1. We are given the function $$f(m,n) = \prod_{k=m}^n \left(1 - \frac{1}{k}\right) = \prod_{k=m}^n \frac{k-1}{k}$$ for integers $m,n$ with $1 < m \le n$. We want to find $$S = \sum_{m=2}^{2025} f(m,2025)$$ and then compute $2S$. 2. Simplify the product $f(m,n)$: $$ f(m,n) = \prod_{k=m}^n \frac{k-1}{k} = \frac{m-1}{m} \times \frac{m}{m+1} \times \frac{m+1}{m+2} \times \cdots \times \frac{n-1}{n} $$ Notice that most terms cancel out. Specifically, all terms from $m$ up to $n-1$ cancel in numerator and denominator, leaving: $$ f(m,n) = \frac{m-1}{n} $$ 3. Substitute $n=2025$ to get: $$ f(m,2025) = \frac{m-1}{2025} $$ 4. Now find $S$: $$ S = \sum_{m=2}^{2025} f(m,2025) = \sum_{m=2}^{2025} \frac{m-1}{2025} = \frac{1}{2025} \sum_{m=2}^{2025} (m-1) $$ 5. Change summation index by letting $j=m-1$, so when $m=2$, $j=1$ and when $m=2025, j=2024$: $$ S = \frac{1}{2025} \sum_{j=1}^{2024} j = \frac{1}{2025} \times \frac{2024 \times 2025}{2} \quad \text{(using the formula } \sum_{j=1}^n j = \frac{n(n+1)}{2} \text{)} $$ 6. Simplify $S$: $$ S = \frac{1}{2025} \times \frac{2024 \times 2025}{2} = \frac{2024}{2} = 1012 $$ 7. Finally, compute $2S$: $$ 2S = 2 \times 1012 = 2024 $$ **Answer:** $2S = 2024$