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Chapter 8: Problem 29

Several terms of a sequence $$\left\\{a_{n}\right\\}_{n=1}^{\infty} \text { are given.}$$ a. Find the next two terms of the sequence. b. Find a recurrence relation that generates the sequence (supply the initial value of the index and the first term of the sequence). c. Find an explicit formula for the nth term of the sequence. $$\\{1,3,9,27,81, \dots\\}$$

Short Answer

Expert verified
Answer: The next two terms are 243 and 729, the recurrence relation is \(a_n = 3a_{n-1}\), and the explicit formula for the nth term is \(a_n = 3^{n-1}\).

Step by step solution

01

Find the next two terms

To find the next two terms, we need to find the constant factor. We can do this by dividing the second term by the first term, i.e., $$3/1 = 3$$, and then the third term by the second term $$9/3 = 3$$. We can see in-between term constant factor is 3. Therefore, the next term will be \(81\times3=243\), and the term after that will be \(243\times3=729\). So the next two terms are 243 and 729.
02

Find the recurrence relation

Since each term in the sequence is obtained by multiplying the previous term by a constant factor, we can write a recurrence relation for the sequence. We have found that the constant factor is 3. Therefore, the recurrence relation is represented as: $$a_n = 3a_{n-1}$$ The sequence starts at index 1, with the first term being 1.
03

Find an explicit formula for the nth term

To find the explicit formula for the nth term, we need to express the sequence as a function of n. This is a geometric sequence because each term is obtained by multiplying the previous term by a constant. Usually, in a geometric sequence, if the first term is \(a_1\), and the constant ratio is r, then the nth term can be represented as: $$a_n = a_1\cdot r^{n-1}$$ In this case, the first term, \(a_1\), is 1, and the constant ratio, r, is 3. So the explicit formula for the nth term of the sequence is: $$a_n = 1 \cdot 3^{n-1}$$ or simply, $$a_n = 3^{n-1}$$

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Key Concepts

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

Recurrence Relation
Understanding the concept of a recurrence relation is crucial when dealing with sequences in mathematics. A recurrence relation describes the way each term in a sequence is formed from its predecessors. In the simple terms of arithmetic we might say that it's a rule that tells us how to find a 'next term' from the 'current term'.

Let's take a closer look at our geometric sequence example \(\{1,3,9,27,81, \dots\}\). To find a pattern, we examine how each term is generated. As we previously observed, each term is obtained by multiplying the term before it by 3. So, if we symbolize any term of the sequence as \(a_n\), we can say that the term that follows it, symbolized as \(a_{n+1}\), is simply \(a_n\) multiplied by 3. This relationship can be succinctly expressed as a recurrence relation: \[a_{n+1} = 3a_n\].

In a sequence, recurrence relations provide a dynamic method of calculation, where you need to know a specific term to find the next one. For this specific sequence, we also define that our starting point, the first term \(a_1\), is 1. So, the pattern is both clear and repetitive: to move forward in the sequence, multiply by 3.
Explicit Formula
The explicit formula, on the other hand, allows us to calculate any term in a sequence without referencing previous terms. It is direct and doesn't necessitate the iterative process that a recurrence relation does. For our sequence, which we've established is a geometric progression, the nth term can be predicted using such a formula.

The explicit formula for a geometric sequence looks like this: \[a_n = a_1 \times r^{(n-1)}\], where \(a_1\) is the first term and \(r\) is the common ratio. In our case, \(a_1 = 1\) and \(r = 3\), so the explicit formula for the nth term is simply \(a_n = 3^{(n-1)}\). This formula is extremely powerful because it allows students to bypass many calculations to find, for instance, the 50th term without calculating the 49 terms before it, which would be a very tedious task using the recurrence relation.
Geometric Progression
A geometric progression (or geometric sequence) is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the 'ratio'. It is this consistent multiplier that defines a sequence as geometric.

In our example \(\{1,3,9,27,81, \dots\}\), we see a clear geometric progression with a common ratio of 3. To see this practically, \(3 / 1 = 9 / 3 = 27 / 9 = 81 / 27 = 3\), cementing that the common ratio is indeed consistent across all terms. Geometric progressions can describe many natural phenomena and are also fundamental in various financial calculations, such as compound interest, making them a valuable concept to understand not just in algebra but in real-world applications as well.

