Determine the least value of n n for which the sum of cubes of the first n n natural numbers is less than 2500 2500 using the following formula: 1 3 + 2 3 + ⋯ + n 3 = [ n ( n + 1 ) 2 ] 2 1^3+2^3+\cdots+n^3=\left\lbrack\frac{n\left(n+1\right)}{2}\right\rbrack^2 1 3 + 2 3 + ⋯ + n 3 = [ 2 n ( n + 1 ) ] 2

Determine the least value of nn for which the sum ...

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