Sequences And Series Question 10
Question 10 - 30 January - Shift 1
Let $\quad \sum _{n=0}^{\infty} \frac{n^{3}((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}=a e+\frac{b}{e}+c$,where $a, b, c \in \mathbb{Z}$ and $e=\sum _{n=0}^{\infty} \frac{1}{n !}$ Then $a^{2}-b+c$ is equal to _____
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Answer: 26
Solution:
Formula: Arithmetico-Geometrical Progression (A.G.P.) (ii)
$\sum _{n=0}^{\infty} \frac{n^{3}((2 n) !)+(2 n-1)(n !)}{(n !)((2 n) !)}$
$=\sum _{n=0}^{\infty} \frac{1}{(n-3) !}+\sum _{n=0}^{\infty} \frac{3}{(n-2) !}$
$+\sum _{n=0}^{\infty} \frac{1}{(n-1) !}+\sum _{n=0}^{\infty} \frac{1}{(2 n-1) !}-\sum _{n=0}^{\infty} \frac{1}{(2 n) !}$
$=e+3 e+e+\frac{1}{2}(e-\frac{1}{e})-\frac{1}{2}(e+\frac{1}{e})$
$=5 e-\frac{1}{e}$
$a^{2}-b+c=26$
