数列an=2n-1数列b1,b2-b1,b3-b2,..,bn-bn-1是首项为1公比为1/2的等比数列,设{Cn}=a

n-b(n-1) = (1/2)^(n-1)
bn - b1 = 1/2 +(1/2)^2+...+(1/2)^(n-1)
bn = 1+1/2 +(1/2)^2+...+(1/2)^(n-1)
= 2( 1- (1/2)^n)
let
S =1.(1/2)^1+2.(1/2)^2+...+n.(1/2)^n (1)
(1/2)S = 1.(1/2)^2+2.(1/2)^3+...+n.(1/2)^(n+1) (2)
(1)-(2)
(1/2)S = (1/2+1/2^2+...+1/2^n) -n.(1/2)^(n+1)
= ( 1- (1/2)^n ) -n.(1/2)^(n+1)
4S =8( 1- (1/2)^n ) -8n.(1/2)^(n+1)
cn = an .bn
= 2(2n-1) .( 1- (1/2)^n)
= 4n-2 + 2(1/2)^n - 4(n.(1/2)^n)
Tn = c1+c2+...+cn
= 2n^2 + 4( 1- (1/2)^n ) -4S
=2n^2 - 4( 1- (1/2)^n ) +8n.(1/2)^(n+1)
=2n^2- 4 + (4n+4).(1/2)^n