对于等差数列除以等比数列前n项和的求解

Sn=1*(1/2)^0+0*(1/2)^1 +(-1)*(1/2)^2+(-2)*(1/2)^3 +……+(2-n)(1/2)^(n-1).①
(1/2)Sn= 1*(1/2)^1 +0*(1/2)^2 +(-1)*(1/2)^3 +(-2)*(1/2)^4+……+(2-n)(1/2)^n.②
①-②
(1/2)Sn=1*(1/2)^0+(-1)*(1/2)^1 +(-1)*(1/2)^2+(-1)*(1/2)^3 +……+(-1)(1/2)^(n-1)+(n-2)(1/2)^n
=1-[ (1/2)^1 +(1/2)^2+(1/2)^3 +……+(1/2)^(n-1) ] + (n-2)(1/2)^n
=1-[1-2^(1-n)] + (n-2)/2^n
=1/2^(n-1) +(n/2 -1)/2^(n-1)
=(n/2)/2^(n-1)
=n/2^n
Sn=2n/2^n = n / 2^(n-1) 即：Sn = n / 2^(n-1)