<p>The variance of first <em>n</em> even natural numbers is:</p>
<p>(A) \(\dfrac{n^2-1}{3}\)</p>
<p>(B) \(n^2 - 1\)</p>
<p>(C) \(\dfrac{n(n+1)}{2}\)</p>
<p>(D) \(\dfrac{(n+1)(2n+1)}{6}\)</p>
Step-by-Step Solution
Key Concept: First n even natural numbers form the sequence 2, 4, 6, ..., 2n. Their variance can be calculated using the formula Var(X) = E(X²) - [E(X)]², where the mean is n+1 and the sum of squares follows a known pattern.
<p><strong>Step 1:</strong> The first n even natural numbers are: 2, 4, 6, ..., 2n</p><p><strong>Step 2:</strong> Calculate mean: Mean = (2 + 4 + 6 + ... + 2n)/n = 2(1 + 2 + 3 + ... + n)/n = 2·n(n+1)/(2n) = n+1</p><p><strong>Step 3:</strong> Calculate E(X²): E(X²) = (4 + 16 + 36 + ... + 4n²)/n = 4(1 + 4 + 9 + ... + n²)/n = 4·n(n+1)(2n+1)/(6n) = 2(n+1)(2n+1)/3</p><p><strong>Step 4:</strong> Apply variance formula: Var(X) = E(X²) - [E(X)]² = 2(n+1)(2n+1)/3 - (n+1)² = (n+1)[2(2n+1)/3 - (n+1)] = (n+1)[(4n+2-3n-3)/3] = (n+1)(n-1)/3 = <strong>(n²-1)/3</strong></p><p>∴ Answer: A</p>
Correct Answer: A