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Chapter 4: Problem 9

For all integers \(n \geq 3\), $$ 4^{3}+4^{4}+4^{5}+\cdots+4^{n}=\frac{4\left(4^{n}-16\right)}{3} $$

Short Answer

Expert verified
Using mathematical induction, we proved the provided formula for all integers \(n \geq 3\). In the base case, we verified that the formula holds true for \(n=3\). For the induction step, we assumed the formula holds for an integer \(n=k\) and proved that it also holds for \(n=k+1\). Thus, for all integers \(n \geq 3\), the given formula for the sum of powers of 4, starting from \(4^3\) up to \(4^n\), is true: \[4^{3}+4^{4}+4^{5}+\cdots+4^{n} = \frac{4\left(4^{n}-16\right)}{3}\]

Step by step solution

01

Base case: Proving the formula for \(n = 3\)

First, we will verify the formula for \(n = 3\). The left-hand side (LHS) of the equation is the sum of powers of 4 from \(4^3\) to \(4^3\), which is just \(4^3 = 64\). The right-hand side (RHS) of the equation for \(n = 3\) is: \[\frac{4\left(4^3-16\right)}{3} = \frac{4\left(64-16\right)}{3} = \frac{4\cdot 48}{3} = 64\] Since LHS = RHS, the formula holds true for \(n = 3\).
02

Induction step: Assuming the formula holds for \(n = k\), prove it for \(n = k + 1\)

Now we will perform the induction step. Let's assume that the formula is true for some integer \(n = k\) (\(k \geq 3\)), \[\sum_{i=3}^{k} 4^i = 4^{3}+4^{4}+4^{5}+\cdots+4^{k} = \frac{4\left(4^{k}-16\right)}{3}\] We want to prove that the formula also holds for \(n = k + 1\): \[\sum_{i=3}^{k+1} 4^i = 4^{3}+4^{4}+4^{5}+\cdots+4^{k} + 4^{k+1} = \frac{4\left(4^{k+1}-16\right)}{3}\] Let's calculate the LHS of this equation by adding \(4^{k+1}\) to our assumed formula: \[\sum_{i=3}^{k+1} 4^i = \frac{4\left(4^{k}-16\right)}{3} + 4^{k+1}\] Now we want to manipulate the expression to match the RHS: \[\frac{4\left(4^{k}-16\right)}{3} + 4^{k+1} = \frac{4\left(4^{k}-16\right)}{3} + \frac{3\cdot 4^{k+1}}{3} = \frac{4\left(4^{k}-16\right) + 3\cdot 4^{k+1}}{3}\] Finally, we will simplify the expression: \[\frac{4\left(4^{k}-16\right) + 3\cdot 4^{k+1}}{3} = \frac{4^{k+1} - 64 + 3\cdot 4^{k+1}}{3} = \frac{4^{k+1} - 64 + 12\cdot 4^{k}}{3} = \frac{4\left(4^{k+1}-16\right)}{3}\] As our last expression matches the RHS of the equation we want to prove, our induction step is complete.
03

Conclusion

Since we have proven the base case for \(n = 3\) and have shown that if the formula holds for an integer \(n = k\), it also holds true for the next integer \(n = k + 1\), the formula is true for all integers \(n \geq 3\). Thus, for all integers \(n \geq 3\), we have: \[4^{3}+4^{4}+4^{5}+\cdots+4^{n} = \frac{4\left(4^{n}-16\right)}{3}\]

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Key Concepts

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

Discrete Mathematics
Discrete mathematics encompasses a wide range of topics that are fundamental to many areas of computer science, economics, and the physical sciences. Central to discrete mathematics is the study of discrete structures, unlike continuous structures found in calculus. Discrete structures include objects such as graphs, sequences, and sets that can be enumerated or arranged in a distinct order.

In the context of series and summation, discrete mathematics deals with sequences of specific terms and their respective sums. The exercise given involves the summation of a series of powers of 4, which is a common type of problem in this field. By approaching such a problem through mathematical induction, a method we cover in more detail in another section, we step through properties of discrete mathematics that highlight the subject's focus on concrete solutions and direct applications to mathematical series.
Series and Summation
In mathematics, a series is the sum of the terms of an infinite or finite sequence. Summation is the operation of adding a sequence of numbers; the result is their sum or total. This concept is widely used across various domains of mathematics, including discrete mathematics and calculus.

Understanding series and summation is crucial when dealing with problems like the one presented. This particular exercise involves a geometric series, where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. Geometric series are particularly amenable to closed-form expressions, as seen in the solution to our exercise. The formula provided gives a straightforward way to calculate the sum, which becomes much more efficient than adding a long list of individual terms. This efficiency is key in computer science and mathematical modeling, where such calculations are often performed.
Proof Techniques
Proof techniques are the tools and methodologies used by mathematicians to establish the truth of mathematical statements. There are various types of proofs, such as direct proof, proof by contradiction, and proof by induction, the latter of which is employed in solving the given exercise.

Mathematical induction is a powerful proof technique commonly used in discrete mathematics to prove statements about integers. It works on the principle of establishing a base case and then proving that if the case is true for a particular integer, it must also be true for the next one. This domino effect ensures that if the initial and induction steps are correct, the statement is proven for an infinite sequence of numbers.

To apply mathematical induction effectively, understanding how to establish the base case and execute the induction step is crucial. The solution's clear demonstration of both steps allows students to grasp the methodical approach required. By mastering induction, students can tackle a wide variety of problems within discrete mathematics and beyond, making it an essential tool in their mathematical arsenal.

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Most popular questions from this chapter

Prove each statement in 8-23 by mathematical induction. \(5^{n}+9<6^{n}\), for all integers \(n \geq 2\).

Suppose \(a[1], a[2], a[3], \ldots, a[m]\) is a one-dimensional array and consider the following algorithm segment: $$ \begin{aligned} &\operatorname{sum}:=0 \\ &\text { for } k:=1 \text { to } m \\ &\text { sum }:=\text { sum }+a[k] \\ &\text { next } k \end{aligned} $$ Fill in the blanks below so that each algorithm segment performs the same job as the one given above. a. \(s u m:=0\) for \(i:=0\) to ______________ \(\operatorname{sum}:=\) ______________ next \(i\) b. \(s u m:=0\) for \(j:=2\) to ______________ sum := ______________ next \(j\)

Write each of \(32-41\) using summation or product notation. $$ \left(2^{2}-1\right) \cdot\left(3^{2}-1\right) \cdot\left(4^{2}-1\right) $$

Use the well-ordering principle to prove that if \(a\) and \(b\) are any integers not both zero, then there exist integers \(u\) and \(v\) such that \(\operatorname{gcd}(a, b)=u a+v b\). (Hint: Let \(S\) be the set of all positive integers of the form \(u a+v b\) for some integers \(u\) and \(v .)\)

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