If $\sum_{i = 1}^n i = \frac{n(n + 1)}{2}$,then $\sum_{i = 1}^n (3i - 2) = $

  • A
    $\frac{n(3n - 1)}{2}$
  • B
    $\frac{n(3n + 1)}{2}$
  • C
    $n(3n + 2)$
  • D
    $\frac{n(3n + 1)}{4}$

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