multivariate power series

$\begin{align*} \fr{1-x-xy}&=\fr{1-x(1+y)} \\ &=\fr{1-x}+\frac{x}{(1-x)^2}y+\frac{x^2}{(1-x)^3}y^2+\frac{x^3}{(1-x)^4}y^3+\cdots \end{align*}$

$\begin{align*} &&= (1&&+x&&+x^2&&+x^3&&+x^4&&+x^5&&+\cdots)&& \\ &&&&+(x&&+2x^2&&+3x^3&&+4x^4&&+5x^5&&+\cdots)y&& \\ &&&&&&+(x^2&&+3x^3&&+6x^4&&+10x^5&&+\cdots)y^2& &\\ &&&&&&&&+(x^3&&+4x^4&&+10x^5&&+\cdots)y^3 &&\\ &&&&&&&&&&+(x^4&&+5x^5&&+\cdots)y^4& &\\ &&&&&&&&&&&&+\cdots&&&& \end{align*}$

$\fr{1-x-xy}=\sum_{n,k=0}^\infty \binom{n}{k} x^n y^k$

“operations on multivariate power series correspond to operations on the (multivariate) coefficient sequence. For example, multiplication by $x$ corresponds to a shift in $n$ and multiplication by $y$ corresponds to shift in $k$ . Our result about the bivariate generating function of the binomial coefficients is therefore just a reformulation of the Pascal triangle relation:”

$\binom{n}{k}-\binom{n-1}{k}-\binom{n-1}{k-1}=0 \quad (n,k>0).$

further reading
The Concrete Tetrahedron