Nested Sums of Integers

Article Quant Q&A · Author: Nobody

Summary

The document derives a closed form for repeatedly summing consecutive integers. It starts with the familiar sum of the first integers, then evaluates the sum of those partial sums by expressing it through the sums of integers and squares. The result is a cubic expression, which the answer generalizes to further iterations using a product formula and a binomial-coefficient identity.

The derivation illustrates how nested sums can be reduced using standard power-sum formulas and Pascal's identity. It gives algebraic steps rather than empirical evidence, and it is a general mathematical result rather than a trading method. The notation in the generalization appears inconsistent about the number of iterations and the binomial-coefficient indices, so those details should be checked before applying the formula beyond the worked case.

Key ideas

The sum of the first consecutive integers has a triangular-number formula.
A second nested sum can be evaluated by splitting the triangular-number expression into sums of integers and squares.
Repeated summation produces a polynomial or binomial-coefficient form.
Pascal's identity can be used to establish relationships between successive nested sums.

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# How can I express this sum in a easier way?


# How can I express this sum in a easier way?












For instance, I know that the sum of the first $101$ natural numbers can be expressed in the following easy computation:

$\sum_{i=1}^{101}i = \frac{101*102}{2}$

One of the questions is: and what about this sum?

$\sum_{i=1}^{101}i + \sum_{i=1}^{100}i + ... + \sum_{i=1}^{1}i = \sum_{i_1=1}^{101}\sum_{i_2=1}^{i_1}i_2$

And specially, what about the $nth$ case, i.e.;

$\sum_{i_1=1}^{101}\sum_{i_2=1}^{i_1}\cdots\sum_{i_n=1}^{i_{n-1}}i_n$

Thanks in advance!

## Answer by Canardini (score 5, accepted)

https://quant.stackexchange.com/a/50010

As you mentioned, we have

$$\sum_{l=0}^{k}{p}=\frac{k(k+1)}{2}$$

You want to know

$$\sum_{k=0}^{n}{\sum_{l=0}^{k}{p}}=\sum_{k=0}^{n}{\frac{k(k+1)}{2}}=\frac{1}{2}\sum_{k=0}^{n}{k^2}+\frac{1}{2}\sum_{k=0}^{n}{k}$$

you know that

$$\sum_{k=0}^{n}{k^2}=\frac{n(n+1)(2n+1)}{6}$$

Therefore $$\sum_{k=0}^{n}{\sum_{l=0}^{k}{p}}=\frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}=\frac{n(n+1)(n+2)}{6}$$

EDIT :

We can generalize it : For $m$ iterations,summing up to $k$, we have

$$Sum(m,k)=\frac{k(k+1)...(k+m)}{(m+1)!}$$

In other words,

$$Sum(m,k)={m+k \choose k-1}$$

$$Sum(m+1,n)=\sum_{k=0}^{n}{{m+k \choose k-1}}$$

Also,

$${m+1 +n \choose n-1}={m+n\choose n-2}+{m+n\choose n-1}$$ $${m+n\choose n-2}={m+n-1\choose n-3}+{m+n-1\choose n-2}$$ We keep doing this, until we get $$Sum(m+1,n)={m+1 +n \choose n-1}$$

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.

Closed Forms for Nested Sums of Consecutive Integers

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