The arithmetic mean of the squares of first n natural numbers is

Options

A. `(n+1)/(6)`
B. `((n+1)(2n+1))/(6)`
C. `(n^(2)-1)/(6)`
D. none of these

Answer

To find the arithmetic mean of the squares of the first \( n \) natural numbers, we can follow these steps:

### Step 1: Identify the squares of the first \( n \) natural numbers
The first \( n \) natural numbers are \( 1, 2, 3, \ldots, n \). Their squares are:
\[ 1^2, 2^2, 3^2, \ldots, n^2 \]

### Step 2: Calculate the sum of the squares
The sum of the squares of the first \( n \) natural numbers can be expressed as:
\[ S = 1^2 + 2^2 + 3^2 + \ldots + n^2 \]
There is a formula for this sum:
\[ S = \frac{n(n + 1)(2n + 1)}{6} \]

### Step 3: Determine the number of observations
The total number of observations (the first \( n \) natural numbers) is \( n \).

### Step 4: Calculate the arithmetic mean
The arithmetic mean (AM) is given by the formula:
\[ \text{AM} = \frac{\text{Sum of observations}}{\text{Number of observations}} \]
Substituting the values we have:
\[ \text{AM} = \frac{S}{n} = \frac{\frac{n(n + 1)(2n + 1)}{6}}{n} \]

### Step 5: Simplify the expression
When we simplify the expression, we can cancel \( n \) in the numerator and denominator:
\[ \text{AM} = \frac{(n + 1)(2n + 1)}{6} \]

### Final Result
Thus, the arithmetic mean of the squares of the first \( n \) natural numbers is:
\[ \text{AM} = \frac{(n + 1)(2n + 1)}{6} \]

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