Question

在数列{an}中,已知a1=4/3,a2=13/9,当n≥2,且n∈N*时,有a（n+1)=4/3an-1/3a(n-1

Answer 1

a(n+1)=4/3a(n) - 1/3a(n-1),

a(n+2)=4a(n+1)/3 - a(n)/3,

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

{b(n)=a(n+1)-a(n)}是首项为b(1)=a(2)-a(1)=13/9-12/9=1/9,公比为1/3的等比数列.

a(n+1)-a(n)=(1/9)(1/3)^(n-1)=1/3^(n+1),

3^(n+1)a(n+1) - 3*3^na(n) = 1,

3^(n+1)a(n+1) = 3*3^na(n) + 1,

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

{3^na(n) + 1/2}是首项为3a(1)+1/2=9/2,公比为3的等比数列.

3^na(n) + 1/2 = (9/2)3^(n-1)=(1/2)3^(n+1),

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

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

4/3 < 1/2 第（2）问有误吧?

若是4/3