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

by DivMit   Last Updated August 14, 2019 05:20 AM

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

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

Hence

$$S_n=$$

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

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

$$=\sum_{r=0}^{n-1}$$

$$=$$

PTDS
August 14, 2019 05:07 AM

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