The standard deviation `sigma` of the first N natural numbers can be obtained using which one of the following formulae?

Options

A. A. `sigma=(N^(2)-1)/(12)`
B. B. `sigma=sqrt(N^(2)-1)/(12)`
C. C. `sigma=sqrt(N-1)/(12)`
D. D. `sigma=sqrt(N^(2)-1)/(6N)`

Answer

To find the standard deviation \( \sigma \) of the first \( N \) natural numbers, we can follow these steps:

### Step 1: Understand the definition of standard deviation
The standard deviation \( \sigma \) is the square root of the variance \( \sigma^2 \). The variance can be calculated using the formula:

\[ \sigma^2 = \frac{1}{N} \sum_{i=1}^{N} x_i^2 - \left( \frac{1}{N} \sum_{i=1}^{N} x_i \right)^2 \]

### Step 2: Identify \( x_i \)
In this case, \( x_i \) represents the first \( N \) natural numbers, which are \( 1, 2, 3, \ldots, N \).

### Step 3: Calculate \( \sum_{i=1}^{N} x_i^2 \)
The sum of the squares of the first \( N \) natural numbers is given by the formula:

\[ \sum_{i=1}^{N} i^2 = \frac{N(N + 1)(2N + 1)}{6} \]

### Step 4: Calculate \( \sum_{i=1}^{N} x_i \)
The sum of the first \( N \) natural numbers is given by the formula:

\[ \sum_{i=1}^{N} i = \frac{N(N + 1)}{2} \]

### Step 5: Substitute into the variance formula
Now, substituting these sums into the variance formula:

\[ \sigma^2 = \frac{1}{N} \left( \frac{N(N + 1)(2N + 1)}{6} \right) - \left( \frac{1}{N} \cdot \frac{N(N + 1)}{2} \right)^2 \]

### Step 6: Simplify the variance
1. The first term simplifies to:

\[ \frac{(N + 1)(2N + 1)}{6} \]

2. The second term simplifies to:

\[ \left( \frac{(N + 1)}{2} \right)^2 = \frac{(N + 1)^2}{4} \]

3. Now substituting back:

\[ \sigma^2 = \frac{(N + 1)(2N + 1)}{6} - \frac{(N + 1)^2}{4} \]

### Step 7: Find a common denominator
The common denominator for 6 and 4 is 12. Thus, we rewrite both fractions:

\[ \sigma^2 = \frac{2(N + 1)(2N + 1)}{12} - \frac{3(N + 1)^2}{12} \]

### Step 8: Combine the fractions
Combine the two fractions:

\[ \sigma^2 = \frac{2(N + 1)(2N + 1) - 3(N + 1)^2}{12} \]

### Step 9: Factor out \( (N + 1) \)
Factoring out \( (N + 1) \):

\[ \sigma^2 = \frac{(N + 1)(2(2N + 1) - 3(N + 1))}{12} \]

### Step 10: Simplify the expression
Now, simplify the expression inside the parentheses:

\[ 2(2N + 1) - 3(N + 1) = 4N + 2 - 3N - 3 = N - 1 \]

Thus, we have:

\[ \sigma^2 = \frac{(N + 1)(N - 1)}{12} \]

### Step 11: Find the standard deviation
Taking the square root to find the standard deviation:

\[ \sigma = \sqrt{\frac{(N + 1)(N - 1)}{12}} = \sqrt{\frac{N^2 - 1}{12}} \]

### Final Answer
The standard deviation \( \sigma \) of the first \( N \) natural numbers is:

\[ \sigma = \sqrt{\frac{N^2 - 1}{12}} \]