Sequences & Series
Summation of Special Series
Grade 11
Question:
<p>The sum to 50 terms of the series \(\dfrac{3}{1^2} + \dfrac{5}{1^2 + 2^2} + \dfrac{7}{1^2 + 2^2 + 3^2} + \cdots\) is</p>
<p>\(\dfrac{100}{17}\)</p>
<p>\(\dfrac{150}{17}\)</p>
<p>\(\dfrac{200}{51}\)</p>
<p>\(\dfrac{50}{17}\)</p>
Step-by-Step Solution
Key Concept: Recognize that the denominator follows the formula for sum of squares: 1² + 2² + ... + n² = n(n+1)(2n+1)/6, and the numerator is 2n+1. This allows telescoping decomposition using partial fractions.
<p><strong>Step 1:</strong> Identify the general term. The nth term has:</p><p>• Numerator: 2n + 1</p><p>• Denominator: 1² + 2² + ... + n² = n(n+1)(2n+1)/6</p><p><strong>Step 2:</strong> Simplify the general term:</p><p>T_n = (2n+1)/(n(n+1)(2n+1)/6) = 6/[n(n+1)]</p><p><strong>Step 3:</strong> Use partial fractions:</p><p>6/[n(n+1)] = 6[1/n - 1/(n+1)]</p><p><strong>Step 4:</strong> Sum the first 50 terms (telescoping series):</p><p>S₅₀ = 6[1/1 - 1/2 + 1/2 - 1/3 + ... + 1/50 - 1/51]</p><p>S₅₀ = 6[1 - 1/51] = 6 × 50/51 = 300/51 = 100/17</p><p>∴ Answer: A</p>
Correct Answer: A