How to prove this binomial identity :
$$ { 2n \choose n } = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$
The left hand side arises while solving a standard binomial problem the right hand side is given in the solution , I checked using induction that this is true but I am inquisitive to prove it in a rather general way.
EDIT: I have gone through all of the answers posted here,I particularly liked Isaac♦
answers after which it was not much difficult for me to figure out something i would rather say an easy and straight algebraic proof, I am posting it here if somebody needs in future :
$$ { 2n \choose n } = \frac{(2n)!}{n! \cdot n!} $$
$$ = \frac{ 1 \cdot 2 \cdot 3 \cdots n \cdot (n+1) \cdots (2n-1)\cdot (2n) }{n! \cdot n!}$$
$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdots (2n)]}{n! \cdot n!} $$
$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [(2.1) \cdot (2.2) \cdot (2.3) \cdots (2.n)]}{n! \cdot n!} $$
$$= \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [ 2^n \cdot (n)! ]}{n! \cdot n!} $$
$$ = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$
I do always welcome your comments 🙂
Best Answer
If you have proved it via induction, the you have proved it. I don't know what you mean by "in a rather general way".
But note that $$1\cdot3\cdot5\cdot\cdots\cdot(2n-1)$$ is just crying out to be rewritten as $$\frac{1\cdot2\cdot3\cdot\cdots\cdot(2n)} {2\cdot4\cdot6\cdot\cdots\cdot(2n)}.$$