Question

已知各项均为正项的数列《an》的前n项和满足sn>1,且sn=1/6(an+1)（an+2）,求an.

Answer 1

S1=a1=(a1+1)(a1+2)/6
6a1=a1^2+3a1+2
a1^2-3a1+2=0
(a1-1)(a1-2)=0
a1=1(Sn>1,n=1时,S1=1,舍去)或a1=2
Sn=(an+1)(an+2)/6
Sn-1=[a(n-1)+1][a(n-1)+2]/6
an=Sn-Sn-1
=(an+1)(an+2)/6-[a(n-1)+1][a(n-1)+2]/6
=(1/6)[an^2+3an+2-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[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-3]=0
an+a(n-1)=0(各项均为正项,等式不成立,舍去）或an-a(n-1)-3=0
an=a(n-1)+3
数列{an}为首项为2,公差为3的等差数列.
an=2+(n-1)*3=3n-1
数列的通项公式为an=3n-1.