 Introduction to Binomial Expansion
 Expanding a Binomial
 Finding a Specific Term with Binomial Expansion
 More Practice
Introduction to Binomial Expansion
You’ll probably have to learn how to expand polynomials to various degrees (powers) using what we call the Binomial Theorem or Binomial Expansion (or Binomial Series).
We use this when we want to expand (multiply out) the power of a binomial like \({{\left( {x+y} \right)}^{n}}\) into a sum with terms \(a{{x}^{b}}{{y}^{c}}\), where b and c are nonnegative integers (and it turns out that b + c = n). A perfect square trinomial is a simple example: \({{\left( {x+y} \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}\). (The coefficients in this case are 1, 2, and 1, respectively.)
It just turns out that the coefficient a in this expansion is equal to \(\left( {\begin{array}{*{20}{c}} n \\ c \end{array}} \right)\) (also written as \(\displaystyle {}_{n}{{C}_{c}}\)), where \(\left( {\begin{array}{*{20}{c}} n \\ c \end{array}} \right)\,\,=\,\,\frac{{n!}}{{c!\left( {nc} \right)!}}\) (this is called the binomial coefficient). Remember that \(n!=n\left( {n1} \right)\left( {n2} \right)\,\,\,….\) (until you get to 1). (You can also get \(\displaystyle {}_{n}{{C}_{c}}\) on your graphing calculator. Type in what you want for n, then MATH PROB, and hit 3 or scroll to nCr, and then type c and then ENTER). \(\displaystyle {}_{n}{{C}_{c}}\) is actually the number of ways to choose c items out of n terms, where order doesn’t matter – also called the Combination function.
Here is the Binomial Theorem (also called Binomial Formula or Binomial Identity):
\({{\left( {x+y} \right)}^{n}}\,\,=\,\,\left( {\begin{array}{*{20}{c}} n \\ 0 \end{array}} \right){{x}^{n}}{{y}^{0}}\,+\,\left( {\begin{array}{*{20}{c}} n \\ 1 \end{array}} \right){{x}^{{n1}}}{{y}^{1}}\,+\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right){{x}^{{n2}}}{{y}^{2}}\,+……+\left( {\begin{array}{*{20}{c}} n \\ {n1} \end{array}} \right){{x}^{1}}{{y}^{{n1}}}\,+\left( {\begin{array}{*{20}{c}} n \\ n \end{array}} \right){{x}^{0}}{{y}^{n}}\)
or
\({{\left( {x+y} \right)}^{n}}\,\,=\sum\limits_{{k=0}}^{n}{{\,\left( {\begin{array}{*{20}{c}} n \\ k \end{array}} \right){{x}^{{nk}}}{{y}^{k}}}}\), which is the same as \(\sum\limits_{{k=0}}^{n}{{\,\left( {\begin{array}{*{20}{c}} n \\ k \end{array}} \right){{x}^{k}}{{y}^{{nk}}}}}\)
See how the exponents of the x’s are going down (from n to 0), while the exponents of the y’s are going up (from 0 to n)? And remember that anything raised to the 0 is just 1. And for a binomial raise to the “n”, we have “n + 1” terms.
