Basic Mathematics & Logarithm
Logarithmic series and telescoping
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 where consecutive logarithms cancel. Rewrite each term using logarithm properties: log(A/B) = log(A) - log(B), then observe how the numerator of one term matches the denominator of the next.
<p><strong>Step 1:</strong> Use logarithm property to split each term:</p><p>$$f(n) = \sum_{r=1}^{n} \left[\log_{10}(9r+1) - \log_{10}(9r-8)\right]$$</p><p><strong>Step 2:</strong> Identify the telescoping pattern. Notice that 9r+1 = 9(r+1)-8:</p><p>For r=1: $\log_{10}(10) - \log_{10}(1)$</p><p>For r=2: $\log_{10}(19) - \log_{10}(10)$</p><p>For r=3: $\log_{10}(28) - \log_{10}(19)$</p><p>⋮</p><p>For r=n: $\log_{10}(9n+1) - \log_{10}(9n-8)$</p><p><strong>Step 3:</strong> After telescoping cancellation, only first and last terms survive:</p><p>$$f(n) = \log_{10}(9n+1) - \log_{10}(1) = \log_{10}(9n+1)$$</p><p><strong>Step 4:</strong> Evaluate at specific values:</p><p>• $f(1) = \log_{10}(10) = 1$ ✓</p><p>• $f(10) = \log_{10}(91)$ ✓</p><p>• $f(100) = \log_{10}(901)$ ✓</p><p>∴ Answer: ACD</p>
Correct Answer: ACD