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

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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

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
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

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
首先,最后两项求和,n(n-2)...2/(n-1)(n-3)...1 + n(n-2)...4/(n-1)(n-3)...3 = [n(n-2)...4/(n-1)(n-3)...3](2/1+1)
=n(n-2)...4/(n-1)(n-3)...5
然后该结果再和前一项求和,n(n-2)...4/(n-1)(n-3)...5+n(n-2)...6/(n-1)(n-3)...5
=[n(n-2)...6/(n-1)(n-3)...5](4+1)=n(n-2)...6/(n-1)(n-3)...7
由此不断往前,可推得结果为n