/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 For all integers \(n \geq 1,2+4+... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 6

For all integers \(n \geq 1,2+4+6+\cdots+2 n=n^{2}+n\)

Short Answer

Expert verified
In this exercise, we use mathematical induction to prove the statement \(2+4+6+\cdots+2n=n^2+n\) for all integers \(n \geq 1\). First, we verify the base case for \(n=1\) and then prove the induction step by assuming the statement holds for \(n=k\) and showing it holds for \(n=k+1\). By successfully proving both the base case and induction step, we conclude that the statement is true for all integers \(n \geq 1\).

Step by step solution

01

Prove the base case

The base case is the smallest integer value for which the statement should hold, which is \(n=1\). We need to verify that \(2+4+6+\cdots+2(1) = 1^2+1\). Since there is only one term in the sum, we simply need to show that \(2(1) = 1^2 + 1\).
02

Prove the induction step

Now, we assume that the statement holds for some integer \(n=k\), meaning that \(2+4+6+\cdots+2k=k^2+k\). We need to show that if the statement holds for \(n=k\), it must also hold for \(n=k+1\). In other words, we want to prove the following equation: \[2+4+6+\cdots+2k+2(k+1)= (k+1)^2+(k+1)\]
03

Substitute the induction assumption

We know from our assumption that \(2+4+6+\cdots+2k=k^2+k\). We substitute this into the left side of the equation in Step 2: \[(k^2+k)+2(k+1) = (k+1)^2+(k+1)\]
04

Expand and simplify

Next, we expand the right side of the equation and simplify: \[\begin{aligned} (k^2+k)+2(k+1) &= (k+1)^2+(k+1) \\ (k^2+k)+2k+2 &= k^2+2k+1+k+1 \\ k^2+3k+2 &= k^2+3k+2 \end{aligned}\] Since we have shown that the equation holds for \(n=k+1\) when it holds for \(n=k\), the induction step is complete.
05

Conclusion

By proving the base case and the induction step, we have shown via induction that the statement holds for all integers \(n \geq 1\): \[2+4+6+\cdots+2n=n^2+n\] This completes the proof.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Proof by Induction
Proof by induction is a powerful mathematical technique often used to establish the truth of an infinite number of cases after testing just a few. It's akin to dominos falling in a sequence; if you can show one falls, and then prove each will cause the next to topple, you can assert all will fall without individually flicking each one.

Inductive proof consists of two critical steps:
  • The Base Case validates the statement for the initial value鈥攍ike pushing the first domino to confirm it falls as expected.
  • The Inductive Step demonstrates that if the statement holds for an arbitrary case鈥攕ay the nth domino鈥攊t will also hold for the next one鈥攖he (n+1)th domino.
Once these steps are achieved, the initial proposition is established as true for all pertinent cases, symbolically demonstrated by the unending cascade of toppling dominos.
Inductive Step
The inductive step in a proof by induction is where the real magic of logic shines. Here, you're not just proving the scenario for one particular instance, but for every single case that follows, by utilising a form of hypothetical reasoning.

In the exercise, the inductive step begins with the assumption that the statement is true for a certain case, when n is equal to k. This presumed-known case serves as the foundation for the subsequent case, n = k + 1. By algebraically formulating and verifying that the result holds when moving from k to k + 1, the gap between the known and the unknown is bridged.

This verification is not merely a hand-waving argument but requires a careful and meticulous journey through algebraic manipulations, ensuring that each side of the equation remains in perfect sync with the rise of k to k + 1. It is only with this safety net of logical reasoning that we can leap from one known truth to an endlessly unfolding path of truths, thus capturing the essence of the inductive step.
Sequence Summation
Sequence summation is essentially the mathematics of adding up an ordered list of numbers. In our context, we're looking at a specific type of sequence: even numbers multiplied by 2, starting from the smallest, 2, and doubling each natural number up to a given limit. For example, the sum of the first three terms in this sequence is 2(1) + 2(2) + 2(3) = 2 + 4 + 6.

In proofs involving sequence summation, it's often not enough to see the numbers; we must manipulate them algebraically to reveal patterns or prove statements. As such, students must become familiar with algebraic expressions that represent sums, the distributive property of multiplication over addition, and the art of simplification.

Summing a sequence can be visualized like stones tossed into a pool鈥攐ne after another, they create ripples that expand and overlap. Each stone is a term in the sequence, the ripples are the expanding sums, and sequence summation is the measurement of the total area affected by all stones, corresponding to the combined effect of all the terms in the summation.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For each positive integer \(n\), let \(P(n)\) be the property \(5^{n}-1\) is divisible by \(4 .\) a. Write \(P(0)\). Is \(P(0)\) true? b. Write \(P(k)\). c. Write \(P(k+0)\). d. In a proof by mathematical induction that this divisibility property holds for all integers \(n \geq 0\), what must be shown in the inductive step?

An L-tromino, or tromino for short, is similar to a domino but is shaped like an L: th. Call a checkerboard that is formed using \(m\) squares on a side an \(m \times m\) checkerboard. If one square is removed from a \(4 \times 4\) checkerboard, the remaining squares can be completely covered by trominos. For instance, a covering for one such board is the following: Use mathematical induction to prove that for any integer \(n \geq 1\), if one square is removed from a \(2^{n} \times 2^{n}\) checkerboard, the remaining squares can be completely covered by trominos.

Prove each statement in 8-23 by mathematical induction. \(\sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}\), for all integers \(n \geq 2\).

"Theorem:" For all integers \(n \geq 1,3^{n}-2\) is even. "Proof (by mathematical induction): Suppose the theorem is true for an integer \(k\), where \(k \geq 1\). That is, suppose that \(3^{k}-2\) is even. We must show that \(3^{k+1}-2\) is even. But $$ \begin{aligned} 3^{k+1}-2 &=3^{k} \cdot 3-2=3^{k}(1+2)-2 \\ &=\left(3^{k}-2\right)+3^{k} \cdot 2 \end{aligned} $$ Now \(3^{k}-2\) is even by inductive hypothesis and \(3^{k} \cdot 2\) is even by inspection. Hence the sum of the two quantities is even (by Theorem 3.1.1). It follows that \(3^{k+1}-2\) is even, which is what we needed to show,"

Use repeated division by 2 to convert (by hand) the integers in \(63-65\) from base 10 to base 2 . $$ 90 $$
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.