数列求和 谁能帮我化简一下

An=(2n-1)*(1/2)^(n-1)=2*n*(1/2)^(n-1)-(1/2)^(n-1);
令Bn=2n*(1/2)^(n-1); Cn=(1/2)^(n-1) ;则An=Bn-Cn; San = Sbn-Scn;
Sbn=B1+B2+B3+B4+.+Bn
1/2Sbn= 1/2B1+1/2B2+1/2B3+.+1/2Bn (错位相减,Bn-1/2B(n-1) = 2*(1/2)^(n-1)=4*(1/2)^n);
1/2Sbn=Sbn-1/2Sbn = 2+{4*(1/2)^2+4*(1/2)^3+.+4*(1/2)^n}-2n*(1/2)^n
=2+(1-(1/2)^(n-1))/(1-1/2)-2n*(1/2)^n=2+2*(1-(1/2)^(n-1))-2n*(1/2)^n
Sbn=8+(1/2)^(n-3)-n*(1/2)^(n-2);
Scn=[1-(1/2)^n]/(1-1/2)=2-(1/2)^(n-1);
San=Sbn-Scn=8+(1/2)^(n-3)-n*(1/2)^(n-2)-(2-(1/2)^(n-1))
=6+(1/2)^(n-1)+(1/2)^(n-3)-n*(1/2)^(n-2)