Binomial Coefficients

Pascal's triangle is a triangular arrangement of the numbers $^kC_r$. Its $(k+1)^{th}$ row consists of $^kC_0,{}^kC_1,\ldots,{}^kC_k$:

Comparing this against the expansions $(a+b)^0=1$, $(a+b)^1=a+b$, $(a+b)^2=a^2+2ab+b^2$, $(a+b)^3=a^3+3a^2b+3ab^2+b^3$ shows the coefficients literally ARE the rows of the triangle.

The building rule. Each row starts and ends with $1$, and every other entry is the sum of the two entries diagonally above it: e.g. $3$ is $1+2$, and $10$ is $4+6$. So Pascal's triangle — and hence the full binomial expansion of $(a+b)^n$ for any $n\in\mathbb N$ — can be built using addition alone, with no multiplication or division at all. Reading off the sixth row, $1,5,10,10,5,1$, against the (unweighted) power-terms $a^5,a^4b,a^3b^2,a^2b^3,ab^4,b^5$ instantly gives …