Medium MCQ +4 / -1 PYQ · JEE Mains 2020

If |x| < 1, |y| < 1 and x $\ne$ y, then the sum to infinity of the following series

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

  1. A ${{x + y - xy} \over {\left( {1 + x} \right)\left( {1 + y} \right)}}$
  2. B ${{x + y - xy} \over {\left( {1 - x} \right)\left( {1 - y} \right)}}$ Correct answer
  3. C ${{x + y + xy} \over {\left( {1 + x} \right)\left( {1 + y} \right)}}$
  4. D ${{x + y + xy} \over {\left( {1 - x} \right)\left( {1 - y} \right)}}$

Solution

(x + y) + (x<sup>2</sup>+xy+y<sup>2</sup>) + (x<sup>3</sup>+x<sup>2</sup>y + xy<sup>2</sup>+y<sup>3</sup>) + .... <br><br>By multiplying and dividing x – y : <br><br>$${{\left( {{x^2} - {y^2}} \right) + \left( {{x^3} - {y^3}} \right) + \left( {{x^4} - {y^4}} \right) + ...} \over {x - y}}$$ <br><br>= $${{\left( {{x^2} + {x^3} + {x^4} + ....} \right) - \left( {{y^2} + {y^3} + {y^4} + ...} \right)} \over {x - y}}$$ <br><br>= ${{{{{x^2}} \over {1 - x}} - {{{y^2}} \over {1 - y}}} \over {x - y}}$ <br><br>= $${{\left( {{x^2} - {y^2}} \right) - xy\left( {x - y} \right)} \over {\left( {1 - x} \right)\left( {1 - y} \right)\left( {x - y} \right)}}$$ <br><br>= ${{x + y - xy} \over {\left( {1 - x} \right)\left( {1 - y} \right)}}$

About this question

Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression

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