∵an=an-1+2n-1（n≥2），∴an-an-1=2n-1．∴an=（an-an-1）+（an-1-an-2）+…+（a2-a1）+a1=（2n-1）+（2n-3）+…+3+1= n(2n?1)+1 2 =n2．故答案为n2．
