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Chapter 1: Problem 17

Prove by induction: (a) \(\frac{1}{1 \cdot 2}+\frac{1}{2+3}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1}\) for \(n \geq 1\) (b) \(\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2}+\cdots+\frac{n}{2^{4}}=2-\frac{n+2}{2^{n}}\) for \(n \geq 1\)

Short Answer

Expert verified
Both statements are proven true for all integers \( n \geq 1 \) by induction.

Step by step solution

01

Base Case Verification

We start with the case \( n = 1 \) and verify both parts: (a) \( \frac{1}{1 \cdot 2} = \frac{1}{2} \) and \( \frac{1}{2} = \frac{1}{2} \) which holds true. (b) \( \frac{1}{2} = 2 - \frac{1 + 2}{2^1} = \frac{1}{2} \), which also holds true.
02

Induction Hypothesis

Assume the statement is true for \( n = k \).(a) Assume \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} = \frac{k}{k+1} \).(b) Assume \( \frac{1}{2} + \frac{2}{2^2} + \cdots + \frac{k}{2^k} = 2 - \frac{k+2}{2^k} \).
03

Induction Step for (a)

Prove it for \( n = k+1 \). We need to show:\[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} = \frac{k+1}{k+2} \]Starting from the hypothesis:\[ \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2) + 1}{(k+1)(k+2)} \]\[ = \frac{k^2 + 2k + 1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2} \]This shows the statement holds for \( n = k+1 \).
04

Induction Step for (b)

Prove it for \( n = k+1 \). We need to show:\[ \frac{1}{2} + \frac{2}{2^2} + \cdots + \frac{k}{2^k} + \frac{k+1}{2^{k+1}} = 2 - \frac{(k+1)+2}{2^{k+1}} \]Starting from the hypothesis:\[ 2 - \frac{k+2}{2^k} + \frac{k+1}{2^{k+1}} \]Combine terms:\[ = 2 - \frac{2(k+2) - (k+1)}{2^{k+1}} \]\[ = 2 - \frac{2k + 4 - k - 1}{2^{k+1}} \]\[ = 2 - \frac{k+3}{2^{k+1}} \]Thus proving the induction step for \( n = k+1 \).

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Key Concepts

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

Discrete Mathematics
Discrete mathematics is a branch of mathematics that deals with distinct and separate values rather than continuous ones. Unlike calculus which involves limits and infinitesimally small differences, discrete mathematics focuses on
  • integers,
  • finite sets,
  • graphs,
  • logic,
  • and structures with distinct and separate elements.

In the context of sequences and series, discrete mathematics provides the tools for analyzing these types of mathematical expressions where values change in specific steps. For example, in our exercise, we look at sequences where each term depends on its position within the series, such as \( rac{1}{n(n+1)}\) in part (a).
Discrete mathematics also includes proof techniques, such as induction, which helps in determining the validity of equations or propositions as they apply to an infinite sequence of events or numbers.
Sequence and Series
A sequence is an ordered list of numbers, and a series is the sum of a sequence of numbers. In mathematics, handling sequences and series involves studying the behaviors and patterns of numbers in a particular order.
Sequences can be arithmetic (where each term increases by a common difference), geometric (where each term is multiplied by a common ratio), or more complex forms like those in our exercise.
In this exercise, part (a) discusses a series formed by the reciprocals of products, represented as \( rac{1}{n(n+1)}\). The sum of this series is shown as \( rac{n}{n+1}\), implying a pattern behind the sequence. Part (b) involves increasing numerators with exponentially increasing denominators, \( rac{n}{2^n}\), which also exhibits a recognizable pattern in its sum.These series are finite because they are considered only up to a certain number of terms, specifically up to the term \( n \). Both sequences illustrate a typical but special case of how terms can be arranged to form a sum or series.
Proof Techniques
Proof techniques are essential methodologies in mathematics used to demonstrate the truth of statements or propositions. Among these techniques, mathematical induction is one of the most powerful and commonly used methods, especially in discrete mathematics.
Induction involves two main steps:
  • Verifying that a statement holds true for an initial value (known as the base case).
  • Assuming the statement for a general case \( n = k \) (called the induction hypothesis) and proving it for the next case \( n = k+1 \).

This stepwise approach shows that since the statement is true for the base case, and assuming it holds for a certain \( n \), it must also hold for any subsequent number, thus making it true for all positive integers.
In our exercise, induction is used to prove complex series summations are accurate across all numbers \( n \) beyond just initially validating it for small numbers like \( n = 1 \). The process provides a robust framework to handle propositions that span infinite or large sets, making it foundational in proof strategies.

