Question

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

• A. $\frac{xy}{\left(x+y-1\right)}$
• B. $\frac{xy}{\left(x+y+1\right)}$
• C. $\frac{xy}{\left(x-y-1\right)}$
• D. $\frac{xy}{\left(x-y+1\right)}$

Answer 1

The correct option is A

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

Explanation of correct option

Step 1: Note the given data

$x=1+a+a^2+.....\infty ,\left|a\right|<1$

$y=1+b+b^2+.......\infty ,\left|b\right|<1$

The series $x$is in geometric progression with $a=1$and $r=a$

The series $y$is in geometric progression with $a=1$and $r=b$

Step 2: Find the sum of infinite terms of a geometric progression

The formula for the sum of infinite numbers in a geometric progression is $S_{\infty }=\frac{a}{1-r}$

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

$\Rightarrow$$x-ax=1$

$\Rightarrow$ $ax=x-1$

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

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

$\Rightarrow$$y-by=1$

$\Rightarrow$ $by=y-1$

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

Multiplying equations (1) and (2)

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

$\Rightarrow$$ab=\frac{\left(x-1\right)\left(y-1\right)}{xy}........\left(3\right)$

Step 3: Find the value of $1+ab+a^2{b}^2+.......\infty$

The series is in geometric progression with $a=1;r=ab$

The formula for the sum of infinite numbers in a geometric progression is $S_{\infty }=\frac{a}{1-r}$

$S_{\infty }=\frac{1}{1-ab}$

$\Rightarrow$$S_{\infty }=\frac{1}{1-\left({\displaystyle \frac{\left(x-1\right)\left(y-1\right)}{xy}}\right)}\left[\because from\left(3\right)\right]$

$\Rightarrow$$S_{\infty }=\frac{xy}{xy-\left(x-1\right)\left(y-1\right)}$

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

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

Hence, the correct option is A.