If log(1 − x + x2) = a1x + a2x2 + a3x3 + ⋯ . Then a3 + a6 + a9 + ⋯ is

Concept

A logarithm of a quadratic can often be simplified by factorising 1 − x + x² = (1 + x³)/(1 + x). Combined with the Maclaurin series log(1 + u) = u − u²/2 + u³/3 − ..., this lets us read off the coefficient aₙ of xⁿ directly. To collect only the coefficients whose index is a multiple of 3, we find a closed form for the general coefficient a(3k) and sum the resulting alternating series using the standard expansion log 2 = 1 − 1/2 + 1/3 − 1/4 + ... .

1. Factorise the argument: since 1 + x³ = (1 + x)(1 − x + x²), we have 1 − x + x² = (1 + x³)/(1 + x). Taking logarithms gives log(1 − x + x²) = log(1 + x³) − log(1 + x).

2. Expand each logarithm as a Maclaurin series: log(1 + x) = x − x²/2 + x³/3 − x⁴/4 + ..., so its coefficient of xⁿ is (−1)^(n−1)/n. Likewise log(1 + x³) = x³ − x⁶/2 + x⁹/3 − ..., so its coefficient of x^(3m) is (−1)^(m−1)/m and it contributes no other powers.

3. Read off the coefficient a(3k). The x^(3k) term receives (−1)^(k−1)/k from log(1 + x³) and minus (−1)^(3k−1)/(3k) from −log(1 + x). Since (−1)^(3k−1) = (−1)^(k−1), this gives a(3k) = (−1)^(k−1)/k − (−1)^(k−1)/(3k) = (−1)^(k−1)(2/(3k)).

4. Sum over all k: a₃ + a₆ + a₉ + ... = Σ_(k≥1) (−1)^(k−1)(2/(3k)) = (2/3)[1 − 1/2 + 1/3 − 1/4 + ...] = (2/3) log 2.

Cross-check: the bracketed alternating harmonic series 1 − 1/2 + 1/3 − ... is the standard expansion of log 2, so the total sum is (2/3) log 2.