The simplest value of $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \cdots + \frac{1}{9 \times 10}$ is

### Concept & Formula This is a standard telescoping series. When the denominator of a fraction is the product of two consecutive integers, it can be expressed as the difference of two adjacent unit fractions. $$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$$ ### Step-by-Step Solution Rewrite each term in the series as a difference using the formula: $$\frac{1}{1 \times 2} = \frac{1}{1} - \frac{1}{2}$$ $$\frac{1}{2 \times 3} = \frac{1}{2} - \frac{1}{3}$$ $$\frac{1}{3 \times 4} = \frac{1}{3} - \frac{1}{4}$$ $$\vdots$$ $$\frac{1}{9 \times 10} = \frac{1}{9} - \frac{1}{10}$$ Sum all these rewritten expressions: $$\text{Sum} = \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \cdots + \left(\frac{1}{9} - \frac{1}{10}\right)$$ Notice that intermediate terms cancel each other out (e.g., $-\frac{1}{2}$ cancels with $+\frac{1}{2}$). The only terms that survive the cancellation are the very first term and the very last term: $$\text{Sum} = 1 - \frac{1}{10}$$ $$= \frac{10 - 1}{10} = \frac{9}{10}$$ ### Exam Strategy & Shortcut For any continuous series of the format $\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \dots + \frac{1}{n \times (n+1)}$, the sum is always equal to $\frac{n}{n+1}$. In this problem, the last term is $\frac{1}{9 \times 10}$, which means $n = 9$. Applying the shortcut gives the answer instantly: $\frac{9}{9+1} = \frac{9}{10}$. ### Common Pitfall Attempting to find a common denominator (LCM) for all the fractions ($2, 6, 12 \dots 90$) will take far too long and is unnecessary. Always look for a telescoping pattern when denominators are products of consecutive numbers. ### Final Answer **Therefore, the correct answer is 9/10.**