The S.D. of the first n natural numbers is

Options

A. `(n+1)/(2)`
B. `sqrt(n(n+1))/(2))`
C. `sqrt((n^(2)+1)/(12))`
D. None of these

Answer

To find the standard deviation (S.D.) of the first n natural numbers, we will follow these steps:

### Step 1: Identify the first n natural numbers
The first n natural numbers are:
\[ 1, 2, 3, \ldots, n \]

### Step 2: Calculate the mean (average)
The mean \( \bar{x} \) of the first n natural numbers can be calculated using the formula:
\[ \bar{x} = \frac{\text{Sum of observations}}{\text{Total number of observations}} = \frac{1 + 2 + 3 + \ldots + n}{n} \]
The sum of the first n natural numbers is given by:
\[ \text{Sum} = \frac{n(n + 1)}{2} \]
Thus, the mean is:
\[ \bar{x} = \frac{\frac{n(n + 1)}{2}}{n} = \frac{n + 1}{2} \]

### Step 3: Calculate the variance
The variance \( \sigma^2 \) is given by the formula:
\[ \sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2 \]
We need to calculate \( \sum x_i^2 \) for the first n natural numbers:
\[ \sum x_i^2 = 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \]
Now, substituting into the variance formula:
\[ \sigma^2 = \frac{\frac{n(n + 1)(2n + 1)}{6}}{n} - \left(\frac{n + 1}{2}\right)^2 \]
This simplifies to:
\[ \sigma^2 = \frac{(n + 1)(2n + 1)}{6} - \frac{(n + 1)^2}{4} \]

### Step 4: Simplify the variance
To simplify, we need a common denominator. The common denominator for 6 and 4 is 12:
\[ \sigma^2 = \frac{2(n + 1)(2n + 1)}{12} - \frac{3(n + 1)^2}{12} \]
This can be combined as:
\[ \sigma^2 = \frac{2(n + 1)(2n + 1) - 3(n + 1)^2}{12} \]
Factoring out \( (n + 1) \):
\[ \sigma^2 = \frac{(n + 1)(2(2n + 1) - 3(n + 1))}{12} \]
Expanding the terms 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 5: Calculate the standard deviation
The standard deviation \( \sigma \) is the square root of the variance:
\[ \sigma = \sqrt{\sigma^2} = \sqrt{\frac{(n + 1)(n - 1)}{12}} = \sqrt{\frac{n^2 - 1}{12}} \]

### Final Result
Thus, the standard deviation of the first n natural numbers is:
\[ \sigma = \sqrt{\frac{n^2 - 1}{12}} \]