Question

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

Answer 1

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