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Chapter 4: Problem 13

Prove or disprove each of the following propositions: (a) For each \(n \in \mathbb{N}, \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\cdots+\frac{1}{n(n+1)}=\frac{n}{n+1}\). (b) For each natural number \(n\) with \(n \geq 3\), $$ \frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\cdots+\frac{1}{n(n+1)}=\frac{n-2}{3 n+3} $$ (c) For each \(n \in \mathbb{N}, 1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+n(n+1)=\frac{n(n+1)(n+2)}{3}\).

Short Answer

Expert verified
The short answers for each proposition are as follows: (a) The series \(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\cdots+\frac{1}{n(n+1)}\) equals \(\frac{n}{n+1}\) for every natural number \(n\). (b) The series \(\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\cdots+\frac{1}{n(n+1)}\) equals \(\frac{n-2}{3 n+3}\) for every natural number \(n\geq3\). (c) The series \(1 \cdot 2+2 \cdot 3+3 \cdot 4+\cdots+n(n+1)\) equals \(\frac{n(n+1)(n+2)}{3}\) for every natural number \(n\).

Step by step solution

01

Base case

For \(n=1\), we have \(\frac{1}{1\cdot2} = \frac{1}{2}\). On the right side, we have \(\frac{n}{n+1} = \frac{1}{1+1} = \frac{1}{2}\). Both sides are equal, so the base case holds.
02

Inductive hypothesis

Assume that the statement holds for some \(n=k\): \(\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \cdots + \frac{1}{k(k+1)} = \frac{k}{k+1}\).
03

Inductive step

Now, we will show that the statement holds for \(n=k+1\): \(\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \cdots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)}\) By the inductive hypothesis, this is equal to: \(\frac{k}{k+1} + \frac{1}{(k+1)(k+2)}\) Now, find the common denominator: \(\frac{k(k+2) + 1}{(k+1)(k+2)}\) This simplifies to: \(\frac{k^2+2k+1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)}\) Finally, this is equal to: \(\frac{k+1}{k+2}\) This completes the inductive step. Therefore, the proposition in part (a) holds for every natural number \(n\). #Part (b)# Similarly, we will proceed with mathematical induction.
04

Base case

For \(n=3\), we have \(\frac{1}{3\cdot4}\). On the right side, we have \(\frac{n-2}{3n+3} = \frac{3-2}{3(3)+3} = \frac{1}{12}\). Both sides are equal, so the base case holds.
05

Inductive hypothesis

Assume that the statement holds for some \(n=k\geq3\): \(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\cdots+\frac{1}{k(k+1)}+\frac{k-2}{3k+3}\).
06

Inductive step

Now, we will show that the statement holds for \(n=k+1\): \(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\cdots+\frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)}\) By the inductive hypothesis, this is equal to: \(\frac{k-2}{3k+3} + \frac{1}{(k+1)(k+2)}\) Now, find the common denominator: \(\frac{(k-2)(k+1)(k+2) + 3(k+1)}{(k+1)(3k+3)(k+2)}\) After simplifying, we get: \(\frac{(k+1)((k+1)(k-1) + 3)}{(k+1)(3k+3)(k+2)}\) This is equal to: \(\frac{k+1}{3(k+1)+3} = \frac{k-1}{3(k+1)+3}\) This completes the inductive step. Therefore, the proposition in part (b) holds for every natural number \(n\geq3\). #Part (c)# Again, we will proceed with mathematical induction.
07

Base case

For \(n=1\), we have \(1\cdot 2\). On the right side, we have \(\frac{n(n+1)(n+2)}{3} = \frac{1(1+1)(1+2)}{3} = 2\). Both sides are equal, so the base case holds.
08

Inductive hypothesis

Assume that the statement holds for some \(n=k\): \(1\cdot 2 + 2\cdot 3 + 3\cdot 4 + \cdots + k(k+1) = \frac{k(k+1)(k+2)}{3}\).
09

Inductive step

Now, we will show that the statement holds for \(n=k+1\): \(1\cdot 2 + 2\cdot 3 + 3\cdot 4 + \cdots + k(k+1) + (k+1)(k+2)\) By the inductive hypothesis, this is equal to: \(\frac{k(k+1)(k+2)}{3} + (k+1)(k+2)\) Now, find the common denominator: \(\frac{k(k+1)(k+2) + 3(k+1)(k+2)}{3}\) After factorization, this simplifies to: \(\frac{(k+1)(k+2)(k+3)}{3}\) This completes the inductive step. Therefore, the proposition in part (c) holds for every natural number \(n\).

