Use generating functions to find the coefficient of $x^{15}$ in $\frac{x^3-5x }{(1-x)^3}$.
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Series expansion: $$-5 x-15 x^2-29 x^3-47 x^4-69 x^5-95 x^6-125 x^7-159 x^8-197 x^9-239 x^{10}-285 x^{11}-335 x^{12}-389 x^{13}-447 x^{14}-509 x^{15}+O\left(x^{16}\right)$$ – David G. Stork May 07 '21 at 19:16
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Thanks for the reply – Someone's name May 07 '21 at 19:21
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Notice that $$\frac{1}{(1-x)^3} = \frac{1}{2}\frac{d^2}{dx^2}\left(\frac{1}{1-x}\right) = \frac{1}{2}\frac{d^2}{dx^2}\left(\sum_{n\geqslant 0}x^n\right)=\sum_{n\geqslant 0}\frac{(n+1)(n+2)}{2}x^n.$$ Write $[x^k]f(x)$ for the coefficient of $x^k$ in the generating function $f$. Then, $$[x^{15}]\left\{\frac{x^3-5x}{(1-x)^3}\right\}=[x^{12}]\left\{\frac{1}{(1-x)^3}\right\}-5[x^{14}]\left\{\frac{1}{(1-x)^3}\right\}=13\cdot7 - 5\cdot 15\cdot8=-509.$$
Alan
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