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?

A similar but slightly different variation of Jack's proof above.
$$\begin{align}
&1\cdot (1)\\
+&2\cdot(1+2)\\
+&3\cdot(1+2+3)\\
+&4\cdot(1+2+3+4)\\
+&\vdots\qquad\vdots \; \quad\ddots\quad \ddots\\
+&n\cdot(1+2+3+\cdots+n)\\
&=\sum_{i=1}^ni\sum_{j=1}^i j=\sum_{i=1}^ni\sum_{j=1}^i {j\choose 1}\\
&=\sum_{i=1}^ni{i+1\choose 2}\\
&=\sum_{i=1}^n\left[(i+2){i+1\choose 2}-2{i+1\choose 2}\right]\\
&=\sum_{i=1}^n\left[3{i+2\choose 3}-2{i+1\choose 2}\right]\\
&=3{n+3\choose 4}-2{n+2\choose 3}\\
&={n+2\choose 3}\left[3\left(\frac{n+3}4\right)-2\right]
\qquad\qquad\text{using }{n+3\choose 4}=\frac{n+3}4{n+2\choose 3}\\
&=\frac{(n+2)(n+1)n}{1\cdot 2\cdot 3}\cdot \left[ \frac{3n+1}4\right]\\
&=\frac1{24}n(n+1)(n+2)(3n+1)\qquad\blacksquare
\end{align}$$

NB - the above makes use of the identity
$$\sum_{i=1}^n{i+m\choose m+1}={n+1+m\choose m+2}$$