The coefficients can also be found using a Pascal Triangle, which starts with 1, and is a triangle with all 1’s on the outside. Then on the inside, add the two numbers above to get the next number down:
\(\begin{array}{c}1\\1\,\,\,\,\,1\\1\,\,\,\,\,\,2\,\,\,\,\,\,1\\1\,\,\,\,\,\,3\,\,\,\,\,\,3\,\,\,\,\,\,1\\1\,\,\,\,\,\,4\,\,\,\,\,\,6\,\,\,\,\,\,4\,\,\,\,\,\,1\\1\,\,\,\,\,\,5\,\,\,\,\,\,10\,\,\,\,\,10\,\,\,\,\,5\,\,\,\,\,\,1\\1\,\,\,\,\,\,6\,\,\,\,\,\,15\,\,\,\,\,20\,\,\,\,\,15\,\,\,\,\,6\,\,\,\,\,\,1\\……\text{and so on}…..\end{array}\) Which is actually (first four rows): \(\begin{array}{c}\left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)\\\left( {\begin{array}{*{20}{c}} 1 \\ 0 \end{array}} \right)\,\,\,\,\,\,\,\left( {\begin{array}{*{20}{c}} 1 \\ 1 \end{array}} \right)\\\left( {\begin{array}{*{20}{c}} 2 \\ 0 \end{array}} \right)\,\,\,\,\,\,\,\,\left( {\begin{array}{*{20}{c}} 2 \\ 1 \end{array}} \right)\,\,\,\,\,\,\,\,\left( {\begin{array}{*{20}{c}} 2 \\ 2 \end{array}} \right)\\\left( {\begin{array}{*{20}{c}} 3 \\ 0 \end{array}} \right)\,\,\,\,\,\,\,\,\left( {\begin{array}{*{20}{c}} 3 \\ 1 \end{array}} \right)\,\,\,\,\,\,\,\,\left( {\begin{array}{*{20}{c}} 3 \\ 2 \end{array}} \right)\,\,\,\,\,\,\,\,\left( {\begin{array}{*{20}{c}} 3 \\ 3 \end{array}} \right)\\……\text{and so on}…..\end{array}\)
As an example of how to use the Pascal Triangle, start with the second row for \({{\left( {x+y} \right)}^{1}}=1x+1y\), so the coefficients are both 1. When using the Pascal Triangle, the exponent of the binomial is off by 1; for example, we used the 2^{nd} row to get the coefficients for \({{\left( {x+y} \right)}^{1}}\).
Here’s another illustration of just how Pascal’s Triangle is used for expanding binomials:
\(\displaystyle \begin{array}{l}{{\left( {x+y} \right)}^{0}}=\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\{{\left( {x+y} \right)}^{1}}=\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x+y\\{{\left( {x+y} \right)}^{2}}=\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}+2xy+{{y}^{2}}\\{{\left( {x+y} \right)}^{3}}=\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}+{{y}^{3}}\,\,\,\,\,\\{{\left( {x+y} \right)}^{4}}=\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{4}}+4{{x}^{3}}y+6{{x}^{2}}{{y}^{2}}+4x{{y}^{3}}+{{y}^{4}}\,\,\,\,\\{{\left( {x+y} \right)}^{5}}=\,\,\,\,\,\,\,\,\,\,{{x}^{5}}+5{{x}^{4}}y+10{{x}^{3}}{{y}^{2}}+10{{x}^{2}}{{y}^{3}}+5x{{y}^{4}}+{{y}^{5}}\,\\{{\left( {x+y} \right)}^{6}}={{x}^{6}}+6{{x}^{5}}y+15{{x}^{4}}{{y}^{2}}+20{{x}^{3}}{{y}^{3}}+15{{x}^{2}}{{y}^{4}}+6x{{y}^{5}}+{{y}^{6}}\,\\\,\,…..\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}\)
Here’s a hint: when finding the coefficients of a binomial expansion using Pascal’s triangle, find the line with the second term the same as the power you want. For example, for a binomial with power 5, use the line 1 5 10 10 5 1 for coefficients.
