Question

Given $x=1+a+a^2+....\infty$ and $y=1+b+b^2....\infty$ where $a$ and $b$ are proper fraction $1+ab+a^2{b}^2+....\infty$ equals to ..........

Answer 1

As a and b are proper fraction so $x$ and $y$ can be calculated as sum of infinite series

Common ratio of series $x$ is $a$ and that of $y$ is $b$

$\therefore x=\frac{1}{1-a},y=\frac{1}{1-b}$

$\Rightarrow a=\frac{x-1}{x},b=\frac{y-1}{y}$

So $1+ab+a^2{b}^2+.......\infty =\frac{1}{1-ab}$

Put values of $a$ and $b$ in above, we get

$=\frac{1}{1-\left(\frac{x-1}{x}\right)\left(\frac{y-1}{y}\right)}$

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