Let ω = e i 2 π / 2015 , evaluate ∑ k = 1 2014 1 1 + ω k + ω 2 k

Question

Let   ω  =      e          i      2      π              /            2015      , evaluate       ∑          k      =      1              2014            1          1      +              ω        k            +              ω                  2          k

Donavan Scott · Accepted Answer

For simplicity I will write n=2015. Since

X^{n} - 1 = \prod_{k = 0}^{n - 1} (X - ω^{k})

we conclude that

\frac{n X^{n - 1}}{X^{n} - 1} = \sum_{k = 0}^{n - 1} \frac{1}{X - ω^{k}}

Substituting

X = j = e^{2 i π / 3}

and

X = \bar{j}

and then subtracting we get

\frac{n j^{n - 1}}{j^{n} - 1} - \frac{n {\bar{j}}^{n - 1}}{{\bar{j}}^{n} - 1} = \sum_{k = 0}^{n - 1} (\frac{1}{j - ω^{k}} - \frac{1}{\bar{j} - ω^{k}}) = (\bar{j} - j) \sum_{k = 0}^{n - 1} \frac{1}{1 + ω^{k} + ω^{2 k}}

Finally, since

n = 2015 \equiv 2 \mod 3

we see that

j^{n - 1} = j

and

j^{n} = j^{2} = \bar{j}

, we get

\frac{1}{- i \sqrt{3}} (\frac{n j}{j^{2} - 1} - \frac{n \bar{j}}{j - 1}) = \frac{1}{3} + \sum_{k = 1}^{n - 1} \frac{1}{1 + ω^{k} + ω^{2 k}}

The final step is easy and we get

\sum_{k = 1}^{n - 1} \frac{1}{1 + ω^{k} + ω^{2 k}} = \frac{2 n - 1}{3} = 1343

This conclusion is valid for every n which is equal to 2(mod3).

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