Answer :

Given the first k cubes of natural numbers and that their sum is 2025,

By formula we have S =

2025 =

Taking square root on both sides, we get = 45

k(k + 1) = 90

k^{2} + k = 90

Or k^{2} + k-90 = 0

Solving the quadratic using the quadratic formula

Where b = 1, a = 1, c = -90

We get,

k = is invalid because it yields a negative k which doesn’t make sense because number of terms in a series cannot be negative.

= 9

The sum S = 1^{3} + 2^{3} + 3^{3} + … + k^{3} = 2025 corresponds to k = 9.

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