \right)}}^{{k1}}}}}\left( {2.5n1.5} \right)\). Try it; it works! 
Express the sum in summation notation:
\(\begin{array}{l}x+\left( {x+s} \right)+\left( {x+2s} \right)\\\,\,\,\,+…+\left( {x+ns} \right)\end{array}\) 
We can see that this is an arithmetic sequence with \(d=s\). The first term is “x”, so we have \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d=x+\left( {n1} \right)\left( s \right)\). The series goes up to \(\left( {x+ns} \right)\), so we need to know what the upper bound (“k“) is. To get this, set the last term, or \(x+ns\,\) to \(x+\left( {k1} \right)\left( s \right)\); we get \(k=n+1\) when we start at \(k=1\).
We can write this series as\(\sum\limits_{{k=1}}^{{n+1}}{{x+\left( {k1} \right)s}}\). (We can also write this series as \(\sum\limits_{{k=0}}^{n}{{x+ks}}\)). Try it; it works! 
Evaluate the double sum:
\(\sum\limits_{{j=2}}^{4}{{\,\left( {\sum\limits_{{n=1}}^{3}{{n+2j}}} \right)}}\)

This is tricky, since it’s a double summation. We will work from right to left; expand the inside summation first: \(\sum\limits_{{n=1}}^{3}{{\left( {n+2j} \right)}}=\left( {1+2j} \right)+\left( {2+2j} \right)+\left( {3+2j} \right)=6+6j\). Now we have \(\sum\limits_{{j=2}}^{4}{{\,\left( {6+6j} \right)}}\). Now expand this outside (leftmost) sum: \(\sum\limits_{{j=2}}^{4}{{\,6+6j}}=\left( {6+6\cdot 2} \right)+\left( {6+6\cdot 3} \right)+\left( {6+6\cdot 4} \right)=72\). 
Geometric Series
Just like we saw that an arithmetic series is the sum of an Arithmetic sequence, Geometric Series are the sum of Geometric Sequences.
For example, for the geometric sequence \({{a}_{n}}=2{{\left( 3 \right)}^{{n1}}}\), for \(n=1\,\,\,\text{to }5\), we have or 2, 6, 18, 54, 162. We would write this series as 2 + 6 + 18 + 54 + 162. Again, another way to write this is to use the summation formula \(\sum\limits_{{k=1}}^{5}{{\,2{{{\left( 3 \right)}}^{{k1}}}}}\). (The generic summation formula is \(\sum\limits_{{k=1}}^{n}{{\,{{a}_{1}}{{{\left( r \right)}}^{{k1}}}}}\)).
Here’s an example of why we might want to use a geometric series. Let’s say we know we are getting a 5% raise for the next five years, and we want to know the total amount of money we will make for these five years. We make $50,000 now.
Our common ratio will be 1.05 (do you see how we multiply this by what we make every year to get our new salary?). So we would need to get the sum of a geometric series of 5 terms (with common ratio 1.05) to see the total amount we’ll make in these five year, or \(\sum\limits_{{k=1}}^{5}{{\,50000{{{\left( {1.05} \right)}}^{{k1}}}}}\).
Finite Geometric Series
For geometric series, we have different formulas, depending on whether the series ends at a certain point (finite), or goes on forever (infinite). It turns out that we may have an infinite geometric series that actually ends up at a number; this is said to converge to this number, since adding each term makes the series get closer to this number. We’ll talk about this below.
Here is the partial sum formula for the finite geometric series (one that doesn’t go to infinity):
This looks really complicated, but it’s really not too bad; if we are solving for the \(n\)th generic term \({{S}_{n}}\), we just plug in the first term (\({{a}_{1}}\)) and \(r\) (common ratio), and our answer has an “\(n\)” in it. If we are solving for an actual number, our answer won’t have an “\(n\)” in it. So it’s actually a little easier than the arithmetic series formula, since we don’t need to find \({{a}_{n}}\) first.
