Question

$\frac{1}{3!}+\frac{2}{5!}+\frac{3}{7!}+...$ is equal to

• A. $\frac{{{e}^{-1}}}{2}$
• B. $e$
• C. $\frac{e}{4}$
• D. $\frac{e}{6}$

Answer 1

Answer: (A)

$\frac{{{e}^{-1}}}{2}$

Answer 2

Let $S=\frac{1}{3!}+\frac{2}{5!}+\frac{3}{7!}+.....$
$\therefore$ ${{T}_{n}}=\frac{n}{(2n+1)!}$
$=\frac{1}{2}\left[ \frac{2n+1-1}{(2n+1)!} \right]=\frac{1}{2}\left[ \frac{1}{(2n)!}-\frac{1}{(2n+1)!} \right]$
$\therefore$ ${{T}_{1}}=\frac{1}{2}\left( \frac{1}{2!}-\frac{1}{3!} \right)$
${{T}_{2}}=\frac{1}{2}\left( \frac{1}{4!}-\frac{1}{5!} \right)$
$\therefore$ $S={{T}_{1}}+{{T}_{2}}+....$
$=\frac{1}{2}\left[ \frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+.....\infty +1-1 \right]$
$=\frac{{{e}^{-1}}}{2}$