/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Evaluate each geometric sum. $... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 17

Evaluate each geometric sum. $$\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+\frac{1}{108}+\dots+\frac{1}{2916}$$

Short Answer

Expert verified
Answer: The sum of the given geometric sequence is $$\frac{4379}{2916}$$.

Step by step solution

01

Identify the initial term, a_1

In the given sequence, the first term is \(\frac{1}{4}\). Hence, \(a_1=\frac{1}{4}\).
02

Identify the common ratio, r

To find the common ratio, divide any term in the sequence by the term immediately preceding it. We can compute the common ratio by dividing the second term by the first term:$$r = \frac{\frac{1}{12}}{\frac{1}{4}} = \frac{1}{3}$$ Thus, the common ratio, r, is \(\frac{1}{3}\).
03

Determine the number of terms, n

The problem states it has a finite number of terms with the last term being \(\frac{1}{2916}\). To find the number of terms in the sequence, let's use the relationship:$$a_n = a_1 \cdot r^{n-1}$$We know the last term is \(\frac{1}{2916}\), and we are solving for n. Substitute the known values:$$\frac{1}{2916} = \frac{1}{4} \cdot \left(\frac{1}{3}\right)^{n-1}$$Solving for n, we get:$$\frac{1}{2916} = \frac{1}{4} \cdot \frac{1}{3^{n-1}}$$$$(4 \cdot 2916) = 3^{n-1}$$$$3^{n-1} = 11664$$Which is a perfect cube. Therefore, $$n-1=6$$$$(n-1)+1=6+1$$Then, the number of terms, n, is 7.
04

Calculate the sum, using the formula

Now we have everything we need to calculate the sum of the geometric sequence, using the formula:$$S_n = \frac{a_1(1 - r^n)}{1 - r}$$Substitute the values we found:a_1 = \frac{1}{4}, r = \frac{1}{3}, and n = 7:$$S_7 = \frac{\frac{1}{4}(1 - \left(\frac{1}{3}\right)^7)}{1 - \frac{1}{3}}$$Compute the sum:$$S_7 = \frac{\frac{1}{4}(1 - \frac{1}{2187})}{\frac{2}{3}}$$$$S_7 = \frac{\frac{4379}{4376}}{\frac{2}{3}}$$$$S_7 = \frac{3}{2} \cdot \frac{4379}{4376}$$Finally, $$S_7 = \frac{4379}{2916}$$ So, the sum of the geometric sequence is $$\frac{4379}{2916}$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Geometric Sequence
Understanding a geometric sequence is foundational in appreciating the beauty of patterns within the realm of mathematics. A geometric sequence, sometimes called a geometric progression, is an ordered set of numbers where each term after the first is found by multiplying the previous term by a constant value, known as the common ratio.

When we look at the given exercise, \(\frac{1}{4}, \frac{1}{12}, \frac{1}{36}, \frac{1}{108}, \dots, \frac{1}{2916}\), we see a pattern where each subsequent term is smaller than the one before it, revealing a system in motion, all connected by the consistent mathematical bond of the common ratio.

What truly excites us about geometric sequences isn't just the repetition, but the capability to forecast future terms, and in some cases, determine the sum of the entire sequence鈥攍eading us to investigate the common ratio more closely.
Common Ratio
The common ratio is the thread that weaves through the fabric of a geometric sequence. It is the multiplier that transitions us from one term to the next.

For instance, in our problem, finding the common ratio means we selected the second term \(\frac{1}{12}\) and divided it by the first term \(\frac{1}{4}\), which elegantly collapses to \(\frac{1}{3}\). We could have chosen any other consecutive terms, and if our sequence is geometric, the ratio would remain unchanged.

The common ratio can be pictured as the steady rhythm in the heart of the sequence, a tool that unlocks the ability to understand how the sequence behaves and expands, and an essential element when it comes to calculating the sum of the series. The common ratio can even define the nature of the sequence, determining whether it's converging or diverging.
Finite Series
A finite series is the summation of a sequence with a definitive end, a story told through numbers that have a clear beginning and conclusion.

