若数列{an}满足a1=2,an+1=an+2n(n∈N*),则数列{an}的通项公式an=_.

问题描述:

若数列{an}满足a1=2,an+1=an+2n(n∈N*),则数列{an}的通项公式an=______.

由an+1=an+2n,得an+1-an=2n,
∴n≥2时,a2-a1=2,a3-a2=4,…,an-an-1=2(n-1),
以上各式相加,得an-a1=

(n-1)(2n-2+2)
2
=n2-n,
∵a1=2,∴an=n2-n+2,
又a1=2适合上式,
∴an=n2-n+2,
故答案为:n2-n+2.