Proof that repeated sum equals binomial formula

Question

Let $s, d$ be positive integers. Can you prove the following general formula for the repeated sum? I developed this problem on my own, but is it a well known result? $$\sum_{i_1 = 0}^s \sum_{i_2 = 0}^{i_1} \sum_{i_2 = 0}^{i_2} \cdots \sum_{i_{d} = 0}^{i_{d-1}} 1 = \binom{s + d}{d}$$ E.g. if $d = 1$ then the sum is $$\sum_{i_1 = 0}^s 1 = s+1 = \binom{s+1}{d}.$$ If $d = 2$ then the sum is $$\sum_{i_1 = 0}^s \sum_{i_2 = 0}^{i_1} 1 = \sum_{i_1 = 0}^s (i_1+1) = \frac{(s+1)(s+2)}{2} = \binom{s+2}{d}.$$

Answer 1

Notice that the sum counts the number of assignments to the variables $i_1,..., i_d$ that satisfy $0\leq i_d \leq i_{d-1}\leq ... \leq i_{2}\leq i_{1}\leq s$. Let's map one of these non-decreasing sequences of $d$ integers (with values between $0$ and $s$) to a string with two symbols, $\uparrow$ and $I$ (the $\uparrow$ represents incrementing our integer and $I$ stands for placing the current integer. A sequence will have $s$ copies of $\uparrow$ and $d$ copies of $I$. We map a string of this nature to a sequence of non-decreasing integers as with the following example:

$II\uparrow I\uparrow \uparrow I I \uparrow$ to $0 0 1 3 3$. We started with $I=0$ and saw $I I$ so we wrote down $0 0$. The first $\uparrow$ incremented $I$ to $1$. We then saw an $I$ and wrote down $1$. Next we saw $\uparrow \uparrow$ which caused us to increment $I$ to 2 and then to $3$. We then saw $II$ and wrote down $33$. The final $\uparrow$ incremented $I$ to $4$, but we never saw an $I$ afterwards to write it down.

The number of such strings is $\binom{d+s}{d}$ because each string contains $s$ increments $\uparrow$'s and $d$ $I$'s.

Answer 2

Yes, this is in essence a couple of well-known results. First, the expression

$$\sum_{i_1=0}^s\sum_{i_2=0}^{i_1}\ldots\sum_{i_d=0}^{i_{d-1}}1\tag{1}$$

just counts $1$ for each $d$-tuple $\langle i_1,i_2,\ldots,i_d\rangle$ of non-negative integers satisfying $$s\ge i_1\ge i_2\ge\ldots\ge i_d\;,$$

so in effect it just counts such $d$-tuples. Let $x_1=s-i_1$, $x_k=i_{k-1}-i_k$ for $k=2,\ldots,d$, and $x_{d+1}=i_d$; clearly the $(d+1)$-tuple $\langle x_1,\ldots,x_{d+1}\rangle$ can be constructed from knowledge of $s$ and $\langle i_1,\ldots,i_d\rangle$, and it’s not hard to see that $s$ and $\langle i_1,\ldots,i_d\rangle$ can be reconstructed from $\langle x_1,\ldots,x_{d+1}\rangle$. What are the possibilities for $\langle x_1,\ldots,x_d\rangle$? It’s clear that the $x_k$ must be non-negative integers, and some thought reveals that the only other restriction on them is that $$x_1+x_2+\ldots+x_{d+1}=s\;.\tag{2}$$

Thus, $(1)$ is simply the number of solutions of $(2)$ in non-negative integers. This is a well-known problem whose answer can be obtained by the so-called stars and bars approach, and the answer turns out to be

$$\binom{s+(d+1)-1}{(d+1)-1}=\binom{s+d}d\;.$$

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It can also be seen as a form of what is sometimes called the hockey stick lemma, though that name is usually applied to the result used to carry the induction step of a proof by induction on $d$ of your result. Specifically, suppose that I know the result for $s$ and $d$; then

$$\sum_{i_1=0}^s\sum_{i_2=0}^{i_1}\ldots\sum_{i_{d+1}=0}^{i_d}1=\sum_{i_1=0}^s\binom{i_1+d}d=\binom{s+d+1}{d+1}\;,$$

giving me the result for $s$ and $d+1$. Here the last step is one of the two identities sometimes called the hockey stick lemma; you can find proofs of it here and at the earlier question linked from there.