An Intriguing Triangle
I recently noticed that the triangle with \(n\)th row and \(k\)th column:
\[\Delta(n, k) = k (n  k)\]…is closely related to the binomial coefficients \(n \choose 3\) and \(n \choose 4\).
Here are the first several rows of the triangle^{1}, along with their sums ∑ and cumulative sums ∑∑:
n 𝚫 ∑ ∑∑
2 1 1 1
3 2 2 4 5
4 3 4 3 10 15
5 4 6 6 4 20 35
6 5 8 9 8 5 35 70
...
(We omit all the zero entries at \(k=0\), \(k=n\), and \(n<2\).)
Theorem 1
Note that the binomial coefficients \({3 \choose 3} = 1, {4 \choose 3} = 4, {5 \choose 3} = 10, {6 \choose 3} = 20, \dots\) look a lot like the ∑ column of the sums of the rows. Sure enough:
\[\sum_{k=1}^n k (n  k) = {n+1 \choose 3}\]Proof
Using the facts^{2} that \(\sum_{k=1}^n k = \frac{n(n+1)}{2}\) and \(\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}\), we have:
\[\sum_{k=1}^n k (n  k) = n \sum_{k=1}^n k  \sum_{k=1}^n k^2 = n \frac{n(n+1)}{2}  \frac{n(n+1)(2n+1)}{6}\] \[= n(n+1) \left( \frac{n}{2}  \frac{2n+1}{6} \right) = n(n+1) \frac{3n  (2n+1)}{6}\] \[= \frac{n(n+1)(n1)}{6} = {n + 1 \choose 3}\] \[\tag*{$\blacksquare$}\]Theorem 2
Note that the binomial coefficients \({4 \choose 4} = 1, {5 \choose 4} = 5, {6 \choose 4} = 15, {7 \choose 4} = 35, \dots\) look a lot like the ∑∑ column of the cumulative sums of the rows. Sure enough:
\[\sum_{m=1}^n \sum_{k=1}^m k(nk) = \sum_{m=1}^n {m+1 \choose 3} = {n+2 \choose 4}\]And in fact, more generally for any \(k\):
\[\sum_{m=0}^n {m \choose k} = {n+1 \choose k+1}\]Proof
(n = 0)
If \(k=0\), \(\sum_{m=0}^0 {m \choose 0} = {0 \choose 0} = 1 = {1 \choose 1}\)
Otherwise, \(k>0\) so \(\sum_{m=0}^0 {m \choose k} = {0 \choose k} = 0 = {1 \choose k+1}\)
(n + 1)
Supposing by induction that \(\sum_{m=0}^{n1} {m \choose k} = {n \choose k+1}\), we have:
\[\sum_{m=0}^n {m \choose k} = {n \choose k} + \sum_{m=0}^{n1} {m \choose k} = {n \choose k} + {n \choose k+1}\]But from Pascal’s triangle we know \({n \choose k} + {n \choose k + 1} = {n+1 \choose k+1}\) \(\tag*{$\blacksquare$}\)
Visual Interpretation
In the context of Pascal’s triangle, \(\sum_{m=0}^n {m \choose k} = {n+1 \choose k+1}\) says that you can sum a diagonal to find the value of the entry below and to the right.
For \(n=3\), \(k=1\) for example, \(\sum_{m=0}^3 {m \choose 1} = 0 + 1 + 2 + 3 = 6 = {4 \choose 2}\) is illustrated by marking the relevant entries of the triangle:
1 +0
1 +1 0
1 +2 1 0
1 +3 3 1 0
1 4 =6 4 1 0

Notice that the triangle is just the antidiagonals of a multiplication table:
1 2 3 4 5 . . . 2 4 6 8 10 3 6 9 12 15 4 8 12 16 20 5 10 15 20 25 . . . . . .

\(\sum_{k=1}^n k\) has many derivations, including this clever visual one.
\(\sum_{k=1}^n k^2\) can be derived by expanding \((k1)^3\) as illustrated here. ↩