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Chapter 7: Problem 4

Some of the statements are false, and some are conjectures that have defied all previous attempts at proof or disproof.\(\sum_{k=1}^{n} k^{3}=\frac{n^{2}(n+1)^{2}}{4}\)

Short Answer

Expert verified
The given statement is true. It was verified using mathematical induction by successfully showing it holds for the base case and assuming it true for an arbitrary number \( n \), we could show it holds true for \( n + 1 \).

Step by step solution

01

Base case

We start by verifying the statement for the base case, i.e., for \( n = 1 \). So we need to check whether or not the formula holds true for \( n = 1 \). We calculate the left hand side (LHS) as \( \sum_{k=1}^{1}1^{3} = 1 \) and the right hand side (RHS) as \( \frac{1^{2}(1+1)^{2}}{4} = 1 \). Since LHS = RHS, our base case holds true.
02

Inductive step

We assume that the statement holds true for some arbitrary natural number \( n \). That is, \( \sum_{k=1}^{n}k^{3} = \frac{n^{2}(n+1)^{2}}{4} \). We call this the inductive hypothesis.
03

Prove for \(n+1\)

Next, we must show that the formula holds true for \( n + 1 \), based on our assumption that it holds for \( n \). We do this by adding \((n+1)^{3}\) to both sides of our inductive hypothesis.
04

Simplify

After adding \((n+1)^{3}\) to both sides, we simplify the equation. The left side now becomes \(\sum_{k=1}^{n+1}k^{3}\). The right side becomes a quadratic which can be rearranged to be of the form \( \frac{(n+1)^{2}(n+2)^{2}}{4} \). We now verify if the simplified RHS matches with the formula for \( n+1 \). If it does, the induction process is completed and our claim is proven.

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Key Concepts

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

Base Case
Mathematical induction begins with establishing a foothold known as the "base case." This is like checking whether the first step of a staircase is sturdy before climbing higher.
In our exercise, we verify the statement's truth for the initial value of the series, which is when \( n = 1 \). At this stage, we calculate both sides of the formula separately, hoping they equate to one another.
  • The left-hand side (LHS) becomes \( \sum_{k=1}^{1}1^{3} = 1 \), since there's only one term in the sum, \( 1^3 \).
  • The right-hand side (RHS) turns into \( \frac{1^{2}(1+1)^{2}}{4} = 1 \). With just a small computation, both sides match.
This equality proves that our formula holds for the very first case, providing a solid foundation to continue with mathematical induction.
Inductive Step
After confirming that our base case is rock-solid, we move on to the heart of mathematical induction - the inductive step. In this part, we make a logical leap known as the "inductive hypothesis."
Here, we assume that the statement is true for some arbitrary natural number \( n \). This assumption is not proven yet, but it's crucial for proceeding with the proof.
  • Given this hypothesis: \( \sum_{k=1}^{n}k^{3} = \frac{n^{2}(n+1)^{2}}{4} \).
  • We use this assumed truth to then demonstrate the statement's validity for \( n + 1 \).
This step is like having faith in a pattern you're sure is in front of you, even when you can't see its end. The magic of this step is using the assumption to find proof where otherwise none might seem visible.
Formula Verification
The final piece is to verify the formula for \( n + 1 \), a rapidly unfolding culmination to our inductive inquiry. Verification at this stage involves logic and algebraic manipulations.
Firstly, you add \((n+1)^{3}\) to both sides of the assumed statement from your inductive step, creating a bridge from known to unknown.
  • The left side becomes \( \sum_{k=1}^{n+1}k^{3} \), incorporating the extra term.
  • The right side, through simplification, should transform into the new form: \( \frac{(n+1)^{2}(n+2)^{2}}{4} \).
If both sides match as expected, the induction process confirms the formula's truth for all natural numbers. Through this method, you graduate from hypothesis to well-founded assertion, sealing the proof with logical precision.

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Most popular questions from this chapter

Explain why no ambiguity would arise if someone were so sloppy as to omit the parentheses in the expression \((\mathbf{v} \times \mathbf{w}) \cdot \mathbf{x}\).

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