已知数列{an}各项均为正数,其前n项和为Sn,且(an+1)^2=4Sn,数列{bn}满足b1=3,b(n+1)=a(
3个回答
(1)(a1+1)^2=4S1=4a1
a1^2-2a1+1=0、(a1-1)^2=0、a1=1.
n>=2时,(an+1)^2-[a(n-1)+1]^2=an^2-a(n-1)^2+2an-2a(n-1)=4Sn-4S(n-1)=4an
[an+a(n-1)][an-a(n-1)]-2[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-2]=0
因为是正项数列,所以an+a(n-1)>0.
即an-a(n-1)-2=0、an-a(n-1)=2.
所以,{an}是首项为1、公差为2的等差数列,an=2n-1,n为正整数.
b(n+1)=a(bn)=2bn-1、b(n+1)-1=2(bn-1).
所以数列{bn-1}是首项为b1-1=2、公比为2的等比数列.
bn-1=2^n、bn=2^n+1,n为正整数.
(2)cn=2^n/[bn*b(n+1)]=2^n/[(2^n+1)(2^(n+1)+1]=1/(2^n+1)-1/[2^(n+1)+1].
c1+c2+c3+…+cn
=1/(2+1)-1/(2^2+1)+1/(2^2+1)-1/(2^3+1)+1/(2^3+1)-1/(2^5+1)+…+1/(2^n+1)-1/[2^(n+1)+1]
=1/(2+1)-1/[2^(n+1)+1]
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