X^2/2^X求和（X=1,2,3...）

Question

X^2/2^X 求和 （X=1,2,3...）

Answer 1

用错位相减法
S=1/2+2^2/2^2+3^2/2^3+……+(n-1)^2/2^(n-1)+n^2/2^n
2S=1+2^2/2+3^2/2^2+4^2/2^3+……+n^2/2^(n-1)
2S-S=1+(2+1)/2+(2*2+1)/2^2+(2*3+1)/2^3+……[2(n-1)+1]/2^(n-1)-n^2/2^n
=1+1/2+1/2^2+1/2^3+……+1/2^(n-1)+1+2/2+3/2^2+……+(n-1)/2^(n-2)-n^2/2^n
=2-1/2^(n-1)-n^2/2^n+[1+2/2+3/2^2+……+(n-1)/2^(n-2)]
T=1+2/2+3/2^2+……+(n-2)/2^(n-3)+(n-1)/2^(n-2)
2T=2+2+3/2+4/2^2+……+(n-1)/2^(n-3)
2T-T=2+1+1/2+1/2^2+……+1/2^(n-3)-(n-1)/2^(n-2)
=2+2-1/2^(n-3)-(n-1)/2^(n-2)
故S=6-1/2^(n-1)-1/2^(n-3)-n^2/2^n-(n-1)/2^(n-2)