If |x|<1, |y|<1 and


If $|x|<1$, $|y|<1$ and $x \neq y$, then the sum to infinity of the following series

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


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

  2. $\frac{x+y-x y}{(1+x)(1+y)}$

  3. $\frac{x+y+x y}{(1+x)(1+y)}$

  4. $\frac{x+y+x y}{(1-x)(1-y)}$

Correct Option: 1,


$|x|<1,|y|<1, x \neq y$

$(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)$........

By multiplying and dividing $x-y$ :

$\frac{\left(x^{2}-y^{2}\right)+\left(x^{3}-y^{3}\right)+\left(x^{4}-y^{4}\right)+\ldots \ldots}{x-y}$

$=\frac{\left(x^{2}+x^{3}+x^{4}+\ldots \ldots\right)-\left(y^{2}+y^{3}+y^{4}+\ldots \ldots\right)}{x-y}$


$=\frac{\left(x^{2}-y^{2}\right)-x y(x-y)}{(1-x)(1-y)(x-y)}$

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

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