The value of $\log_{10} 1\frac{1}{2} + \log_{10} 1\frac{1}{3} + \cdots$ up to $198$ terms is equal to

### Concept & Strategy This question relies on recognizing a telescoping series within a logarithmic addition. When adding logarithms of fractions, the product rule allows you to multiply all the fractions together, leading to massive cancellation. The Product Rule: $$ \log(a) + \log(b) + \log(c) = \log(a \cdot b \cdot c) $$ ### Step-by-Step Solution First, convert all the mixed fractions into improper fractions to see the pattern clearly. Term 1: $1\frac{1}{2} = \frac{3}{2}$ Term 2: $1\frac{1}{3} = \frac{4}{3}$ Term 3: $1\frac{1}{4} = \frac{5}{4}$ ...and so on. Let's determine the 198th term. Notice the pattern in the denominator of the $n$-th term: it is $n + 1$. For the 198th term, the denominator is $198 + 1 = 199$. The numerator is always $1$ greater than the denominator, so the 198th numerator is $200$. Thus, the 198th term is $\frac{200}{199}$. The expression is: $\log_{10}\left(\frac{3}{2}\right) + \log_{10}\left(\frac{4}{3}\right) + \cdots + \log_{10}\left(\frac{200}{199}\right)$ Use the product rule to combine all terms into a single logarithm: $\log_{10}\left( \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4} \times \cdots \times \frac{200}{199} \right)$ Observe the telescoping cancellation. The numerator of each fraction cancels perfectly with the denominator of the next fraction. The $3$s cancel, the $4$s cancel, all the way up to $199$. You are left with only the final numerator and the first denominator: $\log_{10}\left(\frac{200}{2}\right)$ Simplify the argument: $\log_{10}(100)$ Evaluate the logarithm: $\log_{10}(10^2) = 2$ ### Exam Strategy & Shortcut Whenever you see a continuous series of "log(1 + 1/n)" additions, instantly know it's a telescoping product. Write down only the first term's denominator and the last term's numerator. Skip writing out the middle terms completely. $200 / 2 = 100$. The log of $100$ is $2$. You can solve this mentally in 5 seconds. ### Common Pitfall The main trap is incorrectly identifying the final term of the series. Students often see "up to 198 terms" and mistakenly use $198$ in the final fraction (e.g., $199/198$), failing to account for the sequence starting with a denominator of $2$. Always map $n = 1$ to the first term to find the correct algebraic shift. ### Final Answer **Therefore, the correct answer is 2.**