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Chapter 4: Problem 11

\(1^{3}+2^{3}+\cdots+n^{3}=\left[\frac{n(n+1)}{2}\right]^{2}\), for all integers \(n \geq 1\)

Short Answer

Expert verified
Using the Principle of Mathematical Induction, we tested the base case for n = 1 and proved that the formula holds true. Then, we used the induction hypothesis to show that if the formula is true for some integer k, it is also true for k + 1. Thus, we have proven that the given formula \(1^{3}+2^{3}+\cdots+n^{3}=\left[\frac{n(n+1)}{2}\right]^{2}\) is true for all integers n 鈮� 1.

Step by step solution

01

Test the base case (n = 1)

First, we'll test the formula for the base case, where n = 1. If the formula holds true for this case, it means that the sum of the cube of the first integer (1^3) is equal to the square of the sum of the first integer divided by 2. Plugging n = 1 into the formula, we get: \[1^3 = \left[\frac{1(1 + 1)}{2}\right]^2\] Simplifying the equation, we can evaluate the expressions: \[1^3 = \left[\frac{1(2)}{2}\right]^2\] \[1 = \left[1\right]^2\] \[1 = 1\] Since the equation holds true for the base case (n = 1), we can proceed to the induction step.
02

Prove the induction hypothesis

Now, we need to show that if the formula holds true for some positive integer k, it also holds true for k + 1. In other words, we need to prove that: \[1^{3}+2^{3}+\cdots+k^{3}+\left(k+1\right)^{3}=\left[\frac{\left(k+1\right)\left(\left(k+1\right)+1\right)}{2}\right]^{2}\] Let's assume that the formula holds true for k, meaning: \[1^{3}+2^{3}+\cdots+k^{3}=\left[\frac{k\left(k+1\right)}{2}\right]^{2}\] Now, we need to show that the formula also holds for k + 1. We can start by adding \(\left(k+1\right)^{3}\) to both sides of the equation: \[1^{3}+2^{3}+\cdots+k^{3}+\left(k+1\right)^{3}=\left[\frac{k\left(k+1\right)}{2}\right]^{2}+\left(k+1\right)^{3}\] Next, we will factor out \(\left(k+1\right)\) from the right side of the equation: \[1^{3}+2^{3}+\cdots+k^{3}+\left(k+1\right)^{3}=\left(k+1\right)\left[\frac{k^2}{4}+k^2\right]\] Now, we simplify the equation: \[1^{3}+2^{3}+\cdots+k^{3}+\left(k+1\right)^{3}=\left(k+1\right)\left[\frac{k^2+4k^2}{4}\right]\] \[1^{3}+2^{3}+\cdots+k^{3}+\left(k+1\right)^{3}=\left[\frac{\left(k+1\right)\left(k^2+4k^2\right)}{4}\right]\] Finally, we compare this result with the formula for k + 1: \[1^{3}+2^{3}+\cdots+k^{3}+\left(k+1\right)^{3}=\left[\frac{\left(k+1\right)\left(\left(k+1\right)+1\right)}{2}\right]^{2}\] As we can see, the induction hypothesis holds, and we have proven that the formula is true for all integers n 鈮� 1.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Induction Base Case
In mathematical induction, the base case is the starting point of our proof. It's like establishing the first domino in a chain falls, ensuring the rest can follow. In this exercise, the base case tests whether the equation holds true for the smallest possible value of \( n \).

Here, we begin by checking \( n = 1 \). We substitute this into our formula:
  • Cube of 1: \( 1^3 = 1 \)
  • Square of the sum: \( \left[\frac{1(1+1)}{2}\right]^2 = 1^2 = 1 \)
By confirming that both expressions equal 1, we validate the formula at its starting point. With this, we establish a strong foundation, allowing the induction process to move forward.
Inductive Step
The inductive step in mathematical induction is about proving that if a statement holds for some arbitrary positive integer \( k \), then it must also hold for the next integer, \( k + 1 \). This is akin to showing that if one domino falls, then the next one will as well.

We start with the assumption that our given formula holds for \( n = k \):
  • \( 1^{3} + 2^{3} + \cdots + k^{3} = \left[\frac{k(k+1)}{2}\right]^2 \)
We must prove that it holds for \( n = k + 1 \):
  • Add \((k+1)^3\) to both sides of the equation.
  • Simplify the expression, factoring and combining terms.
Through careful manipulation, we show that this results in the cube sum formula for \( n = k + 1 \), demonstrating that every next step follows from its predecessor.
Cube Sum Formula
The cube sum formula is a fascinating expression that represents the sum of cubes of the first \( n \) natural numbers. It's beautifully concise:
  • \( 1^{3} + 2^{3} + \cdots + n^{3} = \left[\frac{n(n+1)}{2}\right]^2 \)
This formula astonishingly states that the sum of cubes is the square of the sum of the first \( n \) numbers. We derive this with examples like calculating the cubes and comparing sums.

Understanding this formula requires practice with numbers, sequences, and algebraic manipulation. The beauty lies in deciphering the symmetrical relationships and recognizing patterns.

As you work through the formula, remember its practicality not just in theoretical mathematics, but in applications that deal with sequences and series.
Proof by Induction
Proof by induction is a powerful mathematical tool used to prove statements about integers. It's similar to a marathon, where validation happens in steps until the finish line.

The Process Involves:
  • Base Case: Verify the statement for an initial value (usually \( n=1 \)). Ensure it holds as a building block for all other cases.
  • Inductive Step: Assuming it's true for \( n=k \), we then prove it for \( n=k+1 \). This is where the heavy lifting happens, involving substitution, algebraic manipulation, and simplification.
Each successful step forms a link in a chain, pulling the proof from one integer to the next. This structured approach not only verifies specific problems like the cube sum, but reinforces many theoretical claims in calculus, combinatorics, and beyond.

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Most popular questions from this chapter

A sequence \(c_{0}, c_{1}, c_{2}, \ldots\) is defined by letting \(c_{0}=3\) and \(c_{k}=\left(c_{k-1}\right)^{2}\) for all integers \(k \geq 1\). Show that \(c_{n}=3^{2^{n}}\) for all integers \(n \geq 0\).

Suppose that \(f_{1}, f_{2}, f_{3}, \ldots\) is a sequence defined as follows: $$ f_{1}=1, f_{k}=2 \cdot f_{(k / 2)} \quad \text { for all integers } k \geq 2 \text {. } $$ Prove that \(f_{n} \leq n\) for all integers \(n \geq 1\).

Prove each statement in 8-23 by mathematical induction. \(5^{n}+9<6^{n}\), for all integers \(n \geq 2\).

Prove each statement in 8-23 by mathematical induction. \(n\left(n^{2}+5\right)\) is divisible by 6 , for each integer \(n \geq 1\).

Compute the summations and products in 19-28 $$ \sum_{k=-1}^{1}\left(k^{2}+3\right) $$
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