已知数列an,a1=2,an+1(n+1是a右下角的注脚）=an+3n+2,则an=多少?

∵a(n+1)＝an+3n+2∴a(n+1)－an＝3n+2则an－a(n-1)=3(n-1) + 2a(n-1)－a(n-2)=3(n-2) + 2a(n-2)－a(n-3)=3(n-3) + 2···a3 － a2 =3×2 + 2a2 － a1 =3×1 + 2全加,得an － a1 =3[1+2+3+……+(n-1)] + 2(n-1)an －...