Prove that the product of 2n consecutive negative integers is divisibl

1.	Prove that the product of 2n consecutive negative integers is divisible by (2n)!
Answer» Let us assume the negative consecutive integers are –1, – 2,......, – (2n) Let M be the product of the negative integers, ⇒ M = ( – 1).( – 2)( – 3)......( – 2n + 1).( – 2n) ⇒ M = ( – 1)2n(1.2.3.......(2n – 1).(2n)) ⇒ M = 1.2.3......……(2n – 1).(2n) We know that, n! = n(n – 1)(n – 2)…………2.1 ⇒ M = (2n)! ⇒ \(\frac{M}{(2n)!}\) = \(\frac{(2n)!}{(2n)!}\) ⇒ \(\frac{M}{(2n)!}\) = 1 ∴ M is divisible by (2n)!.

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

Answer»

Let us assume the negative consecutive integers are –1, – 2,......, – (2n)

Let M be the product of the negative integers,

⇒ M = ( – 1).( – 2)( – 3)......( – 2n + 1).( – 2n)

⇒ M = ( – 1)2n(1.2.3.......(2n – 1).(2n))

⇒ M = 1.2.3......……(2n – 1).(2n)

We know that,

n! = n(n – 1)(n – 2)…………2.1

⇒ M = (2n)!

⇒ \(\frac{M}{(2n)!}\) = \(\frac{(2n)!}{(2n)!}\)

⇒ \(\frac{M}{(2n)!}\) = 1

∴ M is divisible by (2n)!.