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

Answer

Correct option: C.
$\frac{{n(n + 1)(2n + 1)}}{6}$
c
(c) It is nothing but $\sum n^2$ $= \frac{{n(n + 1)(2n + 1)}}{{6}}$

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