Question:

Step 1: Write out the telescoping sum: $f(n) = \log_{10}\prod_{r=1}^{n}\dfrac{9r+1}{9r-8}$. Step 2: The product telescopes as $\dfrac{10}{1} \cdot \dfrac{19}{10} \cdot \dfrac{28}{19} \cdots \dfrac{9n+1}{9n-8} = \dfrac{9n+1}{1} = 9n+1$. Step 3: Therefore $f(n) = \log_{10}(9n+1)$. Step 4: Compute $f(11) = \log_{10}(9 \cdot 11 + 1) = \log_{10}(100) = 2$. Option (a) is correct. Step 5: Compute $f(111) = \log_{10}(9 \cdot 111 + 1) = \log_{10}(1000) = 3$. Option (c) is correct. Step 6: Compute $f(1111) = \log_{10}(9 \cdot 1111 + 1) = \log_{10}(10000) = 4$. Option (d) is correct.