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Chapter 5: Problem 7

Use mathematical induction in Exercises \(3-17\) to prove summation formulae. Be sure to identify where you use the inductive hypothesis. Prove that \(3+3 \cdot 5+3 \cdot 5^{2}+\cdots+3 \cdot 5^{n}=3\left(5^{n+1}-1\right) / 4\) whenever \(n\) is a nonnegative integer.

Short Answer

Expert verified
By mathematical induction, \( 3 + 3 \times 5 + 3 \times 5^2 + \cdots + 3 \times 5^n = \frac{3(5^{n+1} - 1)}{4} \) is true for all nonnegative integers \( n \).

Step by step solution

01

- Base Case

Prove the statement for the base case where \( n = 0 \). Substitute \( n = 0 \) in the given equation: \[ 3 = \frac{3(5^{0+1} - 1)}{4} \] Simplifying, we get: \[ 3 = \frac{3(5 - 1)}{4} = \frac{3 \times 4}{4} = 3 \] This is true, so the base case holds.
02

- Inductive Hypothesis

Assume that the statement holds for some nonnegative integer \( k \). That is, \[ 3 + 3 \times 5 + 3 \times 5^2 + \cdots + 3 \times 5^k = \frac{3(5^{k+1} - 1)}{4} \] This is our inductive hypothesis.
03

- Inductive Step

Prove that the statement holds for \( k+1 \). So we need to show: \[ 3 + 3 \times 5 + 3 \times 5^2 + \cdots + 3 \times 5^k + 3 \times 5^{k+1} = \frac{3(5^{(k+1)+1} - 1)}{4} \] Using the inductive hypothesis, add \( 3 \times 5^{k+1} \) to both sides of the equation: \[ \frac{3(5^{k+1} - 1)}{4} + 3 \times 5^{k+1} \] Rewrite \( 3 \times 5^{k+1} \) as \( \frac{12 \times 5^{k+1}}{4} \), to have a common denominator: \[ \frac{3(5^{k+1} - 1)}{4} + \frac{12 \times 5^{k+1}}{4} = \frac{3(5^{k+1} - 1) + 12 \times 5^{k+1}}{4} \] Simplify the numerator: \[ \frac{3 \times 5^{k+1} - 3 + 12 \times 5^{k+1}}{4} = \frac{15 \times 5^{k+1} - 3}{4} \] Factor out 3: \[ \frac{3(5 \times 5^{k+1} - 1)}{4} = \frac{3(5^{(k+1)+1} - 1)}{4} \] Therefore, the statement holds for \( k+1 \).
04

- Conclusion

By the principle of mathematical induction, the given formula holds for all nonnegative integers \( n \).

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Key Concepts

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

Summation Formulae
To understand how to prove summation formulae, it's important to know the structure of the sequence. In this problem, we are dealing with a geometric series, which means each term can be written as a constant multiplied by a base number raised to a new power each time. In this case, the sequence is represented as: \[ 3 + 3 \times 5 + 3 \times 5^2 + \text{...} + 3 \times 5^n \]. The challenge is to show that the sum of these terms equals the closed form given by the formula: \[ \frac{3(5^{n+1} - 1)}{4} \]. Summation formulae like these simplify the expression of a sequence by providing a single formula, rather than having to add up each term individually.
Base Case
In mathematical induction, the base case is our starting point. This step ensures that the formula we're proving holds for the very first value, typically when \( n = 0 \). For the given sequence, substitute \( n = 0 \) into our formula and the series: \[ 3 = \frac{3(5^{0+1} - 1)}{4} \]. This simplifies to \[ 3 = \frac{3(5 - 1)}{4} = \frac{3 \times 4}{4} = 3 \]. Since both sides of the equation match, we've confirmed that the base case is true. The base case is critical because it serves as the foundation that supports our entire inductive proof.
Inductive Hypothesis
The inductive hypothesis is an assumption that the formula holds true for some arbitrary nonnegative integer \( k \). Formally, we assume: \[ 3 + 3 \times 5 + 3 \times 5^2 + \text{...} + 3 \times 5^k = \frac{3(5^{k+1} - 1)}{4} \]. This assumption allows us to build on it for the next step, which involves proving that if it holds for \( k \), then it must also hold for \( k + 1 \). This step does not prove the general case directly but rather sets the stage to link successive values of \( n \).
Inductive Step
The inductive step is the heart of mathematical induction. Here, we need to prove that if the hypothesis holds for \( k \), then it must also hold for \( k + 1 \). We start with the series sum up to \( 3 \times 5^k \) and add the next term: \[ 3 + 3 \times 5 + 3 \times 5^2 + \text{...} + 3 \times 5^k + 3 \times 5^{k+1} = \frac{3(5^{k+1} - 1)}{4} + 3 \times 5^{k+1} \]. We rewrite \( 3 \times 5^{k+1} \) to have the same denominator, giving: \[ \frac{3(5^{k+1} - 1) + 12 \times 5^{k+1}}{4} \]. Simplifying the numerator, we get: \[ \frac{15 \times 5^{k+1} - 3}{4} = \frac{3(5^{(k+1)+1} - 1)}{4} \]. This confirms that the formula holds for \( k+1 \). By combining the base case, inductive hypothesis, and inductive step, we conclude that the given formula holds for all nonnegative integers \( n \).

