设数列{an}满足a(n+1)=an^2-n*an+1,当a1>=3时,证明对于所有n>=1,有an>=n+2
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设数列{an}满足a(n+1)=an^2-n*an+1,当a1>=3时,证明对于所有n>=1,有an>=n+2设数列{an}满足a(n+1)=an^2-n*an+1,当a1>=3时,证明对于所有n>=
设数列{an}满足a(n+1)=an^2-n*an+1,当a1>=3时,证明对于所有n>=1,有an>=n+2
设数列{an}满足a(n+1)=an^2-n*an+1,当a1>=3时,证明对于所有n>=1,有an>=n+2
设数列{an}满足a(n+1)=an^2-n*an+1,当a1>=3时,证明对于所有n>=1,有an>=n+2
数学归纳法
a1>=3成立
设对于n成立,即an>=n+2,这时证明n+1时也成立即可,
即,证明an+1>=n+1+2即可
即,证明an-n-2>=0即可
即,证明an^2-n*an+1-n-2>=0即可
an^2-n*an+1-n-2在an>=n+2增函数
固有an^2-n*an+1-n-2>=(n+2)^2-n*(n+2)-n-2=n+2>=0,
得证
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