Question

Show that the coefficient of $x^n$ in the expansion of $\frac{x}{(1-x{)}^2-cx}$ is $n\{1+\frac{n^2-1}{\lfloor 3}c+\frac{(n^2-1)(n^2-4)}{\lfloor 5}{c}^2+\frac{(n^2-1)(n^2-4)(n^2-9)}{\lfloor 7}{c}^3+.....\}$.

Answer 1

The co efficient can be expressed as co efficient of $x^{n-1}$ in the expansion of $\frac{1}{{(1-x)}^2-cx}$ or $\frac{1}{{(1-x)}^2}{\{1-\frac{cx}{(1-x)}\}}^{-1}$

Expanding by Binomial Theorem, the last expression becomes,

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

and we have to pick out from the expansions of

${(1-x)}^{-2},{(1-x)}^{-4},{(1-x)}^{-6},....$

co efficient of;-

$=n+\frac{c(n-1)n(n+1)}{3!}+\frac{c^2(n-2)(n-1)n(n+1)(n+2)}{5!}+....$

$=n\{1+\frac{(n^2-1)}{3!}c+\frac{(n^2-1)(n^2-4)}{5!}{c}^2+....\}$