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Please give me the compete solution of the above.

Asked by Balbir 9th October 2017, 8:53 PM
Answered by Expert
Answer:
begin mathsize 16px style fraction numerator 1 over denominator 1 plus straight x end fraction plus fraction numerator 2 over denominator 1 plus straight x squared end fraction plus fraction numerator 4 over denominator 1 plus straight x to the power of 4 end fraction plus............ fraction numerator 2 to the power of 2 straight n end exponent over denominator 1 plus straight x to the power of 2 straight n end exponent end fraction
fraction numerator 1 over denominator 1 minus straight x end fraction plus fraction numerator 1 over denominator 1 plus straight x end fraction equals fraction numerator 2 over denominator 1 minus straight x squared end fraction
fraction numerator begin display style 2 end style over denominator begin display style 1 minus straight x squared end style end fraction plus fraction numerator begin display style 2 end style over denominator begin display style 1 plus straight x squared end style end fraction equals fraction numerator 4 over denominator 1 minus straight x to the power of 4 end fraction
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Hence comma space sum space of space the space straight n space terms space is space fraction numerator 2 to the power of 2 straight n plus 1 end exponent over denominator 1 minus straight x to the power of 2 straight n plus 1 end exponent end fraction minus fraction numerator 1 over denominator 1 minus straight x end fraction
Simplifying space this space as space per space the space option space you space will space get space the space answer. end style
Answered by Expert 18th December 2017, 10:50 AM
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