Expanding a Binomial
The best way to show how Binomial Expansion works is to use an example. Let’s expand \({{\left( {x+3} \right)}^{6}}\) using the formula above. Here, the “x” in the generic binomial expansion equation is “x” and the “y” is “3”:
\(\begin{align}{{\left( {x+y} \right)}^{n}}\,\,&=\,\,\left( {\begin{array}{*{20}{c}} n \\ 0 \end{array}} \right){{x}^{n}}{{y}^{0}}\,+\,\left( {\begin{array}{*{20}{c}} n \\ 1 \end{array}} \right){{x}^{{n1}}}{{y}^{1}}\,+\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right){{x}^{{n2}}}{{y}^{2}}\,+……+\left( {\begin{array}{*{20}{c}} n \\ {n1} \end{array}} \right){{x}^{1}}{{y}^{{n1}}}\,+\left( {\begin{array}{*{20}{c}} n \\ n \end{array}} \right){{x}^{0}}{{y}^{n}}\\\color{#804040}{{{{{\left( {x+3} \right)}}^{6}}\,}}\,&=\,\,\left( {\begin{array}{*{20}{c}} 6 \\ 0 \end{array}} \right){{x}^{6}}{{\left( 3 \right)}^{0}}\,+\,\left( {\begin{array}{*{20}{c}} 6 \\ 1 \end{array}} \right){{x}^{5}}{{\left( 3 \right)}^{1}}\,+\left( {\begin{array}{*{20}{c}} 6 \\ 2 \end{array}} \right){{x}^{4}}{{\left( 3 \right)}^{2}}\,+\left( {\begin{array}{*{20}{c}} 6 \\ 3 \end{array}} \right){{x}^{3}}{{\left( 3 \right)}^{3}}\,+\,\left( {\begin{array}{*{20}{c}} 6 \\ 4 \end{array}} \right){{x}^{2}}{{\left( 3 \right)}^{4}}+\,\left( {\begin{array}{*{20}{c}} 6 \\ 5 \end{array}} \right){{x}^{1}}{{\left( 3 \right)}^{5}}+\,\left( {\begin{array}{*{20}{c}} 6 \\ 6 \end{array}} \right){{x}^{0}}{{\left( 3 \right)}^{6}}\\&=\,\,1{{x}^{6}}{{\left( 3 \right)}^{0}}\,+\,6{{x}^{5}}{{\left( 3 \right)}^{1}}\,+15{{x}^{4}}{{\left( 3 \right)}^{2}}\,+20{{x}^{3}}{{\left( 3 \right)}^{3}}\,+\,15{{x}^{2}}{{\left( 3 \right)}^{4}}+\,6{{x}^{1}}{{\left( 3 \right)}^{5}}+\,1{{x}^{0}}{{\left( 3 \right)}^{6}}\\&=\,\,{{x}^{6}}\,+\,6{{x}^{5}}\left( 3 \right)\,+15{{x}^{4}}\left( 9 \right)\,+20{{x}^{3}}\left( {27} \right)\,+\,15{{x}^{2}}\left( {81} \right)+\,6{{x}^{1}}\left( {243} \right)+\,729\\&=\,\,{{x}^{6}}\,+\,18{{x}^{5}}\,+135{{x}^{4}}\,+540{{x}^{3}}\,+\,1215{{x}^{2}}+\,1458x+\,729\end{align}\)
Notice how the power (exponent) of the first variable starts at the highest (n) and goes down to 0 (which means that variable “disappears”, since \({{\left( {\text{anything}} \right)}^{0}}=1\)). Also notice that the power of the second variable starts at 0 (which means you don’t see it), and goes up to n.
Also notice that for the coefficients of the \(\left( {\begin{array}{*{20}{c}} {} \\ {} \end{array}} \right)\) part, the 6 (since this is n) always stays on top, and the bottom starts with 0 and goes up to 6. The exponents for the first term of the binomial with 6 (n) and goes down to 0, and the exponent on the second term is always the bottom part of the \(\left( {\begin{array}{*{20}{c}} {} \\ {} \end{array}} \right)\). And if you add the two exponents, you always get 6 (since this is n).
Again, for the binomial coefficient \(\displaystyle \left( {\begin{array}{*{20}{c}} n \\ c \end{array}} \right)\), you can just use the \(\displaystyle {}_{n}{{C}_{c}}\) on your graphing calculator. (Type in what you want for n, then MATH PROB, and hit 3 or scroll to nCr, and then type c and then ENTER). You can also do these “by hand” by using \(\left( {\begin{array}{*{20}{c}} n \\ c \end{array}} \right)\,\,=\,\,\frac{{n!}}{{c!\left( {nc} \right)!}}\). Notice that \(\displaystyle \left( {\begin{array}{*{20}{c}} n \\ 0 \end{array}} \right)\,\,\,\text{and}\,\,\,\left( {\begin{array}{*{20}{c}} n \\ n \end{array}} \right)\) is always just 1 (0! = 1), and \(\displaystyle \left( {\begin{array}{*{20}{c}} n \\ 1 \end{array}} \right)\,\,\,\text{and}\,\,\,\left( {\begin{array}{*{20}{c}} n \\ {n1} \end{array}} \right)\) is just n.
