<p>If the sum \(\dfrac{3}{1^2} + \dfrac{5}{1^2+2^2} + \dfrac{7}{1^2+2^2+3^2} + \cdots\) up to 20 terms is equal to \(\dfrac{k}{21}\), then \(k\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the formula for sum of squares: 1² + 2² + ... + n² = n(n+1)(2n+1)/6, then decompose the general term using partial fractions to find a telescoping series.
<p><strong>Step 1: Find the general term</strong></p><p>The numerator follows pattern: 3, 5, 7, ... = 2n + 1</p><p>The denominator is: 1² + 2² + ... + n² = n(n+1)(2n+1)/6</p><p>General term: aₙ = (2n+1)/(n(n+1)(2n+1)/6) = 6/(n(n+1))</p><p><strong>Step 2: Decompose using partial fractions</strong></p><p>6/(n(n+1)) = 6[1/n - 1/(n+1)]</p><p><strong>Step 3: Apply telescoping sum</strong></p><p>S₂₀ = 6[1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/20 - 1/21]</p><p>S₂₀ = 6[1 - 1/21] = 6 · 20/21 = 120/21</p><p><strong>Step 4: Express in form k/21</strong></p><p>120/21 = k/21</p><p>∴ <strong>k = 120</strong></p>
Correct Answer: C