Basic Mathematics & Logarithm
Logarithmic series and summation
Grade 11
Question:
<p>Let \(f(n) = \displaystyle\sum_{r=1}^{n} \log_{10}\!\left(\dfrac{9r+1}{9r-8}\right)\), then:</p>
<p>\(f(11) = 2\)</p>
<p>\(f(11) = -2\)</p>
<p>\(f(111) = 3\)</p>
<p>\(f(1111) = 4\)</p>
Step-by-Step Solution
Key Concept: Recognize this as a telescoping series by expressing each term as a difference of logarithms: log(9r+1) - log(9r-8), where consecutive terms cancel when written out sequentially.
<p><strong>Step 1:</strong> Decompose each term using logarithm properties:</p><p>f(n) = Σ(r=1 to n) [log₁₀(9r+1) - log₁₀(9r-8)]</p><p><strong>Step 2:</strong> Write out the telescoping series:</p><p>= [log₁₀(10) - log₁₀(1)] + [log₁₀(19) - log₁₀(10)] + [log₁₀(28) - log₁₀(19)] + ... + [log₁₀(9n+1) - log₁₀(9n-8)]</p><p><strong>Step 3:</strong> Identify surviving terms after cancellation:</p><p>Most terms cancel, leaving only: log₁₀(9n+1) - log₁₀(1) = log₁₀(9n+1)</p><p><strong>Step 4:</strong> Verify key values:</p><p>• f(1) = log₁₀(10) = 1 ✓</p><p>• f(2) = log₁₀(19)</p><p>• f(n) = log₁₀(9n+1) is increasing and unbounded ✓</p><p>• lim(n→∞) f(n) = ∞ ✓</p><p>∴ Answer: ACD</p>
Correct Answer: ACD