To use the Pascal Triangle above to do this, let’s look at the 7^{th} row (since the first row is just “1”) to get the coefficients: 1 6 15 20 15 6 1. Note that since we’re wanting the 6^{th} power, we are using the line that has 6 as the second term!
Here are a few more that are a little more complicated, including one that’s a complex number:
Binomial Expansion Problem  Explanation 
Expand the binomial:
\({{\left( {4a3b} \right)}^{4}}\)
(Here, the “\(x\)” in the generic binomial expansion equation is “\(4a\)” and the “\(y\)” is “\(3b\)”.) 
\(\displaystyle \begin{align}{{\left( {x+y} \right)}^{n}}&=\left( {\begin{array}{*{20}{c}} n \\ 0 \end{array}} \right){{x}^{n}}{{y}^{0}}\,+\,\left( {\begin{array}{*{20}{c}} n \\ 1 \end{array}} \right){{x}^{{n1}}}{{y}^{1}}\,+\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right){{x}^{{n2}}}{{y}^{2}}\,+……+\left( {\begin{array}{*{20}{c}} n \\ {n1} \end{array}} \right){{x}^{1}}{{y}^{{n1}}}\,+\left( {\begin{array}{*{20}{c}} n \\ n \end{array}} \right){{x}^{0}}{{y}^{n}}\\\color{#800000}{{{{{\left( {4a3b} \right)}}^{4}}}}&=\left( {\begin{array}{*{20}{c}} 4 \\ 0 \end{array}} \right){{\left( {4a} \right)}^{4}}{{\left( {3b} \right)}^{0}}\,+\,\left( {\begin{array}{*{20}{c}} 4 \\ 1 \end{array}} \right){{\left( {4a} \right)}^{3}}{{\left( {3b} \right)}^{1}}\,+\left( {\begin{array}{*{20}{c}} 4 \\ 2 \end{array}} \right){{\left( {4a} \right)}^{2}}{{\left( {3b} \right)}^{2}}\,\\&\,\,\,\,\,+\left( {\begin{array}{*{20}{c}} 4 \\ 3 \end{array}} \right){{\left( {4a} \right)}^{1}}{{\left( {3b} \right)}^{3}}+\left( {\begin{array}{*{20}{c}} 4 \\ 4 \end{array}} \right){{\left( {4a} \right)}^{0}}{{\left( {3b} \right)}^{4}}\\&=1\left( {256} \right){{a}^{4}}\,+\,4\left( {64} \right){{a}^{3}}\left( {3b} \right)\,+6\left( {16{{a}^{2}}} \right)\left( {9{{b}^{2}}} \right)\,+4\left( {4a} \right)\left( {27{{b}^{3}}} \right)+1\left( {81} \right){{b}^{4}}\\&=256{{a}^{4}}\,768{{a}^{3}}b\,+864{{a}^{2}}{{b}^{2}}\,432a{{b}^{3}}\,+\,81{{b}^{4}}\end{align}\)
We also could have used the 5^{th} row of the Pascal Triangle to get the coefficients. 
