Chapter 12.6, Problem 28E

To determine

To find: Expansion of ${\left(2A+B^2\right)}^4$ using Binomial Theorem.

Answer to Problem 28E

  ${\left(2A+B^2\right)}^4=16A^4+32A^3{B}^2+24A^2{B}^4+8A{B}^6+B^8$

Explanation of Solution

Given information:

  ${\left(2A+B^2\right)}^4$

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 $k=0,1,\mathrm{2........}n-1,n$

We know that,

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

  $\begin{array}{l}\Rightarrow \left(\begin{array}{l}4\\ 0\end{array}\right)=\frac{4!}{\left(4-0\right)!0!}=\frac{4!}{\left(4\right)!}=1\\ \Rightarrow \left(\begin{array}{l}4\\ 1\end{array}\right)=\frac{4!}{\left(4-1\right)!1!}=\frac{4!}{\left(3\right)!}=4\\ \Rightarrow \left(\begin{array}{l}4\\ 2\end{array}\right)=\frac{4!}{\left(4-2\right)!2!}=\frac{4!}{\left(2\right)!2!}=\frac{24}{4}=6\\ \Rightarrow \left(\begin{array}{l}4\\ 3\end{array}\right)=\frac{4!}{\left(4-3\right)!3!}=\frac{4!}{\left(1\right)!3!}=\frac{4!}{\left(3\right)!}=4\\ \Rightarrow \left(\begin{array}{l}4\\ 4\end{array}\right)=\frac{4!}{\left(4-4\right)!4!}=\frac{4!}{\left(0\right)!4!}=\frac{4!}{4!}=1\end{array}$

So, therefore

  $\begin{array}{l}{\left(2A+B^2\right)}^4=\left(\begin{array}{l}4\\ 0\end{array}\right){\left(2A\right)}^4{\left(B^2\right)}^0+\left(\begin{array}{l}4\\ 1\end{array}\right){\left(2A\right)}^{4-1}{\left(B^2\right)}^1+\left(\begin{array}{l}4\\ 2\end{array}\right){\left(2A\right)}^{4-2}{\left(B^2\right)}^2+\left(\begin{array}{l}4\\ 3\end{array}\right){\left(2A\right)}^{4-3}{\left(B^2\right)}^3+\left(\begin{array}{l}4\\ 4\end{array}\right){\left(2A\right)}^{4-4}{\left(B^2\right)}^4\\ \Rightarrow {\left(2A+B^2\right)}^4=\left(1\right)16A^4+\left(4\right)8A^3\left(B^2\right)+\left(6\right)4A^2\left(B^4\right)+\left(4\right)2A\left(B^6\right)+\left(1\right)\left(B^8\right)\\ \Rightarrow {\left(2A+B^2\right)}^4=16A^4+32A^3{B}^2+24A^2{B}^4+8A{B}^6+B^8\end{array}$