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Using induction, verify that each equation is true for every positive integer \(n\). $$ \begin{aligned} \frac{1}{2 \cdot 4} &+\frac{1 \cdot 3}{2 \cdot 4 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 8}+\cdots+\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots(2 n+2)} \\ &=\frac{1}{2}-\frac{1 \cdot 3 \cdot 5 \cdots(2 n+1)}{2 \cdot 4 \cdot 6 \cdots(2 n+2)} \end{aligned} $$

Short Answer

Expert verified
Hence, the proof is completed by induction. The given formula holds true for all positive integers \(n\).

Step by step solution

01

Proof Base Case for n=1

To verify the base case, substitute \(n=1\) into both sides of the given equation. We obtain: \[\frac{1}{2*4} = \frac{1}{2} - \frac{1*3}{2*4}\] After simplifying these expressions, it can be seen that both sides are equal, verifying the base case.
02

Inductive Step - Assumption

Now, assume this equation holds for some positive integer \(k\). That is: \[\frac{1}{2 \cdot 4} +\frac{1 \cdot 3}{2 \cdot 4 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 8}+\cdots+\frac{1 \cdot 3 \cdot 5 \cdots(2 k-1)}{2 \cdot 4 \cdot 6 \cdots(2 k+2)} \ =\frac{1}{2}-\frac{1 \cdot 3 \cdot 5 \cdots(2 k+1)}{2 \cdot 4 \cdot 6\cdots(2 k+2)}\]
03

Inductive Step - Proof for n=k+1

To prove by induction, we must show that the given equation holds for \(k+1\) if it is assumed to hold for \(k\). Substitute \(n=k+1\) in the given equation and express the left side of the equation, using the supposed fact that the equation holds for \(n=k\). \[\frac{1}{2 \cdot 4} +\frac{1 \cdot 3}{2 \cdot 4 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 8}+\cdots+\frac{1 \cdot 3 \cdot 5 \cdots(2 k+1)}{2 \cdot 4 \cdot 6 \cdots(2 k+4)} \ = \frac{1}{2}-\frac{1 \cdot 3 \cdot 5 \cdots(2 k+1)}{2 \cdot 4 \cdot 6\cdots(2 k+4)} +\frac{1 \cdot 3 \cdot 5 \cdots(2 k+1)}{2 \cdot 4 \cdot 6 \cdots (2 k+4)} \ = \frac{1}{2}-\frac{1 \cdot 3 \cdot 5 \cdots(2 k+3)}{2 \cdot 4 \cdot 6\cdots (2 k+4)}\] Which is what we wanted to show.

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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 used to prove a statement for every positive integer. The process involves two main steps: the base case and the inductive step.

In the base case, you demonstrate that the statement holds for the initial value (usually when \( n = 1 \) or \( n = 0 \)). This step establishes the foundation of your proof. In our exercise, we substituted \( n = 1 \) into the equation and showed that both sides of the equation are equal, confirming that the base case holds.

The inductive step involves assuming the statement is true for an arbitrary positive integer \( k \). This assumption is called the "inductive hypothesis." The next part of the process is to prove that if the statement holds for \( n = k \), then it must also be true for \( n = k + 1 \). By demonstrating this, you confirm that the domino effect has started, meaning if the statement is true for \( n = k \), it will be true for \( n = k + 1 \), and hence, for all integers following it.

This method is powerful because once both the base case and the induction step are verified, the formula or statement is proven for all positive integers.
Series and Sequences
Series and sequences are fundamental concepts in mathematics, often appearing in many areas such as calculus, algebra, and number theory.

A sequence is an ordered list of numbers that follow a specific pattern. Each number in the sequence is called a term. In our exercise, the sequence on the left side of the equation follows a particular pattern where each term's numerator consists of odd numbers multiplied together, and the denominator consists of corresponding even numbers.

A series, on the other hand, is the sum of the terms of a sequence. In this exercise, we are looking at the sum of a given sequence, starting from the first term \( \frac{1}{2 \cdot 4} \) to the \( n \)-th term \( \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n+2)} \). This creates a cumulative effect as more terms are added.

Understanding the nuances between series and sequences is important for breaking down mathematical problems and seeing the relationships within patterns of numbers. These ideas can have wide-reaching applications, from simple arithmetic to complex mathematical proofs and analyses.
Positive Integers
Positive integers are a fundamental concept that form the backbone of whole numbers greater than zero. These numbers start from 1 and go on indefinitely (1, 2, 3, and so on). They are pivotal in calculations because they most commonly reflect practical counting numbers.

In mathematical induction and other proofs, we often focus on positive integers because they reflect a natural and infinite progression. In our exercise, the equation is asserted to hold for all positive integers \( n \).

Understanding positive integers helps one appreciate the vast number of applications in various mathematical fields. Whether it's counting the quantity of objects in a set or determining sequences and series, positive integers provide a clean, simple context.
  • They start from 1 and are non-negative.
  • They have no fractional or decimal components.
  • They are often used in foundational proofs, especially in methods like mathematical induction.
These numbers make it possible to form concrete conclusions about seemingly infinite problems by leveraging their clear, intuitive nature.

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Most popular questions from this chapter

Suppose that we have two piles of cards each containing \(n\) cards. Two players play a game as follows. Each player, in turn, chooses one pile and then removes any number of cards, but at least one, from the chosen pile. The player who removes the last card wins the game. Show that the second player can always win the game.

Use mathematical induction to prove that $$ \frac{1}{2}+\frac{2}{3}+\cdots+\frac{n}{n+1}<\frac{n^{2}}{n+1} $$ for all \(n \geq 2\). This inequality gives a correct proof of the statement of the preceding exercise.

Deal with plane convex sets. \(A\) plane convex set, subsequently abbreviated to "convex set," is a nonempty set \(X\) in the plane having the property that if \(x\) and \(y\) are any two points in \(X,\) the straight-line segment from \(x\) to \(y\) is also in \(X .\) The following figures illustrate. Prove that if \(X\) and \(Y\) are convex sets and \(X \cap Y\) is nonempty, \(X \cap Y\) is a convex set.

Suppose that \(n>1\) people are positioned in a field (Euclidean plane) so that each has a unique nearest neighbor: Suppose further that each person has a pie that is hurled at the nearest neighbor: A survivor is a person that is not hit by a pie. Give an example to show that if \(n\) is even, there might be no survivor

Let \(S_{1}\) be a subset of 26 cards from an ordinary 52 -card deck, and let \(S_{2}\) be the remaining 26 cards. Suppose that there is a suit in which \(S_{1}\) has at least 9 cards. Prove that there is a suit in which \(S_{2}\) has at least 8 cards. (In the card game bridge, two pairs compete against each other and initially hold 13 cards each. This result says a pair, who between them have a suit of at least 9 cards will have opponents who between them have a suit of at least 8 cards.)

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