Sequences & Series
Infinite geometric series and summation
Grade 11
Question:
<p>Let <em>k</em> be natural number. Defined \( S_k \) as the sum of the infinite geometric series with first term \( (k^2 - 1) \) and common ratio \( \dfrac{1}{k} \), that is \( S_k = \dfrac{k^2-1}{k^0} + \dfrac{k^2-1}{k^1} + \dfrac{k^2-1}{k^2} + \cdots \). The value of \( \displaystyle\sum_{k=1}^{\infty} \dfrac{S_k}{2^{k-1}} \), is:</p>
<p>(a) 20</p>
<p>(b) 18</p>
<p>(c) 16</p>
<p>(d) 14</p>
Step-by-Step Solution
Key Concept: First, find the closed form of S_k using the infinite geometric series formula S = a/(1-r), then evaluate the resulting sum by partial fractions or telescoping.
<p><strong>Step 1:</strong> Find S_k using the geometric series formula. With first term a = k²-1 and common ratio r = 1/k:</p><p>S_k = (k²-1)/(1 - 1/k) = (k²-1)/((k-1)/k) = k(k²-1)/(k-1) = k(k-1)(k+1)/(k-1) = k(k+1)</p><p><strong>Step 2:</strong> Handle k=1 separately: S₁ = 0 (since k²-1 = 0), so it contributes 0 to the sum.</p><p><strong>Step 3:</strong> Now evaluate Σ(k=1 to ∞) [S_k/2^(k-1)] = Σ(k=2 to ∞) [k(k+1)/2^(k-1)] (since k=1 contributes 0)</p><p><strong>Step 4:</strong> Use partial fractions: k(k+1) = k² + k. Rewrite as:</p><p>Σ(k=2 to ∞) [k(k+1)/2^(k-1)] = 2·Σ(k=2 to ∞) [k(k+1)/2^k]</p><p><strong>Step 5:</strong> Use the formulas Σ(k=1 to ∞) k·x^k = x/(1-x)² and Σ(k=1 to ∞) k²·x^k = x(1+x)/(1-x)³ with x = 1/2:</p><p>Σ(k=2 to ∞) k(k+1)/2^k = Σ(k=1 to ∞) k(k+1)/2^k - 2/2 = [sum for k≥1] - 1</p><p>= Σ(k=1 to ∞) k²/2^k + Σ(k=1 to ∞) k/2^k - 1 = (1+1/2)·(1/2)/(1/4) + (1/2)/(1/4) - 1</p><p>= 3/2 · 2 + 2 - 1 = 3 + 2 - 1 = 4</p><p>∴ Answer: 2 × 4 = <strong>8</strong></p>
Correct Answer: A