Question

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

Answer 1

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

Answer 2

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