题目

数学数列题.
在数列an中,a1=1,a(n+1)-(n+1)=2(an-1),bn=an+n
(1)求证bn是等比数列
(2)求数列{(2n-1)/(an+log2bn)}的前n项和Sn.
谢谢.

提问时间：2021-03-13

答案

(1)
a(n+1) -(n+1) = 2(an -1)
a(n+1) =2an +n-1
a(n+1) + (n+1) = 2(an + n )
{an + n } 是等比数列,q=2
bn = an +n 是等比数列
(2)
an + n = 2^(n-1) .( a1+ 1)
= 2^n
an = -n+2^n
bn = an+n = 2^n
let
S = 1.(1/2)^0 +2.(1/2)^1+.+n.(1/2)^(n-1) (1)
(1/2)S = 1.(1/2)^1 +2.(1/2)^2+.+n.(1/2)^n (2)
(1)-(2)
(1/2)S =[ 1+1/2+.+1/2^(n-1)] - n.(1/2)^n
= 2[1- 1/2^n] - n.(1/2)^n
S = 4[1- 1/2^n] - 2n.(1/2)^n
cn ={(2n-1)/(an+logbn)}
= (2n-1)/2^n
= n(1/2)^(n-1) - 1/2^n
Sn = c1+c2+...+cn
= S - 2(1- 1/2^n)
=2[1- 1/2^n] - 2n.(1/2)^n
= 2 - (2n+2) (1/2)^n