Sequences & Series
Summation of Series
Grade 11
Question:
<p>Let \(S_n = \dfrac{1}{1^3} + \dfrac{1+2}{1^3+2^3} + \dfrac{1+2+3}{1^3+2^3+3^3} + \ldots\ldots (n \text{ terms})\), where \(n = 1,2,3,4,\ldots\ldots\), then \(S_n\) is always less than:</p>
<p>1</p>
<p>2</p>
<p>3</p>
<p>4</p>
Step-by-Step Solution
Key Concept: Use the formulas: sum of first k natural numbers = k(k+1)/2 and sum of first k cubes = [k(k+1)/2]², then simplify the general term to find that each term equals 4/(k(k+1)) = 4(1/k - 1/(k+1)), making Sₙ a telescoping series.
<p><strong>Step 1:</strong> Find the general term. The numerator of the kth term is 1+2+...+k = k(k+1)/2. The denominator is 1³+2³+...+k³ = [k(k+1)/2]².</p><p><strong>Step 2:</strong> Simplify: T_k = [k(k+1)/2] / [k(k+1)/2]² = 2/[k(k+1)] = 2(1/k - 1/(k+1)) × 2 = 4(1/k - 1/(k+1))</p><p><strong>Step 3:</strong> Sum telescopes: S_n = 4[(1/1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/(n+1))] = 4[1 - 1/(n+1)] = 4n/(n+1)</p><p><strong>Step 4:</strong> As n → ∞, S_n → 4. For all finite n, S_n < 4 (since n/(n+1) < 1 for all n ≥ 1).</p><p>∴ Answer: B (S_n is always less than 4)</p>
Correct Answer: B