Primal-Trigo Summation cos

Geometry Level 4

$\large{\sum _{ n=1 }^{ 2015 }{ \cos ^{ 5 }{ \left( \frac { n\pi }{ 403 } \right) } } }=\ ?$

The answer is -1.

1 solution

Otto Bretscher
Oct 25, 2015

By symmetry $\cos(5\pi-t)=\cos(\pi-t)=-\cos(t)$ , the summands with indices $n$ and $2015-n$ will cancel out, so that $\sum_{n=1}^{2014}\cos^5\left(\frac{n\pi}{403}\right)=0$ . Adding all the way to $n=2015$ makes the sum $\cos^5(5\pi)=\boxed{-1}$ .

Moderator note:

Nice observation of the symmetry, which allows terms to cancel out. Turns a scary looking problem into a harmless one.

Primal-Trigo Summation cos

The answer is -1.

1 solution

Moderator note:

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