write out the first few terms of the series \\( \\sum _ { n = 0 } ^ { \\infty } \\frac { ( - 1 ) ^ { n } } {…

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

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

Answer

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}).