JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

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Analyzing +4 | -1

|x|<1,|y|<1

and

x \neq y

, then the sum to infinity of the following series:

(x+y)+\left(x^2+x y+y^2\right)+\left(x^3+x^2 y+x y^2+y^3\right)+\ldots

, is

View answer and explanation

Correct option: B

\frac{x+y-x y}{(1-x)(1-y)}

Option B (

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

) is correct because the series can be transformed into a difference of geometric series, leading to this simplified expression.

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