The sum [_n = 1^ 2n^2 + 3n + 4 (2n)! ] is equal to :

Question

The sum $$\sum\limits_{n = 1}^\infty  {{{2{n^2} + 3n + 4} \over {(2n)!}}} $$ is equal to :

Accepted Answer

$\begin{aligned} & \sum_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !} \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{2 n(2 n-1)+8 n+8}{(2 n) !} \\ &= \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{(2 n-2) !}+2 \sum_{n=1}^{\infty} \frac{1}{(2 n-1) !}+4 \sum_{n=1}^{\infty} \frac{1}{(2 n) !} \\ & e=1+1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots . \\ & e^{-1}=1-1+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}+\ldots . \\ & \left(\mathrm{e}+\frac{1}{\mathrm{e}}\right)=2\left(1+\frac{1}{2 !}+\frac{1}{4 !}+\ldots . .\right) \\ & e-\frac{1}{e}=\left(1+\frac{1}{3 !}+\frac{1}{5 !}+\ldots . .\right)\end{aligned}$

Now

$$
\begin{aligned}
& \frac{1}{2}\left(\sum_{n=1}^{\infty} \frac{1}{(2 n-2) !}\right)+2 \sum_{n=1}^{\infty} \frac{1}{(2 n-1) !}+4 \sum_{n=1}^{\infty} \frac{1}{(2 n) !} \\
& =\frac{1}{2}\left[\frac{e+\frac{1}{\mathrm{e}}}{2}\right]+2\left[\frac{\mathrm{e}-\frac{1}{\mathrm{e}}}{2}\right]+4\left[\frac{\mathrm{e}+\frac{1}{\mathrm{e}}-2}{2}\right] \\
& =\frac{\left(\mathrm{e}+\frac{1}{\mathrm{e}}\right)}{4}+e-\frac{1}{\mathrm{e}}+2 \mathrm{e}+\frac{2}{\mathrm{e}}-4 \\
& =\frac{13}{4} e+\frac{5}{4 e}-4
\end{aligned}
$$

Solutions · 1