<s>[INST] Summing up the series $a_{3k}$ where $\log(1-x+x^2) = \sum a_k x^k$ If $\ln(1-x+x^2)  = a_1x+a_2x^2 + \cdots \text{ then } a_3+a_6+a_9+a_{12} + \cdots = $ ?
My approach is to write $1-x+x^2 = \frac{1+x^3}{1+x}$ then expanding the respective logarithms,I got a series (of coefficient) which is nothing but $\frac{2}{3}\ln 2$.But this approach took some time for me (I can't solved it during the test but after the test I solved it)... Any other quick method?
 [/INST] Lemma: Let $f(x) = \sum a_n x^n$ be a power series.  Then
$$\sum a_{3n} x^{3n} = \frac{f(x) + f(\omega x) + f(\omega^2 x)}{3}$$
where $\omega = e^{ \frac{2\pi i}{3} }$.
Proof.  Ignoring convergence it suffices to prove this for a single term, and then it boils down to the identity
$$\frac{1 + \omega^n + \omega^{2n}}{3} = \begin{cases} 1 \text{ if } 3 | n \\\ 0 \text{ otherwise} \end{cases}.$$
This is a special case of the discrete Fourier transform.  Applying the lemma, we readily obtain that the desired sum is
$$\frac{\ln 1 + \ln (1 - \omega + \omega^2) + \ln (1 - \omega^2 + \omega)}{3} = \frac{1}{3} \ln (2 \omega^2 \cdot 2 \omega) = \frac{2}{3} \ln 2$$
where we use the fact that $1 + \omega + \omega^2 = 0$.
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