Evaluate $\sum_{k=1}^n \binom{2k}{2}$ combinatorially How to evaluate the expression $$\displaystyle \sum_{k=1}^n \binom{2k}{2}$$ using a combinatorial argument? 
Sorry I have little clue so cannot provide any working for it. Not homework. Related to previous question.
 A: A simple combinatorial argument of choosing $2$ balls from $2k$ red balls and $1$ blue ball gives us
$$\binom{2k+1}{2} = \binom{2k}{2} + 2k$$
And $$\sum_{k=1}^{2n+1} \binom{k}{2} = \binom{2n+2}{3}$$
Hence
$$2 \sum_{k=1}^{n} \binom{2k}{2} + \sum_{k=1}^{n} 2k = \binom{2n+2}{3}$$
and hence
$$2 \sum_{k=1}^{n} \binom{2k}{2} = \binom{2n+2}{3} - \sum_{k=1}^{n} 2k$$
$$\sum_{k=1}^{n} \binom{2k}{2} = \frac{\binom{2n+2}{3}}{2} - \binom{n+1}{2}$$
(Some of the machinery for this argument has been borrowed from this question in particular the $\displaystyle \sum_{k=1}^{n} \binom{k}{2} = \binom{n+1}{3}$)
A: Well, $\binom{2k}{2} = k(2k-1) = 2k^{2} - k$. So, 
$$ \sum_{k=1}^{n} \binom{2k}{2} = 2\sum_{k=1}^{n} k^{2} - \sum_{k=1}^{n} k = n(n+1) \Big( \frac{2n+1}{3} - \frac{1}{2} \Big) $$
A: The magical question:
$$\displaystyle \sum_{k=1}^n \binom{2k}{2}$$
Substitute with the formula:
$$\displaystyle \sum_{k=1}^n \frac{(2k)!}{(2k-2)!2!}$$
Simplify (we can do this because we know for sure that $2k > 1$):
$$\displaystyle \sum_{k=1}^n \frac{(2k)(2k-1)}{2} = 2\sum_{k=1}^n k^2 - \sum_{k=1}^n k$$
And now:
$$2(\frac{n(n+1)(2n+1)}{6}) - \frac{n(n+1)}{2}$$
$$(n+1)\frac{2n(2n+1)-3n}{6}$$
