Mathematics · 2020
JEE Main · 2 September 2020, Shift 1 · Q58
If |x|<1,|y|<1 and x ≠ y, then the sum to infinity of the following series (x+y)+(x^2+x y+y^2)+(x^3+x^2 y+x y^2+y^3)+…. is:
If $\displaystyle |x|<1,|y|<1$ and $\displaystyle x \neq y$, then the sum to infinity of the following series $\displaystyle (x+y)+\left(x^2+x y+y^2\right)+\left(x^3+x^2 y+x y^2+y^3\right)+\ldots$. is :
Official answer
(3)
$\displaystyle \frac{x+y-x y}{(1-x)(1-y)}$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.