Question

The value of $\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+........$ is

• A. $e^{\frac{1}{2}}$
• B. $e^{-1}$
• C. $e$
• D. $e^{\frac{1}{3}}$

Answer 1

Answer: (B)

Given, $\frac{2}{3 !}+\frac{4}{5 !}+\frac{6}{7 !}+\ldots \infty$
The $n$th term of the above series
$T_{n}=\frac{2 n}{(2 n+1) !}=\frac{(2 n+1)-1}{(2 n+1) !}=\frac{2 n+1}{(2 n+1) !}-\frac{1}{(2 n+1) !}$
$T_{n}=\frac{1}{(2 n) !}-\frac{1}{(2 n+1) !}$
$\therefore T_{1}=\frac{1}{2 !}-\frac{1}{3 !}$
$T_{2}=\frac{1}{4 !}-\frac{1}{5 !}$
$T_{3}=\frac{1}{6 !}-\frac{1}{7 !}$
$\begin{matrix}.&.&.\\ .&.&.\\ .&.&.\end{matrix}$
Hence, $T_{n}=\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\ldots \infty=e^{-1}$