由an+1=an+2n+1得an+1-an=2n+1则an=（an-an-1）+（an-1-an-2）+…+（a3-a2）+a1=[2（n-1）+1]+[2（n-2）+1]+…+（2×2+1）+（2×1+1）+1=2[（n-1）+（n-2）+…+2+1]+（n-1）+1=2× (n?1)n 2 +（n-1）+1=（n-1）（n+1）+1=n2，所以数列{an}的通项公式为an=n2．
