Question

If $x=1+a+a^2+a^3+...$ and $y=1+b+b^2+b^3+....$, then show that $1+ab+a^2{b}^2+a^3{b}^3+...=\frac{xy}{x+y-1}$

Answer 1

Given, $x=1+a+a^2+a^3+...$
$S_{\infty }=\frac{a}{1-r}$
Here, $a=1,r=a$
Therefore, $x=\frac{1}{1-a}$
$\Rightarrow 1-a=\frac{1}{x}$
$\Rightarrow a=1-\frac{1}{x}$
$\Rightarrow y=1+b+b^2+b^3+....$
$\Rightarrow a=1,r=b$
$\Rightarrow y=\frac{1}{1-b}$
$\Rightarrow 1-b=\frac{1}{y}$
$\Rightarrow b=1-\frac{1}{y}$
Taking L.H.S
$1+ab+a^2{b}^2+a^3{b}^2+.....$
$=1+ab+(ab{)}^2+(ab{)}^2+.....$
$=1+ab+(ab{)}^2+(ab{)}^3+.....$
$S_{\infty }=\frac{a}{1-r}$ Here $a=1,r=ab$
$=\frac{1}{1-ab}$
$=\frac{1}{1-(1-\frac{1}{x})(1-\frac{1}{y})}$
$=\frac{1}{1-(1-\frac{1}{y}-\frac{1}{x}+\frac{1}{xy})}$
$=\frac{1}{1-1+\frac{1}{y}+\frac{1}{x}-\frac{1}{xy}}$
$=\frac{1}{\frac{1}{y}+\frac{1}{x}-\frac{1}{xy}}=\frac{1}{\frac{x+y-1}{xy}}$
$=\frac{xy}{x+y-1}$