Question

$x=1+a+a^2+\dots \infty ,y=1+b+b^2+\dots \infty$ then $1+ab+a^2{b}^2+\dots \infty =$

Answer 1

$\begin{array}{c}Wehave\\ x=1+a+a^2+......=\frac{1}{1-a}\\ y=1+b+b^2+.......=\frac{1}{1-b}\\ 1+ab+a^2{b}^2+.......=\frac{1}{1-ab}\\ =\frac{1}{1-\left(1-\frac{1}{x}\right)\left(1-\frac{1}{y}\right)}\\ =\frac{1}{1-\left(1-\frac{1}{x}-\frac{1}{y}+\frac{1}{xy}\right)}\\ =\frac{1}{\frac{1}{x}+\frac{1}{y}-\frac{1}{xy}}\\ =\frac{xy}{y+x-1}\end{array}$

Hence, this is the answer.