$\sum\limits_{m = 1}^n {{m^2}}$ is equal to
- A$\frac{{m(m + 1)}}{2}$
- B$\frac{{m(m + 1)(2m + 1)}}{6}$
- C$\frac{{n(n + 1)(2n + 1)}}{6}$
- D$\frac{{n(n + 1)}}{2}$
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