1. Tardigrade
  2. Question
  3. Mathematics
  4. The sum displaystyle∑n=1∞ (2 n2+3 n+4/(2 n) !) is equal to:

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. The sum $\displaystyle\sum_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !}$ is equal to:

JEE MainJEE Main 2023Integrals

Solution:

$\displaystyle\sum_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !} $
$ \frac{1}{2} \displaystyle\sum_{n=1}^{\infty} \frac{2 n(2 n-1)+8 n+8}{(2 n) !} $
$ \frac{1}{2} \displaystyle\sum_{n=1}^{\infty} \frac{1}{(2 n-2) !}+2 \displaystyle\sum_{n=1}^{\infty} \frac{1}{(2 n-1) !}+4 \displaystyle\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( e +\frac{1}{ e }\right)=2\left(1+\frac{1}{2 !}+\frac{1}{4 !}+\ldots \ldots .\right)$
$ e -\frac{1}{ e }=\left(1+\frac{1}{3 !}+\frac{1}{5 !}+\ldots . .\right)$
Now
$ \frac{1}{2}\left(\displaystyle\sum_{n=1}^{\infty} \frac{1}{(2 n-2) !}\right)+2 \displaystyle\sum_{n=1}^{\infty} \frac{1}{(2 n-1) !}+4 \displaystyle\sum_{n=1}^{\infty} \frac{1}{(2 n) !} $
$ =\frac{1}{2}\left[\frac{e+\frac{1}{e}}{2}\right]+2\left[\frac{e-\frac{1}{e}}{2}\right]+4\left(\frac{e+\frac{1}{e}-2}{2}\right) $
$ =\frac{\left(e+\frac{1}{e}\right)}{4}+e-\frac{1}{e}+2 e+\frac{2}{e}-4$
$ =\frac{13}{4} e+\frac{5}{4 e}-4$