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Most popular questions from this chapter

Let \(S\) be the subset of the set of ordered pairs of integers defined recursively by Basis step: \((0,0) \in S .\) Recursive step: If \((a, b) \in S,\) then \((a+2, b+3) \in S\) and \((a+3, b+2) \in S\) a) List the elements of \(S\) produced by the first five appli- cations of the recursive definition. b) Use strong induction on the number of applications of the recursive step of the definition to show that \(5 | a+b\) when \((a, b) \in S .\) c) Use structural induction to show that \(5 | a+b\) when \((a, b) \in S .\)

Show that it is possible to arrange the numbers \(1,2, \ldots, n\) in a row so that the average of any two of these numbers never appears between them. [Hint: Show that it suffices to prove this fact when \(n\) is a power of \(2 .\) Then use mathematical induction to prove the result when \(n\) is a power of 2.]

Let \(P(n)\) be the statement that a postage of \(n\) cents can be formed using just 4 -cent stamps and 7 -cent stamps. The parts of this exercise outline a strong induction proof that \(P(n)\) is true for all integers \(n \geq 18\) . a) Show that the statements \(P(18), P(19), P(20),\) and \(P(21)\) are true, completing the basis step of a proof by strong induction that \(P(n)\) is true for all integers \(n \geq 18\) . b) What is the inductive hypothesis of a proof by strong induction that \(P(n)\) is true for all integers \(n \geq 18 ?\) c) What do you need to prove in the inductive step of a proof that \(P(n)\) is true for all integers \(n \geq 18 ?\) d) Complete the inductive step for \(k \geq 21\) . e) Explain why these steps show that \(P(n)\) is true for all integers \(n \geq 18\) .

Find the reversal of the following bit strings. a) 0101 b) 11011 c) 100010010111

Deal with values of iterated functions. Suppose that \(f(n)\) is a function from the set of real numbers, or positive real numbers, or some other set of real numbers, to the set of real numbers such that \(f(n)\) is monotonically increasing [that is, \(f(n)< f(m)\) when \(n< m )\) and \(f(n)< n\) for all \(n\) in the domain of \(f . ]\) The function \(f^{(k)}(n)\) is defined recursively by $$f^{(k)}(n)=\left\\{\begin{array}{ll}{n} & {\text { if } k=0} \\\ {f\left(f^{(k-1)}(n)\right)} & {\text { if } k>0}\end{array}\right.$$ Furthermore, let \(c\) be a positive real number. The iterated function \(f_{c}^{*}\) is the number of iterations of \(f\) required to reduce its argument to \(c\) or less, so \(f_{c}^{*}(n)\) is the smallest nonnegative integer \(k\) such that \(f^{k}(n) \leq c\). Let \(f(n)=\sqrt{n} .\) Find a formula for \(f^{(k)}(n) .\) What is the value of \(f_{2}^{*}(n)\) when \(n\) is a positive integer?
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