Problem 25 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The weighted mean of the square of 1st \(n\) natural numbers whose weights are corresponding numbers, equals (A) \(\frac{(n+1)(2 n+1)}{2}\) (B) \(\frac{n(n+1)}{2}\) (C) \(\frac{n+1}{2}\) (D) None of these

Step-by-Step Solution

Answer

The weighted mean is \( \frac{n(n+1)}{2} \), corresponding to option (B).

Step 1: Understand the Problem

We need to find the weighted mean of the squares of the first \(n\) natural numbers, using the corresponding numbers as weights. The options are given, and our task is to identify which option expresses the correct result.

Step 2: Define the Concept of Weighted Mean

The weighted mean is calculated by taking the sum of the product of each value and its corresponding weight, and then dividing by the sum of the weights. Mathematically, it is expressed as:\[ \text{Weighted Mean} = \frac{\sum (x_i \cdot w_i)}{\sum w_i} \]where \(x_i\) are the squares of natural numbers and \(w_i\) are the weights which, in this case, are the natural numbers themselves.

Step 3: Calculate the Weighted Mean

Our values \(x_i\) are the squares of natural numbers, so \(x_1 = 1^2, x_2 = 2^2, \ldots, x_n = n^2\). Therefore, our weighted sum \(\sum (x_i \cdot w_i)\) is:\[ 1^3 + 2^3 + 3^3 + \ldots + n^3 = \left(\frac{n(n+1)}{2}\right)^2 \]The sum of the weights \(\sum w_i\) is the sum of the first \(n\) natural numbers:\[ \sum w_i = 1 + 2 + \ldots + n = \frac{n(n+1)}{2} \]

Step 4: Simplify the Expression

Now substitute into the weighted mean formula:\[ \text{Weighted Mean} = \frac{\left(\frac{n(n+1)}{2}\right)^2}{\frac{n(n+1)}{2}} \]The denominator simplifies the expression:\[ \text{Weighted Mean} = \frac{n(n+1)}{2} \]

Step 5: Choose the Correct Option

The resultant expression \( \frac{n(n+1)}{2} \) corresponds to option (B). Therefore, the correct answer is option (B).

Key Concepts

Natural Numbers

Natural numbers are the most basic numbers that are used for counting and are often learned at an early age. These numbers start from 1 and go on infinitely: 1, 2, 3, 4, and so on. Natural numbers do not include zero, decimals, fractions, or negative numbers. When solving problems in mathematics—such as the calculation of weighted means or sums of sequences—natural numbers play a crucial role.

In the context of our exercise, when we talk about the first \( n \) natural numbers, we mean the sequence: 1, 2, 3, ..., n. These numbers are both the elements in our sequence and the weights for the calculation we are tasked with. Understanding this simple yet foundational set of numbers is key to engaging with many mathematical computations.

Sum of Cubes

The sum of cubes refers to the sum of each of a set of numbers raised to the third power. Mathematically, if you have natural numbers up to \( n \), the sum of cubes is given by:

• \( 1^3 + 2^3 + 3^3 + \,\ldots\, + n^3 \)

An intriguing feature of the sum of cubes is that it is equal to the square of the sum of the first \( n \) natural numbers. This can be expressed with the formula:

• \[ \left( \frac{n(n + 1)}{2} \right)^2 \]

This property is what we use in Step 3 of our original solution where we determine the weighted sum of the cubes for our task. It simplifies the process significantly, allowing us to perform calculations with larger numbers without direct computation of each cube.

Sum of Natural Numbers

The sum of natural numbers is one of the more straightforward arithmetic sequences. To find the sum of the first \( n \) natural numbers—1 through \( n \)—we use the formula:

• \[ 1 + 2 + 3 + \,\ldots\, + n = \frac{n(n + 1)}{2} \]

This formula simplifies the addition process instead of manually summing each individual number in the sequence. Understanding this formula is crucial when calculating the weighted mean in exercises similar to the one we've tackled. It allows us to determine the sum of the weights that are essential to finding the weighted mean efficiently. The result of this sum becomes our denominator in the weighted mean formula, streamlining the calculation significantly.