Prove that for every n \(\in\) \(\mathbb{N}\), \(1^{3}\) + \(2^{3}\) + . . . + \(n^{3}\) = \(\frac{n^{2}(n+1)^{2}}{4}\)

 For every n \(\in\) \(\mathbb{N}\), \(1^{3}\) + \(2^{3}\) + . . . + \(n^{3}\) = \(\frac{n^{2}(n+1)^{2}}{4}\)

Proof By Induction

Base Case: n = 1

\(1^{3}\) = 1

\(\frac{1^{2}(1+1)^{2}}{4}\) = 1

Base case holds!

Induction Hypothesis:

Assume that \(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) = \(\frac{k^{2}(k+1)^{2}}{4}\) is true for some k \(\in\) \(\mathbb{N}\). We will use this statement in our inductive step. 

Inductive Step: 

Suppose P(k) is true, P(k): \(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) = \(\frac{k^{2}(k+1)^{2}}{4}\). Prove then that P(k+1) is true. P(k+1): \(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) = \(\frac{(k+1)^{2}(k+2)^{2}}{4}\) is true for some k \(\in\) \(\mathbb{N}\) 

\(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) = \(\frac{k^{2}(k+1)^{2}}{4}\)

\(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) + \((k+1)^{3}\) = \(\frac{k^{2}(k+1)^{2}}{4}\) + \((k+1)^{3}\)

\(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) + \((k+1)^{3}\) = \(\frac{k^{2}(k+1)^{2} + 4(k+1)^{3}}{4}\)

\(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) + \((k+1)^{3}\) = \(\frac{k^{2}(k+1)^{2} + 4(k+1)^{1}(k+1)^{2}}{4}\)

\(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) + \((k+1)^{3}\) = \(\frac{(k+1)^{2}(k^{2} + 4k + 4)}{4}\)

\(1^{3}\) + \(2^{3}\) + . . . + \(k^{3}\) + \((k+1)^{3}\) = \(\frac{(k+1)^{2}(k+2)^{2}}{4}\)

This result equals the right hand side of P(k+1). We have proven by induction that for every n \(\in\) \(\mathbb{N}\), \(1^{3}\) + \(2^{3}\) + . . . + \(n^{3}\) = \(\frac{n^{2}(n+1)^{2}}{4}\)