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Chapter 4: Problem 11

For the sequence \(a_{1}, a_{2}, \ldots, a_{n}, \ldots,\) assume that \(a_{1}=1, a_{2}=5,\) and that for each \(n \in \mathbb{N}, a_{n+1}=a_{n}+2 a_{n-1}\). Prove that for each natural number \(n\), \(a_{n}=2^{n}+(-1)^{n}\)

Short Answer

Expert verified
The given sequence follows the general formula \(a_n = 2^n + (-1)^n\). This formula holds for all natural numbers n, and we have proved this through the principle of mathematical induction. The general formula was tested for the initial terms, and the inductive step showed the formula still held true for successive terms. Therefore, the general formula \(a_n = 2^n + (-1)^n\) is valid for all natural numbers n.

Step by step solution

01

Identify the general formula for sequence and test for initial terms

The given general formula for the sequence is \(a_n = 2^n + (-1)^n\). Let's verify the formula for initial terms \(a_1\) and \(a_2\). \(a_1 = 2^1 + (-1)^1 = 2 - 1 = 1\) \(a_2 = 2^2 + (-1)^2 = 4 + 1 = 5\) Now, since the formula gives the correct values for the initial terms provided \(a_1\) and \(a_2\), we will proceed to the induction step.
02

Inductive Step

Assume the general formula for a_n holds for n = k and n = k + 1. Then \(a_k = 2^k + (-1)^k\) and \(a_{k+1} = 2^{k+1} + (-1)^{k+1}\) for some integer k. Now, we need to prove that the general formula also holds for n = k+2. We have the recurrence relation \(a_{n+1} = a_n + 2a_{n-1}\). So, to find \(a_{k+2}\), we will use this relation.
03

Step 3:Calculating \(a_{k+2}\)

Using the recurrence relation and our assumptions: \(a_{k+2} = a_{k+1} + 2a_k = (2^{k+1} + (-1)^{k+1}) + 2(2^k + (-1)^k)\). Let's simplify this expression: \(a_{k+2} = 2^{k+1} + (-1)^{k+1} + 2^{k+1} - 2(-1)^k = 2\cdot 2^{k+1} - (-1)^k (2 + 1)\) \(a_{k+2} = 2^{k+2} + (-1)^k\cdot 3\). Now we will check when \(k\) is even and odd.
04

Check for even and odd values of k

When k is even: \(a_{k+2} = 2^{k+2} + 3, \Rightarrow a_{k+2} = 2^{k+2} + (-1)^{k+2}\) When k is odd: \(a_{k+2} = 2^{k+2} - 3, \Rightarrow a_{k+2} = 2^{k+2} + (-1)^{k+2}\) In both cases, the general formula holds true for \(a_{k+2}\). Since we have shown that the general formula holds true for initial terms and if it holds true for n=k and n=k+1, it also holds true for n=k+2, we can conclude that the general formula holds true for all natural numbers n by the principle of mathematical induction. Thus, \(a_n=2^n+(-1)^n\) 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.

Recurrence Relations
Understanding recurrence relations is crucial for solving many types of mathematical problems, particularly those involving sequences. A recurrence relation is a formula that expresses each term of a sequence as a function of its preceding terms. In the context of the exercise, the given relation is \(a_{n+1} = a_n + 2a_{n-1}\), meaning each term is generated by adding the previous term to twice the term before that.

The beauty of recurrence relations lies in their ability to succinctly define complex sequences and the patterns inherent within them. This allows mathematicians to calculate very large terms without enumerating all preceding terms, which can be computationally intensive or practically impossible for large indices. They are particularly prominent in algorithms, computer science, and advanced mathematics like combinatorics.
Sequence and Series
A sequence is essentially a list of numbers generated by a specific mathematical rule, while a series is the sum of a sequence of terms. The distinction is key to not only understanding sequences but also to mastering the domains of calculus and discrete mathematics, where series are used in sums, limits, and other operations.

