已知数列{an}的前n项和为Sn,且对于任意的n∈N*,恒有Sn=2an-n,设bn=log2(an+1)
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已知数列{an}的前n项和为Sn,且对于任意的n∈N*,恒有Sn=2an-n,设bn=log2(an+1)
(1)求证:数列{an+1}是等比数列
(2)求数列{an}、{bn}的通项公式.
(1)求证:数列{an+1}是等比数列
(2)求数列{an}、{bn}的通项公式.
![已知数列{an}的前n项和为Sn,且对于任意的n∈N*,恒有Sn=2an-n,设bn=log2(an+1)](/uploads/image/z/19435252-4-2.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94%E5%AF%B9%E4%BA%8E%E4%BB%BB%E6%84%8F%E7%9A%84n%E2%88%88N%2A%2C%E6%81%92%E6%9C%89Sn%3D2an-n%2C%E8%AE%BEbn%3Dlog2%28an%2B1%29)
Sn=2an - n,S(n-1)=2a(n-1) - (n-1)
Sn-S(n-1)=an=2an - 2 a(n-1)-1 ,2a(n-1)=an -1
an +1= 2 [a(n-1)+1],(an +1)/[a(n-1)+1] = 2
所以(an +1)是公比为2的等比数列.
a1=2a1-1,a1=1,a1 +1=2
an +1= 2*2^(n-1) = 2^n an = 2^n -1
bn=log2(an +1) = log2(2^n) =n
Sn-S(n-1)=an=2an - 2 a(n-1)-1 ,2a(n-1)=an -1
an +1= 2 [a(n-1)+1],(an +1)/[a(n-1)+1] = 2
所以(an +1)是公比为2的等比数列.
a1=2a1-1,a1=1,a1 +1=2
an +1= 2*2^(n-1) = 2^n an = 2^n -1
bn=log2(an +1) = log2(2^n) =n
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