c) Hence, or otherwise, evaluate : \( \begin{aligned}\left(\frac{1}{4}+\frac{3}{4}\right) & +\left(\frac{1}{6}+\frac{3}{6}+\frac{5}{6}\right)+\left(\frac{1}{8}+\frac{3}{8}+\frac{5}{8}+\frac{7}{8}\right)+\cdots+\left(\frac{1}{50}+\frac{3}{50}+\frac{5}{50}+\cdots\right. \\ & \left.+\frac{49}{50}\right)\end{aligned} \)

To evaluate the sum: \[ \left(\frac{1}{4}+\frac{3}{4}\right) + \left(\frac{1}{6}+\frac{3}{6}+\frac{5}{6}\right) + \left(\frac{1}{8}+\frac{3}{8}+\frac{5}{8}+\frac{7}{8}\right) + \cdots + \left(\frac{1}{50}+\frac{3}{50}+\frac{5}{50}+\cdots+\frac{49}{50}\right) \] we can observe the following pattern: 1. **General Structure:** - For each even denominator \( k = 4, 6, 8, \ldots, 50 \), the numerators are the odd numbers less than \( k \). - Specifically, for \( k = 2n \), the numerators are \( 1, 3, 5, \ldots, (2n - 1) \). 2. **Expressing the Sum:** - The sum can be rewritten as: \[ \sum_{n=2}^{25} \left( \sum_{m=1}^{n} \frac{2m-1}{2n} \right) \] - Here, \( n \) ranges from 2 to 25, corresponding to denominators \( k = 4 \) to \( k = 50 \). 3. **Simplifying the Inner Sum:** - The inner sum \( \sum_{m=1}^{n} (2m - 1) \) simplifies to \( n^2 \) because: \[ \sum_{m=1}^{n} (2m - 1) = 2 \sum_{m=1}^{n} m - \sum_{m=1}^{n} 1 = 2 \left( \frac{n(n+1)}{2} \right) - n = n^2 \] - Therefore, each term becomes: \[ \frac{n^2}{2n} = \frac{n}{2} \] 4. **Calculating the Total Sum:** - The total sum is: \[ \sum_{n=2}^{25} \frac{n}{2} = \frac{1}{2} \sum_{n=2}^{25} n \] - The sum \( \sum_{n=2}^{25} n \) is: \[ \sum_{n=1}^{25} n - 1 = \frac{25 \times 26}{2} - 1 = 325 - 1 = 324 \] - Therefore, the total sum is: \[ \frac{324}{2} = 162 \] **Final Answer:** The evaluated sum is **162**.