It can be shown from some rather tedious expansion (or wolframalpha!) that $$\sum_{r=1}^n \binom {2r}2=\frac 16n(n+1)(4n-1)$$
Is it possible to derive this closed form result easily using summation identities of binomial coefficients, with minimal or no algebraic expansion?
The result can also be expressed as $\displaystyle\binom {n+1}2 \frac {4n-1}3$ and resembles the result for the sum of squares. Not that it would necessarily help.
Note also that $\displaystyle\sum_{r=1}^n \binom r2=\binom {n+1}3$.
You can write:$$\binom{2n}{2}=4\binom{n}{2}+\binom{n}{1}$$ This can be seen with a counting argument. The set of pairs of distinct elements in $\{1,2,\dots,2n\}$ contains $n$ pairs of the form $\{2k-1,2k\}$. The paiselrs not of this form can be found by selecting a pair of distinct values, $a,b\in\{1,2,\dots,n\}$ and then choosing any of the four pairs of $\{2a,2b\},\{2a-1,2b\},\{2a,2b-1\},$ or $\{2a-1,2b-1\}$.
Numerically, any polynomial of degree $k$ can be written as:
$$a_k\binom{n}{k}+a_{k-1}\binom{n}{k-1}+\cdots +a_0\binom{n}{0}$$
For your example, $$\binom{2n}{2}=n(2n-1)=2n^2-n = 4\frac{n^2-n}{2}+n=4\binom{n}{2}+\binom{n}{1}.$$