20)If \(1 + 2 + 3 + ... + k = 325\), then find \(1^3 + 2^3 + 3^3 + ... + k^3\).

20)If \(1 + 2 + 3 + ... + k = 325\), then find \(1^3 + 2^3 + 3^3 + ... + k^3\).

Answer: Sum of first k natural numbers = \(\frac{k(k+1)}{2} = 325\)

Sum of cube of first k natural numbers = \(\left(\frac{k(k+1)}{2}\right)^2\)

\(= (325)^2 = 1,05,625\)

\(1^3 + 2^3 + 3^3 + \dots + k^3 = 1,05,625\)