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Most popular questions from this chapter

Which of the following statements are correct? Prove each correct statement. Disprove each incorrect statement by finding a counterexample. (a) \(A\) and \(B\) are disjoint if and only if \(B\) and \(A\) are disjoint. (Read the statement carefully - the order in which the sets are listed might matter') (b) \(A \cup B\) and \(C\) are disjoint if and only if both the following are true: (i) \(A\) and \(C\) are disjoint and (ii) \(B\) and \(C\) are disjoint. (c) \(A \cap B\) and \(C\) are disjoint if and only if both the following are true: (i) \(A\) and \(C\) are disjoint and (ii) \(B\) and \(C\) are disjoint. (d) \(A \cup B\) and \(C\) are disjoint if and only if one of the following is true: (i) \(A\) and \(C\) are disjoint or (ii) \(B\) and \(C\) are disjoint. (e) \(A \cap B\) and \(C\) are disjoint if and only if one of the following is true: (i) \(A\) and \(C\) are disjoint or (ii) \(B\) and \(C\) are disjoint. (f) Let \(U\) be a universal set with \(A, B \subseteq U, A\) and \(B\) are disjoint if and only if \(\bar{A}\) and \(\bar{B}\) are disjoint.

Prove by induction: (a) \(0 \cdot 2^{0}+1 \cdot 2^{1}+2 \cdot 2^{2}+3 \cdot 2^{3}+\cdots+n \cdot 2^{n}=(n-1) 2^{n+1}+2\) for \(n \geq 0\) (b) \(1^{2}+3^{2}+5^{2}+\cdots+(2 n+1)^{2}=(n+1)(2 n+1)(2 n+3) / 3\) for \(n \geq 0\) (c) \(1^{2}-2^{2}+3^{2}+\cdots+(-1)^{n-1} n^{2}=(-1)^{n-1} n(n+1) / 2\) for \(n \geq 0\) (d) \(1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+n \cdot(n+1)=n(n+1)(n+2) / 3\) for \(n \geq 0\) (e) \(1 \cdot 2 \cdot 3+2 \cdot 3 \cdot 4+3 \cdot 4 \cdot 5+\cdots+n \cdot(n+1) \cdot(n+2)=n(n+1)(n+2)$$(n+3) / 4\) for \(n \geq 0\)

Challenge: Exactly where is the mistake in the following proof that all personal computers are the same brand? Let \(\mathcal{T}=\\{n \in \mathbb{N}: n \geq 1\) and in every set of \(n\) personal computers, all the personal computers are the same brand \(\\} .\) Prove by induction that for every natural number \(n\) such that \(n \geq 1\) is in \(T\). (Base step) \(1 \in T\), since, trivially, if a set of personal computers contains only one computer, then every (one) computer in the set has the same brand. (Inductive step) Suppose \(n \in T\). We need to show \(n+1 \in T\). So, let \(P\) be any set of \(n+1\) personal computers. Pick any computer \(c \in P ;\) we need to show that every computer in \(P\) is the same brand as \(c\). So, let \(d\) be any computer in \(P\). If \(d=c\), then, trivially, \(d\) and \(c\) are the same brand. Otherwise, \(c \in P-(d\\} .\) The set \(P-(d)\) contains \(n\) computers, so by inductive hypothesis, all the computers in \(P-(d]\) are the same brand. Furthermore, \(d \in P-\mid c\\},\) and. also by inductive hypothesis, all the computers in \(P-\\{c\\}\) are the same brand. Now, let \(e\) be a computer in both \(P-\mid c\\}\) and \(P-[d\\}\). Then, \(d\) is the same brand as \(e,\) and \(c\) is the same brand as \(e\). Therefore, \(d\) is the same brand as \(c\).

Let \(p\) denote the proposition " Jill plays basketball" and \(q\) denote the proposition "Jim plays soccer." Write out-in the clearest way you can-what the following propositions mean: (a) \(\neg p\) (b) \(p \wedge q\) (c) \(p \vee q\) (d) \(\neg p \wedge q\) (c) \(p \rightarrow q\) (f) \(p \leftrightarrow q\) \((g) \neg q \rightarrow p\)

Show for \(n=0,1,2\) that $$(n+1)(2 n+1)(2 n+3) / 3+(2 n+3)^{2}=(n+2)(2 n+3)(2 n+5) / 3$$
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