MCQ
The sum $\sum \limits_{n=1}^{\infty} \frac{2 n^2+3 n+4}{(2 n) !}$ is equal to :
- A$\frac{11 e }{2}+\frac{7}{2 e }$
- ✓$\frac{13 e }{4}+\frac{5}{4 e }-4$
- C$\frac{11 e }{2}+\frac{7}{2 e }-4$
- D$\frac{13 e }{4}+\frac{5}{4 e }$
$\frac{1}{2} \sum \limits_{ n =1}^{\infty} \frac{2 n (2 n -1)+8 n +8}{(2 n ) !}$
$\frac{1}{2} \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n -2) !}+2 \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n -1) !}+4 \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n ) !}$
$e =1+1+\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}+\ldots \ldots$
$e ^{-1}=1-1+\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}+\ldots \ldots$
$\left( e +\frac{1}{ e }\right)=2\left(1+\frac{1}{2 !}+\frac{1}{4 !}+\ldots \ldots .\right)$
$e -\frac{1}{ e }=\left(1+\frac{1}{3 !}+\frac{1}{5 !}+\ldots . .\right)$
$\text { Now }$
$\frac{1}{2}\left(\sum \limits_{ n =1}^{\infty} \frac{1}{(2 n -2) !}\right)+2 \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n -1) !}+4 \sum \limits_{ n =1}^{\infty} \frac{1}{(2 n ) !}$
$=\frac{1}{2}\left[\frac{ e +\frac{1}{ e }}{2}\right]+2\left[\frac{ e -\frac{1}{ e }}{2}\right]+4\left(\frac{ e +\frac{1}{ e }-2}{2}\right)$
$=\frac{\left( e +\frac{1}{ e }\right)}{4}+ e -\frac{1}{ e }+2 e +\frac{2}{ e }-4$
$=\frac{13}{4} e +\frac{5}{4 e }-4$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.