Question

The sum of the series $\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+...$ to $\infty$ = $\frac{a}{e}$. Find $(a+3{)}^2$.

Answer 1

$S=\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+...$

$t_n=\frac{2n}{(2n+1)!}=\frac{1}{2n!}-\frac{1}{(2n+1)!}$

We know, $\frac{e^x+e^{-x}}{2}$= $\sum _{n=0}^{\infty }\frac{x^{2n}}{\left(2n\right)!}$ and $\frac{e^x+e^{-x}}{2}$= $\sum _{n=0}^{\infty }\frac{x^{2n}}{\left(2n\right)!}$

$\therefore S=\sum _{n=0}^{\infty }{t}_n=\sum _{r=0}^{\infty }\frac{1}{2n!}-\sum _{r=0}^{\infty }\frac{1}{(2n+1)!}=\frac{1}{2}(e+\frac{1}{e})-\frac{1}{2}(e-\frac{1}{e})=\frac{1}{e}$

Therefore, $(a+3{)}^2=16$.... a=1
Ans: 16