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.

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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)]
再问： an的通项公式是2n,
T^T所以麻烦了！请再帮我一下吧！
再答： bn = an . 3^an
= (2n) . 3^(2n)
= 3(2n).3^(2n-1)
Tn =b1+b2+...+bn
= 3∑(i:1->n ) (2i). 3^(2i-1)
consider
1+x^2+x^4+...+x^2n = [x^(2n+2) -1]/(x^2-1)
2x + 4x^3+...+2n.x^(2n-1) =[(x^2-1)( 2n+2)x^(2n+1)- [x^(2n+2)-1](2x)] /(x^2-1)^2
= [2n.x^(2n+3) -(2n+2)x^(2n+1) + 2x ]/(x^2-1)^2
put x=3

∑(i:1->n ) (2i). 3^(2i-1)
= 2.3 + 4.3^3+...+(2n).3^(2n-1)
=64[2n.3^(2n+3) -(2n+2).3^(2n+1) + 6 ]
=64[6+ (16n-2)3^(2n+1)]

Tn =b1+b2+...+bn
= 3∑(i:1->n ) (2i). 3^(2i-1)
=192[6+ (16n-2)3^(2n+1)]

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