Question

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

Answer 1

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