Prove that the product of 2n consecutive negative integers is div… - Vidyadip

Question

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

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Answer

We have,
Product by,
$=\big[(2\text{n}+1)(2\text{n}+3)(2\text{n}+5)....(2\text{n}+\text{r})\big]$
$=\frac{(2\text{n})\big[(2\text{n+1})(2\text{n+3})....(2\text{n}+\text{r})\big]}{(2\text{n}!)}$
$=\frac{(2\text{n}+\text{r})!}{(2\text{n})!}$
Hence r = 2n
$=\frac{(2\text{n}+2\text{n})!}{2\text{n}}$
$=\frac{(4\text{n})!}{(2\text{n})!}$
$=(2\text{n})!$

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