已知数列｛an｝是公差不为零的等差数列，a1=2,且a2,a4,a8成等比数列.求数列{an*3^an}的前n项和.

an =a1+(n-1)d
= 2+(n-1)d
a2,a4,a8成等比数列
(a4)^2 = a2.a8
(2+3d)^2 = (2+d)(2+7d)
2d^2-4d =0
d(d-2)=0
d= 2
an = 2+(n-1)2 = 2n
bn = an . 3^an
= (2n) . 3^(2n)
= 3(2n).3^(2n-1)
Tn =b1+b2+...+bn
= 3∑(i:1->2n ) (2i). 3^(2i-1)
consider
1+x+x^2+...+x^2n = (x^(2n+1) -1)/(x-1)
1+2x+...+(2n)x^(2n-1) = [(x^(2n+1) -1)/(x-1)]'
=[2n.x^(2n+1) -(2n+1)x^n +1]/(x-1)^2

put x=3
∑(i:1-.n ) i. 3^(i-1) = 4[(2n)3^(2n+1) -(2n+1).3^(2n) +1]
=4[1+(4n-1).3^(2n)]

Tn =b1+b2+...+bn
= 3∑(i:1->2n ) (2i). 3^(2i-1)
= 12[1+(4n-1).3^(2n)]