设数列{An}满足An+1=An²－nAn+1,n=1,2,3,···；当A1=2时,求出A2,A3,A4,A

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设数列{An}满足An+1=An²－nAn+1,n=1,2,3,···；当A1=2时,求出A2,A3,A4,A5；
并由此猜测出{An}的一个通项
（要有步骤）

a2=2^2-1*2+1=3
a3=3^2-2*3+1=4
a4=4^2-3*4+1=5
a5=5^2-4*5+1=6
猜测an=n+1
以下用数学归纳法证明
由a1=2=1+1知n=1时an=n+1成立
设n=k(k属于正整数)时an=n+1成立即ak=k+1
则当n=k+1时,因为a(n+1)=an²-n*an+1,
所以a(k+1)=ak²-k*(k+1)+1
=(k+1)²-k*(k+1)+1
=k²+2k+1-k²-k+1
=k+2
综上,an=n+1成立

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