<p>The variance of the first <em>n</em> natural numbers is</p>
<p>\(\dfrac{n^2 - 1}{12}\)</p>
<p>\(\dfrac{n^2 - 1}{6}\)</p>
<p>\(\dfrac{n^2 + 1}{6}\)</p>
<p>\(\dfrac{n^2 + 1}{12}\)</p>
Step-by-Step Solution
Key Concept: Variance = E(X²) - [E(X)]². For first n natural numbers, use the formulas: mean = (n+1)/2 and E(X²) = n(n+1)(2n+1)/6 to find σ² = (n²-1)/12.
<p><strong>Step 1:</strong> For first n natural numbers {1, 2, 3, ..., n}, find the mean:</p><p>Mean = E(X) = (1 + 2 + ... + n)/n = n(n+1)/(2n) = (n+1)/2</p><p><strong>Step 2:</strong> Find E(X²):</p><p>E(X²) = (1² + 2² + ... + n²)/n = [n(n+1)(2n+1)/6]/n = (n+1)(2n+1)/6</p><p><strong>Step 3:</strong> Apply variance formula σ² = E(X²) - [E(X)]²:</p><p>σ² = (n+1)(2n+1)/6 - [(n+1)/2]²</p><p>σ² = (n+1)(2n+1)/6 - (n+1)²/4</p><p><strong>Step 4:</strong> Find common denominator (12):</p><p>σ² = [2(n+1)(2n+1) - 3(n+1)²]/12</p><p>σ² = [(n+1)[2(2n+1) - 3(n+1)]]/12</p><p>σ² = [(n+1)(4n + 2 - 3n - 3)]/12</p><p>σ² = [(n+1)(n-1)]/12 = <strong>(n²-1)/12</strong></p><p>∴ Answer: A</p>
Correct Answer: A