Question

If $x=1+a+a^2+....\infty \mathrm{and}y=1+b+b^2+....\infty$ where a and b are proper fractions, then $1+ab+a^2{b}^2+...\infty =?$

• A. $\frac{xy}{x+y-1}$
• B. $\frac{xy}{x-y}$
• C. $\frac{\left(x^2+y^2\right)}{x-y}$
• D. none of the above

Answer 1

The correct option is A

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

Explanation for the correct answer:

Find the value of $1+ab+a^2{b}^2+...\infty$:

Given,

$$\begin{gathered} x=1+a+a^2+....\infty =\frac{1}{1-a} \\ y=1+b+b^2+....\infty =\frac{1}{1-b} \end{gathered}$$ ; [$\because$ sum of infinite terms in GP is $\frac{a}{1-r}$ , where $a$ is the first term and $r$ is the common ratio.]

From the above equations, we can write

$$\begin{gathered} a=\frac{x-1}{x} \\ b=\frac{y-1}{y} \end{gathered}$$

now,

we can write $1+ab+a^2{b}^2+...\infty$as $\frac{1}{1-ab}$

substitute the values of $a$ and $b$

$\begin{array}{rcl}\frac{1}{1-ab}& =& \frac{1}{1-\left({\displaystyle \frac{x-1}{x}}\right)\left({\displaystyle \frac{y-1}{y}}\right)}\\ & =& \frac{xy}{x+y-1}\end{array}$

Hence, the correct option is A.