For n=1, we have
2!n!(2n)!β=2!!2!β=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+1β
(k+1)!(2k+2)!ββ=2β
2kβ
(k+1)β
k!(2k+2)(2k+1)(2x)!β=2kβ
k!(2k)!ββ
2(k+1)2(k+1)β
(2k+1)β=2kβ
k!(2k)!ββ
(2k+1)where 2kβ
k!(2k)!β,(2k+1) are oddβ
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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