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In the book I'm using, $A(x)$ denotes the formal power series (generating function), $A(x) = \sum a_ix^i$. I'm really stuck on this problem. Thanks for any help.

My attempt after the given hint: $$\begin{align} A(x)&=\sum_{n\geq 0} \binom{n}{2}x^n\\ &=\sum_{n\geq 0} \frac{x^2}{2}\frac{d^2}{dx^2}(x^{n})\\ &=\frac{x^2}{2} \frac{d^2}{dx^2}\sum_{n\geq 0} x^{n}\\ &=\frac{x^2}{2} \frac{d^2}{dx^2}\frac{1}{1-x}\\ &=\frac{x^2}{2} \frac{2}{(1-x)^3} \end{align}$$ Sorry, I'm new to $\rm \LaTeX$.

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Do you mean $a_n=\binom{n}{2}$ and $A(x) =\sum_{n\geq 0}a_nx^n= \sum_{n\geq 0} \binom{n}{2}x^n$?

Then note that $$\binom{n}{2}x^n=\frac{x^2}{2}\cdot n(n-1) x^{n-2}=\frac{x^2}{2}\cdot\frac{d^2}{dx^2}(x^{n})$$ and recall the basic generating function $\sum_{n\geq 0}x^n=\frac{1}{1-x}$.

What is $A(x)$?

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  • $\begingroup$ What is $D^2$ here? $\endgroup$ – davidh Feb 9 at 9:41
  • $\begingroup$ And I believe the question is saying when n is substituted as x, the whole sum turn out to be $n \choose 2$ $\endgroup$ – davidh Feb 9 at 9:42
  • $\begingroup$ $D^2$ is the second derivative with respect to $x$. $\endgroup$ – Robert Z Feb 9 at 9:46
  • $\begingroup$ I don't think so. Otherwise the g.f. is simply $x/2+x^2/2$. $\endgroup$ – Robert Z Feb 9 at 9:48
  • $\begingroup$ Where this question come from? A book? Any web-link? $\endgroup$ – Robert Z Feb 9 at 9:51
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Since $\binom{n}{2}x^n=\frac12 x^2\frac{d^2}{dx^2}x^n$, $$\sum_{n\ge 0}\binom{n}{2}x^n=\frac12 x^2\frac{d^2}{dx^2}\frac{1}{1-x}=\frac{x^2}{(1-x)^3}.$$We can double-check by the binomial theorem: the $x^n$ coefficient is $$\frac{(-1)^n}{(n-2)!}\prod_{j=1}^{n-2}(-2-j)=\frac{n!}{(n-2)!2!}=\binom{n}{2}.$$

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