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

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): \(\frac{1}{2}\left[(\Sigma n)^{2}-\Sigma n^{2}\right]\).

Step 1: Understand the Problem Statement

We are asked to find the sum of all possible products of the first \(n\) natural numbers taken two at a time. This involves selecting two distinct numbers from \(1, 2, \, \ldots, n\) and calculating the sum of their products.

Step 2: Consider the Combinatorial Approach

The number of ways to select two numbers from \(n\) numbers is given by \(\binom{n}{2} = \frac{n(n-1)}{2}\). For each selected pair \((i, j)\), we calculate their product \(ij\). The main focus is forming the expression involving these pairwise products.

Step 3: Sum of the Products Formula Derivation

Let's consider each pair \((i, j)\). The sum of all such products where \(i eq j\) can be expressed as \(\sum_{1 \leq i < j \leq n} ij\). It is better understood by writing out initial terms, for instance: for \(n = 3\), the pairs are \((1,2), (1,3), (2,3)\) with products \(2, 3, 6\), summing up to \(11\). Explore using pairwise sum combinations.

Step 4: Algebraic Expansion Approach

Consider \(\left(\sum_{i=1}^{n} i \right)^2 = (1 + 2 + \ldots + n)^2\). This expands to yield all pair combinations including \(i^2\) terms. The undesirable self-pair terms can be eliminated by \(\sum_{i=1}^{n} i^2\), giving remaining only product terms of interest.

Step 5: Use Correct Formula

The difference from exponent expansion: \[ (\Sigma i)^2 - \Sigma i^2 = 2 \sum_{1 \leq i < j \leq n} ij \]. Thus rearranging for the sum of products gives \[ \frac{1}{2} \left( (\Sigma i)^2 - \Sigma i^2 \right) \]. Must check this formula against given answer choices to confirm it fits format.

Step 6: Confirm Final Expression with Options

Match derived expression \( \frac{1}{2} \left( (\Sigma i)^2 - \Sigma i^2 \right) \) with options. It matches with option (D). Hence, option (D) is the correct choice.

Key Concepts

Combinatorial Approach

The combinatorial approach is a powerful technique for calculating the number of ways to pick two items from a larger set. When dealing with the problem of finding the sum of products of the first \(n\) natural numbers taken two at a time, it's essential to understand how many such products there can be.

In combinatorics, the number of ways to choose two distinct numbers out of \(n\) is represented by the binomial coefficient \(\binom{n}{2}\), which is calculated as \(\frac{n(n-1)}{2}\). This formula arises from choosing an initial number and then selecting a subsequent number to pair it with, avoiding repetition.

• This ensures that each specific pairing, say \((i,j)\), is unique and doesn't repeat with reversed order \((j,i)\).
• Using this approach, the focus is immediately on constructing expressions that involve products of these pairs.
• It ensures that all possible distinct combinations are accounted for.

By understanding this concept, you unlock a foundational step for the algebraic expansion that follows.

Algebraic Expansion

Algebraic expansion involves breaking down expressions to cover all possible terms, including those that are not of immediate interest. When calculating the sum of the pairwise products, expanding the square of the sum of the first \(n\) natural numbers is crucial.

This expression, \(\left(\sum_{i=1}^{n} i\right)^2 = (1 + 2 + \ldots + n)^2\), naturally generates all possible pair combinations, including the square terms \(i^2\). These extra terms needn't be in our sum, so they are carefully eliminated. By subtracting \(\sum_{i=1}^{n} i^2\) from the expanded form, you remove these squares, leaving only the products between distinct numbers.

• The removal of \(i^2\) is key to isolating the terms of interest: the \(ij\) terms.
• This method highlights the power of expansion in simplifying complex expressions involving sums and products.

Understanding this allows you to efficiently compute complicated sums that might otherwise require extensive manual pair-checking.

Pairwise Products

Pairwise products are the backbone of calculating sums of combinations. In the context of finding products from a series of numbers taken two at a time, understanding the nature of these pairwise calculations is vital.

By considering each pair \((i, j)\), the products \(ij\) are unique if \(i eq j\). When you expand an expression like \(\left(\sum_{i=1}^{n} i\right)^2\), all possible combinations—including pairwise products—appear naturally.

• It highlights how pairwise products are the result of every unique \(i\) and \(j\) pairing in the series.
• Summing all \(ij\) terms effectively captures the idea of collective interaction between pairs of numbers in the set.

This concept is structured around forming systematic groupings from which sums develop, making it easier to handle large sets of numbers using established mathematical techniques.