已知各项均为正数的数列{an}的前n项和为Sn,满足S1＞1,且6Sn=（an+1)×(an+2),求通项公式
问题描述：
已知各项均为正数的数列{an}的前n项和为Sn,满足S1＞1,且6Sn=（an+1)×(an+2),求通项公式
答
6Sn=（an+1)×(an+2)=an^2+3an+2
6a1=6S1=a1^2+3a1+2
a1=1或2
又S1=a1>1
a1=2
n>=2时
6Sn=an^2+3an+2
6S(n-1)=a(n-1)^2+3a(n-1)+2
an^2-a(n-1)^2-3an-3a(n-1)=0
(an+a(n-1))(an-a(n-1)-3)=0
又an为正项数列
an+a(n-1)>0
an-a(n-1)-3=0
a1=2
an=3n-1
答
当n=1时.6a1=(a1+1)(a1+2),a1=2或1(因为S1gt;1,则舍)
当n用n-1写时,得:6Sn-1 =(an-1 +1)(an-1 +2),与6Sn=(an+1)(an+2)做差,得:
6an=(an+1)(an+2)-(an-1 +1)(an-1 +2)自己化(移项,分解因式)即得
an-an-1 =3,{an}是以2为首项,3为公差的等差数列.an=3n-1