MCQ
If $\sum\limits_{i = 1}^n {i = \frac{{n(n + 1)}}{2}} $, then $\sum\limits_{i = 1}^n {(3i - 2) = } $
- ✓$\frac{{n(3n - 1)}}{2}$
- B$\frac{{n(3n + 1)}}{2}$
- C$n(3n + 2)$
- D$\frac{{n(3n + 1)}}{4}$
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