Question

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

Short answer

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

The demonstration using algebra is simple and straightforward; therefore let us try the more in-depth combinatorial technique.

Assume we're calculating the number of different ways to choose two integers from the set of integers\({\rm{\{ 1,2, \ldots ,n\} }}\). The solution is\(\left( {\begin{array}{*{20}{c}}{\rm{n}}\\{\rm{2}}\end{array}} \right)\), according to basic knowledge.

So, let's take a different approach. If we pick two numbers \({\rm{i < j}}\)such that\({\rm{i,j}}\), we have \({\rm{(n - 1)}}\) alternatives for the first integer I, namely\({\rm{1,2, \ldots ,n - 1}}\).

The possibilities for\({\rm{j}}\)start at \({\rm{i + 1}}\)and go up to \({\rm{n}}\)for each of these \({\rm{i}}\)options. If we set a counter\({\rm{S}}\)so that it grows by one every time a separate pair of integers \(\left( {{\rm{i,j}}} \right)\)is picked, then\(\sum\limits_{{\rm{i = 1}}}^{{\rm{n - 1}}} {\sum\limits_{{\rm{j = i + 1}}}^{\rm{n}} {\rm{1}} } {\rm{ = S = }}\left( {\begin{array}{*{20}{l}}{\rm{n}}\\{\rm{2}}\end{array}} \right)\).

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