an=(3-2n)(-3分之1)的n次方 求an的前n项和sn

用错位法：
Sn=1*(1/2)^1-1*(1/2)^2-3*(1/2)^3+.+(3-2n)*(1/2)^n
1/2*Sn=1*(1/2)^2-1*(1/2)^3-3*(1/2)^4+.+(3-2n)*(1/2)^(n+1)
上式-下式：
1/2*Sn=1*(1/2)^2-2*[（1/2）^2+(1/2)^3+.(1/2)^n]-(3-2n)*(1/2)^(n+1)
1/2*Sn=1-2*[（1/2）^2+(1/2)^3+.(1/2)^n]-(3-2n)*(1/2)^(n+1)
1/2*Sn=2-1-2*[（1/2）^2+(1/2)^3+.(1/2)^n]-(3-2n)*(1/2)^(n+1)
1/2*Sn=2-2*(1/2)^1-2*[（1/2）^2+(1/2)^3+.(1/2)^n]-(3-2n)*(1/2)^(n+1)
1/2*Sn=2-2*[（1/2）^1+(1/2)^2+.(1/2)^n]-(3-2n)*(1/2)^(n+1)
1/2*Sn=2-2*[1/2-(1/2)^(n+1)]/(1-1/2)-(3-2n)*(1/2)^(n+1)
1/2*Sn=2-2*[1-2(1/2)^(n+1)]-(3-2n)*(1/2)^(n+1)
∴Sn=(-2n+7)/2^n