Question

If $x=1+2+\frac{4}{2!}+\frac{8}{2!}+\frac{16}{2!}+.......$, then $x^{-1}$ is equal to

• A. $e^{-2}$
• B. $e^2$
• C. None of these
• D. $e^{\frac{1}{2}}$

Answer 1

The correct option is A

$e^{-2}$

Explanation for the correct option.

Find the value of $x^{-1}$:

Given: $x=1+2+\frac{4}{2!}+\frac{8}{2!}+\frac{16}{2!}+.......$

$\begin{array}{rcl}x& =& 1+2+\frac{4}{2!}+\frac{8}{2!}+\frac{16}{2!}+.......\\ x& =& 1+\frac{2}{1!}+\frac{2^2}{2!}+\frac{2^3}{2!}+\frac{2^4}{2!}+.......\\ x& =& e^2\mathbf{}\mathbf{}\left[\because e^2=1+\frac{2}{1!}+\frac{2^2}{2!}+\frac{2^3}{2!}+\frac{2^4}{2!}+.......\right]\mathbf{}\mathbf{}\mathbf{}\mathbf{}\end{array}$

Then reciprocal of $x$is:

$x^{-1}=e^{-2}$

Hence, option A is correct answer.