数列An中,A1=1,An+An+1=3^n,Bn=An-1/4乘3的N次方,求证Bn是等比数列,求An前n项和

a(n+1)=3^n-an
bn=an-(1/4)*3^n,
an=bn+(1/4)*3^n,
a(n+1)=b(n+1)+(1/4)*3^(n+1)
an+a(n+1)=bn+(1/4)*3^n+b(n+1)+(1/4)*3^(n+1)=3^n
故得bn+b(n+1)=0,
即b(n+1)/bn=-1,
b1=1-3/4=1/4
{bn}为公比为-1的等比数列,
bn=b1*(-1)^(n-1)=-(1/4)*(-1)^n
(2)an=bn+(1/4)*3^n=(1/4)*[3^n-(-1)^n]
Sn=a1+a2+……+a(n-1)+an
=(1/4)*(3^1+3^2+3^3+……+3^n)-(1/4)[-1+1-1+1+……+(-1)^n]
=(3/8)*(3^n-1)-(1/4)[-1+1-1+1+……+(-1)^n]
当n为奇数时：Sn=(3/8)*(3^n-1)+(1/4)=[3^(n+1)-1]/8
当n为偶数时：Sn=(3/8)*(3^n-1)