<p>The weighted mean of the first \(n\) natural numbers whose weights are equal to the squares of the corresponding numbers is</p>
<p>\(\dfrac{n+1}{2}\)</p>
<p>\(\dfrac{3n(n+1)}{2(2n+1)}\)</p>
<p>\(\dfrac{(n+1)(2n+1)}{6}\)</p>
<p>\(\dfrac{n(n+1)}{2}\)</p>
Step-by-Step Solution
Key Concept: The weighted mean uses weights w_i = i² for each natural number i, requiring careful application of the weighted mean formula: Σ(i·i²)/Σ(i²) = Σ(i³)/Σ(i²), then use standard summation formulas for cubes and squares.
<p><strong>Step 1:</strong> Set up the weighted mean formula where values are {1, 2, 3, ..., n} and weights are {1², 2², 3², ..., n²}</p><p>Weighted Mean = Σ(i·i²)/Σ(i²) = Σi³/Σi²</p><p><strong>Step 2:</strong> Apply standard summation formulas:</p><p>Σi³ = [n(n+1)/2]²</p><p>Σi² = n(n+1)(2n+1)/6</p><p><strong>Step 3:</strong> Substitute and simplify:</p><p>= [n(n+1)/2]² ÷ [n(n+1)(2n+1)/6]</p><p>= [n²(n+1)²/4] × [6/n(n+1)(2n+1)]</p><p>= [n(n+1)·6]/[4(2n+1)]</p><p>= [3n(n+1)]/[2(2n+1)]</p><p>∴ Answer: B (which equals <strong>3n(n+1)/[2(2n+1)]</strong>)</p>
Correct Answer: B