Note: we have to be careful to compute \({{r}^{n}}\) first, before subtracting this from 1, and then multiply this difference by \({{a}_{n}}\), and then divide the result by \(\left( {1r} \right)\). We can’t simply cancel out the \(\left( {1r} \right)\) and the \(\left( {1{{r}^{n}}} \right)\), unless \(n=1\).
Here are some problems:
Geometric Series Problem  Solution 
Find an equation for the nth partial sum \({{S}_{n}}\), and find \({{S}_{6}}\) in this geometric sequence:
\({{a}_{1}}=2,\,\,\,\,r=4\) 
We know the formula for the \(n\)th partial sum of a geometric series is \(\displaystyle {{S}_{n}}=\frac{{{{a}_{1}}\left( {1{{r}^{n}}} \right)}}{{1r}}\), so we have \(\displaystyle {{S}_{n}}=\frac{{2\left( {1{{4}^{n}}} \right)}}{{14}}=\frac{2}{3}\left( {1{{4}^{n}}} \right)\). (The series looks like 2 + 8 + 32 + …., which is \(\sum\limits_{{k=1}}^{n}{{2{{{\left( 4 \right)}}^{{k1}}}}}\)).
To get the sum of the first 6 terms, we have \(\displaystyle {{S}_{6}}=\,\frac{2}{3}\left( {1{{4}^{6}}} \right)=\frac{2}{3}\left( {4095} \right)=2730\). 
Find an equation for the nth partial sum \({{S}_{n}}\):
\(\displaystyle 81+\frac{1}{8}\frac{1}{{64}}…\) 
We can see that this is a geometric series, since the second term/the first = the third term/the second = \(\displaystyle \frac{1}{8}\); this is \(r\), or the common ratio.This series is \(\displaystyle \sum\limits_{{k=1}}^{n}{{8{{{\left( {\frac{1}{8}} \right)}}^{{k1}}}}}\).
The equation for this series’ \(n\)th partial sum is \(\displaystyle {{S}_{n}}=\frac{{8\left( {1{{{\left( {\frac{1}{8}} \right)}}^{n}}} \right)}}{{1\left( {\frac{1}{8}} \right)}}=\frac{{8\left( {1{{{\left( {\frac{1}{8}} \right)}}^{n}}} \right)}}{{\frac{9}{8}}}=\frac{{64}}{9}\left( {1{{{\left( {\frac{1}{8}} \right)}}^{n}}} \right)\). Note that we have to be very careful with the parentheses; in the denominator, we can make the “– –“ into a “\(+\)”, but in the numerator, when the \(\displaystyle \frac{1}{8}\) is in the parentheses, we have to leave it. Try this geometric series for \(n=2\) (\({{S}_{n}}=7\)); it works! 
Find an equation for the nth partial sum \({{S}_{n}}\), and find \({{S}_{3}}\):
\(4{{\left( b \right)}^{{n1}}}\) 
We know from the formula for a geometric sequence that \({{a}_{1}}=4\) and \(r=b\). Since the formula for the \(n\)th partial sum of a geometric series is \(\displaystyle {{S}_{n}}=\frac{{{{a}_{1}}\left( {1{{r}^{n}}} \right)}}{{1r}}\), we have \(\displaystyle {{S}_{n}}=\frac{{4\left( {1{{b}^{n}}} \right)}}{{1b}}\). The series looks like \(4+4b+4{{b}^{2}}+…\), or \(\sum\limits_{{k=1}}^{n}{{4{{{\left( b \right)}}^{{k1}}}}}\).
To get the sum of the first 3 terms, we can factor the sum of two cubes to get:.\(\displaystyle \require{cancel} {{S}_{3}}\,\,=\,\,\frac{{4\left( {1{{b}^{3}}} \right)}}{{1b}}=\frac{{4\cancel{{\left( {1b} \right)}}\left( {1+b+{{b}^{2}}} \right)}}{{\cancel{{1b}}}}=4+4b+4{{b}^{2}}\). It works! 
