已知数列{an}满足{an}=2an-1+1(n≥2)且a1=1,bn=log∨2(a

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已知数列{an}满足{an}=2an-1+1(n≥2)且a1=1,bn=log∨2(a
已知数列{an}满足{an}=2an-1+1(n≥2)且a1=1,bn=log∨2(a2n+1+1),cn=1/b2n-1
求证数列{an+1}为等比数列,并求数列{an}的通项公式 2.求数列{cn}的前项和Sn.看懂了吗

a(n+1) = 2a(n)+ 1
a(n+1)+1 = 2[a(n)+1],
{a(n)+1}是首项为a(1)+1=2,公比为2的等比数列.
a(n)+1 = 2*2^(n-1) = 2^n.
a(n) = 2^n - 1.
b(n) = log_{2}[a(2n+1)+1] = log_{2}[2^(2n+1)] = 2n+1.
c(n) = 1/b(2n-1) = 1/(2(2n-1)+1) = 1/(4n-1)...
s(n)没法求和啊　.
看来,是题目没看懂.
楼主那个,a2n+1+1是a(2n+1)+1,还是[a(n+1)+1]^2啊?另外,那个b2n-1是b(2n-1) 还是[b(n)]^2-1啊.

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