How to deduce that $1\cdot 1 + 2\cdot 1 + 2\cdot 2 + 3\cdot 1+3\cdot 2+3\cdot 3 +...+(n\cdot n) = n(n+1)(n+2)(3n+1)/24$ I know how to reason $$1\cdot2 + 2\cdot3 + 3\cdot4 + n(n-1) = \frac{1}{3}n(n-1)(n+1)$$
However, I'm stuck on proving $$1\cdot1 + (2\cdot1 + 2\cdot2) + (3\cdot1+3\cdot2+3\cdot3) + \cdots +(n\cdot 1+...+n\cdot n) = \frac{1}{24}n(n+1)(n+2)(3n+1).$$
Could somebody shed some light on me?

The combinatorial identity:
$$ \sum_{k=0}^{K}\binom{n+k}{n}=\binom{n+K+1}{n+1}\tag{1}$$
can be easily proved by induction. It gives:
$$\sum_{j=1}^{n-1}j(j+1) = 2\sum_{j=1}^{n-1}\binom{j+1}{2} = 2\binom{n+1}{3} = \frac{(n+1)n(n-1)}{2}\tag{2}$$
as well as:
$$\begin{eqnarray*} \sum_{k=1}^{n}k(1+2+\ldots+k) &=& 2\sum_{k=1}^{n}k\binom{k+1}{2}=2\sum_{k=1}^{n}\binom{k+1}{2}+6\sum_{k=1}^{n}\binom{k+1}{3}\\&=&2\binom{n+2}{3}+6\binom{n+2}{4}.\tag{3}\end{eqnarray*}$$
