Question

Let \(\sum_{n=0}^{\infty }\frac{n^{3}((2n)!+(2n-1)(n!))}{(n!)((2n)!)}=ae+\frac{b}{e}+c\), where $a, b, c \in Z$ and $e=\displaystyle\sum_{n=0}^{\infty} \frac{1}{n !}$ Then $a^2-b+c$ is equal to _______

Answer 1

Correct Answer: 26

The correct answer is 26.
$\sum _{n=0}^{\infty }\frac{n^3((2n)!)+(2n-1)(n!)}{(n!)((2n)!)}$
$=\sum _{n=0}^{\infty }\frac{1}{(n-3)!}+\sum _{n=0}^{\infty }\frac{3}{(n-2)!}+\sum _{n=0}^{\infty }\frac{1}{(n-1)!}+\sum _{n=0}^{\infty }\frac{1}{(2n-1)!}-\sum _{n=0}^{\infty }\frac{1}{(2n)!}$
$=e+3e+e+\frac{1}{2}\left(e-\frac{1}{e}\right)-\frac{1}{2}\left(e+\frac{1}{e}\right)$
$=5e-\frac{1}{e}$
$\therefore a^2-b+c=26$

Answer 2

The correct answer is:
\(\sum \left(\frac{n^3}{n!} + \frac{2n - 1}{(2n)!} \right) = \sum \frac{n^2}{(n - 1)! }+ \sum\frac{1}{(2n - 1)!} - \sum\frac{1}{(2n)!}\rightarrow(1)\)
Consider:
\(∑ \frac{n^2 }{(n - 1)!} = ∑\frac{(n - 1)(n + 1)+1}{(n - 1)! }\)
\(= ∑ \frac{1}{(n - 3)!} + 3∑\frac{1}{(n - 2)!} + ∑\frac{1}{(n - 1)!} = e + 3e + e = 5e\)
\(e+\frac{1}{e}=(1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+.....)+(1-\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+.....)\)
\(e+\frac{1}{e}=2(1+\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+.....)=2\sum\frac{1}{(2n)!}\)
\(e-\frac{1}{e}=2(\frac{1}{1!}+\frac{1}{3!}+\frac{1}{5!}+.....)=2\sum\frac{1}{(2n-1)!}\)
Using in (1):
\(⇒5e+\frac{e-\frac{1}{e}}{2}-\frac{e+\frac{1}{e}}{2}\)
\(5e-\frac{1}{2e}-\frac{1}{2e}=5e-\frac{1}{e}\)
\(=ae+\frac{b}{e}+c\)
\(a^2-b+c=25+1=26\)