If $\alpha = \frac{5}{2! 3} + \frac{5 \cdot 7}{3! 3^2} + \frac{5 \cdot 7 \cdot 9}{4! 3^3} + \ldots$,then $\alpha^2 + 4\alpha$ is equal to

  • A
    $21$
  • B
    $23$
  • C
    $25$
  • D
    $27$

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