2020.11.2

(mathcal{A})

(egin{aligned}sum_{i=0}^negin{Bmatrix} n \ i end{Bmatrix}i!end{aligned})(egin{aligned}sum_{i=0}^negin{Bmatrix} n \ i end{Bmatrix}i!iend{aligned})

(egin{aligned}f(x)&=sum_{n geq 0}frac{x^n}{n!}sum_{i=0}^negin{Bmatrix} n \ i end{Bmatrix}i!\&=sum_{igeq0}sum_{ngeq0}frac{x^n}{n!}egin{Bmatrix} n \ i end{Bmatrix}i!\&=sum_{igeq0}sum_{ngeq0}frac{x^n}{n!}sum_{kgeq0}(-1)^kinom{i}{k}(i-k)^n\&=sum_{igeq0}sum_{k=0}^i(-1)^kinom{i}{i-k}sum_{ngeq0}frac{x^n}{n!}(i-k)^n\&=sum_{igeq0}sum_{k^*=0}^i(-1)^{i-k}inom{i}{k}sum_{ngeq0}frac{(kx)^n}{n!}\&=sum_{igeq0}sum_{k=0}^i(-1)^{i-k}inom{i}{k}(e^x)^k\&=sum_{igeq0}(-1+e^x)^i\&=frac{1}{2-e^x}end{aligned})

同理(g(x)=cfrac{e^x-1}{(2-e^x)^2})

原文地址:https://www.cnblogs.com/herald/p/13915460.html