Evaluate:
\(\sum\limits_{{k=1}}^{5}{{8{{{\left( {\frac{1}{4}} \right)}}^{{k1}}}}}\) 
We know from the summation formula for a geometric sequence that \({{a}_{1}}=8\) and \(\displaystyle r=\frac{1}{4}\). Since the formula for the \(n\)th partial sum of a geometric series is \(\displaystyle {{S}_{n}}=\frac{{{{a}_{1}}\left( {1{{r}^{n}}} \right)}}{{1r}}\), we have \(\displaystyle {{S}_{n}}=\frac{{8\left( {1{{{\left( {\frac{1}{4}} \right)}}^{n}}} \right)}}{{1\frac{1}{4}}}=\frac{{8\left( {1{{{\left( {\frac{1}{4}} \right)}}^{n}}} \right)}}{{\frac{3}{4}}}=\frac{{32}}{3}\left( {1{{{\left( {\frac{1}{4}} \right)}}^{n}}} \right)\). The series looks like \(\displaystyle 8+2+\frac{1}{2}+\frac{1}{8}+\frac{1}{{32}}\).
To get the sum of the first 5 terms, we have \(\displaystyle {{S}_{5}}=\frac{{32}}{3}\left( {1{{{\left( {\frac{1}{4}} \right)}}^{5}}} \right)=10.65625=10\frac{{21}}{{32}}\). 
Infinite Geometric Series
A geometric series is infinite if it’s upper bound (the number on the top of the sigma sign) is infinity, or \(\infty \). So an infinite series would look like this: \(\sum\limits_{{k=1}}^{\infty }{{2{{{\left( {.5} \right)}}^{{k1}}}}}\).
It turns out that the equation for an infinite series is much easier than that of a finite series, but there’s one caveat: we can only get an answer (meaning, the series converges), if the common ratio, or \(r\), is between –1 and 1 (not including –1 and 1). This is the same thing as writing the restriction of \(r\) this way: \(\left r \right<1\). Note that the series diverges if \(\left r \right\ge 1\); we can’t find an answer to the infinite series.
This makes sense since if \(r\) is a fraction, we’ll be adding a term that is getting smaller and smaller, so we should finally arrive at a number if we theoretically go to infinity (called the “limit”).
Here is the formula:
Here are some problems; determine whether the geometric series converges or diverges; if it converges, find the sum:
Infinite Geometric Series Problem 
Solution 
Find the sum, if possible:
\(\displaystyle 2+\frac{2}{3}+\frac{2}{9}+…\)

We can see that this is a geometric series, since the second term/the first = the third term/the second \(\displaystyle =\frac{1}{3}\); this is \(r\), or the common ratio. Since \(\left r \right<1\), the series converges, and we can get the sum!
The equation for this series’ sum is \(\displaystyle {{S}_{\infty }}=\frac{{{{a}_{1}}}}{{1r}}=\frac{2}{{1\frac{1}{3}}}=3\). Try the series in your calculator and you’ll see that it’s correct. 
Find the sum, if possible:
\(\displaystyle 4\frac{1}{3}+3\frac{2}{3}+3+…\) 
Even though this series looks similar to the one above, it is actually an arithmetic series, since the second term – the first = the third term – the second \(\displaystyle \frac{2}{3}\).
Arithmetic series don’t converge (they diverge), so we can’t get the sum. 
Find the sum, if possible:
\(36+12+…\) 
We can see that this is a geometric series, since the second term/the first = the third term/the second \(=2\); this is \(r\), or the common ratio. Since \(\left r \right\ge 1\), the series diverges, and we can’t get the sum! 
Find the sum, if possible:
\(\displaystyle \sum\limits_{{k=1}}^{\infty }{{8{{{\left( {\frac{2}{3}} \right)}}^{{k1}}}}}\) 
We can see that this is a geometric series, since the second term/the first = the third term/the second \(\displaystyle =\frac{2}{3}\); this is \(r\), or the common ratio. Since \(\left r \right<1\), the series converges, and we can get the sum!
