数列{an} a1=4 Sn+Sn+1=5/3 an+1 求An 那些1都是下标

题目

数列{an} a1=4 Sn+Sn+1=5/3 an+1 求An 那些1都是下标

答案

s(n)+s(n+1) = (5/3)a(n+1),s(1)+s(2)=2a(1)+a(2)=(5/3)a(2),2a(1)=(2/3)a(2),a(2)=3a(1)=12.
s(n+1)+s(n+2)=(5/3)a(n+2),
(5/3)[a(n+2)-a(n+1)] = [s(n+1)-s(n)] + [s(n+2)-s(n+1)] = a(n+1) + a(n+2),
(2/3)a(n+2) = (8/3)a(n+1),
a(n+2)=4a(n+1),
{a(n+1)}是首项为 a(2)=12,公比为4的等比数列.
a(n+1)=12*4^(n-1)=3*4^n,
a(1)=4,
n>=2时,a(n)=3*4^(n-1)