Chapter 12.6, Problem 41E

To determine

To find: The term containing $b^8$ in the expansion of ${\left(a+b^2\right)}^{12}$

Answer to Problem 41E

Term containing $b^8$ is $a_5=495a^8{b}^8$

Explanation of Solution

Given information:

  ${\left(a+b^2\right)}^{12}$

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

We know that,

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

Given

  ${\left(a+b^2\right)}^{12}$

Here $n=12$

We need to find the term containing $b^8$

Therefore, from the formula $a_{r+1}=\left(\begin{array}{l}n\\ r\end{array}\right)x^{n-r}{y}^r$

The ${\left(r+1\right)}^{th}$ term is $a_{r+1}=\left(\begin{array}{l}12\\ r\end{array}\right){\left(a\right)}^{12-r}{\left(b^2\right)}^r$

The exponent of $b$ is $8$

  $b^{2r}=b^8$

  $\Rightarrow 2r=8$

  $\Rightarrow r=4$

Substitute the $r=4$

We get

  $\begin{array}{l}{a}_{4+1}=\left(\begin{array}{l}12\\ 4\end{array}\right){\left(a\right)}^{12-4}{\left(b^2\right)}^4\\ \Rightarrow a_5=\left(\begin{array}{l}12\\ 4\end{array}\right)a^8{b}^8\\ \left(\begin{array}{l}12\\ 4\end{array}\right)=\frac{12!}{\left(12-4\right)!4!}=\frac{12!}{8!4!}=\frac{12\times 11\times 10\times 9}{4\times 3\times 2\times 1}=495\\ \Rightarrow a_5=495a^8{b}^8\end{array}$

Therefore,

The term containing $b^8$ in the expansion of ${\left(a+b^2\right)}^{12}$ is

  $a_5=495a^8{b}^8$