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
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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.
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.
![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](/uploads/image/z/18091291-67-1.jpg?t=%E6%95%B0%E5%AD%A6%E6%95%B0%E5%88%97%E9%A2%98.%E5%9C%A8%E6%95%B0%E5%88%97an%E4%B8%AD%2Ca1%3D1%2Ca%28n%2B1%29-%28n%2B1%29%3D2%28an-1%29%2Cbn%3Dan%2Bn%281%29%E6%B1%82%E8%AF%81bn%E6%98%AF%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%282)
(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
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
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