因为an=(2n-1)/2^n=n/2^(n-1)-1/2^n
设数列{n/2^(n-1)}前n项和为Tn,数列{1/2^n}前n项和为Pn,
则Sn=Tn-Pn
Tn=1+2/2+3/2²+4/2³+.+n/2^(n-1)
(1/2)Tn=1/2+2/2²+3/2³+.+(n-1)/2^(n-1)+n/2^n
上两式错项相减得
（1/2)Tn=1+1/2+1/2²+1/2³+.+1/2^(n-1)-n/2^n
(1/2)Tn=2[1-(1/2)^n]-n/2^n
Tn=4-(n+2)/2^(n-1)
Pn=1/2+1/2²+1/2³+.+1/2^n=1-1/2^n
Sn=Tn-pn=4-(n+2)/2^(n-1)-(1-1/2^n) = 3-(2n+3)/2^n