an =a1+(n-1)d= 2+(n-1)da2,a4,a8成等比數(shù)列(a4)^2 = a2.a8(2+3d)^2 = (2+d)(2+7d)2d^2-4d =0d(d-2)=0d= 2an = 2+(n-1)2 = 2nbn = an .3^an= (2n) .3^(2n)= 3(2n).3^(2n-1) Tn =b1+b2+...+bn= 3∑(i:1->2n ) (2i).3^...an的通項公式是2n,
T^T所以麻煩了！請再幫我一下吧！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)]