The sum ∑ n = 1 ∞ 2 n 2 + 3 n + 4 ( 2 n ) ! is equal to :

Given, $\sum _{n=1}^{\infty }\frac{2n^2+3n+4}{\left(2n\right)!}$

$\frac{1}{2}\sum _{n=1}^{\infty }\frac{2\mathrm{n}(2\mathrm{n}-1)+8\mathrm{n}+8}{\left(2\mathrm{n}\right)!}$

$\Rightarrow \frac{1}{2}\sum _{n=1}^{\infty }\frac{1}{(2n-2)!}+2\sum _{n=1}^{\infty }\frac{1}{(2n-1)!}+4\sum _{n=1}^{\infty }\frac{1}{\left(2n\right)!}....\left(1\right)$

We know the expansion of exponential functions as,

${\mathrm{e}}^{\mathrm{x}}=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots \dots \infty$

$\mathrm{e}=1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\dots \dots \infty =\sum _{r=0}^{\infty }\frac{1}{r!}$

${\mathrm{e}}^{-1}=1-1+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}+\dots \dots \infty =\sum _{r=0}^{\infty }\frac{{\left(-1\right)}^r}{r!}$

$\left(\mathrm{e}+\frac{1}{\mathrm{e}}\right)=2\left(1+\frac{1}{2!}+\frac{1}{4!}+\dots \dots .\right)=2\left(\frac{1}{\left(\text{even}\right)!}\right)$

$\mathrm{e}-\frac{1}{\mathrm{e}}=\left(1+\frac{1}{3!}+\frac{1}{5!}+\dots ..\right)=2\left(\frac{1}{\left(\text{odd}\right)!}\right)$

Now from equation $\left(1\right)$,

$\frac{1}{2}\left(\sum _{n=1}^{\infty }\frac{1}{(2n-2)!}\right)+2\sum _{n=1}^{\infty }\frac{1}{(2n-1)!}+4\sum _{n=1}^{\infty }\frac{1}{\left(2n\right)!}$

$=\frac{1}{2}\left[\frac{\mathrm{e}+\frac{1}{\mathrm{e}}}{2}\right]+2\left[\frac{\mathrm{e}-\frac{1}{\mathrm{e}}}{2}\right]+4\left(\frac{\mathrm{e}+\frac{1}{\mathrm{e}}-2}{2}\right)$

$=\frac{\left(\mathrm{e}+\frac{1}{\mathrm{e}}\right)}{4}+\mathrm{e}-\frac{1}{\mathrm{e}}+2\mathrm{e}+\frac{2}{\mathrm{e}}-4$

$=\frac{13}{4}e+\frac{5}{4e}-4$