Problem 40 - A Complete Resource Book in Mathematics for JEE Main

Question

Statement-1: The variance of first \(n\) 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 \(\frac{n(n+1)(2 n+1)}{6}\) (A) Statement-1 is true, Statement-2 is true; Statement- 2 is a correct explanation for Statement- 1 . (B) Statement- 1 is true, Statement-2 is true; Statement- 2 is not a correct explanation for Statement- 1 . (C) Statement- 1 is true, Statement-2 is false. (D) Statement- 1 is false, Statement- 2 is true.

Step-by-Step Solution

Answer

(B) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.

Step 1: Understand the Given Statements

We need to verify whether each statement is correct independently and assess whether Statement-2 explains Statement-1. Statement-1 pertains to the variance of even numbers, while Statement-2 gives formulas for sums and sums of squares of natural numbers.

Step 2: Analyze Statement-1

To find the variance of the first n even natural numbers, start by noting these numbers are 2, 4, 6, ..., 2n. The mean (average) is \( \bar{x} = \frac{2 + 4 + 6 + \ldots + 2n}{n} = \frac{2(1 + 2 + 3 + \ldots + n)}{n} = 2 \cdot \frac{n(n+1)/2}{n} = n+1 \).

Step 3: Calculate Variance for Statement-1

The variance is given by \( \text{Var}(X) = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 \). Calculating requires \( \sum_{i=1}^n x_i^2 = 2^2 + 4^2 + \ldots + (2n)^2 = 4(1^2 + 2^2 + \ldots + n^2) = 4\frac{n(n+1)(2n+1)}{6} \) (from Statement-2).

Step 4: Complete Variance Calculation for Statement-1

Variance calculation using the sum of squares: \( \frac{4\cdot n(n+1)(2n+1)/6 - n(n+1)^2}{n} = \frac{1}{n} \cdot \left(\frac{2n(n+1)(2n+1)}{3} - n(n+1)^2\right)\). Simplifying leads to \( \frac{n^2 - 1}{4} \), confirming Statement-1.

Step 5: Verify Statement-2

Statement-2 provides standard results: the sum \( \frac{n(n+1)}{2} \) and the sum of squares \( \frac{n(n+1)(2n+1)}{6} \) for the first n natural numbers. Both formulas are true.

Step 6: Determine Relationship Between Statements

Statement-2 provides needed formulas for calculations in Statement-1, but does not directly offer explanation for the variance of even numbers. It serves as a calculation tool rather than a conceptual basis.

Key Concepts

Sum of first n natural numbers formula

The concept of the sum of the first \( n \) natural numbers is foundational in mathematics. To find the sum of these numbers, we use a simple and elegant formula:\[S = \frac{n(n+1)}{2}.\]This formula is derived from the pattern we observe when we list out the numbers: 1, 2, 3, ..., \( n \). Individually adding these numbers could be computationally intensive for large \( n \), hence the formula offers an efficient computation method.

• The variable \( n \) represents the number of terms.
• The parentheses in \( n(n+1) \) indicate the product of consecutive integers.
• Dividing by 2 accounts for pairing terms from opposite ends of the sequence.

For example, using \( n = 5 \), the sum is calculated as \( \frac{5(5+1)}{2} = 15 \). Understanding this formula helps us quickly find cumulative totals for any list of consecutive natural numbers.

Sum of squares of first n natural numbers

Finding the sum of squares of the first \( n \) natural numbers results in another important formula:\[S = \frac{n(n+1)(2n+1)}{6}.\]This formula, similar to the previous, is invaluable for efficient computation and is used in various applications, especially in statistical calculations.

• Here, \( n \) still represents the number of terms.
• The formula includes three terms multiplied together: \( n \), \( n+1 \), and \( 2n+1 \).
• The division by 6 normalizes the sum based on the sequence's growth rate.

By using \( n = 3 \) as an example, you compute \( \frac{3(3+1)(2\times3+1)}{6} = 14 \). Each part of the formula helps reduce the complexity involved in manually squaring each number and summing the results.

Variance calculation

Variance is a critical concept in statistics that measures how data points differ from the mean of a dataset. For the first \( n \) even natural numbers, recalculating variance involves several steps:First, understand even natural numbers: 2, 4, 6, ..., \( 2n \).

• Calculate the mean: \( \bar{x} = n+1 \), derived by halving the paired list of even numbers.
• Compute sum of squared deviations from the mean: for even numbers, it's based on the sum of squares formula: \( 4\frac{n(n+1)(2n+1)}{6} \).

The formula for variance \( \text{Var}(X) \) is:\[ \text{Var}(X) = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2,\]where \( x_i \) refers to each even number. For even numbers, variance simplifies to \( \frac{n^2 - 1}{4} \), illustrating the dispersion around the mean. This concise measure of variability is vital for statistical analysis, helping identify how spread out even natural numbers are from their average.