Statement-1: The variance of first n even natural numbers is (n² - 1)/4. Statement-2: The sum of fir

Q: Wait, why do we reject the formula $\frac{n^2-1}{4}$? Is it possible I made a calculation error?

It is a common point of confusion! When you calculate the variance of the first $n$ even natural numbers, you are essentially looking at the variance of the set $\{2, 4, \dots, 2n\}$. Using the property $Var(aX) = a^2 Var(X)$, where $X$ is the set of the first $n$ natural numbers, we know $Var(X) = \frac{n^2-1}{12}$. Since our set is $2X$, the variance must be $2^2 \cdot Var(X) = 4 \cdot \frac{n^2-1}{12} = \frac{n^2-1}{3}$. The denominator is $3$, not $4$. Your intuition is good, but the math confirms the statement is false.

Q: Why can't we just use the standard variance formula for natural numbers directly?

You absolutely can, and it is a brilliant shortcut! The variance of the first $n$ natural numbers is $\frac{n^2-1}{12}$. Since the first $n$ even natural numbers are just the first $n$ natural numbers multiplied by $2$, you can use the scaling property of variance: $Var(kX) = k^2 Var(X)$. Here, $k=2$, so $Var(2X) = 4 \cdot Var(X) = 4 \cdot \frac{n^2-1}{12} = \frac{n^2-1}{3}$. This confirms the result much faster than calculating from scratch.

Q: Is Statement-2 always true, or is there a catch?

Statement-2 is a fundamental pillar of sequences and series. The sum of the first $n$ natural numbers, $\sum i = \frac{n(n+1)}{2}$, and the sum of their squares, $\sum i^2 = \frac{n(n+1)(2n+1)}{6}$, are standard identities derived from the method of differences or mathematical induction. They are universally true for all natural numbers $n \ge 1$. There is no catch here; it is a rock-solid foundation for your calculations.

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 $\frac{n(n+1)(2n+1)}{6}$

Options:

(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.

(C) Statement-1 is false, Statement-2 is true.

(D) Statement-1 is true, Statement-2 is true. Statement-2 is a correct explanation for Statement-1.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are not just solving a problem; we are peeling back the layers of statistical variance. We are presented with two statements, and our mission is to determine their truth values.

Let us embark on this journey by first examining the bedrock of our calculation: Statement-2.

### The Bedrock of Summation

Statement-2 provides us with two of the most iconic identities in mathematics. The sum of the first $n$ natural numbers is given by:

$$ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} $$

The sum of their squares is given by:

$$ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} $$

These are not just formulas; they are the DNA of discrete mathematics. Since these are standard, universally accepted results, we can confidently declare **Statement-2 as True**.

### Defining the Engine

To evaluate Statement-1, we must master the concept of variance. Variance, denoted by $\sigma^2$, is a measure of how spread out our data is.

The most efficient way to compute it is using the formula:

$$ \sigma^2 = E[X^2] - (\bar{x})^2 $$

Here, $E[X^2]$ is the mean of the squares and $\bar{x}$ is the mean of the data set. Think of this as the "average of the squares minus the square of the average."

### The Data Set and the Calculation

Our data set consists of the first $n$ even natural numbers: $X = \{2, 4, 6, \dots, 2n\}$. We can write the general term as $x_i = 2i$ for $i = 1, 2, \dots, n$.

Now, let us calculate the mean, $\bar{x}$:

$$ \bar{x} = \frac{\sum_{i=1}^{n} 2i}{n} = \frac{2}{n} \sum_{i=1}^{n} i $$

Substituting the identity from Statement-2, we get:

$$ \bar{x} = \frac{2}{n} \cdot \frac{n(n+1)}{2} = n+1 $$

Next, we find the mean of the squares, $E[X^2]$:

$$ E[X^2] = \frac{\sum_{i=1}^{n} (2i)^2}{n} = \frac{4}{n} \sum_{i=1}^{n} i^2 $$

Using the identity for the sum of squares, we have:

$$ E[X^2] = \frac{4}{n} \cdot \frac{n(n+1)(2n+1)}{6} = \frac{2(n+1)(2n+1)}{3} $$

### The Final Verdict

Now, we bring it all together into our variance formula:

$$ \sigma^2 = \frac{2(n+1)(2n+1)}{3} - (n+1)^2 $$

To simplify this, we factor out $(n+1)$:

$$ \sigma^2 = (n+1) \left[ \frac{2(2n+1)}{3} - (n+1) \right] $$

Expanding the terms inside the bracket, we get:

$$ \frac{4n+2 - 3n - 3}{3} = \frac{n-1}{3} $$

Thus, the variance is:

$$ \sigma^2 = (n+1) \cdot \frac{n-1}{3} = \frac{n^2-1}{3} $$

Compare this with Statement-1, which claims the variance is $\frac{n^2-1}{4}$. The denominator is clearly different! The statement is a clever trap, but our rigorous derivation reveals the truth: **Statement-1 is False**.

Animated Solution for Mathematics - Statistics: 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 $\frac{n ( n + 1 ) ( 2 n + 1 )}{6}$

Select Answer:

Visualized Solution

Verifying Statement-2

Defining Variance

The Data Set: Even Numbers

Calculating the Mean $(\overset{x}{ˉ})$

Calculating Mean of Squares

Computing the Variance

Final Conclusion

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Bedrock of Summation

Defining the Engine

The Data Set and the Calculation

The Final Verdict

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