设数列{bn}满足bn=n^2/2^(n+1),证明:bn
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设数列{bn}满足bn=n^2/2^(n+1),证明:bn设数列{bn}满足bn=n^2/2^(n+1),证明:bn设数列{bn}满足bn=n^2/2^(n+1),证明:bn设c(n)=b(n)/b(
设数列{bn}满足bn=n^2/2^(n+1),证明:bn
设数列{bn}满足bn=n^2/2^(n+1),证明:bn
设数列{bn}满足bn=n^2/2^(n+1),证明:bn
设c(n)=b(n)/b(n-1)=(1+1/(n-1))^2/2,是个随n增加而递减的数列.
c(2)=2>1,c(3)=1.125>1,c(4)=0.89=5后,c(n)1,则b(n)>b(n-1)；反之b(n)
用数学归纳法证明吧
当n=1时b1=1/2^2=1/4<9/16
假设当n=k时，bk=k^2/2^(k+1)<=9/16成立，即1/2^(k+1)<=9/(16k^2)
则当n=k+1时，b(k+1)=(k+1)^2/2^(k+2)=(k+1)^2/[2*2^(k+1)]
<=(k+1)^2*(9/(32k^2)=9(k+1)^2/(32k^2)=9(k^2+2k...
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用数学归纳法证明吧
当n=1时b1=1/2^2=1/4<9/16
假设当n=k时，bk=k^2/2^(k+1)<=9/16成立，即1/2^(k+1)<=9/(16k^2)
则当n=k+1时，b(k+1)=(k+1)^2/2^(k+2)=(k+1)^2/[2*2^(k+1)]
<=(k+1)^2*(9/(32k^2)=9(k+1)^2/(32k^2)=9(k^2+2k+1)/(32k^2)
当k>3时有2k+1
于是当n=k+1时b(k+1)<=9/16成立
所以对一切n都有bn=n^2/2^(n+1)<=9/16
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