Sequences & Series
Sum of series
Grade 11
Question:
<p>If \(S_n = \dfrac{1}{1^3} + \dfrac{1+2}{1^3+2^3} + \dfrac{1+2+3}{1^3+2^3+3^3} + \cdots + \dfrac{1+2+\cdots+n}{1^3+2^3+\cdots+n^3}\), then \(S_n\) equals:</p>
<p>(1) \(\dfrac{2n}{n+1}\)</p>
<p>(2) \(\dfrac{n}{n+1}\)</p>
<p>(3) \(\dfrac{2(n-1)}{n}\)</p>
<p>(4) \(2 - \dfrac{2}{n+1}\)</p>
Step-by-Step Solution
Key Concept: Use the formulas: sum of first n natural numbers = n(n+1)/2 and sum of first n cubes = [n(n+1)/2]². This transforms each term into a simple fraction that telescopes or simplifies dramatically.
<p><strong>Step 1: Apply summation formulas</strong></p><p>For the r-th 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: Simplify the general term</strong></p><p>T_r = [r(r+1)/2] / [r(r+1)/2]² = 1/[r(r+1)/2] = 2/[r(r+1)]</p><p><strong>Step 3: Apply partial fractions</strong></p><p>2/[r(r+1)] = 2(1/r - 1/(r+1))</p><p><strong>Step 4: Sum using telescoping property</strong></p><p>S_n = 2∑(1/r - 1/(r+1)) from r=1 to n</p><p>= 2[(1/1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/(n+1))]</p><p>= 2[1 - 1/(n+1)]</p><p>= 2n/(n+1)</p><p>∴ Answer: D</p>
Correct Answer: D