作业帮已知函数f(x)=2x/(x+1),数列{an}满足a1=4/5,a(n+1)=f(an),bn=1/an-1.
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已知函数f(x)=2x/(x+1),数列{an}满足a1=4/5,a(n+1)=f(an),bn=1/an-1.
(1)求数列｛bn｝的通项公式
(2)设cn=[an*a(n+1)]/2^（n+2),Tn是数列cn的前n项和,证明：Tn＜1/5
(1)求数列｛bn｝的通项公式
(2)设cn=[an*a(n+1)]/2^（n+2),Tn是数列cn的前n项和,证明：Tn＜1/5
(1)由bn=1/an-1得,
an = 1/(bn + 1),代入a(n+1)=f(an)可得
1/(b(n+1) + 1) = 2/(bn + 1) / [1/(bn + 1) + 1]
化简得b(n+1) = bn/2
又b1 = 1/a1 -1 = 1/4
于是 bn = 1/2^(n+1).
(2)由(1),an = 1/(bn + 1) = 2^(n+1) / [2^(n+1) + 1],
于是cn = 2^(n+1) / [(2^(n+1) + 1)(2^(n+2) + 1)] = 1/[2^(n+1) + 1] - 1/[2^(n+2) + 1]
故Tn = c1+...+cn = 1/5 - 1/[2^(n+2) + 1]
an = 1/(bn + 1),代入a(n+1)=f(an)可得
1/(b(n+1) + 1) = 2/(bn + 1) / [1/(bn + 1) + 1]
化简得b(n+1) = bn/2
又b1 = 1/a1 -1 = 1/4
于是 bn = 1/2^(n+1).
(2)由(1),an = 1/(bn + 1) = 2^(n+1) / [2^(n+1) + 1],
于是cn = 2^(n+1) / [(2^(n+1) + 1)(2^(n+2) + 1)] = 1/[2^(n+1) + 1] - 1/[2^(n+2) + 1]
故Tn = c1+...+cn = 1/5 - 1/[2^(n+2) + 1]
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