∑ n = 0 ∞ n 3 ( ( 2 n ) ! ) + ( 2 n - 1 ) ( n ! ) ( n ! ) ( ( 2 n ) ! ) = a e + b e + c where a …

Given:

$\sum _{n=0}^{\infty }\left[\frac{n^3\left\{\left(2n\right)!\right\}+\left(2n-1\right)n!}{n!\times \left(2n\right)!}\right]$

$=\sum _{n=0}^{\infty }\left[\frac{n^3\left\{\left(2n\right)!\right\}}{n!\times \left(2n\right)!}\right]+\sum _{n=0}^{\infty }\left[\frac{\left(2n-1\right)n!}{n!\times \left(2n\right)!}\right]$

$=\sum _{n=0}^{\infty }\left[\frac{n^3}{n!}\right]+\sum _{n=0}^{\infty }\left[\frac{\left(2n-1\right)}{\left(2n\right)!}\right]$

$=\sum _{n=0}^{\infty }\left[\frac{n\left(n-1\right)\left(n-2\right)+3n\left(n-1\right)+n}{n!}\right]+\sum _{n=1}^{\infty }\left[\frac{1}{\left(2n-1\right)!}\right]-\sum _{n=0}^{\infty }\left[\frac{1}{\left(2n\right)!}\right]$

$=\sum _{n=0}^{\infty }\left[\frac{n\left(n-1\right)\left(n-2\right)}{n!}\right]+\sum _{n=0}^{\infty }\left[\frac{3n\left(n-1\right)}{n!}\right]+\sum _{n=0}^{\infty }\left[\frac{n}{n!}\right]+\left(\frac{e-e^{-1}}{2}\right)-\left(\frac{e+e^{-1}}{2}\right)$

$=\sum _{n=3}^{\infty }\left[\frac{1}{\left(n-3\right)!}\right]+3\sum _{n=2}^{\infty }\left[\frac{1}{\left(n-2\right)!}\right]+\sum _{n=1}^{\infty }\left[\frac{1}{\left(n-1\right)!}\right]-\frac{1}{e}$

$=5e-\frac{1}{e}$

So, $a=5, b=-1, c=0$

Hence, $a^2-b+c=25+1=26$