请问怎样用数学归纳法证明这个数列不等式已知an+1( 指第n+1项 )=an+(an^2)/(n^2),a1=1/3.求证an>1/2 -1/4n .另外我将不等式放缩成更严格的先求证 an>1/2 - 1/5n 后反而好证了,是什么道理呢?
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![请问怎样用数学归纳法证明这个数列不等式已知an+1( 指第n+1项 )=an+(an^2)/(n^2),a1=1/3.求证an>1/2 -1/4n .另外我将不等式放缩成更严格的先求证 an>1/2 - 1/5n 后反而好证了,是什么道理呢?](/uploads/image/z/7646634-18-4.jpg?t=%E8%AF%B7%E9%97%AE%E6%80%8E%E6%A0%B7%E7%94%A8%E6%95%B0%E5%AD%A6%E5%BD%92%E7%BA%B3%E6%B3%95%E8%AF%81%E6%98%8E%E8%BF%99%E4%B8%AA%E6%95%B0%E5%88%97%E4%B8%8D%E7%AD%89%E5%BC%8F%E5%B7%B2%E7%9F%A5an%2B1%28+%E6%8C%87%E7%AC%ACn%2B1%E9%A1%B9+%29%3Dan%2B%28an%5E2%29%2F%28n%5E2%29%2Ca1%3D1%2F3.%E6%B1%82%E8%AF%81an%3E1%2F2+-1%2F4n+.%E5%8F%A6%E5%A4%96%E6%88%91%E5%B0%86%E4%B8%8D%E7%AD%89%E5%BC%8F%E6%94%BE%E7%BC%A9%E6%88%90%E6%9B%B4%E4%B8%A5%E6%A0%BC%E7%9A%84%E5%85%88%E6%B1%82%E8%AF%81+an%3E1%2F2+-+1%2F5n+%E5%90%8E%E5%8F%8D%E8%80%8C%E5%A5%BD%E8%AF%81%E4%BA%86%2C%E6%98%AF%E4%BB%80%E4%B9%88%E9%81%93%E7%90%86%E5%91%A2%3F)
请问怎样用数学归纳法证明这个数列不等式已知an+1( 指第n+1项 )=an+(an^2)/(n^2),a1=1/3.求证an>1/2 -1/4n .另外我将不等式放缩成更严格的先求证 an>1/2 - 1/5n 后反而好证了,是什么道理呢?
请问怎样用数学归纳法证明这个数列不等式
已知an+1( 指第n+1项 )=an+(an^2)/(n^2),a1=1/3.求证an>1/2 -1/4n .另外我将不等式放缩成更严格的先求证 an>1/2 - 1/5n 后反而好证了,是什么道理呢?
请问怎样用数学归纳法证明这个数列不等式已知an+1( 指第n+1项 )=an+(an^2)/(n^2),a1=1/3.求证an>1/2 -1/4n .另外我将不等式放缩成更严格的先求证 an>1/2 - 1/5n 后反而好证了,是什么道理呢?
证明:
当n=1时,a2=a1+(a1^2)/1^2=1/3+1/18=7/18>1/2-1/4=7/28成立
设n=k时,ak+1=ak+(ak^2)/(k^2)>1/2-1/4k成立
当n=k+1时,ak+2=ak+1+(ak+1^2)/(k+1)^2
>1/2-1/4k+(1/2-1/4k)^2/(k+1)^2
=1/2-1/4[k-(1/2-1/4k)^2/(k+1)^2]
=1/2-1/4[(k^3+2k^2+k-1+k-1/4k^2)/(k+1)^2]
>1/2-1/4[(k^3+3k^2+3k+1)/(k+1)^2]
=1/2-1/4[(k+1)^3/(k+1)^2]
=1/2-1/4(k+1)成立
所以an>1/2 -1/4n
n=1,a2=a1+(a1^2)/(1^2)=1/3+1/9=4/9>(1/2-1/4=1/4)
设n=k成立,即ak+1=ak+(ak^2)/(k^2)>1/2-1/4k成立
n=k+1时,
ak+2=ak+1+(ak+1^2)/(k+1)^2
>1/2-1/4k+(1/2-1/4k)/(k+1)^2 //带入n=k时成立的假设
=1/2-1/4[k-(...
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n=1,a2=a1+(a1^2)/(1^2)=1/3+1/9=4/9>(1/2-1/4=1/4)
设n=k成立,即ak+1=ak+(ak^2)/(k^2)>1/2-1/4k成立
n=k+1时,
ak+2=ak+1+(ak+1^2)/(k+1)^2
>1/2-1/4k+(1/2-1/4k)/(k+1)^2 //带入n=k时成立的假设
=1/2-1/4[k-(1/2-1/4k)^2/(k+1)^2]
=1/2-1/4[(k^3+2k^2+k-1+k-1/4k^2)/(k+1)^2]
>1/2-1/4[(k^3+3k^2+3k+1)/(k+1)^2]
=1/2-1/4[(k+1)^3/(k+1)^2]
=1/2-1/4(k+1)成立
即n=k+1时,an>1/2 -1/4n
综上得 n>0且n属于N时,an>1/2 -1/4n 成立
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