for all positive integral., u_(n+1) =u_n (u_(n−1) ^2 −2)−u_n u_n =2 and u_1 =2(1/2) prove that : 3log_2 [u_n ]=2^n −1(−1)^n where [x] is the integral part of x


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Question Number 143474 by Ghaniy last updated on 14/Jun/21

$f o r a l l p o s i t i v e i n t e g r a l .,$ $u_{n + 1} = u_{n} (u_{n - 1}^{2} - 2) - u_{n}$ $u_{n} = 2 a n d u_{1} = 2 \frac{1}{2}$ $p r o v e t h a t : {3 \log}_{2} [u_{n}] = 2^{n} - 1 {(- 1)}^{n}$ $w h e r e [x] i s t h e i n t e g r a l p a r t o f x$
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