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

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

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

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

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

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

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

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Satisfying solution