If log (1 - x + {x^2}) = {a_1}x + {a_2}{x^2} | vedclass

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(B) We know that $\log(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \dots$ for $|z| < 1$.Given $\log(1 - x + x^2) = \log((1+x^3)/(1+x))

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$\frac{2}{3}\log 2$

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(B) We know that $\log(1+z) = z - \frac{z^2}{2} + \frac{z^3}{3} - \frac{z^4}{4} + \dots$ for $|z| Given $\log(1 - x + x^2) = \log((1+x^3)/(1+x)) = \log(1+x^3) - \log(1+x)$.Expanding both terms:$\log(1+x^3) = x^3 - \frac{x^6}{2} + \frac{x^9}{3} - \frac{x^{12}}{4} + \dots$$\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \frac{x^6}{6} + \dots$Subtracting these,the coefficients of $x^{3k}$ are given by the terms in $\log(1+x^3)$ minus the corresponding terms in $\log(1+x)$.Specifically,$a_{3k} = \frac{(-1)^{k-1}}{k} - \frac{(-1)^{3k-1}}{3k} = \frac{(-1)^{k-1}}{k} - \frac{(-1)^{k-1}}{3k} = \frac{(-1)^{k-1}}{k} (1 - 1/3) = \frac{2}{3} \frac{(-1)^{k-1}}{k}$.Thus,$\sum_{k=1}^{\infty} a_{3k} = \sum_{k=1}^{\infty} \frac{2}{3} \frac{(-1)^{k-1}}{k} = \frac{2}{3} (1 - 1/2 + 1/3 - 1/4 + \dots) = \frac{2}{3} \log(2)$.

If $\log (1 - x + {x^2}) = {a_1}x + {a_2}{x^2} + {a_3}{x^3} + \dots$,then ${a_3} + {a_6} + {a_9} + \dots$ is equal to

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