Question

The expression ${\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+...\right)}^2$ will be represented in ascending power of $x$ as

• A. $1+\frac{2^2{x}^2}{2!}+\frac{2^4{x}^4}{4!}+.....$
• B. $1+\frac{{\left(2x\right)}^2}{2!}+\frac{2^2{x}^4}{4!}+....$
• C. $1+\frac{{\left(2x\right)}^2}{2.2!}+\frac{2x^4}{4!}+...$
• D. $1+\frac{{\left(2x\right)}^2}{2.2!}+\frac{{\left(2x\right)}^4}{2.4!}+....$

Answer 1

The correct option is D

$1+\frac{{\left(2x\right)}^2}{2.2!}+\frac{{\left(2x\right)}^4}{2.4!}+....$

Represent the expression as required:

As we know that, $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$

$e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+...$

On adding both the equations,

$\begin{array}{rcl}\left(e^x+e^{-x}\right)& =& \left(1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...\right)+\left(1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+...\right)\\ \left(e^x+e^{-x}\right)& =& 2\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+...\right)\\ \left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+...\right)& =& \left(\frac{e^x+e^{-x}}{2}\right)\\ {\left(1+\frac{x^2}{2!}+\frac{x^4}{4!}+...\right)}^2& =& {\left(\frac{e^x+e^{-x}}{2}\right)}^2\\ & =& \left(\frac{1}{4}\right)\left(e^{2x}+e^{-2x}+2e^x{e}^{-x}\right)\\ & =& \left(\frac{1}{4}\right)\left(1+\frac{\left(2x\right)}{1!}+\frac{{\left(2x\right)}^2}{2!}+\frac{{\left(2x\right)}^3}{3!}+...+1-\frac{\left(2x\right)}{1!}+\frac{{\left(2x\right)}^2}{2!}-\frac{{\left(2x\right)}^3}{3!}+...+2\right)\\ & =& \left(\frac{1}{4}\right)\left(2\left(1+\frac{{\left(2x\right)}^2}{2!}+\frac{{\left(2x\right)}^4}{4!}+...\right)+2\right)\\ & =& \left(\frac{1}{4}\right)\left[4+2\left\{\frac{{\left(2x\right)}^2}{2!}+\frac{{\left(2x\right)}^4}{4!}+....\right\}\right]\\ & =& 1+\frac{{\left(2x\right)}^2}{2.2!}+\frac{{\left(2x\right)}^4}{2.4!}+....\end{array}$

Hence, the given equation can be represented as $1+\frac{{\left(2x\right)}^2}{2.2!}+\frac{{\left(2x\right)}^4}{2.4!}+....$ and therefore option $\left(D\right)$ is correct.