In our exercise example, each term of the sequence is defined using the previous terms, showcasing that sequences can be simple or complex, finite or infinite. And while our focus is on sequences, these often lay the groundwork for the study of series, where the sum of all terms is evaluated, leading to concepts such as convergence and divergence in more advanced mathematics. Recognizing patterns and formulating a sequence are fundamental skills in mathematics that have applications in various fields, like finance for calculating interest or in physics for understanding harmonic patterns.
Proofs in Mathematics
The cornerstone of mathematical certainty is proof, and proofs in mathematics serve as a method to establish the truthfulness of a statement beyond any doubt. Proofs vary from geometric demonstrations to algebraic manipulations, as seen in the use of mathematical induction in the given exercise.

Mathematical induction is a technique that allows us to prove a statement or formula's validity for all natural numbers. It comprises two primary steps: the base step, where the statement is proved for the initial number (usually 1), and the inductive step, where we assume the statement is true for some natural number 'k' and then prove it for 'k+1'. Success in this step requires a thorough understanding of how to manipulate the given relations and often algebraic expansion or factorization. Through such rigorous methodology, we can infer the truth for all numbers within a specific domain, which is exactly what was conducted in the exercise to prove the sequence formula.

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Most popular questions from this chapter

For the sequence \(a_{1}, a_{2}, \ldots, a_{n}, \ldots,\) assume that \(a_{1}=1, a_{2}=1, a_{3}=1,\) and for that each natural number \(n\), $$ a_{n+3}=a_{n+2}+a_{n+1}+a_{n} $$ (a) Compute \(a_{4}, a_{5}, a_{6},\) and \(a_{7}\). (b) Prove that for each natural number \(n\) with \(n>1, a_{n} \leq 2^{n-2}\).

See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) For each natural number \(n\) with \(n \geq 2,2^{n}>1+n\). \mathrm{\\{} \text { Proof. We let } k \text { be a natural number and assume that } 2 ^ { k } > 1 + k \text { . } Multiplying both sides of this inequality by \(2,\) we see that \(2^{k+1}>\) \(2+2 k\). However, \(2+2 k>2+k\) and, hence, $$ 2^{k+1}>1+(k+1) $$ By mathematical induction, we conclude that \(2^{n}>1+n\). (b) Each natural number greater than or equal to 6 can be written as the sum of natural numbers, each of which is a 2 or a 5 . \mathrm{\\{} \text { Proof. We will use a proof by induction. For each natural number } n \text { , } we let \(P(n)\) be, "There exist nonnegative integers \(x\) and \(y\) such that \(n=2 x+5 y "\) 'Since $$ \begin{array}{ll} 6=3 \cdot 2+0 \cdot 5 & 7=2+5 \\ 8=4 \cdot 2+0 \cdot 5 & 9=2 \cdot 2+1 \cdot 5 \end{array} $$ we see that \(P(6), P(7), P(8),\) and \(P(9)\) are true. We now suppose that for some natural number \(k\) with \(k \geq 10\) that \(P(6), P(7), \ldots, P(k)\) are true. Now $$ k+1=(k-4)+5 $$ Since \(k \geq 10\), we see that \(k-4 \geq 6\) and, hence, \(P(k-4)\) is true. So \(k-4=2 x+5 y\) and, hence, $$ \begin{aligned} k+1 &=(2 x+5 y)+5 \\ &=2 x+5(y+1) . \end{aligned} $$ This proves that \(P(k+1)\) is true, and hence, by the Second Principle of Mathematical Induction, we have proved that for each natural number \(n\) with \(n \geq 6\), there exist nonnegative integers \(x\) and \(y\) such that \(n=\) \(2 x+5 y\)

Can each natural number greater than or equal to 4 be written as the sum of at least two natural numbers, each of which is a 2 or a 3 ? Justify your conclusion. For example, \(7=2+2+3,\) and \(17=2+2+2+2+3+3+3\).

