Problem 14 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The sum of all possible products of the first \(n\) natural numbers taken two at a time is (A) \(\frac{1}{2}\left[\Sigma n^{2}-\Sigma n\right]\) (B) \(\frac{1}{2}\left[(\Sigma n)^{2}-\Sigma n\right]\) (C) \(\frac{1}{2}\left[\Sigma n^{2}-\Sigma(n+1)\right]\) (D) \(\frac{1}{2}\left[(\Sigma n)^{2}-\Sigma n^{2}\right]\)

Step-by-Step Solution

Answer

The correct answer is (D).

Step 1: Understanding the Problem

We are asked to find the sum of products for pairs of the first \(n\) natural numbers. For example, if \(n=3\), the numbers are \(1, 2, 3\), and the pairs are \((1,2), (1,3), (2,3)\). We want to find a formula to add all such products for any \(n\).

Step 2: Expressing the Sum of Products

We want the sum \( \sum_{1 \le i < j \le n} i \cdot j \). This is equivalent to taking each possible pair of numbers \(i, j\) where \(1 \leq i < j \leq n\) and multiplying them.

Step 3: Using Combinatorial Identity

The sum of products \( \sum_{1 \le i < j \le n} i \cdot j \) can be calculated using the identity \(\left(\sum_{i=1}^n i\right)^2 = \sum_{i=1}^n i^2 + 2\sum_{1 \le i < j \le n} i \cdot j \). This means we need to use the sums of squares and the total sum.

Step 4: Calculating Total Sum and Sum of Squares

First, compute \( \sum_{i=1}^n i = \frac{n(n+1)}{2} \) and \( \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} \). These are standard formulas for sums of the first \(n\) natural numbers and their squares.

Step 5: Substitute into Identity and Simplify

Substitute \(\sum_{i=1}^n i\) and \(\sum_{i=1}^n i^2\) into the identity:\[\left(\frac{n(n+1)}{2}\right)^2 = \frac{n(n+1)(2n+1)}{6} + 2\sum_{1 \le i < j \le n} i \cdot j\]Simplifying, solve for \(\sum_{1 \le i < j \le n} i \cdot j\):\[2\sum_{1 \le i < j \le n} i \cdot j = \frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6}\]Divide by 2:\[\sum_{1 \le i < j \le n} i \cdot j = \frac{1}{2} \left(\frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6}\right)\]

Step 6: Identifying the Correct Answer

The expression \(\frac{1}{2} \left(\left(\frac{n(n+1)}{2}\right)^2 - \sum_{i=1}^n i^2\right)\) matches option (D), showing it simplifies to this form in select options.

Key Concepts

Sum of Products

In combinatorics, the sum of products is a method used to find the total of all possible products derived from selecting certain elements of a set. Specifically, for this problem, we're interested in the sum of products of pairs of the first \( n \) natural numbers. To better understand, consider the numbers \( 1, 2, 3 \). The products from these pairs are calculated by multiplying each number with the others that follow it, i.e., \( (1 \times 2), (1 \times 3), (2 \times 3) \). The goal is to develop a formula that could compute the sum for any \( n \). The solution involves using a mathematical identity that helps simplify the computation. This identity connects the sum of products to more familiar computations: the sum of first \( n \) natural numbers and the sum of their squares. Combinatorial identities like \( \left(\sum_{i=1}^n i\right)^2 = \sum_{i=1}^n i^2 + 2\sum_{1 \le i < j \le n} i \cdot j \) allow us to relate these sums in a useful way.

Natural Numbers

Natural numbers are the set of positive integers starting from 1. They are the simplest numbers used for counting and ordering. In mathematical problems like this one, we work with the first \( n \) natural numbers, which are \( 1, 2, 3, \, \ldots\, n \). Each of these numbers acts as an individual element that can form products with others when taken two at a time. Understanding this set of numbers is critical, as it lays the foundation for combinatorial problems and exercises. We manipulate these numbers to derive useful formulas or identities that solve problems like the one given, where products of pairs need summing. When you're asked to find products of these numbers, it's crucial to understand the properties that make them 'natural'. They naturally follow a sequential pattern, and this property, among others, makes calculating sums and products systematically straightforward and achievable with precise results.

Formula Derivation

The formula derivation for this problem is rooted in breaking down what seems complex into simpler steps. The derived formula is\[ 2\sum_{1 \le i < j \le n} i \cdot j = \frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6} \]which comes from substituting known formulas of the sums of numbers and their squares into our identity.To simplify this, we start by calculating both \( \sum_{i=1}^n i \), which follows the formula \( \frac{n(n+1)}{2} \), and \( \sum_{i=1}^n i^2 \), using the formula \( \frac{n(n+1)(2n+1)}{6} \).By substituting these into the square of the sum identity \[\left(\frac{n(n+1)}{2}\right)^2 = \sum_{i=1}^n i^2 + 2\sum_{1 \le i < j \le n} i \cdot j \] and rearranging terms, we derive the expression for the sum of products.This systematic approach showcases how understanding basic formulas can be combined to derive complex results efficiently. By dividing our equation by 2, we find the final correct formula, matching one of the given options—specifically, option (D) as per the solution.