Question

If ${\log }_e(1-x+x^2)=a_{1}x+a_2{x}^2+a_3{x}^3+...$, and $n$ is not a multiple of $3$ then $a_n$ is equal to

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

Answer 1

Answer: (B)

$(-1{)}^n\frac{1}{n}$

Since $1-x+x^2=\frac{1+x^3}{1+x}$, 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 ${\log }_e\left(1+x\right)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$ and hence ${\log }_e\left(1+x^3\right)=x^3-\frac{x^6}{2}+\frac{x^9}{3}-\frac{x^{12}}{4}+\cdots$
$\therefore {\log }_e\left(1+x^3\right)-{\log }_e\left(1+x\right)=\left[x^3-\frac{x^6}{2}+\frac{x^9}{3}-\frac{x^{12}}{4}+\cdots \right]-\left[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots \right]$
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 $n$ not multiple of 3, we get $a_n={\left(-1\right)}^n\frac{1}{n}$.