The standard deviation of the first n natural numbers is

Options

A. `(sqrt(n^(2)+1))/(12)`
B. `(n^(2)-1)/(12)`
C. `sqrt((n^(2)-1)/(12))`
D. `(n^(2)+1)/(12)`

Answer

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

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

### Step 2: Calculate the Variance
The variance \( \sigma^2 \) is given by the formula:
\[ \sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2 \]
Where \( \sum x_i^2 \) is the sum of the squares of the first \( n \) natural numbers. The formula for the sum of squares is:
\[ 1^2 + 2^2 + 3^2 + \ldots + n^2 = \frac{n(n + 1)(2n + 1)}{6} \]
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 3: Simplify the Variance Expression
To combine the two terms, we need a common denominator, which is 12:
\[ \sigma^2 = \frac{2(n + 1)(2n + 1)}{12} - \frac{3(n + 1)^2}{12} \]
This leads to:
\[ \sigma^2 = \frac{2(n + 1)(2n + 1) - 3(n + 1)^2}{12} \]
Factoring out \( (n + 1) \):
\[ \sigma^2 = \frac{(n + 1)(2(2n + 1) - 3(n + 1))}{12} \]
Now simplifying 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 4: Calculate the Standard Deviation
The standard deviation \( \sigma \) is the square root of the variance:
\[ \sigma = \sqrt{\sigma^2} = \sqrt{\frac{(n + 1)(n - 1)}{12}} = \frac{\sqrt{n^2 - 1}}{\sqrt{12}} = \frac{\sqrt{n^2 - 1}}{2\sqrt{3}} \]

### Final Answer
Thus, the standard deviation of the first \( n \) natural numbers is:
\[ \sigma = \frac{\sqrt{n^2 - 1}}{2\sqrt{3}} \]

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