Question

If $x = 1 + a + {a^2} + ....\infty {\rm{ }}(a < 1)$ and $y = 1 + b + {b^2} + ....\infty {\rm{ }}(b < 1)$. Then the value of $1 + ab + {a^2}{b^2} + ....\infty$ is
A. $\dfrac{{xy}}{{x + y - 1}}$
B. $\dfrac{{xy}}{{x + y + 1}}$
C. $\dfrac{{xy}}{{x - y - 1}}$
D. $\dfrac{{xy}}{{x - y + 1}}$

Answer 1

Hint: In this question, we must find the sum of the terms of the given infinite series. In this type of the geometric progression, we can find the sum of the series using the appropriate formula.

Formula used: The sum to infinite geometric progression refers to the sum of terms in an infinite geometric progression. The formula of the sum of infinite GP is ${S_\infty } = \dfrac{a}{{1 - r}}$, where $a$ = first term and $r$ = common ratio $ = \dfrac{{{a_n}}}{{{a_{n - 1}}}}$.

Complete step by step solution: The series $x = 1 + a + {a^2} + ....\infty {\rm{ }}(a < 1)$ is in GP with $a = 1$ and $r = a$.
Then, the sum of the infinite geometric progression is calculated by the formula, \({S_\infty } = \dfrac{a}{{1 - r}}\),
\[\begin{array}{l}x = \dfrac{1}{{1 - a}} \Rightarrow x(1 - a) = 1\\1 - a = \dfrac{1}{x} \Rightarrow a = \left( {1 - \dfrac{1}{x}} \right){\rm{ }}....{\rm{(1)}}\end{array}\]
Similarly, the series \(y = 1 + b + {b^2} + ....\infty {\rm{ }}(b < 1)\)is in GP with $a = 1$ and $r = b$.
Then, the sum of the infinite geometric progression is calculated by the formula, \({S_\infty } = \dfrac{a}{{1 - r}}\),
\[\begin{array}{l}y = \dfrac{1}{{1 - b}} \Rightarrow y(1 - a) = 1\\1 - b = \dfrac{1}{y} \Rightarrow b = \left( {1 - \dfrac{1}{y}} \right){\rm{ }}......(2)\end{array}\]
And let \(z = 1 + ab + {a^2}{b^2} + ....\infty \).
In this GP, $a = 1$ and the common ratio is calculated as $r = \dfrac{{{a_n}}}{{{a_{n - 1}}}} = \dfrac{{{a_2}}}{{{a_1}}} = \dfrac{{ab}}{1} = ab$
Then, the sum of the infinite geometric progression is calculated by the formula, ${S_\infty } = \dfrac{a}{{1 - r}}$
$z = \dfrac{1}{{1 - ab}}$
Put the value of $a$ and $b$ in $z$.
  \[\begin{array}{l}1 + ab + {a^2}{b^2} + ....\infty = \dfrac{1}{{1 - ab}}\\\dfrac{1}{{1 - ab}} = \dfrac{1}{{1 - \left( {1 - \dfrac{1}{x}} \right)\left( {1 - \dfrac{1}{y}} \right)}}\\ = \dfrac{1}{{1 - \left( {\dfrac{{x - 1}}{x}} \right)\left( {\dfrac{{y - 1}}{y}} \right)}}\\ = \dfrac{1}{{1 - \left( {\dfrac{{(x - 1)(y - 1)}}{{xy}}} \right)}}\\ = \dfrac{1}{{\left( {\dfrac{{xy - (x - 1)(y - 1)}}{{xy}}} \right)}}\\ = \dfrac{{xy}}{{\left( {xy - (xy - x - y + 1)} \right)}}\\ = \dfrac{{xy}}{{\left( {xy - xy + x + y - 1} \right)}}\\ = \dfrac{{xy}}{{x + y - 1}}\end{array}\]

Thus, Option (A) is correct.

Note: Here the given series is a geometric infinite series. So, we can find the common ratio from this easily and substitute the value in the obtained sum.