用数学归纳法证明“(n+1)(n+2)…(n+n)=2^n·1·3·5…(2n-1)(n∈N*)”时,从n=k到n=k+

n=k时
等式左边为 (k+1)(k+2)...(k+k)
当n=k+1时
等式左边为 [(k+1)+1][(k+1)+2].[(k+1)+k][(k+1)+k+1]
比原来多了 两项[(k+1)+k][(k+1)+k+1] =2(2k+1)(k+1)
但是少了 一项 k+1
所以两式相除得需增加2(2k+1)

等价于乘(2k+1)(2k+2)再除以(k+1)
就是2(2k+1)

n=k时，左边=(k+1)(k+2)……（k+k)
n=k+1时，左边=[(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+2)(k+3)……(k+k)(2k+1)*2(k+1)
=(k+1)(k+2)(k+3)……(k+k)(2k+1)*2