I have a question that I thought of.

There are two boxes.

One box has an infinite number of balls, distributed so that there is $1$ ball labeled $1,$ $2$ balls labeled $2,$ and for any positive integer $n,$ $n$ balls labeled $n.$

Another box has also an infinite number of balls, such that it has $1$ ball for all natural numbers.

How many combinations of balls are there such that the sum of the balls are $100$? (All balls are distinct).

I know this is just a massive sum, but I want to learn how to express this in generating functions.

I know I can represent the first box with $\frac{1}{1-x},$ and the second can be represented as $\frac{x^2}{(1-x)^2}+\frac{x}{1-x}$ (if I applied my arithmo-geometric sequence correctly).

I would need to find the $x^{100}$ term of $\frac{1}{1-x}\left(\frac{x^2}{(1-x)^2}+\frac{x}{1-x}\right),$ but with the powers of $(1-x)$ which don't have a useful expansion, it seems impossible to re-expand it.

How do I manipulate the expression $\frac{1}{(1-x)^n}$ to re-expand it?

  • $\begingroup$ I don't understand. Why are there two separate boxes? Does that have any significance for your question? Could you just have one box with $n+1$ balls labeled* $n$ for each positive integer $n$? Or are you asking two separate questions, one for each box? $\endgroup$ – Zubin Mukerjee Sep 17 '18 at 22:44
  • $\begingroup$ I'm asking one question. The two boxes are relevant since we are looking for the number of ways for the sum to be $100$. I mean, having $n+1$ balls won't mess up the fundamentals, but I have a different example. $\endgroup$ – Jason Kim Sep 17 '18 at 22:47
  • $\begingroup$ Would having just one box with $n+1$ balls labeled $n$ (for each $n$) be the same as your example with two boxes? Or no? $\endgroup$ – Zubin Mukerjee Sep 17 '18 at 22:48
  • $\begingroup$ Oh. Sorry for the late response (I'm still a student so... work.) ... Looking at it, it is basically the same thing as having $n+1$ balls labeled $n$ and drawing $2...$ $\endgroup$ – Jason Kim Sep 18 '18 at 1:23

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