The sum of all proper fractions whose denominators are less than or equal to 100, is

### Concept & Formula This problem requires finding the sum of grouped sequences. A proper fraction has a numerator strictly less than its denominator ($n < d$). By grouping fractions with the same denominator, we can leverage the formula for the sum of the first $n$ natural numbers. ### Step-by-Step Solution * **Given:** Proper fractions $\frac{n}{d}$ where $0 < n < d$ and $d \le 100$. * **Deduction:** For a fixed denominator $d$, the proper fractions are $\frac{1}{d}, \frac{2}{d}, \dots, \frac{d-1}{d}$. The sum of these fractions for a specific $d$ is: $$S_d = \frac{1 + 2 + \dots + (d-1)}{d}$$ Using the sum of arithmetic progression formula $\frac{n(n+1)}{2}$: $$S_d = \frac{\frac{(d-1)d}{2}}{d} = \frac{d-1}{2}$$ * **Calculation:** We need the total sum across all denominators from $d=2$ to $d=100$: $$\text{Total Sum} = \sum_{d=2}^{100} \frac{d-1}{2}$$ Factor out the $\frac{1}{2}$ and expand the summation terms: $$\text{Total Sum} = \frac{1}{2} (1 + 2 + 3 + \dots + 99)$$ Apply the sum formula again for $n=99$: $$\text{Total Sum} = \frac{1}{2} \times \frac{99 \times 100}{2} = \frac{9900}{4}$$ $$\text{Total Sum} = 2475$$ ### Exam Strategy & Shortcut Recognize the underlying pattern immediately: the sum of all proper fractions for any denominator $d$ is exactly $\frac{d-1}{2}$. The total sum for denominators up to $N$ is just half the sum of integers up to $N-1$. $\frac{1}{2} \times \frac{99 \times 100}{2} = 2475$. This is a purely formula-driven shortcut that takes less than 20 seconds. ### Common Pitfall Including $n=d$ (an improper fraction equal to 1) in the sequence or starting the summation at $d=1$ (which has no valid proper fractions and breaks the math logic). Always strictly adhere to the mathematical definition of a proper fraction. ### Final Answer **Therefore, the correct answer is 2475.**