Question

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

Answer 1

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.

Answer 2

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