The following expression, where the process continues indefinitely, is called a continued fraction. Complete parts (a) through (e). \[ 2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\cdots}}} \] a. Show that this expression can be built in steps using the recurrence relation \( \mathrm{a}_{0}=2 \), \( a_{n+1}=2+\frac{1}{a_{n}} \). What are the first four terms of the sequence? Choose the correct answer. A. \( 2, \frac{2}{2+\frac{1}{2}}, \frac{2}{2+\frac{1}{2+\frac{1}{2}}}, \frac{2}{2+\frac{1}{2+\frac{1}{2+1}}} \) B. \[ 2,2+1,2+1+\frac{1}{2+1} \] C. \( 2,2+\frac{1}{2}, 2+\frac{1}{2+\frac{1}{2}}, 2+\frac{1}{2+\frac{1}{2+\frac{1}{2}}} \) D. \( 1, \frac{1}{1+\frac{2}{1}}, \frac{1}{1+\frac{2}{1+\frac{2}{1}}}, \frac{1}{1+\frac{2}{1+\frac{2}{1+2}}} \)

To explore this continued fraction, let's dive into the recurrence relation you provided: Starting with \( a_0 = 2 \): - For \( a_1 \): \[ a_1 = 2 + \frac{1}{a_0} = 2 + \frac{1}{2} = 2 + 0.5 = 2.5 \] - For \( a_2 \): \[ a_2 = 2 + \frac{1}{a_1} = 2 + \frac{1}{2.5} = 2 + 0.4 = 2.4 \] - For \( a_3 \): \[ a_3 = 2 + \frac{1}{a_2} = 2 + \frac{1}{2.4} \approx 2 + 0.4167 \approx 2.4167 \] - For \( a_4 \): \[ a_4 = 2 + \frac{1}{a_3} = 2 + \frac{1}{2.4167} \approx 2 + 0.4130 \approx 2.4130 \] Thus, the first four terms in the sequence using the recurrence relation are approximately \( 2, 2.5, 2.4, 2.4167 \), which corresponds to option C, \( 2, 2 + \frac{1}{2}, 2 + \frac{1}{2 + \frac{1}{2}}, 2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2}}} \). **Here's some more fun stuff:** The concept of continued fractions has been around since ancient times, with mathematicians such as John Wallis and Leonhard Euler delving into their fascinating properties. These fractions represent numbers beautifully and can provide insights into irrational numbers, especially when approximating them through their convergents! In practice, continued fractions are also used in algorithms for numerical methods, especially in computing approximations of irrational numbers. For example, they're handy in calculating square roots or in cryptography, where understanding number systems deeply is crucial. So, next time you're solving for square roots, remember that the continued fraction approach can lead to very effective results!