可得a1=1,a2=2,a3=3.猜an=n对于前k（k＞2）项成立
2S(k+1)=a（k+1)²+k+1
2Sk+2a(k+1)=a(k+1)²+k+1
k²=a（k+1）²-2a（k+1）+1
k=a（k+1）-1
a（k+1）=k+1
证毕