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Most popular questions from this chapter

Use the formal definition of the limit of a sequence to prove the following limits. $$\lim _{n \rightarrow \infty} \frac{n}{n^{2}+1}=0$$

Pick two positive numbers \(a_{0}\) and \(b_{0}\) with \(a_{0}>b_{0},\) and write out the first few terms of the two sequences \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}:\) $$a_{n+1}=\frac{a_{n}+b_{n}}{2}, \quad b_{n+1}=\sqrt{a_{n} b_{n}}, \quad \text { for } n=0,1,2 \dots$$ (Recall that the arithmetic mean \(A=(p+q) / 2\) and the geometric mean \(G=\sqrt{p q}\) of two positive numbers \(p\) and \(q\) satisfy \(A \geq G.)\) a. Show that \(a_{n} > b_{n}\) for all \(n\). b. Show that \(\left\\{a_{n}\right\\}\) is a decreasing sequence and \(\left\\{b_{n}\right\\}\) is an increasing sequence. c. Conclude that \(\left\\{a_{n}\right\\}\) and \(\left\\{b_{n}\right\\}\) converge. d. Show that \(a_{n+1}-b_{n+1}<\left(a_{n}-b_{n}\right) / 2\) and conclude that \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} b_{n} .\) The common value of these limits is called the arithmetic-geometric mean of \(a_{0}\) and \(b_{0},\) denoted \(\mathrm{AGM}\left(a_{0}, b_{0}\right)\). e. Estimate AGM(12,20). Estimate Gauss' constant \(1 / \mathrm{AGM}(1, \sqrt{2})\).

Repeated square roots Consider the expression \(\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}},\) where the process continues indefinitely. a. Show that this expression can be built in steps using the recurrence relation \(a_{0}=1, a_{n+1}=\sqrt{1+a_{n}}\), for \(n=0,1,2,3, \ldots . .\) Explain why the value of the expression can be interpreted as \(\lim _{n \rightarrow \infty} a_{n},\) provided the limit exists. b. Evaluate the first five terms of the sequence \(\left\\{a_{n}\right\\}\) c. Estimate the limit of the sequence. Compare your estimate with \((1+\sqrt{5}) / 2,\) a number known as the golden mean. d. Assuming the limit exists, use the method of Example 5 to determine the limit exactly. e. Repeat the preceding analysis for the expression \(\sqrt{p+\sqrt{p+\sqrt{p+\sqrt{p+\cdots}}},}\) where \(p>0 .\) Make a table showing the approximate value of this expression for various values of \(p .\) Does the expression seem to have a limit for all positive values of \(p ?\)

The prime numbers are those positive integers that are divisible by only 1 and themselves (for example, 2,3,5,7, 11,13, \(\ldots\) ). A celebrated theorem states that the sequence of prime numbers \(\left\\{p_{k}\right\\}\) satisfies \(\lim _{k \rightarrow \infty} p_{k} /(k \ln k)=1 .\) Show that \(\sum_{k=2}^{\infty} \frac{1}{k \ln k}\) diverges, which implies that the series \(\sum_{k=1}^{\infty} \frac{1}{p_{k}}\) diverges.

The Greeks solved several calculus problems almost 2000 years before the discovery of calculus. One example is Archimedes' calculation of the area of the region \(R\) bounded by a segment of a parabola, which he did using the "method of exhaustion." As shown in the figure, the idea was to fill \(R\) with an infinite sequence of triangles. Archimedes began with an isosceles triangle inscribed in the parabola, with area \(A_{1}\), and proceeded in stages, with the number of new triangles doubling at each stage. He was able to show (the key to the solution) that at each stage, the area of a new triangle is \(\frac{1}{8}\) of the area of a triangle at the previous stage; for example, \(A_{2}=\frac{1}{8} A_{1},\) and so forth. Show, as Archimedes did, that the area of \(R\) is \(\frac{4}{3}\) times the area of \(A_{1}\).
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