EXERCISE 7 1. Find the sum of each of the following infinite geometric scries: \( \begin{array}{ll}\text { (a) } 2+\frac{2}{3}+\frac{2}{9}+\ldots & \text { (b) }-64+32-16+\ldots \\ \text { (c) } 24-4+\frac{2}{3}-\ldots & \text { (b) Calculate } \sum_{m=0}^{\infty} 3\left(-\frac{1}{2}\right)^{m} \\ \text { (a) Calculate } \sum_{m=1}^{\infty} 8(2)^{-2 m} & \begin{array}{ll}\text { (c) Calculate } \sum_{i=0}^{\infty} \frac{1}{10^{i}} & \text { (c) } 0,5 \dot{4}\end{array} \\ \text { Convert each of the following recurring decimals to a common fraction by } \\ \text { first writing it as a geometric series. } \\ \text { (a) } 0,2 \dot{23} & \text { (b) } 4,2\end{array} \)

The sum of an infinite geometric series can be found using the formula \( S = \frac{a}{1 - r} \), where \( a \) is the first term and \( r \) is the common ratio. For part (a), \( S = \frac{2}{1 - \frac{1}{3}} = \frac{2}{\frac{2}{3}} = 3 \). For part (b), \( S = \frac{-64}{1 - (-\frac{1}{2})} = \frac{-64}{\frac{3}{2}} = -\frac{128}{3} \). For part (c), \( S = \frac{24}{1 - (-\frac{1}{6})} = \frac{24}{\frac{7}{6}} = \frac{144}{7} \). Now, for the series \( \sum_{m=0}^{\infty} 3\left(-\frac{1}{2}\right)^{m} \), we have \( a = 3 \) and \( r = -\frac{1}{2} \), hence \( S = \frac{3}{1 - (-\frac{1}{2})} = \frac{3}{\frac{3}{2}} = 2 \). Next, for \( \sum_{m=1}^{\infty} 8(2)^{-2 m} \), it's best to rewrite it: \( 8(2)^{-2} + 8(2)^{-4} + \ldots \), where \( a = 2 \) and \( r = \frac{1}{4} \), leading to \( S = \frac{2}{1 - \frac{1}{4}} = \frac{2}{\frac{3}{4}} = \frac{8}{3} \). For \( \sum_{i=0}^{\infty} \frac{1}{10^{i}} \), we have \( a = 1 \) and \( r = \frac{1}{10} \), so \( S = \frac{1}{1 - \frac{1}{10}} = \frac{1}{\frac{9}{10}} = \frac{10}{9} \). To convert the recurring decimal \( 0.2\dot{23} \) into a fraction, denote it as \( x = 0.2223\ldots \). We write it as \( x = 0.2 + 0.0\dot{23} \). The second part can be expressed as a geometric series: \( 0.0\dot{23} = \frac{23}{100} + \frac{23}{10000} + \ldots \). This leads to \( S = \frac{\frac{23}{100}}{1 - \frac{1}{100}} = \frac{\frac{23}{100}}{\frac{99}{100}} = \frac{23}{99} \), and combining yields \( x = 0.2 + \frac{23}{99} = \frac{20}{99} + \frac{23}{99} = \frac{43}{99} \). For \( 4.2 \), write it as \( y = 4 + 0.2\dot{0} \). The recurring part can be treated similarly, resulting in \( 4.2 = 4 + \frac{1}{5} = \frac{21}{5} \). Thus, \( 4.2\ldots = \frac{21}{5} \).