Question

Express each sum using summation notation. \(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\cdots+(-1)^{6}\left(\frac{1}{3^{6}}\right)\)

Short answer

\(\sum_{n=0}^{6} (-1)^n \left(\frac{1}{3^n}\right)\)

Step-by-step solution

Identify the Pattern

Observe the given series: \(1 - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \cdots + (-1)^{6}\left(\frac{1}{3^{6}}\right)\). Notice that this is an alternating series where each term involves a power of \(\frac{1}{3}\).

Write the General Term

The general term of the series can be written as \((-1)^{n}\left(\frac{1}{3^{n}}\right)\). Here, each term’s sign alternates due to \((-1)^{n}\) and the denominator is a power of 3.

Define the Range of the Summation

Identify the range of the summation. The given series starts at \(n=0\) and ends at \(n=6\). This means the summation is over \(n\) going from 0 to 6.

Combine into Summation Notation

Combine the identified components into summation notation to express the sum: \[\sum_{n=0}^{6} (-1)^n \left(\frac{1}{3^n}\right)\]}],

Key concepts

Alternating Series

An alternating series is a series where the terms alternate in sign. In other words, the signs change from positive to negative or vice versa sequentially. This can be observed in the given series:

• First term: 1
• Second term: -1/3
• Third term: 1/9
• Fourth term: -1/27

Notice how the sign changes in each term. The presence of \((-1)^n \) in the general term helps in alternating the signs of the series. When \((-1)^n \) is raised to an even power \(n = 0, 2, 4, 6, \), the term is positive. When it is raised to an odd power \(n = 1, 3, 5\), the term is negative.

General Term

The general term of a series is a formula that represents the terms in the series for a given value of the index of summation. In the given exercise, the general term is represented as \((-1)^n \left(\frac{1}{3^n} \right)\).

• \((-1)^n \) alternates the sign of the terms.
• \(\frac{1}{3^n} \) indicates the sequence follows powers of 3 in the denominator.

For example, when \(n=0\), the term is \(1\). When \(n=1\), the term is \(-\frac{1}{3}\). This pattern continues with \((-1)^n\) alternating the sign and \(\frac{1}{3^n}\) representing the fractions involved.

Range of Summation

The range of summation tells us the starting and ending values of the index for which the summation is considered. For the given series, the index of summation is represented by \(n\) and it ranges from 0 to 6.

This means the summation notation will include values of \(n\) starting from 0, going up to and including 6. \[\textstyle \sum \limits \_{n=0}^{6} \left((-1)^n \frac{1}{3^n} \right)\]Here, \(n=0\) gives the first term, \(n=1\) gives the second term, and so on, until \(n=6\). The formula adds up each term in this range, considering the alternating sign and denominator pattern defined by the general term.