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](/uploads/image/z/14760553-49-3.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E6%BB%A1%E8%B6%B3a%28n%2B1%29%3Dan%5E2-n%2Aan%2B1%2C%E5%BD%93a1%3E%3D3%E6%97%B6%2C%E8%AF%81%E6%98%8E%E5%AF%B9%E4%BA%8E%E6%89%80%E6%9C%89n%3E%3D1%2C%E6%9C%89an%3E%3Dn%2B2)
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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