Question

If ${\log }_e\left(\frac{1}{(1+x+x^2+x^3)}\right)$ be expanded in a series of ascending powers of $x,$

the coefficient of $x^n$ is $-\frac{b}{n}$ if $n$ be odd or of the form $4m+2$ and $\frac{a}{n}$ if $n$ be of the form $4m$. Find the value of $a+b^2$.

Answer 1

We have
${\log }_e\left(\frac{1}{1+x+x^2+x^3}\right)=-{\log }_e(1+x+x^2+x^3)$
$=-{\log }_e(1+x)(1+x^2)$
$=-{\log }_e\left(\right(1+x\left)\right(1+ix\left)\right(1-ix\left)\right)$
(where $i=\sqrt{-1}$)
$=-{\log }_e(1+x)-{\log }_e(1+ix)-{\log }_e(1-ix)$
$=-(x-\frac{x^2}{2}+\frac{x^3}{3}-...+(-1{)}^{n-1}\frac{x^n}{n}+...)-(ix-\frac{(ix{)}^2}{2}+\frac{(ix{)}^3}{3}-...+(-1{)}^{n-1}\frac{(ix{)}^n}{n}+...)+(ix+\frac{(ix{)}^2}{2}+\frac{(ix{)}^3}{3}+...+\frac{(ix{)}^n}{n}+...)$
Hence, coefficient of $x^n$ in ${\log }_e\left(\frac{1}{1+x+x^2+x^3}\right)$
$=-\frac{(-1{)}^{n-1}}{n}-(-1{)}^{n-1}\frac{i^n}{n}+\frac{i^n}{n}$
$=\frac{1}{n}\left[\right(-1{)}^n+(-1{)}^n{i}^n+i^n]$

Case 1: If $n$ be odd
then $(-1{)}^n=-1$
$\therefore$ Coefficient of $x^n$ in ${\log }_e\left(\frac{1}{1+x+x^2+x^3}\right)$
$=\frac{1}{n}(-1-i^n+i^n)$
$=-\frac{1}{n}$
Now if $n=4m+2$
Then coefficient of $x^n$ in ${\log }_e\left(\frac{1}{1+x+x^2+x^3}\right)$
$=\frac{1}{n}\left[\right(-1{)}^n+(-1{)}^n{i}^n+i^n)]$
$=\frac{1}{(4m+2)}\left[\right(-1{)}^{4m+2}+(-1{)}^{4m+2}{i}^{4m+2}+i^{4m+2})]$

$=\frac{1}{(4m+2)}[1+i^2+i^2]$

$=\frac{1}{(4m+2)}[1-1-1]$

$=-\frac{1}{n}=-\frac{b}{n}$

(since $4m+2=n$)
$\Rightarrow b=1$

Case 2: If $n=4m$
Then coefficient of $x^n$ in ${\log }_e\left(\frac{1}{1+x+x^2+x^3}\right)$
$=\frac{1}{n}\left[\right(-1{)}^n+(-1{)}^n{i}^n+i^n)]$

$=\frac{1}{4m}\left[\right(-1{)}^{4m}+(-1{)}^{4m}{i}^{4m}+i^{4m})]$

$=\frac{1}{4m}[1+1+1]$

$=\frac{3}{4m}$

$=\frac{3}{n}=\frac{a}{n}$

($\because n=4m$)

$\Rightarrow a=3$

$\therefore a+b^2=4$