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: Test with a small, concrete case.
Instead of working with the general formula from scratch, pick a small value of $n$, say $n = 3$, and see which option matches.
Step 2: Work out the actual mean for n = 3.
The squares of the first three natural numbers are $1, 4, 9$, and their sum is $14$. So the mean is \[ \text{Mean} = \frac{14}{3} \]
Step 3: Check which option gives this value.
Try $\frac{(n+1)(2n+1)}{6}$ with $n = 3$: \[ \frac{(3+1)(2\times3+1)}{6} = \frac{4 \times 7}{6} = \frac{28}{6} = \frac{14}{3} \] This matches exactly. The other options give $n^2 = 9$, or $\frac{n+1}{2} = 2$, or $\frac{n(n+1)(2n+1)}{6} = 14$, none of which equal $\frac{14}{3}$.
Step 4: Confirm this holds in general.
This matches because the sum of squares up to $n$ is $\frac{n(n+1)(2n+1)}{6}$, and dividing by the $n$ observations cancels one factor of $n$, leaving exactly $\frac{(n+1)(2n+1)}{6}$.
\[ \boxed{\dfrac{(n+1)(2n+1)}{6}} \]