Question

Prove that the product of 2n consecutive negative integers is divisible by (2n)!

Answer 1

Product = [(2n + 1)(2n + 3) (2n + 5)...(2n + r)]
$=\frac{\left(2n\right)!\left[\right(2n+1\left)\right(2n+3)....(2n+1\left)\right]}{\left(2n\right)!}=\frac{\left(2n\right)\left[\right(2n-1\left)\right(2n-2)....4.2(2n+1\left)\right(2n+3\left)\right]}{\left(2n\right)!}=\frac{(2n+r)!}{\left(2n\right)!}$
Hence r = 2n
$=\frac{(2n+2n)!}{2n}=\frac{\left(4n\right)!}{\left(2n\right)!}=\left(2n\right)!$