一道高一数列题,急,已知Sn=1/6(an+1)(an+2) 求an(an>0)

因为Sn=1/6(an+1)(an+2)=1/6(an^2+3an+2)
Sn-1=1/6(an-1^2+3an-1+2)
an=sn-sn-1=1/6(an^2-an-1^2+3an-3an-1)
an^2-an-1^2=3(an+an-1)
(an+an-1)(an-an-1)=3(an+an-1)
因为an>0 an+an-1不等于0
所以 an-an-1=3
公差为3
a1=s1=1/6(a1^2+3a1+2)
(a1-1)(a1-2)=0
所以a1=1或2
an=3n-2或3n-1

6Sn=(an)^2+3a(n)+2
6S(n-1)=[a(n-1)]^2+3a(n-1)+2
两式相减得到
an-a(n-1)=3
等差数列

an=1+3(n-1)=3n-2
或an=2+3(n-1)=3n-1