New Year Problem (5)

Calculus Level 2

What is the value of the expression below?

∑ n = 0 2020 cos ⁡ ( π n 2 3 ) \sum_{n=0}^{2020} \cos \Bigg(\frac{\pi n^2}{3}\Bigg)

− π 2 2 -\frac{\pi^2}{2} π 2 \frac{\pi}{2} − 1 4 -\frac{1}{4} − 1 2 -\frac{1}{2} π 3 \frac{\pi}{3}

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Chew-Seong Cheong
Dec 26, 2019

S = ∑ n = 0 2020 cos ⁡ ( n 2 π 3 ) = cos ⁡ 0 + cos ⁡ π 3 + cos ⁡ 4 3 π + cos ⁡ 9 3 π + cos ⁡ 16 3 π + cos ⁡ 25 3 π + cos ⁡ 36 3 π + cos ⁡ 49 3 π + ⋯ + cos ⁡ 202 0 2 3 π = 1 + 1 2 − 1 2 − 1 − 1 2 + 1 2 ⏟ = 0 + 1 + 1 2 − 1 2 − 1 − 1 2 + 1 2 ⏟ = 0 + ⋯ + 1 + 1 2 − 1 2 − 1 − 1 2 + 1 2 ⏟ = 0 + 1 + 1 2 − 1 2 − 1 − 1 2 = 1 + 1 2 − 1 2 − 1 − 1 2 = − 1 2 \begin{aligned} S & = \sum_{n=0}^{2020} \cos \left(\frac {n^2\pi}3 \right) \\ & = \cos 0 + \cos \frac \pi 3 + \cos \frac 43 \pi + \cos \frac 93 \pi + \cos \frac {16}3 \pi + \cos \frac {25}3 \pi + \cos \frac {36}3 \pi + \cos \frac {49}3 \pi + \cdots + \cos \frac {2020^2}3 \pi \\ & = \small \underbrace{1 + \frac 12 - \frac 12 - 1 - \frac 12 + \frac 12}_{=0} + \underbrace{1 + \frac 12 - \frac 12 - 1 - \frac 12 + \frac 12}_{=0} + \cdots + \underbrace{1 + \frac 12 - \frac 12 - 1 - \frac 12 + \frac 12}_{=0} + 1 + \frac 12 - \frac 12 - 1 - \frac 12 \\ & = 1 + \frac 12 - \frac 12 - 1 - \frac 12 \\ & = \boxed {-\frac 12} \end{aligned}

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