Question

If $n$ is a multiple of $3$ then the coefficient of $x^n$ in the expansion of ${\log }_e(1+x+x^2),\left|x\right|<1,$ is

• A. $-\frac{1}{n}$
• B. $\frac{2}{n}$
• C. $-\frac{2}{n}$
• D. $\frac{1}{n}$

Answer 1

The correct option is C $-\frac{2}{n}$

Since $1+x+x^2=\frac{1-x^3}{1-x}$ for $\left|x\right|<1$, we have ${\log }_e\left(1+x+x^2\right)={\log }_e\left(\frac{1-x^3}{1-x}\right)={\log }_e\left(1-x^3\right)-{\log }_e\left(1-x\right)$.
Also ${\log }_e\left(1-x\right)=-\left[x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots \right]$ and hence ${\log }_e\left(1-x^3\right)=-\left[x^3+\frac{x^6}{2}+\frac{x^9}{3}+\frac{x^{12}}{4}+\cdots \right]$
$\therefore {\log }_e\left(1-x^3\right)-{\log }_e\left(1-x\right)=-\left[x^3+\frac{x^6}{2}+\frac{x^9}{3}+\frac{x^{12}}{4}+\cdots \right]+\left[x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\cdots \right]$
Simplify the R. H. S.:
${\log }_e\left(1+x^3\right)-{\log }_e\left(1+x\right)=x+\frac{x^2}{2}-\frac{2x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}-\frac{2x^6}{6}+\frac{x^7}{7}+\cdots$
Hence for $n$ multiple of 3, we get $a_n=-\frac{2}{n}$.