Taking summation to the fourth power

Geometry Level 4

Let S n = ∑ k = 1 n − 1 2 cos ⁡ 4 ( 2 k − 1 n π ) S_{n} = \displaystyle \sum_{k=1}^{\frac {n-1}2} \cos^{4}\left(\frac{2k-1}{n}\pi \right) . Find S 2019 − S 2003 S_{2019} - S_{2003} .


The answer is 3.

Benny Joseph by Benny Joseph , 20, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Benny Joseph
Apr 27, 2018

S n = 3 n − 8 16 ( for odd n ) ; S_{n} = \dfrac{3n-8}{16} (\text{for odd n}) ;
Hence S 2019 − S 2003 = 3 16 ( 2019 − 2003 ) = 3 S_{2019} - S_{2003} = \dfrac{3}{16}(2019-2003) = 3

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...