Question

If ${\log }_e(1+x+x^2+x^3)$ be expanded in ascending powers of $x$, then the coefficient of $x^n$ is $\frac{a}{n}$ if $n$ is odd or of the form $4m+2$ and $\frac{-b}{n}$ if n is of the form $4m$. Find $a\times b$

Answer 1

${\log }_e(1+x+x^2+x^3)$
$={\log }_e\left[\right(1+x\left)\right(1+x^2\left)\right]$
$={\log }_e(1+x)+{\log }_e(1+x^2)$
$=(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}-\frac{x^6}{6}+\frac{x^7}{7}-\frac{x^8}{8}+...\infty )+(x^2-\frac{x^4}{2}+\frac{x^6}{3}-\frac{x^8}{4}+....\infty )$
$=(x+\frac{x^2}{2}+\frac{x^3}{3})+(-\frac{3x^4}{4}-\frac{3x^8}{8}-.....-\frac{3x^n}{n}+.....\infty )+(\frac{x^5}{5}+\frac{x^6}{6}+.....\frac{x^n}{n}+....\infty )$
So, coefficient of $x^n$ when n is of the form 4m is $-\frac{3}{n}$
and coefficient of $x^n$ when n is of the form 4m+2 or odd is $\frac{1}{n}$
So $a=-3,b=-1$
$a\times b=3$