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.

Answer 1

(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]