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?1/2*n(3n+1).

问题描述:

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?
1/2*n(3n+1).

因为an+1-an=3n+2
所以an-an-1=3n-1
……
a2-a1=3*1+2
上式相加得
an-a1=(3n^2+n-4)/2
所以an=a1+(3n^2+n-4)/2=1/2*n(3n+1)

∵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 -...