Question

Prove that the coefficient of $x^n$ in the expansion of $\log (1+x+x^2)$ is $-\frac{2}{n}$ or $\frac{1}{n}$ according as $n$ is or is not a multiple of 3.

Answer 1

$(1-x^3)=(1-x)(1+x+x^2)$

Taking $\log$ both side ,

$\log (1-x^3)=\log (1-x)+\log (1+x+x^2)$

$\log (1+x+x^2)=\log (1-x^3)-\log (1-x)$

Expansion of $\log (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-...-\frac{x^n}{n}-...$

Expansion of $\log (1-x^3)=-x^3-\frac{x^6}{2}-\frac{x^9}{3}-\frac{x^{12}}{4}-...-\frac{x^{3n}}{n}-...$

Expansion of $\log (1+x+x^2)=x+\frac{x^2}{2}-(1-\frac{1}{3})x^3+\frac{x^4}{4}+\frac{x^5}{5}-\frac{1}{2}(1-\frac{1}{3})x^6...$

Coefficient of $x^n$ in $\log (1+x+x^2)$ when $n$ is not a multiple of $3=\frac{1}{n}$

Coefficient of $x^n$ in $\log (1+x+x^2)$ when $n$ is a multiple of $3$ that is $n=3k$ then $-\frac{3}{n}-\frac{-1}{n}=-\frac{2}{n}$