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1.Derive the generating function for the sequence $$0, 0, 0, 0, 3, 4, 5, 6, . . .$$

2.Derive the generating function for the sequence $$0, 0, −12, 36, −108, 324, .. .$$

So the first function looks like $3x^4 + 4x^5+5x^6...$ = $x^4(3+4x+5x^2...)$ and that looks like the generating function for sequence $0,1,2,3,...$ I assume the answer for the first one is something like $\frac{x^4()}{1-x}$ , what am I missing?

For the second one if I factor $12$ out, ($-x^2+3x^3-9x^4...)$, is it something like $\frac{12x^2}{1+3x}$

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For $|x|<1$ we obtain: $$3x^4+4x^5+...=x^2(3x^2+4x^3+...)=x^2(x^3+x^4+...)'=$$ $$=x^2\cdot\left(\frac{x^3}{1-x}\right)'=\frac{x^4(3-2x)}{(1-x)^2}.$$

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  • $\begingroup$ What's the meaning of notation ' $\endgroup$
    – Rui Yu
    Commented Sep 26, 2017 at 5:25
  • $\begingroup$ I couldn't see how you get the third equation. $\endgroup$
    – Rui Yu
    Commented Sep 26, 2017 at 5:26
  • $\begingroup$ The notation $'$ stands for differentiation: $y'=\frac{dy}{dx}.$ $\endgroup$
    – bof
    Commented Sep 26, 2017 at 5:29
  • $\begingroup$ @Rui Yu It's a derivative. $(x^n)'=nx^{n-1}$. $\endgroup$
    – Michael Rozenberg
    Commented Sep 26, 2017 at 5:29
  • $\begingroup$ I see. Did I do the second question right? $\endgroup$
    – Rui Yu
    Commented Sep 26, 2017 at 5:31

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