Question

The coefficient of $x^n$ in the expansion of ${\log }_e(1+x+x^2)$
(i) When $n$ is a multiple of $3$ is $A$
(ii) When $n$ is not a multiple of $3$ is $B$
Find the value of $n(B-A)$.

Answer 1

We have
${\log }_{e}1+x+x^2={\log }_e\left[\right(1-\omega x\left)\right(1-{\omega }^{2}x\left)\right]$
(where $\omega$ is the cube root of unity)

$={\log }_e\left[\right(1-\omega x\left)\right(1-{\omega }^{2}x\left)\right]$

Using logarithmic series expansion

$=(-(\omega x)-\frac{(\omega x{)}^2}{2}-\frac{(\omega x{)}^3}{3}-...-\frac{(\omega x{)}^n}{n}-...)+(-({\omega }^{2}x)-\frac{({\omega }^{2}x{)}^2}{2}-\frac{({\omega }^{2}x{)}^3}{3}-...-\frac{({\omega }^{2}x{)}^n}{n}-...)$

$=-x(\omega +{\omega }^2)-\frac{x^2}{2}({\omega }^2+{\omega }^4)-\frac{x^3}{3}({\omega }^3+{\omega }^6)-...-\frac{x^n}{n}({\omega }^n+{\omega }^{2n})-...$
$\therefore$ Coefficient of $x^n$ in the expansion of ${\log }_e(1+x+x^2)$
$=-\frac{1}{n}({\omega }^n+{\omega }^{2n})$
(i) If $n$ is a multiple of $3$

$n=3m$ say $m\in I$
$\therefore -\frac{1}{n}({\omega }^n+{\omega }^{2n})=-\frac{1}{n}({\omega }^{3m}+{\omega }^{6m})=-\frac{1}{n}(1+1)=-\frac{2}{n}=A$

(ii) If $n$ is not a multiple of $3$
$n=3m+1$

$\therefore -\frac{1}{n}({\omega }^n+{\omega }^{2n})=-\frac{1}{n}({\omega }^{3m+1}+{\omega }^{6m+2})=-\frac{1}{n}(\omega +{\omega }^2)=\frac{1}{n}=B$
$\Rightarrow n(B-A)=n(\frac{1}{n}+\frac{2}{n})=3$

Ans: 3.