Zero solvers challenge 2

Define $g(x) = \frac{2}{e}f(x+1) = \frac{2}{e}\frac{e^{x+1}}{(x+1)^3} = \frac{2}{(x+1)^3}e^x$ and note that $g^{(100)}(0) = \frac{2}{e}f^{(100)}(1) = \frac{2}{e}\cdot \frac{ke}{2} = k.$

Then, using $\frac{2}{(x+1)^3} = \sum_{n=0}^\infty (-1)^n (n+2)(n+1)x^n \\ e^x = \sum_{n=0}^\infty \frac{1}{n!}x^n,$ we may write $\begin{aligned} g(x) &= \left(\sum_{n=0}^\infty (-1)^n (n+2)(n+1)x^n\right)\cdot \left(\sum_{n=0}^\infty \frac{1}{n!}x^n\right) \\ &= \sum_{n=0}^\infty \left(\sum_{r=0}^n (-1)^r (r+2)(r+1) \cdot \frac{1}{(n-r)!}\right)x^n \\ &= \sum_{n=0}^\infty \frac{1}{n!} \left(\sum_{r=0}^n \frac{n!}{r!(n-r)!} (-1)^r (r+2)(r+1)r!\right)x^n \\ &= \sum_{n=0}^\infty \frac{1}{n!} \left(\sum_{r=0}^n \binom{n}{r} (-1)^r (r+2)!\right)x^n \end{aligned}$

Therefore, $k = g^{(100)}(0) = \sum_{r=0}^{100} \binom{100}{r} (-1)^r (r+2)!$