How to calculate sum of product of first $n$ natural numbers taken three at a time$?$

Question

How to calculate sum of product of $n$ natural numbers taken three at a time$?$

There is a post on stackexchange but in that post it was assumed that the original poster knew how to solve double or triple summation. Moreover it was for $k$ taken at a time.

In my opinion, we can write the expression of our answer as $$\sum_{i = 1}^{n}\sum_{j = 1}^{n - i}\sum_{h = 1}^{n - j - i} i\cdot(i+j)\cdot(i + j +h)$$ but how to solve it$?$

Is there any general way of doing so$?$

Any help is greatly appreciated.

What I want to say is this

Suppose we take first $5$ natural numbers and then we take any three natural numbers lying in between $1$ and $5$ and multiply them (both $1$ and $5$ included). Then there will be $\binom{5}{3}$ ways of doing this. And we also add the product of each way. Then how will we calculate it$?$ $$ $$

We obtain for $k=3$ \begin{align*} \sum_{i=0}^{n}&\sum_{j=0}^{n-i}\sum_{k=0}^{n-i-j}i(i+j)(i+j+k)\\ &=\sum_{i=0}^{n}\sum_{j=0}^{n-i}\sum_{k=0}^{n-i-j}(i^3+2i^2j+i^2k+ij^2+ijk)\\ &=\sum_{i=0}^{n}\sum_{j=0}^{n-i}\sum_{k=0}^{n-i-j}(i^3+4i^2j+ijk)\tag{7}\\ &=\sum_{i=0}^{n}i^3\sum_{j=0}^{n-i}1\sum_{k=0}^{n-i-j}1 +4\sum_{i=0}^{n}i^2\sum_{j=0}^{n-i}j\sum_{k=0}^{n-i-j}1\tag{8}\\ &\qquad+\sum_{i=0}^{n}i\sum_{j=0}^{n-i}j\sum_{k=0}^{n-i-j}k\\ \end{align*}

This person is first distributing the summation over variables and then solving$?$ im having problem here. How to do it? Is there any general way$?#

Answer 1

Surprised that no one gave you a complete answer. Your posting clearly showed work.

Your method of indexing seems unnatural to me. Of the $~\displaystyle \binom{n}{3}~$ pertinent unordered triples, I will assume that $i > j > k$. So, you have ordered triples of the form $(i,j,k).$

This means that :

• $i$ can run from $3$ through $n$, inclusive.
• For any value of $i$, you have that $j$ can run from $2$ through $(i-1)$ inclusive.
• For any value of $j$, you have that $k$ can run from $1$ through $(j-1)$ inclusive.

Tools:

• $~\displaystyle \sum_{r=1}^s r = \frac{s^2}{2} + \frac{s}{2} ~: ~s \in \Bbb{Z^+}.$

• $~\displaystyle \sum_{r=1}^s r^2 = \frac{s^3}{3} + \frac{s^2}{2} + \frac{s}{6} ~: ~s \in \Bbb{Z^+}.$

• $~\displaystyle \sum_{r=1}^s r^3 = \frac{s^4}{4} + \frac{s^3}{2} + \frac{s^2}{4} ~: ~s \in \Bbb{Z^+}.$

• $~\displaystyle \sum_{r=1}^s r^4 = \frac{s^5}{5} + \frac{s^4}{2} + \frac{s^3}{3} - \frac{s}{30}~: ~s \in \Bbb{Z^+}.$

• $~\displaystyle \sum_{r=1}^s r^5 = \frac{s^6}{6} + \frac{s^5}{2} + \frac{5s^4}{12} - \frac{s^2}{12}~: ~s \in \Bbb{Z^+}.$

For what it's worth, the above formulas may be derived using nothing more than Algebra, Mathematical Induction, and a basic knowledge of Pascal's Triangle. This means that the derivation can avoid Bernoulli Numbers, Calculus, or Faulhaber's formula.

See, for example, this answer.

