The following expression, where the process continues indefinitely, is called a continued fraction. Complete parts (a) through (e). \( 2+\frac{1}{2+\frac{1}{2+\frac{1}{2+\cdots}}} \) b. Evaluate the first five terms of the sequence \( \left\{a_{n}\right\} \). \( a_{0}=2, a_{1}=\square, a_{2}=\square, a_{3}=\square, a_{4}=\square \) (Type integers or decimals rounded to three decimal places as needed.)
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To evaluate the first five terms of the sequence \( \{a_n\} \), we'll recursively define \( a_n \) based on the continued fraction provided. Starting with \( a_0 \): - \( a_0 = 2 \) Now for the subsequent terms: - \( a_1 = 2 + \frac{1}{a_0} = 2 + \frac{1}{2} = 2 + 0.5 = 2.5 \) - \( a_2 = 2 + \frac{1}{a_1} = 2 + \frac{1}{2.5} = 2 + 0.4 = 2.4 \) - \( a_3 = 2 + \frac{1}{a_2} = 2 + \frac{1}{2.4} \approx 2 + 0.4167 \approx 2.4167 \) - \( a_4 = 2 + \frac{1}{a_3} \approx 2 + \frac{1}{2.4167} \approx 2 + 0.4132 \approx 2.4132 \) So, the values for \( a_n \) rounded to three decimal places are: - \( a_0 = 2 \) - \( a_1 \approx 2.500 \) - \( a_2 \approx 2.400 \) - \( a_3 \approx 2.417 \) - \( a_4 \approx 2.413 \) In summary, the sequence is: \( a_{0} = 2, a_{1} = 2.500, a_{2} = 2.400, a_{3} = 2.417, a_{4} = 2.413 \)
