Chapter 12.6, Problem 33E

To determine

To find:Middle term in the expansion of ${\left(x^2+1\right)}^{18}$ using Binomial Theorem.

Answer to Problem 33E

Middle term = $a_{10}=48620x^{18}$

Explanation of Solution

Given information:

  ${\left(x^2+1\right)}^{18}$

Calculation:

According to the Binomial theorem, expansion of ${\left(x+y\right)}^n$ with non-negative integer $n$ is:

  ${\left(x+y\right)}^n=\left(\begin{array}{l}n\\ 0\end{array}\right)x^n{y}^0+\left(\begin{array}{l}n\\ 1\end{array}\right)x^{n-1}{y}^1+\mathrm{.............}+\left(\begin{array}{l}n\\ n-1\end{array}\right)x^1{y}^{n-1}+\left(\begin{array}{l}n\\ n\end{array}\right)x^0{y}^n$

Where $n\ge 0$ , and $\left(\begin{array}{l}n\\ k\end{array}\right)$ denotes the binomial coefficient for $k=0,1,\mathrm{2........}n-1,n$

The ${\left(r+1\right)}^{th}$ term of the binomial expansion of ${\left(x+y\right)}^n$ is $a_{r+1}=\left(\begin{array}{l}n\\ r\end{array}\right)x^{n-r}{y}^r$

If $n$ is even, then number of terms that is $n+1$ is odd

Middle term = ${\left(\frac{n}{2}+1\right)}^{th}$ term.

If $n$ is odd then number of terms that is $n+1$ is even.

Then

Middle term 1 = ${\left(\frac{n+1}{2}\right)}^{th}$ term

Middle term 2 = ${\left(\frac{n+1}{2}+1\right)}^{th}$ term

We know that,

  $\left(\begin{array}{l}n\\ k\end{array}\right)=\frac{n!}{\left(n-k\right)!k!}$

Given

  ${\left(x^2+1\right)}^{18}$

Here $n=18$

Middle term = $r+1$ = $\left(\frac{n}{2}+1\right)$ = $\left(\frac{18}{2}+1\right)=9+1=10$

Substitute $r=9$

We get

  $\begin{array}{l}{a}_{9+1}=\left(\begin{array}{l}18\\ 9\end{array}\right){\left(x^2\right)}^{18-9}{\left(1\right)}^9\\ \Rightarrow a_{10}=\left(\begin{array}{l}18\\ 9\end{array}\right){\left(x^2\right)}^9\\ \left(\begin{array}{l}18\\ 9\end{array}\right)=\frac{18!}{\left(18-9\right)!9!}=\frac{18\times 17\times 16\times \mathrm{....}\times 10}{9!}=48620\\ \Rightarrow a_{10}=48620x^{18}\end{array}$

Therefore

Middle term of expansion of ${\left(x^2+1\right)}^{18}$ is

  $a_{10}=48620x^{18}$