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 展开
(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 展开
2个回答
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解: (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]<1/5.
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]<1/5.
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①解:
bn = 1/2^(n+1);
由bn=1/an-1
得:
an = 1/(bn + 1)
又a(n+1)=f(an)
带入化简得:
b(n+1) = bn/2
b1 = 1/a1 -1 = 1/4
=>
bn = 1/2^(n+1).
②证明:
由①
an = 1/(bn +1)
= 2^(n+1) / [2^(n+1) +1]
∴n = 2^(n+1) / [(2^(n+1) +1)(2^(n+2) +1)]
= 1/[2^(n+1) + 1] - 1/[2^(n+2) +1]
=>
Tn = c1+...+cn
= 1/5 - 1/[2^(n+2) +1]
<1/5
bn = 1/2^(n+1);
由bn=1/an-1
得:
an = 1/(bn + 1)
又a(n+1)=f(an)
带入化简得:
b(n+1) = bn/2
b1 = 1/a1 -1 = 1/4
=>
bn = 1/2^(n+1).
②证明:
由①
an = 1/(bn +1)
= 2^(n+1) / [2^(n+1) +1]
∴n = 2^(n+1) / [(2^(n+1) +1)(2^(n+2) +1)]
= 1/[2^(n+1) + 1] - 1/[2^(n+2) +1]
=>
Tn = c1+...+cn
= 1/5 - 1/[2^(n+2) +1]
<1/5
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