$\sum_{k=1}^{\infty} \frac{1}{k !} \left(\sum_{n=1}^k 2^{n-1}\right)$ is equal to

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

The value of the infinite series $\frac{1^{2}+2^{2}}{3 !} + \frac{1^{2}+2^{2}+3^{2}}{4 !} + \frac{1^{2}+2^{2}+3^{2}+4^{2}}{5 !} + \dots$ is:

$1 + \frac{2^3}{2!} + \frac{3^3}{3!} + \frac{4^3}{4!} + \dots \infty =$ (in $e$)

The sum of the series $\frac{1}{1 \times 2} + \frac{1 \times 3}{1 \times 2 \times 3 \times 4} + \frac{1 \times 3 \times 5}{1 \times 2 \times 3 \times 4 \times 5 \times 6} + \dots \infty$ is

The money invested in a company is compounded continuously. ₹ $400$ invested today becomes ₹ $800$ in $6$ years. What will it become at the end of $33$ years? (Given $\sqrt{2} \approx 1.4142$)