Chapter 12.6, Problem 25E

To determine

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

Answer to Problem 25E

  $x^4+8x^{3}y+24x^2{y}^2+32x{y}^3+16y^4$

Explanation of Solution

Given information:

  ${\left(x+2y\right)}^4$

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

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