Prove the following


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

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

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

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

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

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

Correct Option: , 3


$\mathrm{S}=(x+y)+\left(x^{2}+y^{2}+x y\right)+\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)+\ldots \infty$

$=\frac{1}{x-y}\left[\left(x^{2}-y^{2}\right)+\left(x^{3}-y^{3}\right)+\left(x^{4}-y^{4}+\ldots \infty\right)\right]$

$=\frac{1}{x-y}\left[\frac{x^{2}}{1-x}-\frac{y^{2}}{1-y}\right]=\frac{(x-y)(x+y-x y)}{(x-y)(1-x)(1-y)}$ $\left[\because S_{\infty}=\frac{a}{1-r}\right]$

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


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