Question

Let $\sum _{n=0}^{\infty }\frac{n^3((2n)!)+(2n-1)(n!)}{(n!)((2n)!)}=ae+\frac{b}{e}+c$, where $a,b,c\in Z$ and $e=\sum _{n=0}^{\infty }\frac{1}{n!}$ Then $a^2-b+c$ is equal to ___

Answer 1

Answer: 26

$\sum _{n=0}^{\infty }\frac{n^3((2n)!)+(2n-1)(n!)}{(n!)((2n)!)}$
$=\sum _{n=0}^{\infty }\frac{1}{(n-3)!}+\sum _{n=0}^{\infty }\frac{3}{(n-2)!}+\sum _{n=0}^{\infty }\frac{1}{(n-1)!}+\sum _{n=0}^{\infty }\frac{1}{(2n-1)!}-\sum _{n=0}^{\infty }\frac{1}{(2n)!}$
$=e+3e+e+\frac{1}{2}\left(e-\frac{1}{e}\right)-\frac{1}{2}\left(e+\frac{1}{e}\right)$
$=5e-\frac{1}{e}$
$\therefore a^2-b+c=26$