Question

If $X=1+a+a^2+....\infty$, where |a| < 1 and $y=1+b+b^2+.....\infty$ where |b| <1,

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

Answer 1

We have, x = 1+ $a+a^2+....\infty$

$\therefore x=\frac{1}{1-a}[\because \text{sum of infinite GP is}{S}_{\infty }=\frac{a}{1-r}]$

$1-a=\frac{1}{x}\Rightarrow a=1-\frac{1}{x}....\left(i\right)$

$y=1+b+b^2+....\infty \Rightarrow y=\frac{1}{1-b}$

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

$1+ab+a^2{b}^2+....\infty =\frac{1}{1-ab}[\because S_{\infty }-\frac{1}{1-r}]$

$1+ab+a^2{b}^2+....\infty =\frac{1}{1-(1-\frac{1}{x})(1-\frac{1}{y})}=\frac{1}{1-\frac{(x-1)(y-1)}{xy}}$

$=\frac{xy}{xy-xy+x+y-1}=\frac{xy}{x+y-1}$