设这个式子的和为S
那么 S = 1/2 +2/2^2 +3/2^3 +4/2^4 +...+n/2^n
所以 2S = 1 +2/2 +3/2^2 +4/2^3 +5/2^4 +...+n/2^(n-1)
两式相减：S = 2S-S = 1 +(2-1)/2 +(3-2)/2^2 +(4-3)/2^3 +(5-4)/2^4 +...+(n-n+1)/2^(n-1) -n/2^n
所以 S = 1/2^0 +1/2^1 +1/2^2 +1/2^3 +1/2^(n-1) -n/2^n
所以 S = [1 -(1/2)^n]/(1 -1/2) -n/2^n
S = 2 -2/2^n -n/2^n = 2 -(n+2)/2^n
即 原式 = 2 -(n+2)/2^n