Question

If $|x|<1,|y|<1$ and $x\ne y,$ then the sum to infinity of the following series $(x+y)+(x^2+xy+y^2)+(x^3+x^{2}y+x{y}^2+y^3)+......\infty$

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

Answer 1

The correct option is B $\frac{x+y-xy}{(1-x)(1-y)}$

$(x+y)+(x^2+xy+y^2)+(x^3+x^{2}y+x{y}^2+y^3)+......\infty$

$=\frac{1}{(x-y)}\{(x^2-y^2)+(x^3-y^3)+(x^4-y^4)+....\infty \}$ $=\frac{\frac{x^2}{1-x}-\frac{y^2}{1-y}}{x-y}$

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

$=\frac{(x^2-y^2)-xy(x-y)}{(1-x)(1-y)(x-y)}=\frac{((x+y)-xy(x-y))}{(1-x)(1-y)(x-y)}$

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