Proving the Formula for the Sum of Cubes Using Mathematical Induction

One of the fundamental techniques in mathematics, particularly when dealing with sequences and series, is the method of mathematical induction. This technique is used to prove statements or formulas that hold for all natural numbers n. In this article, we will delve into a specific problem: proving the formula for the sum of the cubes of the first n natural numbers.

Mathematically, the formula we aim to prove is:

(sum_{k1}^{n} k^3 frac{n^2(n 1)^2}{4})

To prove this, we will use the principle of mathematical induction, which involves two steps: the base case and the induction step.

Base Case

First, let's verify that the formula holds for the base case n 1.

(sum_{k1}^{1} k^3 1^3 1)

According to the formula,

(frac{1^2(1 1)^2}{4} frac{4}{4} 1)

Hence, the formula holds for n 1.

Induction Hypothesis

Next, we assume that the formula is true for some arbitrary natural number m. This is known as the induction hypothesis:

(sum_{k1}^{m} k^3 frac{m^2(m 1)^2}{4})

Induction Step

We now need to prove that if the formula holds for m, it also holds for m 1.

Let's start by considering the sum of the cubes from 1 to m 1 and show that it satisfies the formula:

(sum_{k1}^{m 1} k^3 sum_{k1}^{m} k^3 (m 1)^3)

Using the induction hypothesis,

(sum_{k1}^{m 1} k^3 frac{m^2(m 1)^2}{4} (m 1)^3)

Let's simplify the right-hand side:

(sum_{k1}^{m 1} k^3 frac{m^2(m 1)^2}{4} frac{4(m 1)^3}{4})

( frac{m^2(m 1)^2 4(m 1)^3}{4})

( frac{(m 1)^2(m^2 4(m 1))}{4})

( frac{(m 1)^2(m^2 4m 4)}{4})

( frac{(m 1)^2(m 2)^2}{4})

This shows that the formula holds for m 1. Therefore, by the principle of mathematical induction, the formula is true for all natural numbers n in (mathbb{Z}^ ).

Thus, we have shown that the formula (sum_{k1}^{n} k^3 frac{n^2(n 1)^2}{4}) is valid for all natural numbers n.