用数学归纳法证明4n/(n+1)≤(2n)!/(n!)^2n为大于1的整数
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用数学归纳法证明4n/(n+1)≤(2n)!/(n!)^2n为大于1的整数
用数学归纳法证明4n/(n+1)≤(2n)!/(n!)^2
n为大于1的整数
用数学归纳法证明4n/(n+1)≤(2n)!/(n!)^2n为大于1的整数
说明:此题n为大于等于的整数也是成立的
证明:(1)当n=1时,∵4n/(n+1)=4*1/(1+1)=2
(2n)!/(n!)^2=(2*1)!/(1!)^2=2
∴4n/(n+1)≤(2n)!/(n!)^2成立
当n=2时,∵4n/(n+1)=4*2/(2+1)=8/3
(2n)!/(n!)^2=(2*2)!/(2!)^2=6
∴4n/(n+1)≤(2n)!/(n!)^2成立
(2)假设当n=k (k>2)时,4k/(k+1)≤(2k)!/(k!)^2成立
当n=k+1时,∵(2n)!/(n!)^2=(2(k+1))!/((k+1)!)^2
=[(2k+2)(2k+1)(2k)!]/[(k+1)^2*(k!)^2]
=[(2k)!/(k!)^2]*[2(k+1)(2k+1)/(k+1)^2]
≥[4k/(k+1)]*[2(k+1)(2k+1)/(k+1)^2]
=[4(k+1)/(k+2)]*[(2k)(k+2)(2k+1)/(k+1)^3]
=[4(k+1)/(k+2)]*[(2k)/(k+1)]*[(k+2)/(k+1)]*[(2k+1)/(k+1)]
>[4(k+1)/(k+2)]*1*1*1
=[4(k+1)/(k+2)]
=4n/(n+1)
∴4n/(n+1)≤(2n)!/(n!)^2成立
故综合(1)和(2),由数学归纳法得4n/(n+1)≤(2n)!/(n!)^2 (n为大于等于的整数)成立.