Sum of all possible products formed from a set of consecutive integers

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Homework Statement

Say I have a set of consecutive integers up to N, for example up to 5, the set 1,2,3,4,5. I need to know a general formula for the value of the sum of all products that can be formed from a certain number of elements taken from such a set. For example with the 5 set, say I want 3 numbers; I'll get 1.2.3+1.2.4+1.2.5+1.3.4+1.3.5+1.4.5+2.3.4 etc etc. I'd like to find a general formula for any set size and any length of product. Spent most of today looking for one, and from what I did it looks like a nice formula may not exist, but you never know.

Homework Equations

The Attempt at a Solution

Let N be the number of elements in the set and S be the number of terms per product. The furthest I got was something like

$\sum_{i=1}^{N+1-S}i(\sum_{j=i+1}^{N+2-S}j(\sum_{k=j+1}^{N+3-S}k(...\sum_{z=y+1}^{N}z))...)$

The final sum (over z here) can be done easily, it gives something depending on y, then the sum over y can be done, and so on, but it very rapidly gets completely out of hand. Any alternative ways of looking at the problem would be greatly appreciated.

 

[mighty2000]

Thank you for posting your question about finding a general formula for the sum of all products that can be formed from a set of consecutive integers up to N. This is an interesting problem and I can understand your frustration in trying to find a suitable formula.

After considering your approach, I believe there may be a simpler way to approach this problem. Let us consider the example you provided with the set {1,2,3,4,5} and choosing 3 numbers to form products. We can write out all the possible products as follows:

1.2.3 + 1.2.4 + 1.2.5 + 1.3.4 + 1.3.5 + 1.4.5 + 2.3.4 + 2.3.5 + 2.4.5 + 3.4.5

Notice that each term in this sum can be written as the product of the smallest number in that term and the sum of the remaining two numbers. For example, 1.2.3 = 1(2+3) and 1.2.4 = 1(2+4). Therefore, we can rewrite the sum as:

1(2+3+4+5) + 2(3+4+5) + 3(4+5) + 4(5)

Now, let's generalize this for any set size N and number of terms per product S. We can write the sum as follows:

\sum_{i=1}^{N-S+1} i \left(\sum_{j=i+1}^{N-S+2} j \left(\sum_{k=j+1}^{N-S+3} k \left(...\sum_{z=N-(S-1)}^{N} z \right) \right) \right)

Using the same logic as before, we can rewrite this as:

\sum_{i=1}^{N-S+1} i \left(\sum_{j=i+1}^{N-S+2} j \left(\sum_{k=j+1}^{N-S+3} k \left(...(N-S+1) + (N-S+2) + ... + N \right) \right) \right)

This can be simplified to:

\sum_{i=1}^{N-S+1} i \left