Notation for nested sums of arbitrary degree?

Question

e.g. 3rd degree nesting:

$$\sum_{a=1}^{10}\sum_{b=1}^a\sum_{c=1}^bc$$

I may need to nest up to degree 10.

Answer at Notation for nested sigmas (summations) seems sub-optimal.

Answer 1

Short Answer

This is just repeated iterations of the Hockey-stick Identity.

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Quick Overview of the Hockey-stick Identity

Here's my attempt at a more approachable description:

Notice that for each of the colors, the sum of all of the numbers in the diagonal is the last number on the bottom that goes off in the opposite direction. For instance, if we look at the yellow, we see that $1+6+21+56 = 84$.

The proof for this is very intuitive if you begin with Pascal's identity (as in, you know how Pascal's Triangle numbers are obtained). $$ \begin{align} 84 &= 56 + 28 \\\\ &= 56 + (21 + 7) \\\\ &= 56 + 21 + (6 + 1) \end{align} $$ There are lots of cool/elegant counting proofs for this identity too, but we'll skip those for now.

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Summations

The numbers of Pascal's Triangle are also known as binomial coefficients, which are computed as: $$\binom{n}{k}$$ where $n$ is the row of Pascal's Triangle, and $k$ is the row element number (both of which start at $0$, not $1$).

So now we can start pulling apart at the summations from the inside out and see how this is related:

$$\begin{align} \sum_{c=1}^{b}c &= 1 + 2 + 3 + 4 + \cdots + b \\\\ \sum_{c=1}^{b}c &= \binom{1}{1} + \binom{2}{1} + \binom{3}{1} + \binom{4}{1} + \cdots + \binom{b}{1} \\\\ \sum_{c=1}^{b}c &= \binom{b+1}{2} \\\\ \end{align}$$

Now we can apply the same idea to the next summation layer:

$$\begin{align} \sum_{b=1}^a\sum_{c=1}^bc &= \sum_{b=1}^a \binom{b+1}{2} \\\\ \sum_{b=1}^a\sum_{c=1}^bc &= \binom{2}{2} + \binom{3}{2} + \binom{4}{2} + \binom{5}{2} + \cdots + \binom{a+1}{2} \\\\ \sum_{b=1}^a\sum_{c=1}^bc &= \binom{a+2}{3} \\\\ \end{align}$$

Finally:

$$\begin{align} \sum_{a=1}^{10}\sum_{b=1}^a\sum_{c=1}^bc &= \sum_{a=1}^{10} \binom{a+2}{3} \\\\ \sum_{a=1}^{10}\sum_{b=1}^a\sum_{c=1}^bc &= \binom{3}{3} + \binom{4}{3} + \binom{5}{3} + \binom{6}{3} + \cdots + \binom{12}{3} \\\\ \sum_{a=1}^{10}\sum_{b=1}^a\sum_{c=1}^bc &= \binom{13}{4} = 715 \\\\ \end{align}$$

So in general, given the format of nested summations you have, if the number of summations you have is $k$ and the upper limit is $n$, then the sum is $$\binom{n+k}{k+1}$$.