Problem 66 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

Statement-1 : The variance of first \(\mathrm{n}\) even natural numbers is \(\frac{n^{2}-1}{4}\). Statement-2 : The sum of first \(n\) natural numbers is \(\frac{n(n+1)}{2}\) and the sum of squares of first \(n\) natural numbers is \(\frac{n(n+1)(2 n+1)}{6}\). [2009] (a) Statement- 1 is true, Statement- 2 is true. Statement- 2 is not a correct explanation for Statement-1. (b) Statement- 1 is true, Statement- 2 is false.

Step-by-Step Solution

Answer

Statement-1 is true. Statement-2 is true but does not explain Statement-1.

Step 1: Understand What Each Statement Asserts

Statement-1 talks about finding the variance of the first \( n \) even natural numbers, while Statement-2 provides formulas for the sum of the first \( n \) natural numbers and the sum of their squares. Begin by examining how these might relate to the problem at hand.

Step 2: Formulate the Sequences

The first \( n \) even natural numbers are \( 2, 4, 6, \, ..., \, 2n \). The formula to find these numbers is \( 2k \) where \( k \) ranges from \( 1 \) to \( n \).

Step 3: Calculate Mean of Even Numbers

The mean \( \bar{x} \) is calculated as \( \bar{x} = \frac{\sum_{k=1}^{n} 2k}{n} = \frac{2(1 + 2 + 3 + \, ... \, + n)}{n} \). Using the formula from Statement-2, \( 1 + 2 + \, ... \, n = \frac{n(n+1)}{2}, \) so the mean becomes \( \frac{2 \times \frac{n(n+1)}{2}}{n} = n+1 \).

Step 4: Determine the Variance Formula

Variance \( \sigma^2 = \frac{1}{n}\sum_{k=1}^{n}(x_k - \bar{x})^2 \). Substitute \( x_k \) and \( \bar{x} \), and simplify each term \((2k - (n+1))^2 = 4k^2 - 4k(n+1) + (n+1)^2\).

Step 5: Use Provided Formulas to Expand

Expand \( \sum_{k=1}^{n}(2k - (n+1))^2 \) using the expressions: \( \sum_{k=1}^{n}4k^2, \sum_{k=1}^{n}4k(n+1), \) and \( n(n+1)^2 \). Substitute the sum of squares formula from Statement-2 into the sum: \( 4\sum_{k=1}^{n}k^2 = 4\times\frac{n(n+1)(2n+1)}{6} \).

Step 6: Simplify and Evaluate the Variance

After substitution, simplify to arrive at \( \sigma^2 = \frac{n(2n+1) - 2n(n+1) + n(n+1)^2}{3n}\). Reducing this, you reach the variance as \( \frac{n^2-1}{4} \), confirming Statement-1.

Step 7: Determine Correctness of Statement-2

Statement-2 offers formulas for sums and squares which are standard and useful in derivations for variances, but does not provide a direct explanation as to why the variance formula holds as in Statement-1. It merely assists in calculations without directly proving variance results.

Key Concepts

Even Natural Numbers

Even natural numbers are simply numbers that are divisible by 2. They form a sequence starting from 2 running through numbers like 4, 6, 8, \, ... all the way to a number like 2n, where n represents the last term you are interested in. Understanding this set of numbers and how they progress is important in many areas of mathematics. For example, in this exercise, knowing that the sequence of first n even natural numbers is structured as \(2, 4, 6, \ldots, 2n\) allows us to manipulate and calculate properties like mean and variance more efficiently.
Finding even numbers is as simple as multiplying 2 with the natural numbers. Here's a quick example list:

• 2, which is \(2 \times 1\)
• 4, which is \(2 \times 2\)
• 6, which is \(2 \times 3\)
• ...and so on up to \(2n\).

This understanding makes it possible to tackle a wide variety of mathematical calculations involving even numbers.

Sum of Natural Numbers

The sum of the first n natural numbers is a fundamental formula in mathematics, expressed as \(\frac{n(n+1)}{2}\). This expression arises when you add the sequence \(1, 2, 3, 4, \ldots, n\).

Here’s the intuition behind this formula: if you list numbers from 1 through n, and then list them in reverse order and add corresponding pairs, each pair sums to \(n + 1\). For instance, with numbers 1 to n:

• Pair the first and the last number: \(1 + n = n + 1 \)
• Pair the second and second last: \(2 + (n-1) = n + 1 \)

This pattern continues throughout the list. Since there are n numbers, you will have \(\frac{n}{2}\) pairs (assuming n is even or handles the middle number separately if odd), each adding to \(n + 1\). Therefore, the sum becomes:

• \(\frac{n}{2} \times (n+1) = \frac{n(n+1)}{2}\)

Understanding this formula helps in simplifying various calculations where sums of sequences are involved.

Sum of Squares of Natural Numbers

The sum of the squares of the first n natural numbers is written by the formula \(\frac{n(n+1)(2n+1)}{6}\). Though it looks complex at first glance, it is derived from applying the simplification principles of polynomial expressions.
Here’s a quick explanation of the sequence you would typically calculate:

• The sequence begins with \(1^2, 2^2, 3^2, 4^2, \ldots, n^2\).

When expanded, each term in the sequence is squared and gradually increases in size as you progress from 1 to n. Why do we need this formula? It’s often used in statistical calculations like finding variance, as seen when expanding expressions to simplify results.

Consider using this formula in practical calculations:

• If you substitute a specific n value into the formula, you get a direct and efficient way to find the sum of squares without listing and adding each term individually.

Understanding and applying this formula lays the groundwork for more advanced operations involving variances, as it directly relates to terms squared and their relationships within a sequence.