<s>[INST] Find the sum to n terms $$S=1^2+3^2+6^2+10^2+15^2+.......$$
My attempt is as follows:
$$T_n=\left(\frac{n\cdot\left(n+1\right))}{2}\right)^2$$
$$T_n=\frac{n^4+n^2+2\cdot n^3}{4}$$
$$S=\frac{1}{4}\cdot\sum_{n=1}^{n}\left(n^4+n^2+2\cdot n^3\right)$$ 
Now to solve this one has to calculate $\sum_{n=1}^{n}n^4$ which will be a very lengthy process, is there any shorter method to solve this question?
By the way I calculated $\sum_{n=1}^{n}n^4$ and it came as $\dfrac{\left(n\right)\cdot\left(n+1\right)\cdot\left(2\cdot n+1\right)\cdot\left(3\cdot n^2+3\cdot n-1\right)}{30}$, then I substituted this value into the original equation.
Then I got final answer as $\dfrac{\left(n\right)\cdot\left(n+1\right)\cdot\left(n+2\right)\cdot\left(3\cdot n^2+6\cdot n+1\right)}{60}$
But it took me a very long time to calculate all of this, is there any shorter way to solve this problem?
 [/INST] You may use the hockey-stick identity, like in the computation of $\sum_{n=1}^{N}n^m$. We have
$$ \binom{n}{2}^2 = 6\binom{n}{4}+12\binom{n}{3}+7\binom{n}{2}+\binom{n}{1} $$
hence
$$\begin{eqnarray*} \sum_{n=1}^{N}\binom{n}{2}^2 &=& 6\binom{N+1}{5}+12\binom{N+1}{4}+7\binom{N+1}{3}+\binom{N+1}{2}\\&=&\color{red}{\frac{N(N+1)(N+2)(3N^2+6N+1)}{60}}.\end{eqnarray*}$$
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