已知数列AN的各项均为正数,且前N项和满足6Sn=an^2+3an+2,求数列通项公式

6Sn=an^2+3an+2
6S(n-1)=a(n-1)^2+3a(n-1)+2
6Sn-6S(n-1)=6an=an^2+3an+2-a(n-1)^2-3a(n-1)-2
6an=an^2+3an-a(n-1)^2-3a(n-1)
a(n-1)^2+3a(n-1)=an^2-3an
a(n-1)^2+3a(n-1)+9/4=an^2-3an+9/4
(a(n-1)+3/2)^2=(an-3/2)^2
所以
an-3/2=a(n-1)+3/2 或an-3/2=-a(n-1)-3/2
an-a(n-1)=6 或 an+a(n-1)=0(舍去)
所以是等差数列,公差d=6
6Sn=an^2+3an+2
6S1=6a1=a1^2+3a1+2
a1^2-3a1+2=0
(a1-2)(a1-1)=0
a1=2 或a1=1
所以an=a1+(n-1)d=2+(n-1)*6=6n-4
或 an=a1+(n-1)d=1+(n-1)*6=6n-5