数列【an】满足(n+1)an+1=2(n+2)an+3n2+9n+6,a1=6,求通项公式

(n+1)a(n+1)=2(n+2)an+3n²+9n+6
(n+1)a(n+1)=2(n+2)an+3(n+1)(n+2)
等式两边同除以(n+1)(n+2)
a(n+1)/(n+2)=2an/(n+1) +3
a(n+1)/(n+2) +3=2an/(n+1)+6=2[an/(n+1)+3]
[a(n+1)/(n+2) +3]/[an/(n+1)+3]=2,为定值
a1/(1+1) +3=6/2 +3=6
数列{an/(n+1) +3}是以6为首项,2为公比的等比数列
an/(n+1) +3=6×2^(n-1)=3×2ⁿ
an/(n+1)=3×2ⁿ-3
an=(n+1)(3×2ⁿ-3)=3(n+1)(2ⁿ-1)
n=1时,a1=3×(1+1)(2-1)=6,同样满足通项公式
数列{an}的通项公式为an=3(n+1)(2ⁿ-1)

式子没打错吗？ 3n2是什么?

先把括号打开 化简成nan+3an+3n2+9n+6=0
然后 (n+3)an=-3n2-9n-6
等式右边是 -3（n+1）（n+2）
an=-3（n+1）（n+2）/(n+3)