I am trying to solve the recurrence: $ a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}} $ but here is a problem for me. After few steps I have this: $ a_n^2 = a_{n-1}\cdot a_{n-2} $ and I don't now what to do further. I can solve a recurrence like that $ a_{n+2} + a_{n+1} - a_n = 5 \cdot 2^n, $ but I can't find any information about this case (when I have some degree or
square root in a recurrence).
How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$?
6
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discrete-mathematics
recurrence-relations
1 Answers
13
Hint: Let $b_n=\log a_n$ and solve a recurrence for $b_n$.
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0@Buga1234 Those two equations are the same - it yields the same values for $\lambda$. – 2012-12-01