If $7 = 5 + \frac{1}{7}(5 + \alpha) + \frac{1}{7^{2}}(5 + 2\alpha) + \frac{1}{7^{3}}(5 + 3\alpha) + \cdots \infty$, then the value of $\alpha$ is:\\ (1) $\frac{6}{7}$\\ (2) 6\\ (3) $\frac{1}{7}$\\ (4) 1