The provided infinite series has a common ratio of \( \frac{1}{7} \) and can be represented as: \[ 7 = 5 + \frac{1}{7} (5 + \alpha) + \frac{1}{7^2} (5 + 2\alpha) + \cdots. \] This is a geometric series. The value of \( \alpha \) is determined by formulating the sum of the series and solving for \( \alpha \).
Final Answer: \( \alpha = \frac{6}{7} \).
\[ 5m \sum_{r=m}^{2m} T_r \text{ is equal to:} \]