I've been staring at this identity which appears in my textbook for a while. Plugging in numbers I can verify that this is true, however I have no idea how this was determined or proved? Is there a nice result about binomial coefficients that gives this identity?
The only thing that comes to mind is a creative application of Pascal's identity, but even then I'm getting the result.
On
Recall that $\displaystyle{a \choose b} = \frac{a \cdot (a-1) \cdots (a-b+1)}{b!}$
The denominator of $\displaystyle{2n \choose n}$ is $n!$.
The denominator of $\displaystyle{2(n-1) \choose n-1}$ is $(n-1)!$.
So $\displaystyle\frac{1}{n} {2(n-1) \choose n-1}$ has the same denominator as ${2n \choose n}$.
The numerator of $\displaystyle{2n \choose n}$ is $2n\cdots (n+1)$.
The numerator of $\displaystyle{2(n-1) \choose n-1}$ is $(2n-2)\cdots n$.
So $\displaystyle\frac{(2n)(2n-1)}{n}{2(n-1) \choose n-1}$ has the same numerator as ${2n \choose n}$.
$$2(2n-1)\frac1n\binom{2n-2}{n-1}=\frac{(2n)(2n-1)}{n^2}\frac{(2n-2)!}{(n-1)!(n-1)!} =\frac{(2n)!}{n!n!}=\binom{2n}n.$$