Summation of $arccosleft(frac{n^2+r^2+r}{sqrt{(n^2+r^2+r)^2+n^2}}right)$












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I found this question in a book, and cannot solve it.



I have to find the the sum $$S_n=sum_{r=0}^{n-1} arccosleft(frac{n^2+r^2+r}{sqrt{(n^2+r^2+r)^2+n^2}}right)$$



I tried converting this to $arctan(frac{n}{n^2+r^2+r})$ which seemed the most possible way of solving this but can't convert this into a difference of two terms which would help in telescoping the sum.



So my question is:




Am I on the right track or do I need to change my approach completely?
Any help would be highly appreciated.











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    1












    $begingroup$


    I found this question in a book, and cannot solve it.



    I have to find the the sum $$S_n=sum_{r=0}^{n-1} arccosleft(frac{n^2+r^2+r}{sqrt{(n^2+r^2+r)^2+n^2}}right)$$



    I tried converting this to $arctan(frac{n}{n^2+r^2+r})$ which seemed the most possible way of solving this but can't convert this into a difference of two terms which would help in telescoping the sum.



    So my question is:




    Am I on the right track or do I need to change my approach completely?
    Any help would be highly appreciated.











    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      I found this question in a book, and cannot solve it.



      I have to find the the sum $$S_n=sum_{r=0}^{n-1} arccosleft(frac{n^2+r^2+r}{sqrt{(n^2+r^2+r)^2+n^2}}right)$$



      I tried converting this to $arctan(frac{n}{n^2+r^2+r})$ which seemed the most possible way of solving this but can't convert this into a difference of two terms which would help in telescoping the sum.



      So my question is:




      Am I on the right track or do I need to change my approach completely?
      Any help would be highly appreciated.











      share|cite|improve this question











      $endgroup$




      I found this question in a book, and cannot solve it.



      I have to find the the sum $$S_n=sum_{r=0}^{n-1} arccosleft(frac{n^2+r^2+r}{sqrt{(n^2+r^2+r)^2+n^2}}right)$$



      I tried converting this to $arctan(frac{n}{n^2+r^2+r})$ which seemed the most possible way of solving this but can't convert this into a difference of two terms which would help in telescoping the sum.



      So my question is:




      Am I on the right track or do I need to change my approach completely?
      Any help would be highly appreciated.








      summation trigonometric-series






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      edited Dec 17 '18 at 17:46









      user10354138

      7,4322925




      7,4322925










      asked Dec 17 '18 at 17:19









      user35508user35508

      2,9392623




      2,9392623






















          1 Answer
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          $begingroup$

          Hint: compute $$tanleft(arctan{frac{r+1}{n}}-arctan{frac{r}{n}}right).$$






          share|cite|improve this answer









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          • $begingroup$
            Ah ,That was it...Thanks,I got it
            $endgroup$
            – user35508
            Dec 17 '18 at 17:35











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          6












          $begingroup$

          Hint: compute $$tanleft(arctan{frac{r+1}{n}}-arctan{frac{r}{n}}right).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah ,That was it...Thanks,I got it
            $endgroup$
            – user35508
            Dec 17 '18 at 17:35
















          6












          $begingroup$

          Hint: compute $$tanleft(arctan{frac{r+1}{n}}-arctan{frac{r}{n}}right).$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Ah ,That was it...Thanks,I got it
            $endgroup$
            – user35508
            Dec 17 '18 at 17:35














          6












          6








          6





          $begingroup$

          Hint: compute $$tanleft(arctan{frac{r+1}{n}}-arctan{frac{r}{n}}right).$$






          share|cite|improve this answer









          $endgroup$



          Hint: compute $$tanleft(arctan{frac{r+1}{n}}-arctan{frac{r}{n}}right).$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 17:32









          MindlackMindlack

          4,920211




          4,920211












          • $begingroup$
            Ah ,That was it...Thanks,I got it
            $endgroup$
            – user35508
            Dec 17 '18 at 17:35


















          • $begingroup$
            Ah ,That was it...Thanks,I got it
            $endgroup$
            – user35508
            Dec 17 '18 at 17:35
















          $begingroup$
          Ah ,That was it...Thanks,I got it
          $endgroup$
          – user35508
          Dec 17 '18 at 17:35




          $begingroup$
          Ah ,That was it...Thanks,I got it
          $endgroup$
          – user35508
          Dec 17 '18 at 17:35


















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