Buga1234
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 Dec1 comment How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$? is it the linear recurrence? Will be right if I'll do something like this $2\lambda^2 - \lambda - 1 = 0$ or I must to do this $\lambda^2 - 1/2 \lambda - 1/2 = 0$ Nov28 comment How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$? yeap... it must be $2b_n = b_{n-1} + b_{n-2}$ Nov28 awarded Scholar Nov28 awarded Supporter Nov28 accepted How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$? Nov28 comment How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$? oh, thank you very much! Now I got it, I think... $a_n = 10^{b_n}$, $10^{b_n} = 10^{b_{n-1}} * 10^{b_{n-2}}$, $10^{b_n} = 10^{b_{n-1} + b_{n-2}}$, $b_n = b_{n-1} + b_{n-2}$ is this a rigth way? Nov28 comment How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$? :) Now I've the same question. Where did you get that $e$ ? Nov28 comment How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$? $$b_n=\log a_n , a_n^{b_n} = k ?$$ What do you mean? How it works? Nov28 awarded Student Nov28 asked How to solve this recurrence relation $a_{n+2} = \sqrt{a_{n+1}\cdot a_{n}}$?