Question

已知数列{an}中,a1=a>0,an+1=f(an),其中f(x)=3x/1—+2x.(1)求a2,a3,a4.(2)

Answer 1

a(n+1)=3a(n)/[1+2a(n)],

若a(n+1)=0,则,a(n)=0, ..., a(1)=0,与a(1)=a>0矛盾.

因此,a(n)不为0.

1/a(n+1)=[1+2a(n)]/[3a(n)] = (1/3)[1/a(n)] + 2/3 = (1/3)[1/a(n)] + 1 - 1/3,

1/a(n+1) - 1 = (1/3)[1/a(n) - 1],

{1/a(n) - 1}是首项为1/a(1)-1=[1/a-1],公比为(1/3)的等比数列.

1/a(n) - 1 = [1/a - 1](1/3)^(n-1) = (1-a)/[a*3^(n-1)],

1/a(n) = 1 + (1-a)/[a*3^(n-1)] = [1-a + a*3^(n-1)]/[a*3^(n-1)],

a(n) = [a*3^(n-1)]/[1-a+a*3^(n-1)]

a(2)=[3a]/[1-a+3a]=(3a)/(1+2a),

a(3)=[9a]/[1-a+9a]=(9a)/(1+8a),

a(4)=[27a]/[1-a+27a]=(27a)/(1+26a)