JEE PYQ: Sequence And Series Question 19
Question 19 - 2020 (02 Sep Shift 1)
If $|x| < 1$, $|y| < 1$ and $x \neq y$, then the sum to infinity of the following series $(x + y) + (x^2 + xy + y^2) + (x^3 + x^2y + xy^2 + y^3) + \ldots$ is:
(1) $\frac{x + y - xy}{(1+x)(1+y)}$
(2) $\frac{x + y + xy}{(1+x)(1+y)}$
(3) $\frac{x + y - xy}{(1-x)(1-y)}$
(4) $\frac{x + y + xy}{(1-x)(1-y)}$
Show Answer
Answer: (3)
Solution
$= \frac{1}{x-y}\left[(x^2 - y^2) + (x^3 - y^3) + (x^4 - y^4) + \ldots\right]$
$= \frac{1}{x-y}\left[\frac{x^2}{1-x} - \frac{y^2}{1-y}\right] = \frac{(x-y)(x+y-xy)}{(x-y)(1-x)(1-y)}$
$= \frac{x + y - xy}{(1-x)(1-y)}$
BETA
