Question

If $\left|x\right|<1$, $\left|y\right|<1$, then the sum of infinity of the following series: $\left(x+y\right)+\left(x^2+xy+y^2\right)+\left(x^3+x^{2}y+x{y}^2+y^3\right)+......$ is

• A. $\frac{x+y+xy}{\left(1-x\right)\left(1-y\right)}$
• B. $\frac{x+y-xy}{\left(1-x\right)\left(1-y\right)}$
• C. $\frac{x+y+xy}{\left(1+x\right)\left(1+y\right)}$
• D. $\frac{x+y-xy}{\left(1+x\right)\left(1+y\right)}$

Answer 1

The correct option is B

$\frac{x+y-xy}{\left(1-x\right)\left(1-y\right)}$

The explanation for the correct option

The given series: $\left(x+y\right)+\left(x^2+xy+y^2\right)+\left(x^3+x^{2}y+x{y}^2+y^3\right)+......+\infty$

$$\begin{gathered} =\frac{\left(x-y\right)\left(x+y\right)+\left(x-y\right)\left(x^2+xy+y^2\right)+\left(x-y\right)\left(x^3+x^{2}y+x{y}^2+y^3\right)+......+\infty }{\left(x-y\right)} \\ =\frac{\left(x^2-y^2\right)+\left(x^3-y^3\right)+\left(x^4-y^4\right)+.....+\infty }{\left(x-y\right)} \\ =\frac{\left(x^2+x^3+x^4+......+\infty \right)-\left(y^2+y^3+y^4+......+\infty \right)}{\left(x-y\right)} \end{gathered}$$

General formula: The sum of an infinite geometric progression with first term $a$ and common ratio $(r<1)$ can be given by $\frac{a}{1-r}$.

It is given that $\left|x\right|<1$, $\left|y\right|<1$.

Thus, $\left(x+y\right)+\left(x^2+xy+y^2\right)+\left(x^3+x^{2}y+x{y}^2+y^3\right)+......+\infty =\frac{{\displaystyle \frac{x^2}{1-x}}-{\displaystyle \frac{y^2}{1-y}}}{x-y}$

$$\begin{gathered} \Rightarrow \frac{{\displaystyle \frac{x^2\left(1-y\right)-y^2\left(1-x\right)}{\left(1-x\right)\left(1-y\right)}}}{x-y} \\ =\frac{\left(x^2-y^2\right)-\left(x^{2}y-x{y}^2\right)}{\left(x-y\right)\left(1-x\right)\left(1-y\right)} \\ =\frac{\left(x-y\right)\left(x+y\right)-xy\left(x-y\right)}{\left(x-y\right)\left(1-x\right)\left(1-y\right)} \\ =\frac{x+y-xy}{\left(1-x\right)\left(1-y\right)} \end{gathered}$$

Hence, (B) is the correct option.