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Let $f(i,j)$ denote the sum of the products of the pertinent ordered triples, where $i > j$, and where $k$ runs from $1$ through $(j-1).$

That is, $~\displaystyle f(i,j) = \sum_{k=1}^{j-1} i \times j \times k ~: i \in \{3,4,\cdots,n\}, ~j \in \{2,3,\cdots,i-1\}.$

Let $g(i)$ denote $~\displaystyle \sum_{j=2}^{i-1} f(i,j).$

Let $H$ denote $~\displaystyle \sum_{i=3}^n g(i).$

Then, the goal is to calculate $H$.

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For $i,j$ fixed,

$$f(i,j) = [i \times j] \times \sum_{k=1}^{j-1} k$$

$$= [i \times j] \times \frac{(j-1)^2}{2} + \frac{j-1}{2}$$

$$= [i \times j] \times \frac{1}{2} \times \left[ ~\left(j^2 - 2j + 1 ~\right) + (j - 1) ~\right]$$

$$= [i \times j] \times \frac{1}{2} \times \left[ ~j^2 - j ~\right]$$

$$\frac{i}{2} \times \left( ~j^3 - j^2 ~\right).$$

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For $i$ fixed:

$$g(i) = \sum_{j=2}^{i-1} f(i,j) = \sum_{j=2}^{i-1} \left[ ~\frac{i}{2} \times \left( ~j^3 - j^2 ~\right) ~\right].$$

There is a mild shortcut here. Note that when $j=1$, the expression $(j^3 - j^2) = 0.$

Therefore,

$$g(i) = \sum_{j=1}^{i-1} \left[ ~\frac{i}{2} \times \left( ~j^3 - j^2 ~\right) ~\right]$$

$$ = \frac{i}{2} \times \sum_{j=1}^{i-1} ~j^3 \\ - \frac{i}{2} \times \sum_{j=1}^{i-1} ~j^2$$

$$= \frac{i}{2} \times ~\left[ ~\frac{(i-1)^4}{4} + \frac{(i-1)^3}{2} + \frac{(i-1)^2}{4} ~\right] \\ - \frac{i}{2} ~\left[ ~\frac{(i-1)^3}{3} + \frac{(i-1)^2}{2} + \frac{(i-1)}{6} ~\right]$$

$$= \frac{i}{2} \times ~\left[ ~\frac{(i-1)^4}{4} + \frac{(i-1)^3}{6} - \frac{(i-1)^2}{4} - \frac{i-1}{6} ~\right] $$

$$= \frac{i}{2} \times \frac{1}{12} \times \\ \left\{ \\ ~~[3i^4 - 12i^3 + 18i^2 - 12i + 3] \\ + [ 2i^3 - 6i^2 + 6i -2] \\ - [3i^2 - 6i + 3] \\ - [2i - 2] \\ ~\right\}$$

$$= \frac{1}{24} \times [3i^5 - 10i^4 + 9i^3 - 2i^2].$$

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$$H = \sum_{i=3}^n g(i) = \sum_{i=1}^n g(i) - \sum_{i=1}^2 g(i).$$

Again, there is a mild shortcut here. Note that $~\displaystyle \sum_{i=1}^2 g(i) = 0.$

Therefore,

$$H = \sum_{i=1}^n g(i) \\ = \frac{1}{24} \times \sum_{i=1}^n [3i^5 - 10i^4 + 9i^3 - 2i^2] \\ = \frac{1}{24} \times \\ \left\{ \\ ~~\left[\frac{n^6}{2} + \frac{3n^5}{2} + \frac{5n^4}{4} - \frac{n^2}{4}\right] \\ - \left[2n^5 + 5n^4 + \frac{10n^3}{3} - \frac{n}{3}\right] \\ + \left[\frac{9n^4}{4} + \frac{9n^3}{2} + \frac{9n^2}{4}\right] \\ - \left[\frac{2n^3}{3} + n^2 + \frac{n}{3}\right] \\ \right\}$$

$$= \frac{1}{24} \times \left[\frac{n^6}{2} - \frac{n^5}{2} - \frac{3n^4}{2} + \frac{n^3}{2} + n^2\right].$$