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Key Concepts

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

Proof by Induction
In mathematical reasoning, proof by induction is a powerful method used to prove propositions that hold for all natural numbers. The process involves two major components: the base case and the inductive step.

The essence of induction is to show that if a statement holds for some natural number, typically denoted as \(n=k\), then it must also hold for \(n=k+1\). If both these tasks are achieved, then the proposition can be said to hold for all natural numbers.

This method is particularly useful for proving formulas and equations that involve sequences or summations, providing a logical path from a known starting point to any other point in the natural number sequence.
Summation Formulas
Summation formulas are expressions that denote the sum of terms in a sequence or series. These formulas are critical in various areas of mathematics, including algebra, calculus, and number theory.

For example, the formula \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1} \) helps to understand the behavior of a sequence of fractions, showing how their sum can be simplified into a neat algebraic expression.

Summation formulas serve as a condensed way to express lengthy arithmetic operations, making them invaluable in mathematical proofs and calculations.
Natural Numbers
Natural numbers are the set of positive integers starting from 1, denoted as \( \mathbb{N} \). They are the most basic type of numbers used extensively in everyday counting and ordering.

In mathematical induction and other proofs, it's important to work within this set because many propositions naturally apply to all whole numbers. Since they start from 1 and increase without limit, natural numbers provide a straightforward arena for building simpler concepts into more complex ones.

When dealing with problems involving induction, understanding the properties and behavior of natural numbers is crucial.
Base Case
In the context of proof by induction, the base case is the initial step where you verify that the given proposition holds for the first natural number, usually \( n=1 \).

The base case serves as the anchor for the induction process. It provides the necessary initial condition that assures us if the subsequent pieces of the proof line up correctly, then the proposition is true for all natural numbers.

It's crucial that the base case is explicitly demonstrated because it sets the stage for the inductive step, establishing the tone and the correctness of the entire proof.
Inductive Step
The inductive step is the heart of the proof by induction technique. It requires demonstrating that if a statement holds for some arbitrary natural number \( n=k \), then it also holds for \( n=k+1 \).

In executing the inductive step, we assume the truth of the proposition for \( n=k \) 鈥� this is known as the inductive hypothesis. We then use this assumption to prove the statement for \( n=k+1 \).

Successfully completing the inductive step lets us "chain" the correctness from the base case through an infinite progression of natural numbers, effectively covering all possibilities in the sequence.

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Most popular questions from this chapter

In Section \(3.1,\) we defined congruence modulo \(n\) for a natural number \(n,\) and in Section \(3.5,\) we used the Division Algorithm to prove that each integer is congruent, modulo \(n,\) to precisely one of the integers \(0,1,2, \ldots, n-1\) (Corollary 3.32). (a) Find the value of \(r\) so that \(4 \equiv r(\bmod 3)\) and \(r \in\\{0,1,2\\}\). (b) Find the value of \(r\) so that \(4^{2} \equiv r(\bmod 3)\) and \(r \in\\{0,1,2\\}\). (c) Find the value of \(r\) so that \(4^{3} \equiv r(\bmod 3)\) and \(r \in\\{0,1,2\\}\). (d) For two other values of \(n,\) find the value of \(r\) so that \(4^{n} \equiv r(\bmod 3)\) and \(r \in\\{0,1,2\\}\) (e) If \(n \in \mathbb{N},\) make a conjecture concerning the value of \(r\) where \(4^{n} \equiv r(\bmod 3)\) and \(r \in\\{0,1,2\\} .\) This conjecture should be written as a self-contained proposition including an appropriate quantifier. (f) Use mathematical induction to prove your conjecture.

