For n=1, we have
2!n!(2n)!=2!!2!=1 is odd
Hence the result is true for n=1 (base case)
Let it be true for n=K.
Now for n=k+1
2k+1⋅(k+1)!(2k+2)!=2⋅2k⋅(k+1)⋅k!(2k+2)(2k+1)(2x)!=2k⋅k!(2k)!⋅2(k+1)2(k+1)⋅(2k+1)=2k⋅k!(2k)!⋅(2k+1)where 2k⋅k!(2k)!,(2k+1) are odd
Hence the result is true for n=k+1.
Thus the result is true for all +ve integers by mathematical induction.
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