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



Answers 1


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
PTDS
August 14, 2019 05:07 AM

Related Questions


Updated December 10, 2017 22:20 PM

Updated April 03, 2019 09:20 AM

Updated June 02, 2017 02:20 AM

Updated May 24, 2017 12:20 PM

Updated October 12, 2017 07:20 AM