由
an+1+an−1
an+1−an+1=n可得an+1+an-1=nan+1-nan+n
∴(1-n)an+1+(1+n)an=1+n
∴an+1=
n+1
n−1an−
n+1
n−1=
1
n−1(an−1)×(n+1)
∴
an+1
n+1=
1
n−1an−
1
n−1=
n
n−1•
an
n−
n
n−1•
1
n
∴
an+1
n+1−1=
n
n−1(
an
n−1)
∴
1
n(
an+1
n+1−1)=
1
n−1(
an
n− 1)
∴{
1
n−1(
an
n −1)}为常数列
而
1
n−1(
an
n−1)=
1
2−1• (
a2
2−1)=2
an=[2(n-1)+1]n=2n2-n
当n=1时,
6+a1−1
6−a1+1=1可得a1=1适合上式
故答案为:2n2-n