Chapter 10, Problem 14CR

To determine

To calculate:

The binomial expansion of given expression.

Answer to Problem 14CR

The binomial expansion of given expression of ${\left(2x+y^2\right)}^4$ is $=16x^4+32x^3{y}^2+24x^2{y}^4+8x{y}^6+y^8$ .

Explanation of Solution

Given information:

Expression:

  ${\left(2x+y^2\right)}^4$

Formula used:

Binomial theorem:

  ${\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i$

Where,

  $a=$ first term

  $b=$ second term

Calculation:

Consider expression ${\left(2x+y^2\right)}^4$

Apply the binomial theorem for ${\left(2x+y^2\right)}^4$

  ${\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i$

Where,

  $a=2x,b=y^2$

So, the simplification are as follows,

  $\begin{array}{l}{\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\begin{array}{c}n\\ i\end{array}\right)}{a}^{\left(n-i\right)}{b}^i\\ {\left(2x+y^2\right)}^4={\displaystyle {\sum }_{i=0}^4\left(\begin{array}{c}4\\ i\end{array}\right)}{\left(2x\right)}^{\left(4-i\right)}{\left(y^2\right)}^i\end{array}$

The expand summation are as follows:

  $=\frac{4!}{0!\left(4-0\right)!}{\left(2x\right)}^4{\left(y^2\right)}^0+\frac{4!}{1!\left(4-1\right)!}{\left(2x\right)}^3{\left(y^2\right)}^1+\frac{4!}{2!\left(4-2\right)!}{\left(2x\right)}^2{\left(y^2\right)}^2+\frac{4!}{!\left(4-3\right)!}{\left(2x\right)}^1{\left(y^2\right)}^3+\frac{4!}{4!\left(4-0\right)!}{\left(2x\right)}^0{\left(y^2\right)}^4$

By simplifying each summation, the expansion is as follows,

  $=16x^4+32x^3{y}^2+24x^2{y}^4+8x{y}^6+y^8$