Question:
If $|x|<1$, $|y|<1$ and $x \neq y$, then the sum to infinity of the following series
$(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots \ldots$
Correct Option: 1,
Solution:
$|x|<1,|y|<1, x \neq y$
$(x+y)+\left(x^{2}+x y+y^{2}\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)$........
By multiplying and dividing $x-y$ :
$\frac{\left(x^{2}-y^{2}\right)+\left(x^{3}-y^{3}\right)+\left(x^{4}-y^{4}\right)+\ldots \ldots}{x-y}$
$=\frac{\left(x^{2}+x^{3}+x^{4}+\ldots \ldots\right)-\left(y^{2}+y^{3}+y^{4}+\ldots \ldots\right)}{x-y}$
$=\frac{\frac{x^{2}}{1-x}-\frac{y^{2}}{1-y}}{x-y}$
$=\frac{\left(x^{2}-y^{2}\right)-x y(x-y)}{(1-x)(1-y)(x-y)}$
$=\frac{x+y-x y}{(1-x)(1-y)}$