If |x| < 1, |y| < 1 and x ≠ y, then the sum to infinity of the following series (x + y) + (x2 + xy + y2 ) + …

Question

If |x| < 1, |y| < 1 and x $\neq$ y, then the sum to infinity of the following series (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + ..... is : 

(1) $x+y+xy\over(1-x)(1-y)$

(2) $x+y-xy\over(1-x)(1-y)$

(3) $x+y-xy\over(1-x)(1+y)$

(4) $x+y+xy\over(1+x)(1+y)$

Answer 1

Ans. (2) $x+y-xy\over(1-x)(1-y)$

Sol. (x + y) + (x² + xy + y²) + (x³ + x²y + xy² + y³) + ........

$={1\over x-y}\left({x^2\over1-x}-{y^2\over1-y}\right)$

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

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

Comments

Satisfying solution