TITLE: Why is Pascal's Triangle called a Triangle?
QUESTION [19 upvotes]: Pascal's triangle extends infinitely downwards, meaning in only has two sides. Why then is it called a triangle if a triangle, by definition must have three sides?

REPLY [37 votes]: I find Pascal's original form greatly preferable. Slightly modernized:
$$
\begin{array}{ll|llllllllll}
\label{my-label}
  &   & j & → &    &    &     &     &    &    &   &   \\
  &   & 0 & 1 & 2  & 3  & 4   & 5   & 6  & 7  & 8 & 9 \\ \hline
i & 0 & 1 & 1 & 1  & 1  & 1   & 1   & 1  & 1  & 1 & 1 \\
↓ & 1 & 1 & 2 & 3  & 4  & 5   & 6   & 7  & 8  & 9 &   \\
  & 2 & 1 & 3 & 6  & 10 & 15  & 21  & 28 & 36 &   &   \\
  & 3 & 1 & 4 & 10 & 20 & 35  & 56  & 84 &    &   &   \\
  & 4 & 1 & 5 & 15 & 35 & 70  & 126 &    &    &   &   \\
  & 5 & 1 & 6 & 21 & 56 & 126 &     &    &    &   &   \\
  & 6 & 1 & 7 & 28 & 84 &     &     &    &    &   &   \\
  & 7 & 1 & 8 & 36 &    &     &     &    &    &   &   \\
  & 8 & 1 & 9 &    &    &     &     &    &    &   &   \\
  & 9 & 1 &   &    &    &     &     &    &    &   &  
\end{array}
$$
This way, it's defined for all $(i, j) \in \mathbb{N}^2$. The identities are nicer, too:
$$T_{i,j} = T_{j, i}$$
$$T_{i,j} = \frac{(i+j)!}{i!j!}$$
$$T_{i+1,j+1} = T_{i+1,j} + T_{i,j+1}$$
Row sum:
$$\sum_{i+j=n}T_{i,j} = 2^n$$
The "hockey stick":
$$T_{i,j+1} = \sum_{0 \leq k \leq i}{T_{k,j}}$$
And finally, for fans of the binomial coefficient:
$$T_{i,j} = \binom{i+j}{i} = \binom{i+j}{j}$$
Of course, the higher-dimensional forms are easy too:
$$T_{\hat{x}} = \frac{(\sum_{x \in \hat{x}}{x})!}{\prod_{x \in \hat{x}}{(x!)}}$$
$$T_{\hat{x}} = T_{\sigma(\hat{x})}$$
$$T_{\hat{x}} = T_{x_1 - 1, x_2, ..., x_n} + T_{x_1, x_2 - 1, ..., x_n} + ... + T_{x_1, x_2, ..., x_n-1}$$