Second Column and Diagonal of Pascal's Triangle consist of Triangular Numbers

From ProofWiki
Jump to navigation Jump to search

Theorem

The $2$nd column and $2$nd diagonal of Pascal's triangle consists of the set of triangular numbers.


Proof

Recall Pascal's triangle:

$\begin{array}{r|rrrrrrrrrr} n & \binom n 0 & \binom n 1 & \binom n 2 & \binom n 3 & \binom n 4 & \binom n 5 & \binom n 6 & \binom n 7 & \binom n 8 & \binom n 9 & \binom n {10} & \binom n {11} & \binom n {12} \\ \hline 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 2 & 1 & 2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 3 & 1 & 3 & 3 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 4 & 1 & 4 & 6 & 4 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 5 & 1 & 5 & 10 & 10 & 5 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 6 & 1 & 6 & 15 & 20 & 15 & 6 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 7 & 1 & 7 & 21 & 35 & 35 & 21 & 7 & 1 & 0 & 0 & 0 & 0 & 0 \\ 8 & 1 & 8 & 28 & 56 & 70 & 56 & 28 & 8 & 1 & 0 & 0 & 0 & 0 \\ 9 & 1 & 9 & 36 & 84 & 126 & 126 & 84 & 36 & 9 & 1 & 0 & 0 & 0 \\ 10 & 1 & 10 & 45 & 120 & 210 & 252 & 210 & 120 & 45 & 10 & 1 & 0 & 0 \\ 11 & 1 & 11 & 55 & 165 & 330 & 462 & 462 & 330 & 165 & 55 & 11 & 1 & 0 \\ 12 & 1 & 12 & 66 & 220 & 495 & 792 & 924 & 792 & 495 & 220 & 66 & 12 & 1 \\ \end{array}$


By definition, the entry in row $n$ and column $m$ contains the binomial coefficient $\dbinom n m$.

Thus the $2$nd column contains all the elements of the form $\dbinom n 2$.


The $m$th diagonal consists of the elements in column $n - m$.

Thus the $m$th diagonal contains the binomial coefficients $\dbinom n {n - m}$.

By Symmetry Rule for Binomial Coefficients:

$\dbinom n {n - m} = \dbinom n m$

Thus the $2$nd diagonal also contains the binomial coefficients $\dbinom n 2$.


By Binomial Coefficient with Two: Corollary, the triangular numbers are precisely those numbers of the form $\dbinom n 2$.

Hence the result.

$\blacksquare$


Sources