Problem 9 - Precalculus: Mathematics for Calculus

Question

Use mathematical induction to prove that the formula is true for all natural numbers \(n\) $$1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\frac{n^{2}(n+1)^{2}}{4}$$

Step-by-Step Solution

Answer

The formula is proven true for all natural numbers \(n\) by mathematical induction.

Step 1: Base Case

First, check if the formula holds for the smallest natural number, which is 1. Substituting \(n = 1\) in the formula, we have:\[1^3 = \frac{1^2(1+1)^2}{4}\]Simplifying the right-hand side, we get:\[1 = \frac{1 \cdot 4}{4} = 1\]Since both sides equal 1, the base case is valid.

Step 2: Inductive Hypothesis

Assume that the formula holds true for some natural number \(k\). That is, assume:\[1^3 + 2^3 + 3^3 + \, \cdots \, + k^3 = \frac{k^2(k+1)^2}{4}\]We will use this assumption to show that the formula also holds for \(k+1\).

Step 3: Inductive Step

Consider the sum including the next term \((k+1)^3\):\[1^3 + 2^3 + 3^3 + \, \cdots \, + k^3 + (k+1)^3\]Using the inductive hypothesis, this can be rewritten as:\[\frac{k^2(k+1)^2}{4} + (k+1)^3\]

Step 4: Simplifying the Inductive Step

Factor out \((k+1)^2\) from the equation:\[\frac{k^2(k+1)^2}{4} + (k+1)^3 = (k+1)^2\left(\frac{k^2}{4} + (k+1)\right)\]Simplify the expression inside the bracket:\[\frac{k^2}{4} + k + 1 = \frac{k^2 + 4k + 4}{4} = \frac{(k+2)^2}{4}\]Thus, the equation becomes:\[\frac{(k+1)^2(k+2)^2}{4}\]

Step 5: Conclusion of Inductive Step

This shows that if the formula holds for \(n = k\), then it also holds for \(n = k + 1\), since we have:\[1^3 + 2^3 + 3^3 + \, \cdots \, + k^3 + (k+1)^3 = \frac{(k+1)^2(k+2)^2}{4}\]Since the base case is true and the inductive step is verified, by mathematical induction, the formula is true for all natural numbers \(n\).

Key Concepts

Sum of Cubes Formula

The sum of cubes formula is a fascinating equation that allows us to express the sum of the cubes of the first \(n\) natural numbers in a concise form. Specifically, the formula is given by: \[1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}\] This formula provides a shortcut to calculate the sum of the cubes without having to manually add each individual cube.

• When you cube a number, you multiply it by itself twice. For example, \(2^3 = 2 \times 2 \times 2 = 8\).
• Adding cubes one by one for large \(n\) can be tedious, which is why this formula is so helpful.
• This formula is elegant in its simplicity, turning a complex sum into a neat function of \(n\).

Understanding this formula gives insight into the deeper patterns of numbers and can be a wonderful example of mathematical beauty.

Base Case

The base case in mathematical induction is the starting point of your proof. It involves verifying that the given formula works for the smallest possible value of \(n\), which is typically \(n = 1\).

• In our sum of cubes exercise, we plug \(1\) into the equation to ensure both sides equal.
• The calculation needs to show that \(1^3 = \frac{1^2(1+1)^2}{4}\), and simplifying confirms both sides equal \(1\).
• This base step is crucial because it establishes the formula's validity at least for the smallest number, confirming the process can begin.

Thus, confirming the base case step establishes a foundation on which the rest of the proof is built.

Inductive Hypothesis

The inductive hypothesis is the assumption step in the induction process. Here, you assume the formula holds true for some arbitrary natural number \(k\). This unconditional acceptance of the formula at \(k\) serves as a stepping stone for proving it at \(k + 1\).

• For our formula, the inductive hypothesis is assuming \(1^3 + 2^3 + \cdots + k^3 = \frac{k^2(k+1)^2}{4}\).
• While assuming the hypothesis, you are not actually proving it for \(k\) here, but just accepting it as true.
• It's a bridge that connects the base step and the eventual proof for the next element.

The inductive hypothesis is fundamental as it allows the next step, the inductive step, to be executed.

Inductive Step

The inductive step is the crux where the real magic of mathematical induction happens. At this stage, you prove that if the formula is true for \(n = k\) (from the inductive hypothesis), then it must be true for \(n = k + 1\).

• To do this, consider the sum for \(k+1\): \(1^3 + 2^3 + \cdots + k^3 + (k+1)^3\).
• By using the inductive hypothesis, rewrite this as: \(\frac{k^2(k+1)^2}{4} + (k+1)^3\).
• The goal is to show this equals \(\frac{(k+1)^2(k+2)^2}{4}\), thereby proving the formula works for \(k+1\).

This step verifies that the truth extends from \(k\) to \(k+1\).

• If successful, it concludes that since the base case holds and one true case leads to the next, the formula is valid for all natural numbers.
• It's akin to verifying a domino effect: if one falls (the base case), and the proof ensures that each one pushes the next (inductive step), they all fall (the formula holds for all \(n\)).

Mastering the inductive step is key to proving many mathematical theories and problem statements effectively.