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Special Series in Sequences and Series

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Question

$If a_{n + 1} = \frac{1}{1 - a_{n}} for n \geq 1 and a_{3} = a_{1} then |{(a_{2001})}^{2001}| is$

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Solution

$\begin{array}{l} a_{2} = \frac{1}{1 - a_{1}}, a_{3} = \frac{1}{1 - a_{2}} = \frac{1}{1 - \frac{1}{1 - a_{1}}} = \frac{1 - a_{1}}{- a_{1}} \\ Since a_{3} = a_{1} \to - a_{1}^{2} = 1 - a_{1} \\ \Rightarrow a_{1}^{2} - a_{1} + 1 = 0 \Rightarrow a_{1} = - w_{,} - w^{2} \\ a_{5} = \frac{1}{1 - a_{4}} = \frac{1}{1 - \frac{1}{1 - a_{3}}} = \frac{a_{3} - 1}{a_{3}} \end{array}$
$\begin{array}{l} ∴ continuing in their way, we obtain \\ a_{1} = a_{3} = a_{5} = \dots = a_{2001} \Rightarrow {(a_{2001})}^{2001} = (- w)^{2001} = - 1 \\ \Rightarrow |{(a_{2001})}^{2001}| = 1 \end{array}$

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