Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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).

share|cite|improve this question
up vote 13 down vote accepted

Hint: Let $b_n=\log a_n$ and solve a recurrence for $b_n$.

share|cite|improve this answer
$$ b_n=\log a_n , a_n^{b_n} = k ? $$ What do you mean? How it works? – Buga1234 Nov 28 '12 at 19:13
Where did you get $k$? If $b_n=\log a_n$ then $e^{b_n}=a_n$. – Thomas Andrews Nov 28 '12 at 19:18
:) Now I've the same question. Where did you get that $ e $ ? – Buga1234 Nov 28 '12 at 19:24
You don't need to use natural logarithms, you can use any logarithm base you want with this problem. If you define $b_n=\log_{10} a_n$ then you know that $10^{b_n}=a_n$. – Thomas Andrews Nov 28 '12 at 19:58
yeap... it must be $ 2b_n = b_{n-1} + b_{n-2} $ – Buga1234 Nov 28 '12 at 20:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.