数列{an}的前n项和为Sn=3an+2 设bn=n 求数列{an·bn}的和Tn

an=Sn-S(n-1)
=3an+2-3a(n-1)-2
an=3/2a(n-1)
a1=3a1+2
a1=-1
an=(-1)*(3/2)^(n-1)
anbn=-n*(3/2)^(n-1)
Tn=-1(3/2)^0-2(3/2)^1-3(3/2)^2-n*(3/2)^(n-1)
3/2Tn=-(3/2)^1-2(3/2)^2-(n-1)*(3/2)^(n-1)-n*(3/2)^n
Tn-3/2Tn=-(3/2)^0-(3/2)^1-(3/2)^2-*(3/2)^(n-1)+n*(3/2)^n
=-1*(1-(3/2)^n)/(1-3/2)+n*(3/2)^n
-1/2Tn=2*(1-(3/2)^n)+n*(3/2)^n
Tn=-4*(1-(3/2)^n)-2n*(3/2)^n