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Chapter 13: Problem 55

Expand each sum. \(\sum_{k=0}^{n-1} \frac{1}{3^{k+1}}\)

Short Answer

Expert verified
The expanded sum is \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots + \frac{1}{3^n}\).

Step by step solution

01

Title - Understand the Summation

The given summation is \(\sum_{k=0}^{n-1} \frac{1}{3^{k+1}}\). This indicates that we need to calculate the sum of the terms \(\frac{1}{3^{k+1}}\) for values of \(k\) starting from \(0\) and going up to \((n-1)\).
02

Title - Write Out the First Few Terms

Substitute \(k = 0, 1, 2, \ldots, n-1\) into the term \(\frac{1}{3^{k+1}}\). The first few terms are: \(\frac{1}{3^1}, \frac{1}{3^2}, \frac{1}{3^3},\ \ldots,\ \frac{1}{3^n}\).
03

Title - Expand the Summation

Expand the summation notation into an explicit sum: \(\frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots + \frac{1}{3^n}\).
04

Title - Conclusion

The expanded form of the summation \(\sum_{k=0}^{n-1} \frac{1}{3^{k+1}}\) is \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots + \frac{1}{3^n}\).

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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 series with a constant ratio between successive terms. In the given exercise, the ratio between each term is \(\frac{1}{3}\). Here's how you can recognize a geometric series:
  • Each term is obtained by multiplying the previous term by a fixed, non-zero number called the common ratio.
  • In our sum \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots + \frac{1}{3^n}\), the common ratio \(\frac{1}{3}\). So, each term is \(\frac{1}{3}\) times the previous term.
To write it out more formally, if \{a_1, a_2, a_3, \ldots\}\ is a geometric series, then:
\{a, ar, ar^2, ar^3, \ldots\}\ where 'a' is the first term and 'r' is the common ratio.
series expansion
Series expansion involves writing out the terms of a series to make summing them up easier. For the sum \(\frac{1}{3^1} + \frac{1}{3^2} + \frac{1}{3^3} + \ldots + \frac{1}{3^n}\), you deal with each term individually.
To expand it, we substitute each value of k from 0 to (n-1) into \(\frac{1}{3^{k+1}}\):
  • For k=0: \(\frac{1}{3^{0+1}} = \frac{1}{3}\).
  • For k=1: \(\frac{1}{3^{1+1}} = \frac{1}{9}\).
  • For k=2: \(\frac{1}{3^{2+1}} = \frac{1}{27}\).
This process continues until the (n-1) term \(\frac{1}{3^n}\). This step-by-step expansion helps you sum the series more easily.
algebraic notation
Algebraic notation helps in representing series in a concise form using symbols and variables. In the initial exercise, the series is written as: \(\sum_{k=0}^{n-1} \frac{1}{3^{k+1}}\).
Here's how the algebraic notation works:
  • The summation symbol \(\sum\) indicates that you are summing up terms.
  • The variable 'k' is called the index of summation, starting at 0 and ending at \((n-1)\).
  • The expression \(\frac{1}{3^{k+1}}\) is the general term being summed.
Algebraic notation makes it easier to write and manipulate series, particularly when dealing with long or complex sums.
finite series
A finite series is a series that has a definite number of terms. Unlike an infinite series that goes on forever, a finite series stops after a certain number of terms.
In the exercise, the series is finite because it stops at the term \(\frac{1}{3^n}\).
  • Finite series are simpler to sum because of their limited number of terms.
  • You can use specific formulas to easily sum them.
For a finite geometric series of the form \(\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots + \frac{1}{3^n}\), the sum is given by: \(\frac{a(1-r^n)}{1-r}\), where 'a' is the first term and 'r' is the common ratio. Using this, you can quickly find the sum rather than adding each term individually.

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Most popular questions from this chapter

Which of the following choices, \(A\) or \(B\), results in more money? A: To receive \(\$ 1000\) on day \(1, \$ 999\) on day \(2, \$ 998\) on day \(3,\) with the process to end after 1000 days B: To receive \(\$ 1\) on day \(1, \$ 2\) on day \(2, \$ 4\) on day 3 , for 19 days

Determine whether the given sequence is arithmetic, geometric, or neither. If the sequence is arithmetic, find the common difference; if it is geometric, find the common ratio. If the sequence is arithmetic or geometric, find the sum of the first 50 terms. $$ \left\\{\left(\frac{5}{4}\right)^{n}\right\\} $$

Use the Principle of Mathematical Induction to show that the given statement is true for all natural numbers \(n\). $$ 1 \cdot 2+3 \cdot 4+5 \cdot 6+\cdots+(2 n-1)(2 n)=\frac{1}{3} n(n+1)(4 n-1) $$

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Find the value of the determinant: $$\left|\begin{array}{rrr}3 & 1 & 0 \\ 0 & -2 & 6 \\ 4 & -1 & -2\end{array}\right|$$

Show that $$ 1+2+\cdots+(n-1)+n=\frac{n(n+1)}{2} $$ [Hint: Let $$ \begin{array}{l} S=1+2+\cdots+(n-1)+n \\ S=n+(n-1)+(n-2)+\cdots+1 \end{array} $$Add these equations. Then $$ 2 S=[1+n]+[2+(n-1)]+\cdots+[n+1] $$
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