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

Question

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 $\rm \LaTeX$.

Answer 1

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 $A(x)$?

Answer 2

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