Question

Second central moment (Variance) for the first $n$ natural numbers is ____.

• A. $\frac{n(n+1)}{2}$
• B. $\frac{n(n+1)(2n+1)}{6}$
• C. $\frac{n^2+1}{12}$
• D. $\frac{n^2-1}{12}$

Hint:

Memorize variance of first $n$ natural numbers: $\frac{n^2-1}{12}$.

Answer 1

The Correct Option is D

Concept: Variance of first $n$ natural numbers.
Step 1: Mean of first $n$ natural numbers.
\[ \bar{x} = \frac{n+1}{2} \]
Step 2: Variance formula.
\[ \sigma^2 = \frac{1}{n} \sum x^2 - (\bar{x})^2 \]
Step 3: Use standard results.
\[ \sum x^2 = \frac{n(n+1)(2n+1)}{6} \]
Step 4: Substitute.
\[ \sigma^2 = \frac{(n+1)(2n+1)}{6} - \left(\frac{n+1}{2}\right)^2 \] \[ = \frac{(n+1)}{12} (2(2n+1) - 3(n+1)) \] \[ = \frac{(n+1)}{12}(4n+2 -3n -3) = \frac{(n+1)(n-1)}{12} \] \[ = \frac{n^2 - 1}{12} \]
Hence, the variance is $\frac{n^2 - 1{12}$.