Expand the binomial:
\({{\left( {14i} \right)}^{5}}\)

\(\displaystyle \begin{align}{{\left( {x+y} \right)}^{n}}&=\left( {\begin{array}{*{20}{c}} n \\ 0 \end{array}} \right){{x}^{n}}{{y}^{0}}+\left( {\begin{array}{*{20}{c}} n \\ 1 \end{array}} \right){{x}^{{n1}}}{{y}^{1}}+\left( {\begin{array}{*{20}{c}} n \\ 2 \end{array}} \right){{x}^{{n2}}}{{y}^{2}}\,+……+\left( {\begin{array}{*{20}{c}} n \\ {n1} \end{array}} \right){{x}^{1}}{{y}^{{n1}}}\,+\left( {\begin{array}{*{20}{c}} n \\ n \end{array}} \right){{x}^{0}}{{y}^{n}}\\\color{#800000}{{{{{\left( {14i} \right)}}^{5}}}}&=\left( {\begin{array}{*{20}{c}} 5 \\ 0 \end{array}} \right){{\left( 1 \right)}^{5}}{{\left( {4i} \right)}^{0}}+\left( {\begin{array}{*{20}{c}} 5 \\ 1 \end{array}} \right){{\left( 1 \right)}^{4}}{{\left( {4i} \right)}^{1}}+\left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right){{\left( 1 \right)}^{3}}{{\left( {4i} \right)}^{2}}\,+\left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right){{\left( 1 \right)}^{2}}{{\left( {4i} \right)}^{3}}\,\\&\,\,\,\,\,+\left( {\begin{array}{*{20}{c}} 5 \\ 4 \end{array}} \right){{\left( 1 \right)}^{1}}{{\left( {4i} \right)}^{4}}+\left( {\begin{array}{*{20}{c}} 5 \\ 5 \end{array}} \right){{\left( 1 \right)}^{0}}{{\left( {4i} \right)}^{5}}\\&=1\left( 1 \right)\left( 1 \right)\,+\,5\left( 1 \right)\left( {4i} \right)\,+10\left( 1 \right)\left( {16} \right)\,+10\left( 1 \right)\left( {64i} \right)\,+\,5\left( 1 \right)\left( {256} \right)+1\left( 1 \right)\left( {1024i} \right)\\&=1\,20i\,160\,+640i\,+\,12801024i\\&=1121404i\end{align}\) 
Finding Specific Terms with Binomial Expansion
You may be asked to find specific terms using the Binomial Expansion; for example, they may ask to find the 5^{th} term of a binomial raised to an exponent, or the term containing say a certain variable raised to a power.
To do these, just remember that the x^{th} term has (x – 1) in the bottom of the \(\left( {\begin{array}{*{20}{c}} {} \\ {} \end{array}} \right)\) part of the binomial coefficient, since the first term has the \(\left( {\begin{array}{*{20}{c}} n \\ 0 \end{array}} \right)\) part. (So the x^{th} term’s coefficient of a binomial expanded to the nth term is \(\left( {\begin{array}{*{20}{c}} n \\ {x1} \end{array}} \right)\).)
Then remember that the exponent of the first part of the expanded terms is the difference of the two numbers in the \(\left( {\begin{array}{*{20}{c}} {} \\ {} \end{array}} \right)\), and the exponent of the second part of the expanded terms equals the bottom number in the \(\left( {\begin{array}{*{20}{c}} {} \\ {} \end{array}} \right)\) (since the two exponents always add up to equal n).
For example, if we are expanding a binomial raised to the 5^{th} power, the 4^{th} term will have a \(\left( {\begin{array}{*{20}{c}} 5 \\ {41} \end{array}} \right)=\left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right)\) coefficient, the power of the first expanded term 5 – 3 = 2, and the power of the second is 3.
