作者existfor (余歪歪)
看板Math
标题Re: [其他] 数学归纳法证明
时间Wed Jan 19 22:50:23 2011
n=1 => 1/2 <= 1/2
设当n=k时成立
则(1/2)(3/4).....((2k-1)/2k)<=1/(3k+1)^(1/2)
目标是证n=k+1也成立
则(1/2)(3/4).....((2k-1)/2k)*((2k+1)/(2k+2))
<=1/(3k+1)^(1/2)*((2k+1)/(2k+2))
=(2k+1)/(3k+1)^(1/2)*(2k+2)
=(2k+1)/2*[(3k+1)(k^2+2k+1)]^(1/2)
=(2k+1)/(12k^3+28k^2+20k+4)^(1/2)
=1/(3k+4)^(1/2)
=1/[3(k+1)+1]^(1/2)
by数学归纳法.....................................DONE.
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