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
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