The equation for this series’ sum is \(\displaystyle {{S}_{\infty }}=\frac{{{{a}_{1}}}}{{1r}}=\frac{8}{{1\left( {\frac{2}{3}} \right)}}=\frac{8}{{\frac{5}{3}}}=\frac{{24}}{5}\). 
Find the sum, if possible:
\(\sum\limits_{{k=1}}^{\infty }{{.5{{{\left( {1.2} \right)}}^{{k1}}}}}\) 
We can see that this is a geometric series, since the second term/the first = the third term/the second \(=1.2\); this is \(r\), or the common ratio. Since \(\left r \right\ge 1\), the series diverges, and we can’t get the sum! 
The sum to infinity of a geometric series is 9. The second term is 2. Find \(r\), the common ratio.  Since we only know the second term, let’s first use the Finite Geometric Sequences formula. Then we’ll have to set up a system with the Infinite Geometric Series formula.
\(\displaystyle {{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}};\,\,\,\,{{a}_{2}}={{a}_{1}}{{\left( r \right)}^{{21}}}={{a}_{1}}r=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{S}_{\infty }}=\frac{{{{a}_{1}}}}{{1r}}=9\) We now have two equations and two unknowns, so we can solve for \({{a}_{1}}\) and \(r\): \(\displaystyle {{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}};\,\,\,\,{{a}_{2}}=2={{a}_{1}}{{\left( r \right)}^{{21}}};\,\,\,\,{{a}_{1}}r=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{S}_{\infty }}=\frac{{{{a}_{1}}}}{{1r}}=9\) \(\displaystyle \begin{array}{c}9\left( {r{{r}^{2}}} \right)=2;\,\,\,\,9{{r}^{2}}9r+2=0;\,\,\,\,\left( {3r1} \right)\left( {3r2} \right)=0\,\\\text{Answer 1: }r=\frac{1}{3}\,;\,\,{{a}_{1}}=\frac{2}{r}=6\\\text{Answer 2: }r=\frac{2}{3}\,;\,\,{{a}_{1}}=\frac{2}{r}=3\end{array}\) Try the answers; they work! 
Summary of Formulas for Sequences and Series
Here is a summary of the main formulas for Sequences and Series:
\(\begin{array}{l}{{a}_{1}}=\,\,\text{first term}\\{{a}_{n}}=\,\,n\text{th term}\end{array}\) 
Arithmetic \(d=\) common difference 
Geometric \(r=\) common ratio 
Sequences 
\({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\)
(Recursive: \({{a}_{n}}=a,\,\,\,\,\,{{a}_{n}}={{a}_{{n1}}}+d\))

\({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\)
(Recursive: \({{a}_{n}}=a,\,\,\,\,\,{{a}_{n}}=r{{a}_{{n1}}}\) ) 
Finite Series (Sum) 
\(\displaystyle {{S}_{n}}=\frac{n}{2}\left( {{{a}_{1}}+{{a}_{n}}} \right)\,\,\,=\,\,\,\frac{n}{2}\left( {2{{a}_{1}}+\left( {n1} \right)d} \right)\)

\(\displaystyle {{S}_{n}}=\frac{{{{a}_{1}}\left( {1{{r}^{n}}} \right)}}{{1r}}\) 
Infinite Series (Sum) 
Not Applicable 
\(\displaystyle {{S}_{\infty }}=\frac{{{{a}_{1}}}}{{1r}},\,\,\,\,\,\,\,\left r \right<1\)

Applications of Sequences and Series
Sequence and Series are very useful in many applications; in fact, with geometric sequences, especially when we’re dealing with growth or decay (like with money), we’ll see that they equations look a lot like some of the exponential equations we worked with here in the Exponential Functions section.
The main thing to remember about word problems with sequences and series is that when we want an amount for a single thing, such as a particular row, year, for example, we use a sequence. When we want to know a total amount, such as money or rows, we want to use a series (which is a sum).
Here are some problems:
Problem:
A stadium with 75 rows has 10 seats in the first row, 15 in the second, and so on. How many seats are in the top row? How many total seats are in the stadium?
