当n为正偶数,求证n/(n-1)+n(n-2)/(n-1)(n-3)+...+n(n-2).2/(n-1)(n-3)..
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当n为正偶数,求证n/(n-1)+n(n-2)/(n-1)(n-3)+...+n(n-2).2/(n-1)(n-3)...1=n
网上能找到一个回答、但我看不懂.求详解
网上能找到一个回答、但我看不懂.求详解
可利用归纳法证明
n=2时,2/1=2,成立
假设n=2k时,k为正整数,结论成立
则n=2k+2时,有
(2k+2)/(2k+1)+(2k+2)(2k)/[(2k+1)(2k-1)]+...+(2k+2)(2k).2/[(2k+1)(2k-1)...1]
=[(2k+2)/(2k+1)]{1+2k/(2k-1)+...+2k(2k-2)...2/[(2k-1)(2k-3)...1]}
由归纳假设知 2k/(2k-1)+...+2k(2k-2)...2/[(2k-1)(2k-3)...1]=2k
∴上式=[(2k+2)/(2k+1)](1+2k)
=2k+2,由此知n=2k+2时也成立
∴对任意正偶数n,该结论均成立
n=2时,2/1=2,成立
假设n=2k时,k为正整数,结论成立
则n=2k+2时,有
(2k+2)/(2k+1)+(2k+2)(2k)/[(2k+1)(2k-1)]+...+(2k+2)(2k).2/[(2k+1)(2k-1)...1]
=[(2k+2)/(2k+1)]{1+2k/(2k-1)+...+2k(2k-2)...2/[(2k-1)(2k-3)...1]}
由归纳假设知 2k/(2k-1)+...+2k(2k-2)...2/[(2k-1)(2k-3)...1]=2k
∴上式=[(2k+2)/(2k+1)](1+2k)
=2k+2,由此知n=2k+2时也成立
∴对任意正偶数n,该结论均成立
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