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an+1=2+an\large a_{n+1} = \sqrt{2+a_n} an+1=2+an
Consider the recurrence relation ana_nan above for n=1,2,3,…n = 1,2, 3, \ldots n=1,2,3,… and a1=3a_1 = \sqrt3a1=3.
Evaluate log2[cos−1(a20162)] \log_2 \left [ \cos^{-1} \left ( \dfrac{a_{2016}}2 \right) \right ] log2[cos−1(2a2016)].
If this number can be expressed as log2π−log2a−b \log_2 \pi - \log_2 a - blog2π−log2a−b, where aaa and bbb are positive integers with aaa odd, submit your answer as a+ba+ba+b.
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