已知各项均为正数的数列｛an}的前n项和为Sn,且6Sn=（an+1)(an+2),n为正整数,求an

已知各项均为正数的数列｛an}的前n项和为Sn,且6Sn=（an+1)(an+2),n为正整数,求an
已知各项均为正数的数列｛an}的前n项和为Sn,且6Sn=（an+1)(an+2),n为正整数,求an

已知各项均为正数的数列｛an}的前n项和为Sn,且6Sn=（an+1)(an+2),n为正整数,求an
an = Sn - Sn-1,利用这个 将 6Sn=（an+1)(an+2) 变形后可以得
（an + an-1）（an - an-1 -3）=0
由于各项为正,故 an - an-1 = 3
则 an为公差为3 的等比数列,再由 6Sn=（an+1)(an+2)可求得 a1 =1 或 2
则 an = 3n - 2 或 3n - 1

6Sn=an^2+3an+2
6Sn-1=an-1^2+3an-1+2
6(Sn-Sn-1)=(an+an-1)(an-an-1)+3(an-an-1)
6an=(an+an-1)(an-an-1)+3(an-an-1)
0=(an+an-1)(an-an-1)+3(an-an-1)-6an
0=(an+an-1)(an-an-1)-3(an+an-1)
3=an-an-1
3=an-1-an-2
...............
3=a2-a1
3(n-1)=an-a1
an=a1+3(n-1)