<p>Find the sum of the infinite series \(1 + \left(1 + \dfrac{1}{5}\right)\left(\dfrac{1}{2}\right) + \left(1 + \dfrac{1}{5} + \dfrac{1}{5^2}\right)\left(\dfrac{1}{2^2}\right) + \cdots\)</p>
Step-by-Step Solution
Key Concept: Recognize that each term has the form of a geometric series sum multiplied by a power of 1/2. The coefficient of (1/2)^n is the sum of a geometric series: 1 + 1/5 + 1/5² + ... + 1/5^n = (1-(1/5)^(n+1))/(4/5). Reorganize the double sum by separating the geometric series from the power-of-2 denominator.
<p><strong>Step 1:</strong> Write the general term. The n-th term (n ≥ 0) is:</p><p>T_n = (1 + 1/5 + 1/5² + ... + 1/5^n) · (1/2)^n</p><p><strong>Step 2:</strong> Use the geometric series formula for the bracketed part:</p><p>1 + 1/5 + ... + 1/5^n = (1 - (1/5)^(n+1))/(1 - 1/5) = 5/4 · (1 - (1/5)^(n+1))</p><p><strong>Step 3:</strong> Rewrite the sum by separating:</p><p>S = Σ(n=0 to ∞) [5/4 · (1 - (1/5)^(n+1))] · (1/2)^n</p><p>= (5/4)Σ(n=0 to ∞) (1/2)^n - (5/4)Σ(n=0 to ∞) (1/5)^(n+1) · (1/2)^n</p><p><strong>Step 4:</strong> Evaluate each sum:</p><p>First sum: Σ(1/2)^n = 1/(1-1/2) = 2</p><p>Second sum: Σ(1/5)^(n+1) · (1/2)^n = (1/5)Σ(1/10)^n = (1/5) · 1/(1-1/10) = (1/5) · (10/9) = 2/9</p><p><strong>Step 5:</strong> Combine:</p><p>S = (5/4) · 2 - (5/4) · (2/9) = 5/2 - 5/18 = 45/18 - 5/18 = 40/18 = 20/9</p><p>∴ Answer: <strong>20/9</strong></p>
Correct Answer: 20/9