Find the remainder when $P= \displaystyle\sum_{n=3}^{42} \binom{\binom{n}{2}}{2}$ is divided by 10.
I went down the obvious, brute force method of writing $\binom{k}{2}=\frac{k(k-1)}{2}$ and ended up with $$\frac 18 \sum_{n=3}^{42}(n^4-2n^3+3n^2-2n)$$ and applied the power sum formulas. The algebra got really messy.
Can anyone provide a faster calculation of the above method or some other innovative procedure? Geometric or combinatorial interpretations like the first amswer of Combinatorial Geometric proof of $\binom{\binom{n}{2}}{3} > \binom{\binom{n}{3}}{2}$ are also appreciated.
Hint
$$P_k= \displaystyle\sum_{n=3}^{k} \binom{\binom{n}{2}}{2}=\frac{1}{40} (k-2) (k-1) k (k+1) (k+2)$$ could help