Question

The sum of the series $\sum _{n=1}^{\infty }\frac{n^2+6n+10}{(2n+1)!}$ is equal to :

• A. $\frac{41}{8}e+\frac{19}{8}{e}^{-1}-10$
• B. $-\frac{41}{8}e+\frac{19}{8}{e}^{-1}-10$
• C. $\frac{41}{8}e-\frac{19}{8}{e}^{-1}-10$
• D. $\frac{41}{8}e+\frac{19}{8}{e}^{-1}+10$

Answer 1

The correct option is C $\frac{41}{8}e-\frac{19}{8}{e}^{-1}-10$

$\sum _{n=1}^{\infty }\frac{n^2+6n+10}{(2n+1)!}$
Put $2n+1=r,$ where $r=3,5,\mathrm{7....}$
$\Rightarrow n=\frac{r-1}{2}\therefore \frac{n^2+6n+10}{(2n+1)!}=\frac{{\left(\frac{r-1}{2}\right)}^2+3r-3+10}{r!}=\frac{r^2+10r+29}{4(r!)}$
Now
$\sum _{r=3,5,\mathrm{7...}}\frac{r(r-1)+11r+29}{4(r!)}=\frac{1}{4}\sum _{r=3,5,\mathrm{7...}}(\frac{1}{(r-2)!}+\frac{11}{(r-1)!}+\frac{29}{r!})=\frac{1}{4}\{(\frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}....)+11(\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}.....)+29(\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+....)\}=\frac{1}{4}\{\frac{e-\frac{1}{e}}{2}+11\left(\frac{e+\frac{1}{e}-2}{2}\right)+29\left(\frac{e-\frac{1}{e}-2}{2}\right)\}=\frac{1}{8}\{e-\frac{1}{e}+11e+\frac{11}{e}-22+29e-\frac{29}{e}-58\}=\frac{1}{8}\{41e-\frac{19}{e}-80\}$
$\frac{41}{8}e-\frac{19}{8}{e}^{-1}-10$