Solution:
We can see that an arithmetic sequence models this situation (since we are adding 5 seats to each successive row), and when we are adding up all the seats in the stadium, we are dealing with an arithmetic series.
To get how many seats are in the top row, we have \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\), where \({{a}_{1}}=10\), and \(d=5\). We have \({{a}_{{75}}}=10+\left( {751} \right)5=380\), so there are 380 seats in the top row.
To get the total number of seats, we’ll use \(\displaystyle {{S}_{n}}=\frac{n}{2}\left( {{{a}_{1}}+{{a}_{n}}} \right)\), or \({{S}_{{75}}}=\frac{{75}}{2}\left( {10+380} \right)=14625\), so there are a total of 14625 seats. See how useful these equations could be, if you had to put in an order in for these seats?
Problem:
For the first year of a new job, you make $60,000. You get an annual raise each year of 3%. Find out how much you’ll make in your fifth year, and also the total amount of money you will make in 5 years.
Solution
Let’s first put some real numbers in to see what type of sequence or series this is. If you make $60,000 the first year, and you get a 3% raise for the second year, you will make \(60000 + (.03)(60000)\), or $61,800. (This is the same as multiplying 60000 by 1.03). To get how much you’ll make in the third year, multiply $61,800 by 1.03 again. So we have a geometric sequence with \(r=1.03\).
To find out how much money you’ll make in the fifth year, we’ll use the equation \({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\), with \(n=5\), \({{a}_{1}}=600000\), and \(r=1.03\). We’ll have \({{a}_{5}}=60000{{\left( {1.03} \right)}^{{51}}}=67530.53\). So you will make $67,530.53 in year 5. This makes sense, since we know it needs to be more than $60,000.
To find out how much you will make in the first 5 years, we need the partial sum formula for a geometric series, which is \(\displaystyle {{S}_{n}}\,\,=\,\,\frac{{{{a}_{1}}\left( {1{{r}^{n}}} \right)}}{{1r}}\). We have \(\displaystyle {{S}_{5}}\,\,=\,\,\frac{{60000\left( {1{{{\left( {1.03} \right)}}^{5}}} \right)}}{{11.03}}=318548.15\). So you’ll make $318,548.15 in the first 5 years!
Problem:
A pendulum swings through an arc (distance of the curve) of 5 feet initially. For the second swing, the length of the arc is .85 the length of the previous swing, and so on. On which swing will the length of the arc be less than 2 feet?
Solution:
We can see that we have a geometric sequence here, since we are multiplying the length of the arc by .85 with each swing. Since the initial swing is 5 feet, we have \({{a}_{n}}={{a}_{1}}{{\left( r \right)}^{{n1}}}\), or \({{a}_{n}}=5{{\left( {.85} \right)}^{{n1}}}\).
Since the question asks when (which swing) the length of the arc will be less than 2 feet, we need to use the equation \(5{{\left( {.85} \right)}^{{n1}}}=2\) to see what \(n\) is when the swing of the arc is 2 feet. The best way to solve this is to use logs (you could also use “guess and check”); we have \(5{{\left( {.85} \right)}^{{n1}}}=2;\,\,\,{{\left( {.85} \right)}^{{n1}}}=\frac{2}{5};\,\,\,\left( {n1} \right)\ln \left( {.85} \right)=\ln \left( {\frac{2}{5}} \right);\,\,n\approx 6.6\).
So when the length of the arc is 2, it will be after the 6^{th} swing, since \(n\) has to be an integer. So the swing will have an arc length of less than 2 on the 7^{th} swing. Tricky!
Problem:
Lyndal wants to run a total of at least 78 miles from all her runs before she starts school in the fall. She runs 2 miles the first day, 2.2 miles, the second, 2.4 miles the third, and so on, to make sure she doesn’t overheat. What’s the minimum number of days she has to run in the summer to make sure she runs a total of 78 miles?
