$1^{3} + 2^{3} + \ldots + n^{3}$ equals

$1^{3} + 2^{3} + \ldots + n^{3}$ equals
  1. $\left(\dfrac{n(n+1)}{2}\right)^{2}$
  2. $\dfrac{n(n+1)}{2}$
  3. $n^{2}$
  4. $\dfrac{n(n+1)(2n+1)}{6}$

Solution

Famous identity: sum of first $n$ cubes $=$ square of sum of first $n$ naturals.

Asked in: IMO

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