Question

If $a,b$ are positive numbers such that
$x=1+a+a^2+.....\infty (a<1)$;
$y=1+b+b^2....\infty (b<1)$;
then $1+ab+a^2{b}^2+....\infty =$

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

Answer 1

Answer: (A)

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

Given,

$x=1+a+a^2+.......\infty$

or, $x=\frac{1}{1-a}$ [ Since $a<1$].........(1).

Also given,

$y=1+b+b^2+........\infty$

or, $y=\frac{1}{1-b}$ [ Since $b<1$].......(2).

Now,

$x+y-1=\frac{1-ab}{(1-a)(1-b)}$........(3).[ Using (1), (2)]

Now

$\frac{xy}{x+y-1}=\frac{1}{1-ab}$......(4) [ using (3)}

Now,

$1+ab+a^2{b}^2+.......\infty$

$=\frac{1}{1-ab}$ [ As $a<1$ and $b<1$ then $ab<1$]

$=\frac{xy}{x+y-1}$ [ From (4)]