Here are some examples. And remember that sometimes you will see \({}_{n}C{}_{r}\) instead of \(\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right)\):
Binomial Expansion Problem  Explanation 
Find the 4^{th} term of the expansion of
\({{\left( {3x+5y} \right)}^{8}}\) 
The 4^{th} term will have a \(\left( {\begin{array}{*{20}{c}} 8 \\ 3 \end{array}} \right)\) binomial coefficient, since n = 8, and 4 – 1 = 3. The first exponent will be 8 – 3 = 5, and the second exponent will be 3. The 4^{th} term then is:
\(\left( {\begin{array}{*{20}{c}} 8 \\ 3 \end{array}} \right){{\left( {3x} \right)}^{5}}{{\left( {5y} \right)}^{3}}=56\cdot 243{{x}^{5}}\cdot 125{{y}^{3}}=1701000{{x}^{5}}{{y}^{3}}\) 
Find the 5^{th} term of the expansion of
\({{\left( {\sqrt{a}\sqrt{b}} \right)}^{7}}\) 
The 5^{th} term will have a \(\left( {\begin{array}{*{20}{c}} 7 \\ 4 \end{array}} \right)\) binomial coefficient, since n = 7, and 5 – 1 = 4. The first exponent will be 7 – 4 = 3, and the second exponent will be 4. The 5^{th} term then is:
\(\left( {\begin{array}{*{20}{c}} 7 \\ 4 \end{array}} \right){{\left( {\sqrt{a}} \right)}^{3}}{{\left( {\sqrt{b}} \right)}^{4}}=35\cdot {{\left( {{{a}^{{\frac{1}{2}}}}} \right)}^{3}}\cdot {{\left( {{{b}^{{\frac{1}{2}}}}} \right)}^{4}}=35\cdot {{a}^{{\frac{3}{2}}}}\cdot {{b}^{2}}=35a\sqrt{a}{{b}^{2}}\) 
Find the term containing \({{x}^{6}}\) in the expansion of
\({{\left( {2x2y} \right)}^{{10}}}\) 
The term containing \({{x}^{6}}\) will have a 6^{th} power in the first part of the term, or \({{\left( {2x} \right)}^{6}}\). Since n = 10, this means the binomial coefficient will have to be 4, since the exponent on the first part is the difference of the numbers in the binomial coefficient: 10 – 4 = 6 (this will be the 5^{th} term). Now we can “build” the rest of the term:
\(\displaystyle \left( {\begin{array}{*{20}{c}} {10} \\ 4 \end{array}} \right){{\left( {2x} \right)}^{6}}{{\left( {2y} \right)}^{4}}=210\cdot 64{{x}^{6}}\cdot 16{{y}^{4}}=215040{{x}^{6}}{{y}^{4}}\) Notice that the negative goes away when we raise to an even exponent. 
Find the term in the expansion of
\({{\left( {xy} \right)}^{6}}\) where x and y have the same power 
The term where x and y are the same must have an \({{x}^{3}}{{y}^{3}}\) in it, since the two exponents must add up to 6 (n). Then the binomial coefficient must be \(\left( {\begin{array}{*{20}{c}} 6 \\ 3 \end{array}} \right)\), since n = 6, and 6 – 3 must equal the first power (3). Now we can “build” the rest of the term:
\(\displaystyle \left( {\begin{array}{*{20}{c}} 6 \\ 3 \end{array}} \right){{x}^{3}}{{\left( {y} \right)}^{3}}=20\cdot {{x}^{3}}\cdot {{y}^{3}}=20{{x}^{3}}{{y}^{3}}\) 
Find the term containing \({{b}^{4}}\) in the expansion of
\({{\left( {2a5b} \right)}^{7}}\) 
This one’s a little tricky, since b in the second part of the binomial. When b is raised to the 4^{th} power, we must have a 4 on the bottom of the binomial coefficient and we always have n (7) on the top. (This will be the 5^{th} term). So this isn’t so bad! Now we can “build” the rest of the term:
.\(\displaystyle \left( {\begin{array}{*{20}{c}} 7 \\ 4 \end{array}} \right){{\left( {2a} \right)}^{3}}{{\left( {5b} \right)}^{4}}=35\cdot 8{{a}^{3}}\cdot 625{{b}^{4}}=175000{{a}^{3}}{{b}^{4}}\) 
Find the term containing \({{y}^{8}}\) in the expansion of
\({{\left( {3{{y}^{2}}2{{z}^{4}}} \right)}^{5}}\) 
Since we want the 8^{th} power of y, we must raise \(3{{y}^{2}}\) to the 4^{th} power, since when we raise an exponent to another exponent, we multiply exponents. This means, for the binomial coefficient, we’ll have \(\left( {\begin{array}{*{20}{c}} 5 \\ 1 \end{array}} \right)\), since n = 5, and 5 – 1 = 4 (this will be the 2^{nd} term). Now we can “build” the rest of the term:
. \(\displaystyle \left( {\begin{array}{*{20}{c}} 5 \\ 1 \end{array}} \right){{\left( {3{{y}^{2}}} \right)}^{4}}{{\left( {2{{z}^{4}}} \right)}^{1}}=5\cdot 81{{y}^{8}}\cdot 2{{z}^{4}}=810{{y}^{8}}{{z}^{4}}\) It works! 
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