Question

设数列{an}的前n项和为Sn，且对任意正整数n都有2Sn=n(an+3),

（1）求证：数列{an}为等差数列
谢谢

Answer 1

2a(1)=2s(1)=a(1)+3,a(1)=3,

2a(n+1)=2s(n+1)-2s(n)=(n+1)[a(n+1)+3]-n[a(n)+3],
(n-1)a(n+1)=na(n)+3,
a(n+1)/n=a(n)/(n-1)+3/[n(n-1)] = a(n)/(n-1) + 3/n - 3/(n-1),
[a(n+1)-3]/n = [a(n)-3]/(n-1) = ... = [a(2)-3]/(2-1)=a(2)-3,
a(n)-3=[a(2)-3](n-1),
a(n)=3 + [a(2)-3](n-1).
{a(n)}为首项为a(1)=3,公差为[a(2)-3]的等差数列.