Intuition for sum of triangular numbers and significance for $3\choose{k}$ Specific question: according to my calculation based on Timbuc's answer to this question,
$$\sum_{k=0}^n\frac{k(k+1)}{2}=\frac{n(n+1)(2n+4)}{12} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; =\frac{n(n+1)(n+2)}{6}$$
[Edit: RHS simplified based on suggestion from herbsteinberg.]
If this is right, is there an intuitive or geometric proof of this?
Background and motivation: I'm trying to relate the concepts of integration and summation using reasoning which I find simple and intuitive. 
As part of that, I'm trying to understand the process of summing sequences in this rotated section of Pascal's triangle: 
$(0)\;\;01\;\;01\;\;01\;\;01\;\;01\;\;01\;\;...\frac{k^0}{0!} \\ (1)\;\;01\;\;02\;\;03\;\;04\;\;05\;\;06\;\;...\frac{k+0}{1!} \\ (2)\;\;01\;\;03\;\;06\;\;10\;\;15\;\;21\;\;...\frac{k(k+1)}{2!} \\ (3)\;\;01\;\;04\;\;10\;\;20\;\;35\;\;56\;\;...\frac{k(k+1)(k+2)}{3!}$
I see that summing each line seems to increase the degree of the expression by $1$ and I can imagine that rearrangement, simplification and/or a limit process could later reduce these terms to $\frac{k^2}{2}$, $\frac{k^3}{3}$ etc., but for now I'm interested in how/why each line has the exact expression it does.
Line $(1)$ is just counting. 
Line $(2)$ I can picture and understand as in Fig. 1 below.
Fig. 1

Line $(3)$ [Edited to add the following, which may start to answer my question] I can picture and understand as a stepped version of the right-angled tetraga in Fig. 2 below.
Fig. 2

 A: Intuitively, you are walking the pascal triangle, in a zigzag way.
$$\binom{n}{k} + \binom{n}{k-1} = \binom{n+1}{k}$$
For your example, $\large \sum_{k=0}^{n}{k(k+1)\over 2} = \sum_{k=1}^{n}\binom{k+1}{2} = \binom{n+2}{3}$
$$\large\begin{matrix}
\binom{2}{2} = \binom{3}{3} = 1 \cr 
\binom{3}{3} + \binom{3}{2} = \binom{4}{3} = 4 \cr
\binom{4}{3} + \binom{4}{2} = \binom{5}{3} = 10 \cr
\binom{5}{3} + \binom{5}{2} = \binom{6}{3} = 20 \cr
\cdots \cr
\Large\binom{n+1}{3} + \binom{n+1}{2} = \binom{n+2}{3}
\end{matrix}$$
A: Inspired by Sum of Consecutive Squares Formula Proof
$$S_n = \sum_{k=1}^n\binom{k+1}{2}$$
$$3S_4 = \begin{matrix}
1 && 1 && 4\cr
12 && 21 && 33\cr
123 &+& 321 &+& 222\cr
1234 &&4321 && 1111\cr
\end{matrix}$$
$3S_4 = \begin{matrix}
6\cr
66\cr
666\cr
6666 \end{matrix}$
$6S_4 = \begin{matrix}
66666\cr
66666\cr
66666\cr
66666 \end{matrix}$
$$S_4 = {4(4+1)(4+2) \over 6}$$
$$→ S_n = {n(n+1)(n+2) \over 6}$$
A: Each row can be written as a sum of combinatorial numbers. This is what is known as the Hockey Stick Identity
A nice way of imaginining the intuition behind these identities is understanding a combinatorial proof for the hockey stick identity
