<s>[INST] finding the coefficient of $x^{14}$ in the expression: $\frac{5x^2-x^4}{(1-x)^3}$ I have a homework question which requires me to find the coefficient of $x^{14}$ in the expression: $\dfrac{5x^2-x^4}{(1-x)^3}$
I have not figured out a way to do this (I believe this is because my algebra is weak).
This is a question in a combinatorics class and has to do with Generating Functions if that is any help.
Help is greatly appreciated. Thanks, Jason
 [/INST] Worth mention is the use of partial fractions to derive differential operator representations, viz.
$$\rm\displaystyle \frac{5-x^2}{(1-x)^3}\ =\ \frac{4}{(1-x)^3}- \frac{2}{(1-x)^2} - \frac{1}{1-x}\ =\ (2\ D^2 + 2\ D - 1)\ \bigg(\frac{1}{1-x}\bigg),\quad D\ =\ \frac{d}{dx} $$
Consider a general $2\:$nd-order constant coefficient differential operator applied to $\rm\:1/(1-x)\:.$
$\rm\qquad\displaystyle \frac{2\:a\:+b+c-(b+2\:c)\ x + c\ x^2}{(1-x)^3}\ =\ (a\ D^2 + b\ D + c)\ \bigg(\!\!\frac{1}{1-x}\bigg)$
$\rm\displaystyle\phantom{\qquad \frac{2\:a\:+b+c-(b+2\:c)\ x + c\ x^2}{(1-x)^3}}\ =\ (a\ D^2 + b\ D + c)\ \sum_{k\:=\:0}^{\infty}\ x^k $
$\rm\displaystyle\phantom{\qquad \frac{2\:a\:+b+c-(b+2\:c)\ x + c\ x^2}{(1-x)^3}}\ =\ \sum_{k\:=\:0}^{\infty}\ ((k+2)\:(k+1)\:a+(k+1)\:b+c)\ x^k$
Thus $\rm\ a = 2 = b,\ c = -1\ $ yields
$$\rm \frac{5-x^2}{(1-x)^3}\: =\ \sum_{k\:=\:0}^\infty\ (2\:(k+3)\:(k+1)-1)\ x^k $$  
Hence we find that the coefficient of $\rm\:x^{12}\:$ equals $\ 2\cdot 15\cdot 13 -1\ =\ 389\:.$
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