Question

If $x=1+a+a^2+...\infty ,(|a|<1),y=1+b+b^2+...\infty ,(|b|<1),$ then $z=1+ab+a^2{b}^2+a^3{b}^3+...\infty$ is

• A. x+y-1xy
• B. 2xyx+y-1
• C. xyx+y-1
• D. x+y-12xy

Answer 1

The correct option is C xyx+y-1

We want to find $z$ in this problem. It is going to be in terms of 1 and $b.$

But the options are in terms of $x$ and $y.$

We can find $a$ and $b$ in terms of $x$ and $y$ from the first two infinite series.

$x=\frac{1}{1-a}$ ... (1)

$y=\frac{1}{1-b}$ ... (2)

$z=\frac{1}{1-ab}$ ... (3)

From (1), $1-a=\frac{1}{x}$

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

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

Similarly, $b=\frac{y-1}{y}$

$\therefore z=\frac{1}{1-\frac{(x-1)}{x}\frac{(y-1)}{y}}$

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

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