If x < 1 ,   y < 1 and x ≠ 1 , then the sum to infinity of the following series x + y + x…

Let $S=\left(x+y\right)+\left(x^2+xy+y^2\right)+\left(x^3+x^{2}y+x{y}^2+y^3\right)+.....$

Multiply $\left(x-y\right)$ both sides

$\Rightarrow \left(x-y\right)S=\left(x-y\right)(\left(x+y\right)+\left(x^2+xy+y^2\right)$$+\left(x^3+x^{2}y+x{y}^2+y^3\right)+...)$

$\Rightarrow \left(x-y\right)S=\left(x-y\right)\left(x+y\right)+\left(x-y\right)\left(x^2+xy+y^2\right)+\left(x-y\right)\left(x^3+x^{2}y+x{y}^2+y^3\right)+.....$

$\Rightarrow \left(x-y\right)S=\left(x^2-y^2\right)+\left(x^3-y^3\right)+\left(x^4-y^4\right)+.....$

$\Rightarrow \left(x-y\right)S=\left(x^2+x^3+x^4+....\right)-\left(y^2+y^3+y^4+.....\right)$

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

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