$1^{3} + 2^{3} + \ldots + n^{3}$ equals
$1^{3} + 2^{3} + \ldots + n^{3}$ equals
- $\left(\dfrac{n(n+1)}{2}\right)^{2}$
- $\dfrac{n(n+1)}{2}$
- $n^{2}$
- $\dfrac{n(n+1)(2n+1)}{6}$
Solution
Famous identity: sum of first $n$ cubes $=$ square of sum of first $n$ naturals.
Asked in: IMO
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