Question

Show that \(\sum\limits_{{\rm{i = 1}}}^{{\rm{n - 2}}} {\sum\limits_{{\rm{j = i + 1}}}^{{\rm{n - 1}}} {\sum\limits_{{\rm{k = j + 1}}}^{\rm{n}} {\rm{1}} } } {\rm{ = }}\left( {\begin{array}{*{20}{c}}{\rm{n}}\\{\rm{3}}\end{array}} \right)\) if \({\rm{n}}\) is an integer with\({\rm{n}} \ge {\rm{3}}\).

Short answer

It is shown that \(\sum\limits_{{\rm{i = 1}}}^{{\rm{n - 2}}} {\sum\limits_{{\rm{j = i + 1}}}^{{\rm{n - 1}}} {\sum\limits_{{\rm{k = j + 1}}}^{\rm{n}} {\rm{1}} } } {\rm{ = }}\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{3}}\end{array}} \right)\) is an integer.

Step-by-step solution

Introduction

An integer is a number from the set of negative and positive numbers, including zero that has no decimal or fractional element.

Explanation

Let’s take three separate integers \({\rm{(i,j,k)}}\)from \({\rm{\{ 1,2, \ldots ,n\} }}\)such that \({\rm{i < j < k}}\)and count the diverse number of ways in which it may be done, and equate that with the known numerical quantity: \(\sum\limits_{{\rm{i = 1}}}^{{\rm{n - 2}}} {\sum\limits_{{\rm{j = i + 1}}}^{{\rm{n - 1}}} {\sum\limits_{{\rm{k = j + 1}}}^{\rm{n}} {\rm{1}} } } {\rm{ = }}\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{3}}\end{array}} \right)\).

Therefore, it is shown that \(\sum\limits_{{\rm{i = 1}}}^{{\rm{n - 2}}} {\sum\limits_{{\rm{j = i + 1}}}^{{\rm{n - 1}}} {\sum\limits_{{\rm{k = j + 1}}}^{\rm{n}} {\rm{1}} } } {\rm{ = }}\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{3}}\end{array}} \right)\) is an integer.