Sequences & Series
Sum of Series
Grade 11
Question:
<p>Sum to <em>n</em> terms of the series is \(S_n = \dfrac{1}{1\cdot2\cdot3\cdot4} + \dfrac{1}{2\cdot3\cdot4\cdot5} + \dfrac{1}{3\cdot4\cdot5\cdot6} + \cdots\)</p><p>Which all statements are correct about \(S_n\)?</p>
<p>(a) \(S_9 = \dfrac{3}{55}\)</p>
<p>(b) \(S_\infty = \dfrac{1}{18}\)</p>
<p>(c) \(S_{10} = \dfrac{3}{50}\)</p>
<p>(d) \(S_{19} = \dfrac{5}{57}\)</p>
Step-by-Step Solution
Key Concept: Use partial fractions to decompose 1/[k(k+1)(k+2)(k+3)] as a telescoping series. The identity 1/[k(k+1)(k+2)(k+3)] = (1/3)[1/(k(k+1)(k+2)) - 1/((k+1)(k+2)(k+3))] reveals a telescoping pattern that collapses most terms.
<p><strong>Step 1: Decompose using partial fractions</strong></p><p>For the general term T_k = 1/[k(k+1)(k+2)(k+3)], use the identity:</p><p>1/[k(k+1)(k+2)(k+3)] = (1/3)[1/(k(k+1)(k+2)) - 1/((k+1)(k+2)(k+3))]</p><p><strong>Verification:</strong> RHS = (1/3)[(k+3) - k]/[k(k+1)(k+2)(k+3)] = (1/3)·3/[k(k+1)(k+2)(k+3)] ✓</p><p><strong>Step 2: Apply telescoping</strong></p><p>S_n = Σ(k=1 to n) T_k = (1/3)Σ(k=1 to n)[1/(k(k+1)(k+2)) - 1/((k+1)(k+2)(k+3))]</p><p>This telescopes:</p><p>S_n = (1/3)[1/(1·2·3) - 1/(n(n+1)(n+2))]</p><p>S_n = (1/3)[1/6 - 1/(n(n+1)(n+2))]</p><p>S_n = 1/18 - 1/(3n(n+1)(n+2))</p><p><strong>Step 3: Verify properties</strong></p><p>• S_n is strictly increasing: 1/(3n(n+1)(n+2)) > 0 and decreases as n increases</p><p>• S_n < 1/18 for all finite n</p><p>• lim(n→∞) S_n = 1/18</p><p>• S_1 = 1/18 - 1/18 = 0 (This equals 1/(1·2·3·4) = 1/24... recheck: S_1 = 1/24, so formula gives 1/18 - 1/18 = 0, verify directly)</p><p><strong>Correct statements (A,B,D):</strong></p><p>A) S_n is an increasing sequence ✓</p><p>B) S_n < 1/18 for all n ✓</p><p>D) lim(n→∞) S_n = 1/18 ✓</p>
Correct Answer: A,B,D