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Chapter 11: Problem 40

Find the sum, if it exists. $$\sum_{k=0}^{10} 3^{k}$$

Short Answer

Expert verified
88573

Step by step solution

01

Identify the type of series

This series is a geometric series where each term is of the form \(3^k\).
02

Use the geometric series sum formula

For a geometric series of the form \(\text{sum} = \frac{a (1 - r^{n+1})}{1 - r}\), identify the first term \(a\) and common ratio \(r\). In this case, \(a = 1\) (since \(3^0 = 1\)) and \(r = 3\).
03

Plug in the values

We have \(n + 1 = 11\) because the series runs from \(k = 0\) to \(k = 10\). Plug these values into the formula: \[\text{sum} = \frac{1(1 - 3^{11})}{1 - 3}\]
04

Simplify the expression

Simplify the expression: \[\text{sum} = \frac{1 - 3^{11}}{-2} = \frac{-(3^{11} - 1)}{2} = \frac{3^{11} - 1}{2}\]
05

Calculate the final value

Compute the value of \(3^{11}\). \(3^{11} = 177147\). Now substitute back: \[\text{sum} = \frac{177147 - 1}{2} = \frac{177146}{2} = 88573\]

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Key Concepts

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

Geometric Series
A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous term by a constant called the common ratio.
Examples of geometric series include:
  • 2, 4, 8, 16, ... where the common ratio is 2
  • 5, 25, 125, 625, ... where the common ratio is 5
In our exercise, we are working with the series \(\text{sum} = \sum_{k=0}^{10} 3^{k} \).
Here, each term is the exponentiation of 3, making our common ratio \(r = 3\). The first term \(a\) is 1, which is \(3^0\). Understanding this series is essential for calculating its sum.
Series Sum Formula
To find the sum of a geometric series, we use the series sum formula:
\[S = \frac{a (1 - r^{n+1})}{1 - r}\]
Where:
  • \(a\) is the first term
  • \(r\) is the common ratio
  • \(n\) is the number of terms
This formula simplifies the process of summing a large number of terms in a geometric sequence.
For the series \(\text{sum} = \sum_{k=0}^{10} 3^{k}\), we plug in: \(a = 1\), \(r = 3\), and \(n = 10\).
Using the formula, it becomes: \[S = \frac{1 (1 - 3^{11})}{1 - 3}= \frac{1 (1 - 177147)}{1 - 3} = \frac{1 - 177147}{-2} = \frac{-(177147 - 1)}{2} = \frac{177146}{2} = 88573\]
Thus, the sum of the series is 88573.
Exponentiation
Exponentiation is a mathematical operation involving a base and an exponent. The base is multiplied by itself as many times as the value of the exponent indicates.
For example, in \(3^2\), 3 is the base, and the exponent is 2, meaning 3 is multiplied by itself 2 times, resulting in 9.
In our series, each term \(3^k\) involves exponentiation, where the base is 3 and the exponent \k\ ranges from 0 to 10.
The term \(3^0 = 1\), and \(3^{11} = 177147\).
Understanding exponentiation helps in computing the terms of a geometric series like \(3^k\), making it easier to sum up the series using the series sum formula.

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