JEE PYQ: Sequence And Series Question 16
Question 16 - 2021 (26 Feb Shift 2)
The sum of the series $\sum_{n=1}^{\infty} \frac{n^2+6n+10}{(2n+1)!}$ is equal to:
(1) $\frac{11}{2}e + \frac{11}{2}e^{-1} - 10$
(2) $-\frac{11}{2}e + \frac{11}{2}e^{-1} - 10$
(3) $\frac{11}{2}e - \frac{11}{2}e^{-1} - 10$
(4) $\frac{41}{8}e + \frac{19}{8}e^{-1} + 10$
Show Answer
Answer: (3)
Solution
Put $2n + 1 = r$, where $r = 3, 5, 7, \ldots$
$n = \frac{r-1}{2}$
$\frac{n^2 + 6n + 10}{(2n+1)!} = \frac{\frac{r^2+10r+29}{4}}{r!}$
Sum computed using $e$ series expansions gives $\frac{11}{2}e - \frac{11}{2}e^{-1} - 10$
BETA
