Question

If $x=1+a+a^2+a^3+.....\infty$ and $y=1+b+b^2+b^3+.....\infty$; prove that $1+ab+a^2{b}^2+....\infty =\frac{xy}{x+y-1}$.

Answer 1

$X=1+a+a^2+a^3+.....\infty \to \left(1\right)$

$y=1+b+b^2+b^3+.....\infty \to \left(2\right)$

$1+ab+a^2{b}^2+....\infty =?\to \left(3\right)$

$S_{\infty }=\frac{a}{1-r}=\frac{1}{1-a}=x$ for equation (1)

$S_{\infty }=\frac{1}{1-b}=y$ for equation (2)

so 1 - a $=\frac{1}{x}$ so a $=1-\frac{1}{x}$

& b $=1\frac{-1}{y}$

for equation (3):-

$S_{\infty }=\frac{1}{1-ab}=\frac{1}{1-(1-\frac{1}{x})(1-\frac{1}{y})}=\frac{1}{1-1+\frac{1}{y}+\frac{1}{x}-\frac{1}{xy}}$

$S_{\infty }=\frac{xy}{x+y-1}\Rightarrow$ Hence Proved