Please give me the compete solution of the above.

Asked by Balbir | 9th Oct, 2017, 08:53: PM

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 Sneha shidid | 18th Dec, 2017, 10:50: AM