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

The correct answer is 26.
n=0∑∞​(n!)((2n)!)n3((2n)!)+(2n−1)(n!)​
=n=0∑∞​(n−3)!1​+n=0∑∞​(n−2)!3​+n=0∑∞​(n−1)!1​+n=0∑∞​(2n−1)!1​−n=0∑∞​(2n)!1​
=e+3e+e+21​(e−e1​)−21​(e+e1​)
=5e−e1​
∴a2−b+c=26