Question

If n is a multiple of $3$, show that $1-\frac{n-3}{2}+\frac{(n-4)(n-5)}{\lfloor 3}-\frac{(n-5)(n-6)(n-7)}{\lfloor 4}+.....$

$+(-1{)}^{r-1}\frac{(n-r-1)(n-r-2)....(n-2r+1)}{\lfloor r}+......,$ is equal to $\frac{3}{n}$ or $-\frac{1}{n}$, according as n is odd or even.

Answer 1

$\Rightarrow$Here, $\frac{n-3}{2}$ is the co efficient of $x^{n-4}$ in $\frac{1}{2}{(1-x)}^{-2}$

$\frac{(n-5)(n-4)}{3!}$ is the co efficient of $x^{n-6}$ in $\frac{1}{3}{(1-x)}^{-3}$

$\frac{(n-7)(n-6)(n-5)}{4!}$ is the co efficient of $x^{n-8}$ in $\frac{1}{4}{(1-x)}^{-4}$

and so on

Hence $S=$ the co efficient of $x^n$ in

$x^2{(1-x)}^{-1}-\frac{1}{2}{x}^4{(1-x)}^{-2}+\frac{1}{3}{x}^6{(1-x)}^{-3}-\frac{1}{4}{x}^8{(1-x)}^{-4}+....$

=the co efficient of $x^n$ in $\log \{1+x^2{(1-x)}^{-1}\}$

But $1+\frac{x^2}{1-x}=\frac{(1-x+x^2)}{(1-x)}\frac{(1+x)}{(1+x)}=\frac{1+x^3}{1+x^2}$

$\therefore S=$ the co efficient of $x^n$ in $\log (1+x^3)-\log (1-x^2)$

If $n=6r$, the co efficient of $x^n$ is $-\frac{1}{2r}$ from the first series and $\frac{1}{3r}$ from the second series.

$\therefore S=-\frac{1}{2r}+\frac{1}{3r}=-\frac{1}{6r}=-\frac{1}{n}$

If $n=6r+3$, the co efficient of $x^n$ is $\frac{1}{2r+1}$ from first series and $0$ from the second.

$S=\frac{1}{2r+1}=\frac{3}{n}$