Finding a Specific Term

There may be instances when we want to identify a certain term in the expansion of $\displaystyle{ (x+y) }^{ n }$. It is straightforward to identify the terms where $n$ is an integer with a low value. You might multiply each binomial out to identify the coefficients, or you might use Pascal's triangle. However, what if $n$ has a greater value? How would you identify a particular term of ${(3x-4)}^{12}$? Luckily, there is a formula that can be used to calculate the terms in such situations.

Let's go through a few expansions of binomials, in order to consider any patterns that are present in the terms.

$\displaystyle {(a+b)}^{1}=a+b$

$\displaystyle {(a+b)}^{2}={a}^{2}+2ab+{b}^{2}$

$\displaystyle {(a+b)}^{3}={a}^{3}+3{a}^{2}b+3a{b}^{2}+{b}^3$

$\displaystyle {(a+b)}^{4}={a}^{4}+4{a}^{3}b+6{a}^{2}{b}^{2}+4a{b}^{3}+{b}^4$

A few things should be noticed:

• The number of terms is one more than $n$ (the exponent).
• The power of $a$ starts with $n$ and decreases by $1$ each term.
• The power of $b$ starts with $0$ and increases by $1$ each term.
• The sum of the exponents in each term adds up to $n$.  
• The coefficients of the first and last terms are both $1$ and they follow Pascal's triangle.

If the expansion is short, such as:

$\displaystyle \begin{aligned} {(x+2)}^{3}&={x}^{3}+2{x}^{2}{2}^{1}+2{x}^{1}{2}^{2}+{2}^{3}\\ &={x}^{3}+4{x}^{2}+8{x}+8 \end{aligned}$

Then it is easy to find a particular term. This becomes difficult and time consuming when the expansion is large. There is, luckily, a shortcut for identifying particular terms of longer expansions. The following formula yields the $r$th term in the expansion:

$\displaystyle { \begin{pmatrix} n \\ r-1 \end{pmatrix} }{ a }^{ n-(r-1) }{ b }^{ r-1 }$

Recall that the combination formula provides a way to calculate $\begin{pmatrix} n \\ k \end{pmatrix}$:

$\displaystyle {\begin{pmatrix} n \\ k \end{pmatrix}=\frac{n!}{(n-k)!k! }}$

Example: Find the fifth term of ${(3x-4)}^{12}$

Note that the value of $n = 12$ in this case. Because we are looking for the fifth term, we use $r=5$. Plugging these values into the formula, we have:

$\displaystyle { \begin{pmatrix} 12 \\ 5-1 \end{pmatrix} }{ (3x) }^{ 12-(5-1) }{ (-4) }^{ 5-1 }$

$\displaystyle { \begin{pmatrix} 12 \\ 4 \end{pmatrix} }{ (3x)}^{ 8 }{ (-4) }^{ 4 }$

Remember to evaluate $\begin{pmatrix} 12 \\ 4 \end{pmatrix}$ using the combination formula:

$\displaystyle \begin{aligned} \frac{n!}{(n-k)!k! }&=\frac{12!}{(12-4)!4! }\\ &=495 \end{aligned}$

Subbing in $\begin{pmatrix} 12 \\ 4 \end{pmatrix}=495$ in the formula, we have: 

$\displaystyle 495{ (3x)}^{ 8 }{ (-4) }^{ 4 }$

When the power is applied to the terms, the result is:

$\displaystyle 495\cdot 6561{x}^{8} \cdot 256 =831409920{x}^{8}$

Thus, the fifth term of ${(3x-4)}^{12}$ is $831409920{x}^{8}$.