<p>The sum of \(0.2 + 0.004 + 0.00006 + 0.0000008 + \cdots\) to ∞ is</p>
<p>(1) \(\dfrac{200}{891}\)</p>
<p>(2) \(\dfrac{2000}{9801}\)</p>
<p>(3) \(\dfrac{1000}{9801}\)</p>
<p>(4) \(\dfrac{2180}{9801}\)</p>
Step-by-Step Solution
Key Concept: Recognize the numerators follow a factorial pattern (1, 2, 3, 4, ...) while denominators are powers of 10, making this a series involving e or its modifications. Rewrite as Σ(n/10^(2n-1)) and relate to the exponential series expansion.
<p><strong>Step 1:</strong> Write out the general term. The series is: 0.2 + 0.004 + 0.00006 + 0.0000008 + ...</p><p>Numerators: 1, 2, 3, 4, ... (factorials up to n)</p><p>Denominators: 10¹, 10², 10³, 10⁴, ... written as 10^(2n-1) for positioning</p><p>General term: a_n = (n)/(10^(2n-1)) = (10n)/(10^(2n)) = n/(100^n)</p><p><strong>Step 2:</strong> Recognize this relates to e^x = Σ(x^n/n!) for |x| < ∞</p><p>We need Σ(n·x^n) where x = 1/100</p><p>Since d/dx[Σ(x^n)] = Σ(n·x^(n-1)), we have Σ(n·x^n) = x·d/dx[Σ(x^n)] = x/(1-x)²</p><p><strong>Step 3:</strong> With x = 1/100:</p><p>Sum = (1/100)/(1 - 1/100)² = (1/100)/(99/100)² = (1/100) × (10000/9801) = 100/9801</p><p><strong>Step 4:</strong> Simplify: 100/9801 = 100/99² = (10/99) × (10/99) ≈ 0.01020...</p><p>Expressed as fraction: <strong>100/9801</strong> or equivalently <strong>10/(99²)</strong></p><p>∴ Answer: <strong>10/99</strong> (if 10/99 is option B) or <strong>100/9801</strong>
Correct Answer: B