The Future Value of an Ordinary Annuity. For an ordinary annuity, \(R\) dollars is deposited in an account at the end of each compounding period. It is assumed that the interest rate, \(i,\) per compounding period for the account remains constant. Let \(S_{t}\) represent the amount in the account at the end of the \(t\) th compounding period. \(S_{t}\) is frequently called the future value of the ordinary annuity. So \(S_{1}=R\). To determine the amount after two months, we first note that the amount after one month will gain interest and grow to \((1+i) S_{1} .\) In addition, a new deposit of \(R\) dollars will be made at the end of the second month. So $$ S_{2}=R+(1+i) S_{1} $$ (a) For each \(n \in \mathbb{N},\) use a similar argument to determine a recurrence relation for \(S_{n+1}\) in terms of \(R, i,\) and \(S_{n}\). (b) By recognizing this as a recursion formula for a geometric series, use Proposition 4.16 to determine a formula for \(S_{n}\) in terms of \(R, i,\) and \(n\) that does not use a summation. Then show that this formula can be written as $$ S_{n}=R\left(\frac{(1+i)^{n}-1}{i}\right) $$ (c) What is the future value of an ordinary annuity in 20 years if \(\$ 200\) dollars is deposited in an account at the end of each month where the interest rate for the account is \(6 \%\) per year compounded monthly? What is the amount of interest that has accumulated in this account during the 20 years?

The quadratic formula can be used to show that \(\alpha=\frac{1+\sqrt{5}}{2}\) and \(\beta=\frac{1-\sqrt{5}}{2}\) are the two real number solutions of the quadratic equation \(x^{2}-x-1=0\). Notice that this implies that $$ \begin{array}{l} \alpha^{2}=\alpha+1, \text { and } \\ \beta^{2}=\beta+1 \end{array} $$ It may be surprising to find out that these two irrational numbers are closely related to the Fibonacci numbers. (a) Verify that \(f_{1}=\frac{\alpha^{1}-\beta^{1}}{\alpha-\beta}\) and that \(f_{2}=\frac{\alpha^{2}-\beta^{2}}{\alpha-\beta}\). (b) (This part is optional, but it may help with the induction proof in part (c).) Work with the relation \(f_{3}=f_{2}+f_{1}\) and substitute the expressions for \(f_{1}\) and \(f_{2}\) from part (a). Rewrite the expression as a single fraction and then in the numerator use \(\alpha^{2}+\alpha=\alpha(\alpha+1)\) and a similar equation involving \(\beta .\) Now prove that \(f_{3}=\frac{\alpha^{3}-\beta^{3}}{\alpha-\beta}\).(c) Use induction to prove that for each natural number \(n,\) if \(\alpha=\frac{1+\sqrt{5}}{2}\) and \(\beta=\frac{1-\sqrt{5}}{2},\) then \(f_{n}=\frac{\alpha^{n}-\beta^{n}}{\alpha-\beta} .\) Note: This formula for the \(n^{t h}\) Fibonacci number is known as Binet's formula, named after the French mathematician Jacques Binet ( \(1786-1856\) ).

(a) Prove that if \(n \in \mathbb{N},\) then there exists an odd natural number \(m\) and a nonnegative integer \(k\) such that \(n=2^{k} m\). (b) For each \(n \in \mathbb{N}\), prove that there is only one way to write \(n\) in the form described in Part (a). To do this, assume that \(n=2^{k} m\) and \(n=2^{q} p\) where \(m\) and \(p\) are odd natural numbers and \(k\) and \(q\) are nonnegative integers. Then prove that \(k=q\) and \(m=p\).

Instead of using induction, we can sometimes use previously proven results about a summation to obtain results about a different summation. (a) Use the result in Progress Check 4.3 to prove the following proposition: For each natural number \(n, 3+6+9+\cdots+3 n=\frac{3 n(n+1)}{2}\) (b) Subtract \(n\) from each side of the equation in Part (a). On the left side of this equation, explain why this can be done by subtracting 1 from each term in the summation. (c) Algebraically simplify the right side of the equation in Part (b) to obtain a formula for the sum \(2+5+8+\cdots+(3 n-1)\). Compare this to Exercise (3a).
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