Prove that $\frac{(2n)!}{(n!)^2}-1$ is divisible by $(2n+1)$

Question

Prove that $$\frac{(2n)!}{(n!)^2}-1$$ is divisible by $(2n+1)\;,$ Where $n\in \mathbb{N}$ and $n>1$

$\bf{My\; Try::}$ Let $$S = \frac{(2n)!}{n!^2}-1 = \frac{2^n(2n-1)(2n-3)\cdot \cdot ........\cdot 3 \cdot 2 \cdot 1}{n!}-1$$

Now How can we prove that $2^n(2n-1)(2n-3)\cdot \cdot \cdot .... 3\cdot 2 \cdot 1$ is divisible by $n!$ and odd number

Help Required, Thanks

See http://math.stackexchange.com/questions/1459278/is-72-36-1-divisible-by-73 — lab bhattacharjee, Jan 30 '16 at 13:13

score 3 · Answer 1 · answered Jan 30 '16 at 12:14

3

$\dfrac{(2n)!}{(n!)^2} - 1 = \dbinom{2n}{n}-1$

This is the count of ways to arrange $n$ discrete objects into two equal piles, excluding a particular arrangement.

$\begin{array}{l:l:l} n & \binom{2n}n-1 & 2n+1 \\ 1 & 1 & 3 \\ 2 & 5 & 5 \\ 3 & 19 & 7 \\ 4 & 69 & 9 \\ \vdots & \vdots & \vdots \end{array}$

Only occasionally is $\binom{2n}n -1$ divisible by $2n+1$

answered Jan 30 '16 at 12:14

Graham Kemp

I was struggling to prove this by induction. Guess it was because it was false. – Saikat Jan 30 '16 at 13:25

1 Answers1