2020.11.2

\(\mathcal{A}\)

求\(\begin{aligned}\sum_{i=0}^n\begin{Bmatrix} n \\ i \end{Bmatrix}i!\end{aligned}\)，\(\begin{aligned}\sum_{i=0}^n\begin{Bmatrix} n \\ i \end{Bmatrix}i!i\end{aligned}\)

\(\begin{aligned}f(x)&=\sum_{n \geq 0}\frac{x^n}{n!}\sum_{i=0}^n\begin{Bmatrix} n \\ i \end{Bmatrix}i!\\&=\sum_{i\geq0}\sum_{n\geq0}\frac{x^n}{n!}\begin{Bmatrix} n \\ i \end{Bmatrix}i!\\&=\sum_{i\geq0}\sum_{n\geq0}\frac{x^n}{n!}\sum_{k\geq0}(-1)^k\binom{i}{k}(i-k)^n\\&=\sum_{i\geq0}\sum_{k=0}^i(-1)^k\binom{i}{i-k}\sum_{n\geq0}\frac{x^n}{n!}(i-k)^n\\&=\sum_{i\geq0}\sum_{k^*=0}^i(-1)^{i-k}\binom{i}{k}\sum_{n\geq0}\frac{(kx)^n}{n!}\\&=\sum_{i\geq0}\sum_{k=0}^i(-1)^{i-k}\binom{i}{k}(e^x)^k\\&=\sum_{i\geq0}(-1+e^x)^i\\&=\frac{1}{2-e^x}\end{aligned}\)

同理\(g(x)=\cfrac{e^x-1}{(2-e^x)^2}\)