Chapter 11 - Infinite Sequences and Series - 11.2 Exercises - Page 736: 55

$1.5342424242=\frac{5063}{3300}$

$1.5342424242$ $=1.53 + \frac{42}{10^{4}}+\frac{42}{10^{6}}+\frac{42}{10^{8}}+...$ $=1.53 + \Sigma^{\infty}_{n=0} \frac{42}{10^{4}}(\frac{1}{10^{2}})^{n}$ $a=\frac{42}{104}$ and $r=\frac{1}{10^{2}}$ $1.5342424242 = 1.53 + \frac{\frac{42}{10^{4}}}{1-\frac{1}{10^{2}}}$ $=\frac{153}{100}+\frac{42}{9900}$ $=\frac{15189}{9900}$ $=\frac{5063}{3300}$