# Determinants in Pascal's Triangle

#### Tony Foster November 6, 2016

Starting with the sequences of figurate numbers in Pascal's triangle, we reveal, via determinants, triangular numbers as exponents with base $2.$

Below are several examples with $\displaystyle T_{n}=\sum_{k=1}^nk=\frac{n(n+1)}{2},$ the $n^{\text{th}}$ triangular number.

\displaystyle \begin{align} &\left|\begin{array}{cc}1&1\\1&3\end{array}\right|=2=2^1=2^{T_1}\\ \,\\ &\left|\begin{array}{ccc}1&1&1\\1&3&6\\1&5&15\end{array}\right|=8=2^{1+2}=2^{T_2}\\ \,\\ &\left|\begin{array}{cccc}1&1&1&1\\1&3&6&10\\1&5&15&35\\1&7&28&84\end{array}\right|=64=2^{1+2+3}=2^{T_3}\\ \,\\ &\left|\begin{array}{ccccc}1&1&1&1&1\\1&3&6&10&15\\1&5&15&35&70\\1&7&28&84&210\\1&9&45&165&495\end{array}\right|=1024=2^{1+2+3+4}=2^{T_4}\\ \,\\ &\left|\begin{array}{cccccc}1&1&1&1&1&1\\1&3&6&10&15&21\\1&5&15&35&70&126\\1&7&28&84&210&462\\1&9&45&165&495&1287\\1&11&66&286&1001&3003\end{array}\right|=1024=2^{1+2+3+4+5}=2^{T_5}\\ \end{align}

The twitter user @mathforge has pointed out a paper Determinants of Matrices from Pascal's Triangle by Andrew Stacey with a much more general result.

Write Pascal's triangle in left-justified form, starting with $0\text{s}:$

$\left|\begin{array}{cccccc}1&0&0&0&0&\ldots\\1&1&0&0&0&\ldots\\1&2&1&0&0&\ldots\\1&3&3&1&0&\ldots\\1&4&6&4&1&\ldots\\&\ldots\end{array}\right|$

The table may also be expanded upwards using $\displaystyle {n \choose i}=-(1)^{i}{i-n-1\choose i},$ for negative $n.$ Then we have the following statement (Corollary 3.1 from the article):

Let $k\in\mathbb{N}$ and $l_1,\ldots,l_k,n_1,\ldots,n_k\in\mathbb{Z}.$ Let $B$ be the $(k+1)\times (k+1)$ matrix

$\displaystyle B=\left(\begin{array}{ccccc}1&n_1&{n_2\choose 2}&\ldots&{n_k\choose k}\\ 1&n_1+l_1&{n_2+l_2\choose 2}&\ldots&{n_k+l_k\choose k}\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ 1&n_1+l_1k&{n_2+l_2k\choose 2}&\ldots&{n_k+l_kk\choose k}\end{array}\right),$

then $\displaystyle \det B=\prod_{i=1}^{k}l_i^{\,i}.$ In particular, if $l_i=1$ for all $i$ then $\det B=1$ and if $l_i=l$ for all $i$ then $\displaystyle \det B=l^{\frac{k(k+1)}{2}}.$