Solution:
We know this will be a Partial Sum problem, since we want the total of all her runs to equal 78. We also see that the sequence is Arithmetic, since she is adding .2 miles to each run (\(d=.2\)). From this, we know that \({{a}_{n}}={{a}_{1}}+\left( {n1} \right)d\), or \({{a}_{n}}=2+\left( {n1} \right).2=.2n+1.8\).
We’ll use the following partial sum formula, with \(\displaystyle {{S}_{n}}=78:\,\,\,{{S}_{n}}=\frac{n}{2}\left( {{{a}_{1}}+{{a}_{n}}} \right)\), so \(\displaystyle 78=\frac{n}{2}\left[ {2+\left( {.2n+1.8} \right)} \right]\).
It’s a little tricky to solve for \(n\), since we’ll actually end up with a quadratic. It’s easiest to convert to fractions and multiply by common denominator, and then factor (we could have also used the Quadratic Formula):
\(\begin{align}78&=\frac{n}{2}\left[ {2+\left( {.2n+1.8} \right)} \right]=\frac{n}{2}\left( {.2n+3.8} \right)\\78&=\frac{n}{2}\left( {\frac{n}{5}+\frac{{19}}{5}} \right)=\frac{{{{n}^{2}}}}{{10}}+\frac{{19}}{{10}}\\78\cdot 10&=\left( {\frac{{{{n}^{2}}}}{{\cancel{{10}}}}+\frac{{19}}{{\cancel{{10}}}}} \right)\cdot \cancel{{10}}\\780&={{n}^{2}}+19\\{{n}^{2}}+19780&=0\\\left( {n20} \right)\left( {n+39} \right)&=0\\n&=20\,\,\,\,\,\,\,\text{(can }\!\!’\!\!\text{ t use }\,39)\end{align}\)
Lyndal must run 20 days to make sure she runs a total of 78 miles in the summer.
Problem:
How much (what fraction) of the square will eventually be shaded if the shading pattern (in the shapes of squares) is continued indefinitely from the top left to the bottom right?
Solution:
It looks like the first shaded square is one quarter of the whole square, and the second shaded square is one quarter of that, and so on. Since the shading theoretically goes on indefinitely, this will be a infinite geometric series, with \(\displaystyle {{a}_{1}}=\frac{1}{4}\), and \(\displaystyle r=\frac{1}{4}\) (see how second/first \(\displaystyle =\frac{1}{4}\) )? So we have \(\displaystyle {{S}_{\infty }}=\frac{{{{a}_{1}}}}{{1r}}=\frac{{\left( {\frac{1}{4}} \right)}}{{1\left( {\frac{1}{4}} \right)}}=\frac{{\left( {\frac{1}{4}} \right)}}{{\left( {\frac{3}{4}} \right)}}=\frac{1}{3}\).
Sequences and Sums on the Graphing Calculator
We can use the TI Graphing Calculator to create sequences and determine the sum of sequences (series).
We can use SEQ and SUM in the Catalog list, or in the 2^{nd} STAT (LIST) OPS 5 or seq and 2^{nd} STAT MATH 5 or sum, as in the following. Note that the calculator will either have you fill in the steps or (if you have an older operating system), you have to enter the arguments on one line.
You first enter the expression, using the X,T,θ,n button, the variable (hit X,T,θ,n again), starting and ending values of the variable, and (optional) 1 if you want all the values – which you probably will.
With the newer operating systems, when you use sum, you can get a summation symbol where you can enter the variable, the beginning and ending values, and the expression, like in the last screen. (You can also use MATH summation (MATH 0) or the shortcut ALPHA WINDOW 2 to do this.)
Note that if you want the sum of an infinite series, you can put a large number in the upper bound, such as 999, and you should arrive at the solution, or pretty close!
You can even put a sequence in a LIST, like \({{L}_{1}}\), by using the seq command and then using STO and then the name of the list:
Learn these rules, and practice, practice, practice!
For Practice: Use the Mathway widget below to try a Sequences and Series problem. Click on Submit (the blue arrow to the right of the problem) and click on Identify the Sequence to see the answer.
You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.
If you click on Tap to view steps, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).
On to Binomial Expansion – you are ready!