Question

求级数∑n^2/(n!*2^n)的和

Answer 1

∑n^2/(n!*2^n)=∑(n^2/n!)*(1/2)^n=∑n(n-1)/n!*(1/2)^n+∑n/n!*(1/2)^n

e^x=∑(1/n!)*x^n

求导得：

e^x=∑n/n!*x^(n-1) e^x=∑n(n-1)/n!*x^(n-2)

令x=1/2得

∑n(n-1)/n!*(1/2)^n=1/2*e^(1/2)

∑n/n!*(1/2)^n=1/4*e^(1/2)

原式=3/4*e^(1/2)