JEE Main 2021 (Online) 26th February Evening ShiftQuestion 1
The sum of the series $\sum_{n=1}^{\infty}\frac{n^{2}+6n+10}{(2n+1)!}$ is equal to :
Sequence and Series · Logarithmic and Exponential Series, and Recursive Sequences
Solution
Formulas used. (1) $\sum_{j\ \text{odd}}\frac1{j!}=\sinh1=\frac{e-e^{-1}}{2}$, $\ \sum_{j\ \text{even}}\frac1{j!}=\cosh1=\frac{e+e^{-1}}{2}$(2) $\frac{m(m-1)}{m!}=\frac1{(m-2)!}$, $\ \frac{m}{m!}=\frac1{(m-1)!}$Step 1 - Substitute $m=2n+1$. $n=\tfrac{m-1}{2}$: $$n^2+6n+10=\frac{m^2+10m+29}{4}=\frac{m(m-1)+11m+29}{4}.$$ Step 2 - Split (sum over odd $m\ge3$): $$S=\frac14\left[\sum\frac1{(m-2)!}+11\sum\frac1{(m-1)!}+29\sum\frac1{m!}\right].$$ Step 3 - Each part. - $m-2=1,3,5,\dots$: $\sinh1$ - $m-1=2,4,6,\dots$: $\cosh1-1$ - $m=3,5,7,\dots$: $\sinh1-1$ Step 4 - Add. $$S=\frac14\left[30\sinh1+11\cosh1-40\right]=\frac14\left[15\left(e-e^{-1}\right)+\frac{11}{2}\left(e+e^{-1}\right)-40\right]=\frac{41}{8}e-\frac{19}{8}e^{-1}-10 .$$ Answer: (D)