Question

Select the mean of squares of first n natural numbers :

• A. \(n^2\)
• B. \(\frac{(n+1)(2n+1)}{6}\)
• C. \(\frac{n+1}{2}\)
• D. \(\frac{n(n+1)(2n+1)}{6}\)

Hint:

If you forget the formula during the exam, plug in \(n = 1\).
The first natural number is 1, and its square is 1. Its mean must be 1.
Checking Option (B): \(\frac{(1+1)(2\cdot1+1)}{6} = \frac{2 \times 3}{6} = 1\).
Checking Option (D): \(\frac{1(1+1)(2\cdot1+1)}{6} = 1\).
To differentiate, try \(n = 2\). The squares are 1 and 4, so mean is \(\frac{1+4}{2} = 2.5\).
Option (B) for \(n=2\) gives: \(\frac{(2+1)(2\cdot2+1)}{6} = \frac{3 \times 5}{6} = 2.5\). This confirms Option (B) is correct.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question is from the topic of Statistics and Mathematical Induction/Series.
We are asked to find the arithmetic mean of the squares of the first \(n\) natural numbers.

Step 2: Key Formula or Approach:
The mean of \(n\) observations is calculated as:
\[ \text{Mean} = \frac{\text{Sum of all observations}}{\text{Total number of observations}} \]
The sum of squares of the first \(n\) natural numbers is given by the standard formula:
\[ \sum_{i=1}^{n} i^2 = 1^2 + 2^2 + 3^2 + \dots + n^2 = \frac{n(n+1)(2n+1)}{6} \]

Step 3: Detailed Explanation:
Let the observations be the squares of the first \(n\) natural numbers:
\[ 1^2, 2^2, 3^2, \dots, n^2 \]
The total number of observations is \(n\).
Using the sum formula:
\[ \text{Sum} = \frac{n(n+1)(2n+1)}{6} \]
Now, compute the mean:
\[ \text{Mean} = \frac{\text{Sum}}{n} \]
Substitute the expression for the sum:
\[ \text{Mean} = \frac{\frac{n(n+1)(2n+1)}{6}}{n} \]
Cancel \(n\) from the numerator and denominator:
\[ \text{Mean} = \frac{(n+1)(2n+1)}{6} \]

Step 4: Final Answer:
The mean of the squares of the first \(n\) natural numbers is \(\frac{(n+1)(2n+1)}{6}\).