Question

$x=1+a+a^2+....\infty ,y=1+b+b^2+....\infty ,\left|ab\right|<1,$ then $1+ab+a^2{b}^2+....\infty =$

• A. $\frac{xy}{x+y-1}$
• B. $\frac{x+y-1}{yx}$
• C. $\frac{xy}{x+y+1}$
• D. $\frac{x+y+1}{xy}$

Answer 1

Answer: (A)

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

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