根据题意：
S(n)=1/2+2/2²+3/2³+……++(n-1)/[2^(n-1)]+n/(2^n)
(1/2)S(n)=1/2²+2/2³+3/(2^4)+……++(n-1)/(2^n)+n/[2^(n+1)]
上面两式相减,得
(1/2)S(n)=1/2+1/2²+1/2³+……+1/(2^n)-n/[2^(n+1)]
=1-1/(2^n)-n/[2^(n+1)]
=1-(n+2)/[2^(n+1)]
所以
S(n)=2-(n+2)/(2^n)