（1）由an+2-an+1=an+1-an+2可得：数列an+1-an为等差数列，且首项a1-a0=2-0=2，公差为2（3分）∴an-an-1=（a1-a0）+2（n-1）=2+2（n-1）=2n（4分）∴an＝a1+(a2−a1)+(a3−a2)++(an−an−1)＝2+4+6++2n＝n(2+2n)2＝n...