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

Question

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

Answer 1

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

Hence

$\displaystyle S_n=$

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

$\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.