JEE Mains · Maths · STD 11 - 8. sequence and series
If \(|x|<1,|y|<1\) and \(x \neq y,\) then the sum to infinity of the following series \((x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots .\)
- A \(\frac{x+y-x y}{(1-x)(1-y)}\)
- B \(\frac{x+y-x y}{(1+x)(1+y)}\)
- C \(\frac{x+y+x y}{(1+x)(1+y)}\)
- D \(\frac{x+y+x y}{(1-x)(1-y)}\)
Answer & Solution
Correct Answer
(A) \(\frac{x+y-x y}{(1-x)(1-y)}\)
Step-by-step Solution
Detailed explanation
\(|x|<1,|y|<1, x \neq y\) \((x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)\) \(+\ldots \ldots \ldots\) By multiplying and dividing \(x-y:\) \(\frac{\left(x^{2}-y^{2}\right)+\left(x^{3}-y^{3}\right)+\left(x^{4}-y^{4}\right)+\ldots \ldots}{x-y}\)…
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