$$S_n = \sum_{r=0}^{n-1} \cos^{-1} ( \frac{n^2+r^2+r}{\sqrt{n^4 +r^4+2r^3+2n^2 r^2+2n^2r+n^2+r^2}})$$ find$S_{100}$

Now the denominator of this expression is to big and confusing.I don't know how can we resolve it, because of such big expression in limited amount of time. It needs to be telescopic of some kind, but I don't know how to make it. Help please


Notice that $\displaystyle n^4 +r^4+2r^3+2n^2 r^2+2n^2r+n^2+r^2 = (n^2 + r^2 + r)^2 + n^2$


$\displaystyle S_n=$

$\displaystyle \sum_{r=0}^{n-1} \cos^{-1} ( \frac{n^2+r^2+r}{\sqrt{(n^2 + r^2 + r)^2 + n^2}})$

$\displaystyle =\sum_{r=0}^{n-1} \tan^{-1} \frac{n}{n^2+r^2+r} $

$\displaystyle =\sum_{r=0}^{n-1} \tan^{-1} \frac{\frac{1}{n}}{1 + \frac{r^2+r}{n^2}} $

(Divide numerator and denominator by $n^2$)

$\displaystyle =\sum_{r=0}^{n-1} \tan^{-1} \frac{\frac{r+1}{n} - \frac{r}{n}}{1 + \frac{r+1}{n}\frac{r}{n}}$

$\displaystyle =\sum_{r=0}^{n-1} \left (\tan^{-1} \frac{r+1}{n} - \tan^{-1} \frac{r}{n} \right)$

$\displaystyle = \frac{\pi}{4}$

Please check the calculations.

  • 1
    $\begingroup$ The expression is $(n^2+r^2+r)^2+n^2$ $\endgroup$ – RandomAspirant Aug 14 at 5:12
  • $\begingroup$ Thank You so so much $\endgroup$ – RandomAspirant Aug 14 at 5:54
  • $\begingroup$ Think it should be $n^2$ in the denominator before the explanation in parentheses. Just a typo I think. $\endgroup$ – Mike Aug 14 at 8:11
  • $\begingroup$ @Mike Fixed... Thanks a lot for pointing out. $\endgroup$ – PTDS Aug 14 at 12:48

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