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Chapter 4: Problem 7

For \(n \in \mathbf{Z}^{+}\), use mathematical induction to show that $$ \sum_{i=1}^{n} i^{4}=\frac{n(n+1)(2 n+1)\left(3 n^{2}+3 n-1\right)}{30} $$

Short Answer

Expert verified
The formula \(\sum_{i=1}^{n} i^{4}=\frac{n(n+1)(2 n+1)(3 n^{2}+3 n-1)}{30}\) is proven valid for any positive integer \(n\) by mathematical induction.

Step by step solution

01

Base Case Verification

The base case is when \(n=1\). Plugging \(n=1\) into the formula gives: \(1^{4}=\frac{1(1+1)(2(1)+1)(3(1)^{2}+3(1)-1)}{30}\) which simplifies to: \(1=1\). So the formula holds for the base case.
02

Inductive Hypothesis

The inductive hypothesis is that the formula holds for \(n=k\). That is, assume \(\sum_{i=1}^{k} i^{4}=\frac{k(k+1)(2k+1)(3k^{2}+3 k-1)}{30}\) is true.
03

Inductive Step

To complete the proof, it needs to show that if the formula holds for \(n=k\), then it also holds for \(n=k+1\). That is, need to show that \(\sum_{i=1}^{k+1} i^{4}=\frac{(k+1)(k+2)(2(k+1)+1)(3(k+1)^{2}+3 (k+1)-1)}{30}\). Start by adding \((k+1)^{4}\) to both sides of the equation from the inductive hypothesis. Simplify the right side and show that it's equivalent to the formula for \(n=k+1\).
04

Conclusion of the Proof

Since the formula holds for the base case and if it holds for \(n=k\), it also holds for \(n=k+1\), it is proven by mathematical induction that the formula is valid for all positive integer \(n\).

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Key Concepts

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

Base Case
In mathematical induction, the base case is the starting point of the proof. It serves as the foundation to verify that the formula works for the smallest possible value of the variable involved. Typically, this is the simplest scenario one can consider. Let's see how it applies to the sum of powers of integers:
  • The base case is generally established by showing that the formula is true when the variable equals the smallest positive integer, usually 1.
  • For the given problem, this means plugging in 1 for \(n\) in the equation \(\sum_{i=1}^{n} i^{4}=\frac{n(n+1)(2n+1)(3n^{2}+3n-1)}{30}\).
  • The equation simplifies to \(1^{4}=\frac{1(1+1)(2(1)+1)(3(1)^2+3(1)-1)}{30}\), leading us to \(1=1\).
By confirming the validity of the equation in this simple setting, the base case is verified, and the first step in proving our formula is complete.
Inductive Hypothesis
The inductive hypothesis is the assumption that forms the bridge to prove the respective case for the next integer. It plays a key role in the method of mathematical induction. Here's how it fits in:
  • We assume that the statement is true for some arbitrary positive integer \(k\).
  • In our exercise, it means assuming that the formula holds: \(\sum_{i=1}^{k} i^{4}=\frac{k(k+1)(2k+1)(3k^{2}+3k-1)}{30}\).
  • This assumption is what we call the inductive hypothesis and it allows us to proceed to the next step of the induction process.
Assuming this step is crucial because it helps establish that if it holds for \(n=k\), then the result can be generalized further.
Inductive Step
The inductive step is where the magic of mathematical induction comes into action. Now, it's time to prove that the formula also holds for the next integer, \(k+1\), using our inductive hypothesis:
  • The goal is to show that if the formula works for \(n=k\), then it needs to be valid for \(n=k+1\) as well.
  • To do this, we need to add \((k+1)^4\) to both sides of the equation: \(\sum_{i=1}^{k+1} i^{4} = \sum_{i=1}^{k} i^{4} + (k+1)^{4}\).
  • Substituting the inductive hypothesis for \(\sum_{i=1}^{k} i^{4}\) simplifies this to match the expression for \(n=k+1\).
Successfully performing this step demonstrates that our assumption holds for \(n=k\) leads naturally to the truth of \(n=k+1\), further solidifying the proof.
Sum of Powers of Integers
The sum of powers of integers is a fascinating series that this problem explores. Specifically, we examine the sum of the fourth powers of integers. Here's what makes it interesting:
  • Finding a closed formula for sums like \(\sum_{i=1}^{n} i^{4}\) can significantly simplify analysis and calculations.
  • These sums have applications in various areas of mathematics, including number theory and algebra.
  • The formula \(\sum_{i=1}^{n} i^{4} = \frac{n(n+1)(2n+1)(3n^{2}+3n-1)}{30}\) offers a compact way to calculate these sums without direct computation.
Understanding such formulas is crucial as they provide insight and tools for deeper mathematical investigations.

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Most popular questions from this chapter

Two hundred coins numbered 1 to 200 are put in a row across the top of a cafeteria table. Two hundred students are assigned numbers (from 1 to 200 ) and are asked to turn over certain coins. The student assigned number 1 is supposed to turn over all the coins. The student assigned number 2 is supposed to turn over every other coin, starting with the second coin. In general, the student assigned the number \(n\), for each \(1 \leq n \leq 200\), is supposed to turn over every \(n\)th coin, starting with the \(n\)th coin. a) How many times will the 200 th coin be turned over? b) Will any other coin(s) be turned over as many times as the 200 th coin? c) Will any coin be turned over more times than the 200 th coin?

For \(n \in \mathbf{Z}^{+}\)define the sum \(s_{n}\) by the formula $$ s_{n}=\frac{1}{2 !}+\frac{2}{3 !}+\frac{3}{4 !}+\cdots+\frac{(n-1)}{n !}+\frac{n}{(n+1) !} . $$ a) Verify that \(s_{1}=\frac{1}{2}, s_{2}=\frac{5}{6}\), and \(s_{3}=\frac{23}{24}\). b) Compute \(s_{4}, s_{5}\), and \(s_{6}\). c) On the basis of your results in parts (a) and (b), conjecture a formula for the sum of the terms in \(s_{n}\). d) Verify your conjecture in part (c) for all \(n \in \mathbf{Z}^{+}\) by the principle of finite induction.

For \(n \in \mathbf{Z}^{+}\), prove each of the following by mathematical induction: a) \(5 \mid\left(n^{5}-n\right)\) b) \(6 \mid\left(n^{3}+5 n\right)\)

If \(p\) is any prime, prove that \(\sqrt[3]{p}\) is irrational.

Define the sequence of rational numbers \(a_{0}, a_{1}, a_{2}, a_{3}, \ldots\), recursively by 1) \(a_{0}=0, a_{1}=1\); and, 2) For \(n \geq 2, a_{n}=\frac{a_{n-1}+(n-1) a_{n-2}}{n}\). For all \(n \in \mathbf{N}\) prove that \(0 \leq a_{n} \leq 1 .\)
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