Question

write out the first few terms of the series \\( \sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { 8 ^ { n } } \\). what is the series sum?
the first term is 1 (type an integer or a simplified fraction.)
the second term is \\( \frac { 2 } { 3 } \\) (type an integer or a simplified fraction.)

Explanation:

Step1: Find the first few terms

For \(n = 0\), \(a_{0}=\frac{(-1)^{0}}{8^{0}}=\frac{1}{1}=1\).
For \(n = 1\), \(a_{1}=\frac{(-1)^{1}}{8^{1}}=-\frac{1}{8}\).
For \(n = 2\), \(a_{2}=\frac{(-1)^{2}}{8^{2}}=\frac{1}{64}\).
For \(n = 3\), \(a_{3}=\frac{(-1)^{3}}{8^{3}}=-\frac{1}{512}\).

Step2: Recognize the series as a geometric series

The general form of a geometric series is \(\sum_{n = 0}^{\infty}ar^{n}\), where \(a\) is the first - term and \(r\) is the common ratio.
In the series \(\sum_{n=0}^{\infty}\frac{(-1)^{n}}{8^{n}}=\sum_{n = 0}^{\infty}a r^{n}\), we have \(a = 1\) and \(r=-\frac{1}{8}\).

Step3: Use the formula for the sum of an infinite geometric series

The formula for the sum of an infinite geometric series is \(S=\frac{a}{1 - r}\), when \(\vert r\vert<1\).
Substitute \(a = 1\) and \(r=-\frac{1}{8}\) into the formula:
\(S=\frac{1}{1-(-\frac{1}{8})}=\frac{1}{1 + \frac{1}{8}}=\frac{1}{\frac{9}{8}}=\frac{8}{9}\).

Answer:

The first few terms are \(1,-\frac{1}{8},\frac{1}{64},-\frac{1}{512},\cdots\). The sum of the series is \(\frac{8}{9}\).