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$$a_1 = 1; a_2 = 9; a_{n+2} = \frac{a_{n+1}a_n}{6a_n - 9a_{n+1}}$$

I need to find non-recurring formula for $a_n$. Is there any good way to do this? The only one comes to mind is to guess the formula and then prove it using mathematical induction.

Thanks in advance!

I've got the result and it looks like this: $a_n = \frac{-3*2^{n-1} + 2^{2n - 1} + 1}{3}$ but I really don't like this way and would love to know how to solve this properly.

Mishael
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3 Answers3

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You formula for $a_n$ seems wrong, is it $\frac{1}{a_n}$? Take inverse, one has $$ \frac{1}{a_{n+2}}=6\frac{1}{a_{n+1}}-9\frac{1}{a_n}. $$ there is a standard way to solve this (http://en.wikipedia.org/wiki/Recurrence_relation#Linear_homogeneous_recurrence_relations_with_constant_coefficients).

Ma Ming
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Let $b_n = \dfrac1{a_n}$. We then have $$b_1 = 1; b_2 = \dfrac19 \text{ and } b_{n+2} = -9b_n + 6b_{n+1} \implies b_{n+2} - 3b_{n+1} = 3(b_{n+1} - 3b_{n+2})$$ This gives us $$b_{n+2} - 3 b_{n+1} = 3^{n+1}(b_1 - 3b_2) = 2 \cdot 3^n$$ I will leave the rest to you to complete the details.

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Hint:

$$a_{n+2}=\frac{a_{n+1}a_n}{6a_n-9a_{n+1}}$$ $$6a_na_{n+2}-9a_{n+1}a_{n+2}=a_{n+1}a_n$$ Now divide by $a_na_{n+1}a_{n+2}$ to get: $$\frac{1}{a_{n+2}}-\frac{6}{a_{n+1}}+\frac{9}{a_n}=0$$ This is reccurence relation is easy to solve

Amr
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