｛an}是正数列,前n项和Sn满足：4Sn=(an-1)(an+3),则通项公式an=?

Question

｛an}是正数列,前n项和Sn满足：4Sn=(an-1)(an+3),则通项公式an=?

Answer 1

a(n)>0.
4s(n)=[a(n)-1][a(n)+3]
4a(1)=[a(1)-1][a(1)+3],
0=[a(1)]^2-2a(1)-3=[a(1)-3][a(1)+1],a(1)=3.
4a(n+1)=4s(n+1)-4s(n)=[a(n+1)-1][a(n+1)+3]-[a(n)-1][a(n)+3],
[a(n)-1][a(n)+3]=[a(n+1)-3][a(n+1)+1],
[a(n+1)]^2-2[a(n+1)] - 3 = [a(n)]^2 + 2a(n) - 3,
0=[a(n+1)]^2 - [a(n)]^2 - 2[a(n+1)+a(n)] =[a(n+1)+a(n)][a(n+1)-a(n)-2],
0=a(n+1)-a(n)-2,
a(n+1)=a(n)+2,
{a(n)}是首项为a(1)=3,公差为2的等差数列.
a(n)=3+2(n-1)=2n+1