Question

The sum of the series $1+\frac{1}{1!}.\frac{1}{4}+\frac{1.3}{2!}.{\left(\frac{1}{4}\right)}^2+\frac{\mathrm{1.3.5}}{3!}.{\left(\frac{1}{4}\right)}^3+....$ to $\infty$ is

• A. $\sqrt{2}$
• B. $2$
• C. $\frac{1}{\sqrt{2}}$
• D. none of these

Answer 1

The correct option is A $\sqrt{2}$

Considering an infinite expansion of $(1+x{)}^n$
We get

$1+nx+\frac{n(n-1)}{2!}{x}^2...\infty$
Comparing coefficients, we get
$nx=\frac{1}{4}$
$\Rightarrow \frac{n(n-1)}{2}{x}^2=\frac{3}{2!.16}$
$\Rightarrow nx(nx-x)=\frac{3}{16}$
$\Rightarrow \frac{1}{4}-x=\frac{3}{4}...(nx=\frac{1}{4})$
$\Rightarrow x=\frac{-1}{2}$
Therefore $n=\frac{-1}{2}$
Hence sum of the above series will be
$(1-\frac{1}{2}{)}^{-\frac{1}{2}}$
$=\sqrt{2}$