求数列(2的n次方)分之n的前n项和

运用错位相减法：
an=n/2^n
Sn=a1+a2+a3+……+an=1/2^1+2/2^2+3/2^3+……+n/2^n
(1/2)Sn=1/2a1+1/2a2+……+1/2an=1/2^2+2/2^3+3/2^4+……+n/2^(n+1)
Sn-(1/2)Sn=1/2^1+1/2^2+1/2^3+……+1/2^n-n/2^(n+1)
(1/2)Sn=1/2*(1-(1/2)^n)/(1/2)-n/2^(n+1)
Sn=2-1/2^(n-1)-n/2^n=2-(n+2)/2^n.