Why is it true that $\sum_{n \geq 0} (1+x)^n y^n = \frac{1}{1-y(1+x)}$?

Question

Let $B_n = (1 + x)^n = \sum_k {{n}\choose{k}}x^n$.

Then it is true that for any $y$ and any $n$:

$$\sum_{n \geq 0}B_n x y^n = \sum_{n \geq 0}\sum_k {{n}\choose{k}} x^k y^k = \sum_{n \geq 0} (1+x)^n y^n = \frac{1}{1-y(1+x)}$$

The first two of these assertions follow from the formula. But where does the third one, the fraction, come from?

For reference, this is from page 16 of generatingfunctionology.

Answer 1

Use: $1+x+x^2+...=\frac{1}{1-x}$ if $|x|<1$

Answer 2

Provided $|y(1+x)|<1$, the series is Geometric. Use a formula.