The sum of the series sum[n=1 to ∞](n^2+6n+10)/(2n+1)! is equal to:

Question

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

(1) \(\frac{41}{8}e+\frac{19}{8}e^{-1}-10\)

(2) \(\frac{41}{8}e-\frac{19}{8}e^{-1}-10\)

(3) \(\frac{41}{8}e+\frac{19}{8}e^{-1}+10\)

(4) \(-\frac{41}{8}e+\frac{19}{8}e^{-1}-10\)

Answer 1

Correct option is (2) \(\frac{41}{8}e-\frac{19}{8}e^{-1}-10\)

\(T_n=\frac{n^2+6n+10}{(2n+1)!}\)\(=\frac{4n^2+24n+40}{4.(2n+1)!}\)

\(=\frac{(2n+1)^2+20n+39}{4.(2n+1)!}\)