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

Given, $\mathop{∑}_{n = 1}^{\infty}\dfrac{2{n}^{2} + 3n + 4}{( 2n ) !}$

$\dfrac{1}{2}\mathop{∑}_{n = 1}^{\infty}\dfrac{2n ( 2n - 1 ) + 8n + 8}{( 2n ) !}$

$⇒ \dfrac{1}{2}\mathop{∑}_{n = 1}^{\infty}\dfrac{1}{( 2n - 2 ) !} + 2\mathop{∑}_{n = 1}^{\infty}\dfrac{1}{( 2n - 1 ) !} + 4\mathop{∑}_{n = 1}^{\infty}\dfrac{1}{( 2n ) !} . . . . ( 1 )$

We know the expansion of exponential functions as,

${e}^{x} = 1 + \dfrac{x}{1 !} + \dfrac{{x}^{2}}{2 !} + \dfrac{{x}^{3}}{3 !} + \dfrac{{x}^{4}}{4 !} + \ldots \ldots \infty$

$e = 1 + 1 + \dfrac{1}{2 !} + \dfrac{1}{3 !} + \dfrac{1}{4 !} + \ldots \ldots \infty = \mathop{∑}_{r = 0}^{\infty}\dfrac{1}{r !}$

${e}^{- 1} = 1 - 1 + \dfrac{1}{2 !} - \dfrac{1}{3 !} + \dfrac{1}{4 !} + \ldots \ldots \infty = \mathop{∑}_{r = 0}^{\infty}\dfrac{{\left(- 1\right)}^{r}}{r !}$

\left(e + \dfrac{1}{e}\right) = 2\left(1 + \dfrac{1}{2 !} + \dfrac{1}{4 !} + \ldots \ldots .\right) = 2\left(\dfrac{1}{\left(\text{even}\right) !}\right)$</p><p>$e - \dfrac{1}{e} = \left(1 + \dfrac{1}{3 !} + \dfrac{1}{5 !} + \ldots . .\right) = 2\left(\dfrac{1}{\left(\text{odd}\right) !}\right)$</p><p>Now from equation \left(1\right)$,

$\dfrac{1}{2}\left(\mathop{∑}_{n = 1}^{\infty}\dfrac{1}{( 2n - 2 ) !}\right) + 2\mathop{∑}_{n = 1}^{\infty}\dfrac{1}{( 2n - 1 ) !} + 4\mathop{∑}_{n = 1}^{\infty}\dfrac{1}{( 2n ) !}$

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

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

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