If $y = \frac{3}{4} + \frac{3 \cdot 5}{4 \cdot 8} + \frac{3 \cdot 5 \cdot 7}{4 \cdot 8 \cdot 12} + \ldots \infty$,then:

  • A
    $y^2 - 2y + 5 = 0$
  • B
    $y^2 + 2y - 7 = 0$
  • C
    $y^2 - 3y + 4 = 0$
  • D
    $y^2 + 4y - 6 = 0$

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