Question

试求sn=1+3/2^2+4/2^3+```+n/2^n-1+n+1/2^n 求过程

Answer 1

Sn = 2/2 + 3/2^2 + 4/2^3 + …… + n/2^(n-1) + (n+1)/2^n
(1/2)Sn = 2/2^2+ 3/2^3 + …… + n/2^n + (n+1)/2^(n+1)
两式相减得
(1/2)Sn= 1 + [ 1/2^2+ 1/2^3 +……+ 1/2^n ] - (n+1)/2^(n+1)
= 1/2+ [ 1/2 + 1/2^2+ 1/2^3 +……+ 1/2^n ] - (n+1)/2^(n+1)
= 1/2 + [(2^n - 1)/2^n] - (n+1)/2^(n+1)
= [2^(n+1) + 2^n - n - 3]/[2^(n+1)]
= (3×2^n - n - 3)/[2^(n+1)]

Answer 2

sn=2/2+3/2^2+4/2^3+```+n/2^(n-1)+(n+1)/2^n
0.5sn=2/2^2+3/2^3+4/2^4+...+(n-1)/2^(n-1)+n/2^n+(n+1)/2^(n+1)

错位相减
0.5Sn=1-(n+1)/2^(n+1)+[1/2^(n+1)-1/4]/(1/2-1)
Sn=3-(n+1)/2^n-1/2^(n-1)

Answer 3

Sn/2=2/2^2+3/2^3+…+(n+1)/2^(n+1)
Sn-Sn/2=1+1/2^2+1/2^3+…1/2^n-1/2^(n+1)
Sn=3-3/2^n