The series produced by our sequence is indeed finite as it concludes with the term \(\frac{1}{2916}\). Unlike its counterpart, the infinite series, which stretches toward infinity, a finite series allows for the calculation of an exact sum which we can pinpoint with precision. This creates a satisfying sense of completeness, as we can neatly package the entirety of the series into a single, finite value.

Dealing with a finite series is reassuring but calculating its sum requires a formula that accounts for the finite nature and utilizes both the common ratio and the number of terms in the sequence.
Sum of Geometric Series
The climax of our journey through a geometric sequence is the discovery of its sum. The sum of a geometric series is the total value obtained when all terms of the sequence are added together.

To find this sum, particularly for a finite series, we use a special formula: \(S_n = \frac{a_1(1 - r^n)}{1 - r}\), where \(S_n\) is the sum of the first n terms, \(a_1\) is the first term, \(r\) is the common ratio, and \(n\) is the number of terms in the series.

In our exercise, we applied this formula beautifully after determining the sequence's characteristics, reaching the sum of \(\frac{4379}{2916}\). The formula is potent, drawing upon the proven consistency of the common ratio and the bounds of the finite series to deliver a single, encompassing value.

Through each step, from recognizing the sequence, identifying the common ratio, ensuring the series is finite, and applying the formula, the sum reflects the harmony of these interconnected concepts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Marie takes out a \(\$ 20,000\) loan for a new car. The loan has an annual interest rate of \(6 \%\) or, equivalently, a monthly interest rate of \(0.5 \% .\) Each month, the bank adds interest to the loan balance (the interest is always \(0.5 \%\) of the current balance), and then Marie makes a \(\$ 200\) payment to reduce the loan balance. Let \(B_{n}\) be the loan balance immediately after the \(n\) th payment, where \(B_{0}=\$ 20,000\). a. Write the first five terms of the sequence \(\left\\{B_{n}\right\\}\). b. Find a recurrence relation that generates the sequence \(\left\\{B_{n}\right\\}\). c. Determine how many months are needed to reduce the loan balance to zero.

An infinite product \(P=a_{1} a_{2} a_{3} \ldots,\) which is denoted \(\prod_{k=1}^{\infty} a_{k}\) is the limit of the sequence of partial products \(\left\\{a_{1}, a_{1} a_{2}, a_{1} a_{2} a_{3}, \ldots\right\\} .\) Assume that \(a_{k}>0\) for all \(k\) a. Show that the infinite product converges (which means its sequence of partial products converges) provided the series \(\sum_{k=1}^{\infty} \ln a_{k}\) converges. b. Consider the infinite product $$P=\prod_{k=2}^{\infty}\left(1-\frac{1}{k^{2}}\right)=\frac{3}{4} \cdot \frac{8}{9} \cdot \frac{15}{16} \cdot \frac{24}{25} \cdots$$ Write out the first few terms of the sequence of partial products, $$P_{n}=\prod_{k=2}^{n}\left(1-\frac{1}{k^{2}}\right)$$ (for example, \(P_{2}=\frac{3}{4}, P_{3}=\frac{2}{3}\) ). Write out enough terms to determine the value of \(P=\lim _{n \rightarrow \infty} P_{n}\) c. Use the results of parts (a) and (b) to evaluate the series $$\sum_{k=2}^{\infty} \ln \left(1-\frac{1}{k^{2}}\right)$$

Consider the following situations that generate a sequence. a. Write out the first five terms of the sequence. b. Find an explicit formula for the terms of the sequence. c. Find a recurrence relation that generates the sequence. d. Using a calculator or a graphing utility, estimate the limit of the sequence or state that it does not exist. Radioactive decay A material transmutes \(50 \%\) of its mass to another element every 10 years due to radioactive decay. Let \(M_{n}\) be the mass of the radioactive material at the end of the \(n\) th decade, where the initial mass of the material is \(M_{0}=20 \mathrm{g}\)

It can be proved that if a series converges absolutely, then its terms may be summed in any order without changing the value of the series. However, if a series converges conditionally, then the value of the series depends on the order of summation. For example, the (conditionally convergent) alternating harmonic series has the value $$ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots=\ln 2 $$ Show that by rearranging the terms (so the sign pattern is \(++-\) ), $$ 1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots=\frac{3}{2} \ln 2 $$

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.
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Probability and Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.