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