/*! 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 39 Find the sum of each infinite ge... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 39

Find the sum of each infinite geometric series. $$3+\frac{3}{4}+\frac{3}{4^{2}}+\frac{3}{4^{3}}+\cdots$$

Short Answer

Expert verified
The sum of the infinite geometric series is 4.

Step by step solution

01

Identify the First Term and the Common Ratio

The first term \(a\) of the series is 3 and the common ratio \(r\) is \(\frac{3}{4}\). This can be seen because each succeeding term after 3 is obtained by multiplying the preceding term by \(\frac{1}{4}\), for example, \(\frac{3}{4}=\frac{3}{4^{1}}=3 \times \(\frac{1}{4}\), \(\frac{3}{4^{2}}=3 \times \(\frac{1}{4}\)^{2}\), and so on.
02

Substitute the values into the Series Sum Formula

To compute the sum of the series, you substitute \(a = 3\) and \(r = \frac{1}{4}\) into the series sum formula \(S=\frac{a}{1-r}\). This gives: \(S=\frac{3}{1-\frac{1}{4}}\).
03

Solve the Expression

Then you simplify the expression on the right-hand side of the equation to find the sum \(S\). The computation is as follows: \(S=\frac{3}{1-\frac{1}{4}} = \frac{3}{\frac{3}{4}} = \frac{3}{1} \times \frac{4}{3} = 4.

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!

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

Show that $$ \left(\begin{array}{l}n \\\r\end{array}\right)+\left(\begin{array}{c}n \\\r+1\end{array}\right)=\left(\begin{array}{l}n+1 \\\r+1 \end{array}\right) $$ Hints: $$ \begin{aligned}&(n-r) !=(n-r)(n-r-1) !\\\&(r+1) !=(r+1) r !\end{aligned} $$

If \(f(x)=x^{4},\) find \(\frac{f(x+h)-f(x)}{h}\) and simplify.

Evaluate the given binomial coefficient. $$ \left(\begin{array}{l}8 \\\3\end{array}\right) $$

Write the first three terms in each binomial expansion, expressing the result in simplified form. $$ \left(x^{2}+1\right)^{17} $$

Use the Binomial Theorem to expand and then simplify the result: \(\left(x^{2}+x+1\right)^{3}\). [ Hint: Write \(x^{2}+x+1\) as \(\left.x^{2}+(x+1)\right]\)
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Calculus

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.