已知数列｛An｝的前n项和为Sn,且6Sn=（2n+3）An+n〖n∈N*〗〈1〉求数列｛An｝的通项公式.

Question

已知数列｛An｝的前n项和为Sn,且6Sn=（2n+3）An+n〖n∈N*〗 〈1〉求数列｛An｝的通项公式.

Answer 1

6s(n)=(2n+3)a(n)+n,
6a(1)=6s(1)=5a(1)+1,a(1)=1.
6s(n+1)=(2n+5)a(n+1)+n+1,
6a(n+1)=6s(n+1)-6s(n)=(2n+5)a(n+1)+1 - (2n+3)a(n),
(2n-1)a(n+1) = (2n+3)a(n) - 1 = (2n+3)a(n) + [(2n-1)-(2n+3)]/4,
(2n-1)[a(n+1)-1/4] = (2n+3)[a(n)-1/4],
(2n-1)(2n+1)[a(n+1)-1/4] = (2n+1)(2n+3) [a(n)-1/4],
[a(n+1)-1/4]/[(2n+1)(2n+3)] = [a(n)-1/4]/[(2n-1)(2n+1)],
{[a(n)-1/4]/[(2n-1)(2n+1)]}是常数列.
[a(n)-1/4]/[(2n-1)(2n+1)]=[a(1)-1/4]/[(2-1)(2+1)]=(3/4)/3=1/4,
a(n)=[(2n-1)(2n+1)+1]/4 = n^2