设数列{an}的各项都是正数,a1=1,(an+1)/(an+1+1)=an+1/2an,bn=an2+an.(1)求数列{bn}的通项公式;(2)求
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![设数列{an}的各项都是正数,a1=1,(an+1)/(an+1+1)=an+1/2an,bn=an2+an.(1)求数列{bn}的通项公式;(2)求](/uploads/image/z/3987945-9-5.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%90%84%E9%A1%B9%E9%83%BD%E6%98%AF%E6%AD%A3%E6%95%B0%2Ca1%3D1%2C%28an%2B1%29%2F%28an%2B1%2B1%29%3Dan%2B1%2F2an%2Cbn%3Dan2%2Ban.%281%29%E6%B1%82%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%9B%282%29%E6%B1%82)
设数列{an}的各项都是正数,a1=1,(an+1)/(an+1+1)=an+1/2an,bn=an2+an.(1)求数列{bn}的通项公式;(2)求
设数列{an}的各项都是正数,a1=1,(an+1)/(an+1+1)=an+1/2an,bn=an2+an.(1)求数列{bn}的通项公式;(2)求
设数列{an}的各项都是正数,a1=1,(an+1)/(an+1+1)=an+1/2an,bn=an2+an.(1)求数列{bn}的通项公式;(2)求
1...
(An+1)/{A(n+1)+1}=A(n+1)/2An
∴{A(n+1)^2+A(n+1)}/{(An)^2+An}=2
∵Bn=(An)^2+An ∴B(n+1)={A(n+1)^2+A(n+1)} ∴B(n+1)/Bn=2 既Bn为公比2的等比数列 B1=(a1)^2+a1=2
所以Bn=2*2^(n-1)=2^n
2...
∵ Bn...
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1...
(An+1)/{A(n+1)+1}=A(n+1)/2An
∴{A(n+1)^2+A(n+1)}/{(An)^2+An}=2
∵Bn=(An)^2+An ∴B(n+1)={A(n+1)^2+A(n+1)} ∴B(n+1)/Bn=2 既Bn为公比2的等比数列 B1=(a1)^2+a1=2
所以Bn=2*2^(n-1)=2^n
2...
∵ Bn=(An)^2+An Bn=2^n ∴(An)^2+An=2^2 解得An=√(2^n+1/4)-1/2
3...(1+An)A(n+1)
={√(2^n+1/4)+1/2}*{√(4^n+1/4)-1/2}
>{√(2^n+1/4)+1/2}*{√(2^n+1/4)-1/2}
=2^n
∴1/(1+An)A(n+1)<1/2^n
∴原式
< 1/2^1+1/2^2+……+1/2^n
设原式=Cn=1/2^1+1/2^2+……+1/2^n
则2Cn=1+1/2^1+1/2^2+……+1/2^(n-1)
2Cn-Cn=1-1/2^n
即Cn=1-1/2^n<1
∴原式<1
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