Problem 18 - Precalculus with Limits

Question

In Exercises 11 - 24, use mathematical induction to prove the formula for every positive integer \( n \). \( 1^3 + 2^3 + 3^3 + 4^3 + \cdots + n^3 = \dfrac{n^2\left(n + 1\right)^2}{4} \)

Step-by-Step Solution

Answer

The mathematical induction proves that the given formula \(1^3 + 2^3 + 3^3 + 4^3 + \cdots + n^3 = \dfrac{n^2\left(n + 1\right)^2}{4}\) is correct for all positive integers \(n\). This is verified by proving it for the base case (\(n = 1\)), making an assumption that it is true for \(n = k\), and then proving that it is true for the next integer \(k + 1\).

Step 1: Base Case Test

Check the base case for \(n = 1\): if the formula is correct, \(1^3 = \dfrac{1^2\left(1 + 1\right)^2}{4}\), which simplifies to \(1 = 1\). So, the base case is valid.

Step 2: Inductive Step

Assume the formula holds true for \(n = k\), so \(1^3 + 2^3 + 3^3 + 4^3 + \cdots + k^3 = \dfrac{k^2\left(k + 1\right)^2}{4}\). This is called the induction hypothesis.

Step 3: Inductive Conclusion

Now prove that the formula also holds for \(n = k + 1\). Therefore, \(1^3 + 2^3 + 3^3 + 4^3 + \cdots + k^3 + (k+1)^3 = \dfrac{(k+1)^2\left((k + 1) + 1\right)^2}{4}\). Now to prove this, apply the assumed formula for \(n = k\), then \(1^3 + 2^3 + 3^3 + 4^3 + \cdots + k^3 = \dfrac{k^2\left(k + 1\right)^2}{4}\), so the equation to prove becomes \( \dfrac{k^2\left(k + 1\right)^2}{4} + (k+1)^3 = \dfrac{(k+1)^2\left((k + 1) + 1\right)^2}{4}\). Simplify the equation and make sure both sides balance out.

Key Concepts

Inductive Step

The heart of mathematical induction is the inductive step, which is a method of proving that if we assume a statement to be true for some integer, then that statement must also be true for the next integer. In our exercise, the inductive step involves assuming the formula for the sum of cubes works perfectly for an abstract but fixed number, say k.

In concrete terms, if we assume that 1³ + 2³ + ... + k³ equals \(\dfrac{k^2(k + 1)^2}{4}\), then for the inductive step, we must show this relationship holds when we add the next cube, (k+1)³, to our summation. This step is critical because it bridges our assumption (the truth for k) with the conclusion we wish to reach (the truth for k+1).

To enhance understanding, visualize this step as a domino effect. If we can knock down the first domino (our base case) and ensure each domino can knock down the next (the inductive step), then all subsequent dominos will fall, proving our statement for all integers.

Inductive Conclusion

The inductive conclusion is what we arrive at after performing the inductive step successfully. It binds the truth of our initial assumption to a universal truth for all integers. Following the step-by-step solution provided, our task was to establish that if our formula holds for a number k, it must also hold for k+1.

This confirmation is completed by showing the left and right sides of our proposed equation for k+1 are indeed equal when we add (k+1)³ to our assumed sum. Simplifying the resulting expression and reaching the form \(\dfrac{(k+1)^2((k+1)+1)^2}{4}\) confirms that our formula is valid for k+1, and thus by the principle of mathematical induction, for all positive integers. This brings us to our end goal: with every step validated, we've proven the sum of cubes formula.

Summation of Cubes

The summation of cubes formula is a fascinating result in number theory. It encapsulates the idea that the addition of the cubes of the first n natural numbers is equal to the square of the sum of those n numbers, divided by four. In our exercise, the formula presented is 1³ + 2³ + 3³ + ... + n³ = \(\dfrac{n^2(n + 1)^2}{4}\).

Understanding this formula is not just about memorization but about appreciating the beautiful pattern from the sum of cubes to a squared quantity. Students often find it helpful to look at specific examples, like the sum of the first three cubes 1³ + 2³ + 3³ = 36, which then relates to (3*4/2)² = 36, representing a special relationship in mathematics called a perfect square. Moreover, knowing this formula empowers one to calculate a sum of a long series of cubes without direct addition, saving time and enhancing computational efficiency.