# Find a generating function for which $A(n)={n \choose 2}$

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 $$\rm \LaTeX$$.

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 $$A(x)$$?

• What is $D^2$ here? – davidh Feb 9 at 9:41
• And I believe the question is saying when n is substituted as x, the whole sum turn out to be $n \choose 2$ – davidh Feb 9 at 9:42
• $D^2$ is the second derivative with respect to $x$. – Robert Z Feb 9 at 9:46
• I don't think so. Otherwise the g.f. is simply $x/2+x^2/2$. – Robert Z Feb 9 at 9:48
• Where this question come from? A book? Any web-link? – Robert Z Feb 9 at 9:51

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 $$\frac{(-1)^n}{(n-2)!}\prod_{j=1}^{n-2}(-2-j)=\frac{n!}{(n-2)!2!}=\binom{n}{2}.$$