Problem 21 - A Complete Resource Book in Mathematics for JEE Main

Question

The reciprocal of the weighted mean of first \(n\) natural numbers whose weights are equal to the squares of the corresponding numbers is (A) \(\frac{2(2 n+1)}{3 n(n+1)}\) (B) \(\frac{3 n(n+1)}{n(2 n+1)}\) (C) \(\frac{3 n(n+1)}{2 n+1}\) (D) None of these

Step-by-Step Solution

Answer

The correct option is (A) \( \frac{2(2n+1)}{3n(n+1)} \).

Step 1: Understanding the Problem

We need to find the reciprocal of the weighted mean of the first \(n\) natural numbers. The weights are the squares of the corresponding numbers \(1^2, 2^2, 3^2, \ldots, n^2\).

Step 2: Formula for Weighted Mean

Recall that the weighted mean is given by \( \frac{\sum w_i x_i}{\sum w_i} \), where \( x_i \) are the values (natural numbers in our case) and \( w_i \) are the weights (squares of the numbers).

Step 3: Calculate Weighted Mean Numerator

The numerator of the weighted mean is \( \sum_{i=1}^{n} i \cdot i^2 = \sum_{i=1}^{n} i^3 \), which is the sum of cubes of the first \(n\) natural numbers.

Step 4: Calculate Sum of Cubes

Use the formula for the sum of cubes: \( \sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2 \).

Step 5: Calculate Weighted Mean Denominator

The denominator is \( \sum_{i=1}^{n} i^2 \), which is the sum of squares of the first \(n\) natural numbers.

Step 6: Calculate Sum of Squares

Use the formula for the sum of squares: \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \).

Step 7: Weighted Mean Formula

The weighted mean is \( \frac{\left(\frac{n(n+1)}{2}\right)^2}{\frac{n(n+1)(2n+1)}{6}} \). Simplifying this expression gives \( \frac{3n(n+1)}{2n+1} \).

Step 8: Finding the Reciprocal

The reciprocal of the weighted mean is \( \frac{2n+1}{3n(n+1)} \).

Step 9: Choosing the Correct Option

The reciprocal we computed matches with option (A), which is \( \frac{2(2n+1)}{3n(n+1)} \).

Key Concepts

Reciprocal

Understanding the concept of "reciprocal" can make a big difference in solving mathematical problems, especially when dealing with fractions and ratios. When you find the reciprocal of a number, you are essentially flipping the number over. For example, the reciprocal of a fraction like \( \frac{a}{b} \) is \( \frac{b}{a} \). This is important because multiplying a number by its reciprocal will always give you 1.

In the context of a weighted mean, after you calculate it, taking the reciprocal involves flipping the expression to simplify further or achieve the desired form. When dealing with complex problems with weighted means, recognizing when to use a reciprocal can be pivotal to reaching the solution efficiently.

Natural Numbers

Natural numbers are the simplest set of numbers that we often use in counting and ordering. These include all positive integers starting from 1, such as 1, 2, 3, and so on. Natural numbers are foundational not just in everyday mathematics, but also in advanced concepts.

In problems involving sequences or series, the term "natural numbers" indicates that we're only considering positive whole numbers. This is crucial because many formulas, like those for the sum of cubes or squares, assume we're working with these specific numbers to provide valid solutions.

Squares of Numbers

When we speak of the squares of numbers, we are referring to multiplying a number by itself. For instance, the square of 3 is \(3^2 = 9\). Squares of numbers appear frequently in various mathematical contexts, including geometry and algebra.

In weighted mean problems, squares can serve as weights. This means each natural number is multiplied by its square, influencing how much it contributes to the overall mean. Recognizing and calculating squares accurately is key when these weights are given by such values.

Sum of Cubes

The sum of cubes refers to the total when each number in a sequence is cubed and those cubes are added together. For the first \( n \) natural numbers, the sum of cubes is given by the formula \( \sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2 \). This formula expresses a fascinating symmetry, highlighting how cubes relate to squares.

This formula often simplifies problems related to volume and geometric progression. Recognizing expressions as a sum of cubes can make problems appear less daunting, allowing for streamlined calculations by using established formulas.

Sum of Squares

The sum of squares is an important mathematical concept that appears often in statistics, physics, and beyond. It is the sum of each number in a sequence squared. For example, the sum of squares of the first \(n\) natural numbers is \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \).

This particular formula is essential in calculating variances and standard deviations in statistics. It also simplifies the solving of problems that involve quadratic expressions. Understanding how to compute and use the sum of squares is crucial for solving weighted mean calculations, as seen in many mathematical exercises and practical applications.