More fun in 2015, Part 7

The key is that $\begin{aligned} (1+w^k+w^{2k})(1+w^{3k}+w^{6k} + \cdots + w^{2013k})(1-w^k) &= 1-w^{2016k} \\ &= 1-w^k, \end{aligned}$ and $1-w^k \ne 0$ , so the sum is $\sum_{k=1}^{2014} \sum_{n=0}^{671} w^{3nk} = 2014 + \sum_{n=1}^{671} \sum_{k=1}^{2014} w^{3nk}$

But $\sum_{k=1}^{2014} \zeta^k$ is $-1$ if $\zeta$ is a nontrivial $2015$ th root of unity (same argument as above; multiply by $1-\zeta$ to get $\zeta-\zeta^{2015} = \zeta-1$ ).

So the answer is $2014 + \sum_{n=1}^{671} -1 = \fbox{1343}$ .

This solution is not elegant at all, but I would still share it, because this is how I did it.

Firstly, $z^2+z+1=(z-e^{\frac{2i\pi}3})(z-e^{\frac{4i\pi}3})$ .

Therefore, by Partial Fractions - Linear Factors , $\dfrac1{1+z+z^2}=-\frac i{\sqrt3}\left(\dfrac1{z-e^{\frac{2i\pi}3}}-\dfrac1{z-e^{\frac{4i\pi}3}}\right)$ .

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$z^{2015}=1$ is the equation that generates all the " $\omega$ "s.

Note that $z=1$ is an extra solution that I will deal with later.

Then, $(z+e^{\frac{2i\pi}3})^{2015}=1$ would produce "( $\omega-e^{\frac{2i\pi}3}$ )".

Then, $\left(\dfrac1z+e^{\frac{2i\pi}3}\right)^{2015}=1$ would produce " $\left(\dfrac1{\omega-e^{\frac{2i\pi}3}}\right)$ ".

Multiply by $e^{\frac{4i\pi}3}z^{2015}$ to produce $(e^{\frac{4i\pi}3}+z)^{2015}=e^{\frac{4i\pi}3}z^{2015}$ .

Sum of roots by Vieta's Formula $= \dfrac{-2015e^{\frac{4i\pi}3}}{1-e^{\frac{4i\pi}3}}=\dfrac{\sqrt3}3(2015)(i)$

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Similarly, the sum of roots of the other part $\left(\dfrac1{z-e^{\frac{4i\pi}3}}\right)$ is $-\dfrac{\sqrt3}3(2015)(i)$ .

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Therefore, $\displaystyle\sum_{k=1}^{2015}\frac1{1+\omega_k+\omega_k^2}=-\frac i{\sqrt3}\left(\frac{\sqrt3}3\right)(2015)[i-(-i)]=\frac{4030}3$ .

Note that it is summed to 2015 instead of 2014

Then, the required answer is $\dfrac{4030}3-\dfrac13=1343$ .

I computed it for smaller values $n$ in the place of 2015 (and $n-1$ in the place of 2014), and noticed a pattern. I claim that:

• if $n \equiv 0 \pmod{3}$ , the sum does not exist (one of the denominators is zero)

• if $n \equiv 1 \pmod{3}$ , the sum is $\frac{n-1}{3}$

• if $n \equiv 2 \pmod{3}$ , the sum is $\frac{2n-1}{3}$

In this case we have $n = 2015 \equiv 2 \pmod{3}$ so the sum is $\frac{2*2015-1}{3}=1343$ . I don't have a proof of this claim but I'm working on it.

I didn't feel like working out the math as Patrick & Kenny did. But note that e^(2πi/2015) is simply the unit vector for the angle 2π/2015. w^k is the unit vector for 2πk/2015, and similarly for w^2k. So I put that in a spreadsheet, 2014 rows calculating each part of the sum, then summed it. Much quicker than working the theoretical aspect.

In general, for $w=e^{\frac{2\pi i}{3k+2}}$ , that is, $w$ being the smallest non-trivial $(3k+2)\mbox{-}th$ root of unity, the result of the sum is $2k+1$ .

In fact the result holds for $w$ being any non-trivial $(3k+2)\mbox{-}th$ root of unity.

Enjoyable Roots-of-Unity problem, Sir Otto! Happy Holidays to you!