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

  • A
    $1$
  • B
    $\frac{6}{7}$
  • C
    $6$
  • D
    $\frac{1}{7}$

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