<p>Let \(S_n = \dfrac{1}{1^3} + \dfrac{1+2}{1^3+2^3} + \dfrac{1+2+3}{1^3+2^3+3^3} + \cdots + \dfrac{1+2+\cdots+n}{1^3+2^3+\cdots+n^3}\). If \(100S_n = n\), then \(n\) is equal to</p>
Step-by-Step Solution
Key Concept: Use the formulas: sum of first n natural numbers = n(n+1)/2 and sum of cubes = [n(n+1)/2]². This transforms each term into (n(n+1)/2) / [n(n+1)/2]² = 2/(n(n+1)), which telescopes via partial fractions.
<p><strong>Step 1:</strong> Identify formulas for numerators and denominators.</p><p>For the rth term: Numerator = 1+2+···+r = r(r+1)/2</p><p>Denominator = 1³+2³+···+r³ = [r(r+1)/2]²</p><p><strong>Step 2:</strong> Simplify the rth term.</p><p>tᵣ = [r(r+1)/2] / [r(r+1)/2]² = 1/[r(r+1)/2] = 2/(r(r+1))</p><p><strong>Step 3:</strong> Apply partial fractions.</p><p>2/(r(r+1)) = 2[1/r - 1/(r+1)]</p><p><strong>Step 4:</strong> Compute the telescoping sum.</p><p>Sₙ = Σᵣ₌₁ⁿ 2[1/r - 1/(r+1)]</p><p>= 2[(1/1 - 1/2) + (1/2 - 1/3) + ··· + (1/n - 1/(n+1))]</p><p>= 2[1 - 1/(n+1)] = 2n/(n+1)</p><p><strong>Step 5:</strong> Solve 100Sₙ = n.</p><p>100 · 2n/(n+1) = n</p><p>200n/(n+1) = n</p><p>200n = n(n+1)</p><p>200 = n+1</p><p>n = 199</p><p>∴ Answer: D (199)</p>
Correct Answer: D