# 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 triangle1, 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 facts2 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:

$\sum_{k=1}^n k (n - k) = n \sum_{k=1}^n k - \sum_{k=1}^n k^2 = n \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6}$ $= n(n+1) \left( \frac{n}{2} - \frac{2n+1}{6} \right) = n(n+1) \frac{3n - (2n+1)}{6}$ $= \frac{n(n+1)(n-1)}{6} = {n + 1 \choose 3}$ $\tag*{\blacksquare}$

## 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:

$\sum_{m=1}^n \sum_{k=1}^m k(n-k) = \sum_{m=1}^n {m+1 \choose 3} = {n+2 \choose 4}$

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 $$\sum_{m=0}^{n-1} {m \choose k} = {n \choose k+1}$$, we have:

$\sum_{m=0}^n {m \choose k} = {n \choose k} + \sum_{m=0}^{n-1} {m \choose k} = {n \choose k} + {n \choose k+1}$

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

1. 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
.              .
.                 .
.                    .


It’s also known as OEIS A003991

2. $$\sum_{k=1}^n k$$ has many derivations, including this clever visual one.

$$\sum_{k=1}^n k^2$$ can be derived by expanding $$(k-1)^3$$ as illustrated here