证明 (n+1)(n+2)(n+3)...(2n-1)2n=2n*1*3*5***(2n-3)(2n-1)

你好
证明：
n=1时,n+1=2
(2^1)*1=2,等式成立.
假设当n=k(k为自然数,且k>=1)时等式成立.
即
(k+1)(k+2)...(k+k)=(2^k)*1*3*...*(2k-1)
则当n=k+1时,
(k+1+1)(k+1+2)...(k+1+k-1)(k+1+k)(k+1+k+1)
=(k+2)(k+3)...(k+k)(2k+1)(2k+2)
=(k+1)(k+2)...(k+k)(2k+1)(2k+2)/(k+1)
=(k+1)(k+2)...(k+k)(2k+1)2
=(2^k)*1*3*...*(2k-1)*(2k+1)*2
=[2^(k+1)]*1*3*...*[2(k+1)-1]
等式也成立.
综上,(n+1)(n+2)(n+3)+.+(n+n)=(2^n)*1*3*.(2n-1)
等式成立.

原题是:求证: (n+1)(n+2)(n+3)...(2n-1)2n=(2^n)·1·3·5·...·(2n-3)(2n-1),n∈N*

用分析法证明：
要证 (n+1)(n+2)(n+3)...(2n-1)2n=(2^n)·1·3·5·...·(2n-3)(2n-1),n∈N*
只需证 1·2·3·...·n·(n+1)(n+2)(n+3)...(2n-1)(2n)
=1·2·3·...·n·(2^n)·1·3·5·...·(2n-3)(2n-1),n∈N*
只需证 1·2·3·...·n·(n+1)(n+2)(n+3)...(2n-1)(2n)
=(2·4·6·...·2n)·(1·3·5·...·(2n-3)(2n-1)),n∈N*
因左边=(2n)!
右边=(2·4·6·...·2n)·(1·3·5·...·(2n-3)(2n-1))
=1·2·3·...·(2n-3)(2n-2)(2n-1)(2n)
=(2n)!
即1·2·3·...·n·(n+1)(n+2)(n+3)...(2n-1)(2n)
=(2·4·6·...·2n)·(1·3·5·...·(2n-3)(2n-1)),n∈N*成立
所以 (n+1)(n+2)(n+3)...(2n-1)2n=(2^n)·1·3·5·...·(2n-3)(2n-1),n∈N*

希望能帮到你！