In calculus, it can be shown that $$ \begin{array}{l} \int \sin ^{2} x d x=\frac{x}{2}-\frac{1}{2} \sin x \cos x+c \quad \text { and } \\ \int \cos ^{2} x d x=\frac{x}{2}+\frac{1}{2} \sin x \cos x+c \end{array} $$ Using integration by parts, it is also possible to prove that for each natural number \(n,\) $$ \begin{aligned} \int \sin ^{n} x d x &=-\frac{1}{n} \sin ^{n-1} x \cos x+\frac{n-1}{n} \int \sin ^{n-2} x d x \text { and } \\ \int \cos ^{n} x d x &=\frac{1}{n} \cos ^{n-1} x \sin x+\frac{n-1}{n} \int \cos ^{n-2} x d x \end{aligned} $$ (a) Determine the values of $$ \int_{0}^{\pi / 2} \sin ^{2} x d x $$ and $$ \int_{0}^{\pi / 2} \sin ^{4} x d x $$ (b) Use mathematical induction to prove that for each natural number \(n\), $$ \begin{aligned} \int_{0}^{\pi / 2} \sin ^{2 n} x d x &=\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots(2 n)} \frac{\pi}{2} \text { and } \\ \int_{0}^{\pi / 2} \sin ^{2 n+1} x d x &=\frac{2 \cdot 4 \cdot 6 \cdots(2 n)}{1 \cdot 3 \cdot 5 \cdots(2 n+1)} \end{aligned} $$ These are known as the Wallis sine formulas. (c) Use mathematical induction to prove that $$ \begin{aligned} \int_{0}^{\pi / 2} \cos ^{2 n} x d x &=\frac{1 \cdot 3 \cdot 5 \cdots(2 n-1)}{2 \cdot 4 \cdot 6 \cdots(2 n)} \frac{\pi}{2} \quad \text { and } \\ \int_{0}^{\pi / 2} \cos ^{2 n+1} x d x &=\frac{2 \cdot 4 \cdot 6 \cdots(2 n)}{1 \cdot 3 \cdot 5 \cdots(2 n+1)} \end{aligned} $$ These are known as the Wallis cosine formulas.

Evaluation of proofs See the instructions for Exercise (19) on page 100 from Section 3.1 . (a) Let \(f_{n}\) be the \(n^{\text {th }}\) Fibonacci number, and let \(\alpha\) be the positive solution of the equation \(x^{2}=x+1 .\) So \(\alpha=\frac{1+\sqrt{5}}{2}\). For each natural number \(n, f_{n} \leq \alpha^{n-1}\) Proof. We will use a proof by mathematical induction. For each natural number \(n,\) we let \(P(n)\) be, " \(f_{n} \leq \alpha^{n-1}\) " We first note that \(P(1)\) is true since \(f_{1}=1\) and \(\alpha^{0}=1\). We also notice that \(P(2)\) is true since \(f_{2}=1\) and, hence, \(f_{2} \leq \alpha^{1}\). We now let \(k\) be a natural number with \(k \geq 2\) and assume that \(P(1)\), \(P(2), \ldots, P(k)\) are all true. We now need to prove that \(P(k+1)\) is true or that \(f_{k+1} \leq \alpha^{k}\). Since \(P(k-1)\) and \(P(k)\) are true, we know that \(f_{k-1} \leq \alpha^{k-2}\) and \(f_{k} \leq \alpha^{k-1}\). Therefore, $$ \begin{aligned} f_{k+1} &=f_{k}+f_{k-1} \\ f_{k+1} & \leq \alpha^{k-1}+\alpha^{k-2} \\ f_{k+1} & \leq \alpha^{k-2}(\alpha+1) . \end{aligned} $$ We now use the fact that \(\alpha+1=\alpha^{2}\) and the preceding inequality to obtain $$ \begin{array}{l} f_{k+1} \leq \alpha^{k-2} \alpha^{2} \\ f_{k+1} \leq \alpha^{k} \end{array} $$ This proves that if \(P(1), P(2), \ldots, P(k)\) are true, then \(P(k+1)\) is true. Hence, by the Second Principle of Mathematical Induction, we conclude that or each natural number \(n, f_{n} \leq \alpha^{n-1}\).
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