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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.

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try it –  belisarius Apr 28 '13 at 16:22

2 Answers 2

up vote 14 down vote accepted

Hint: Let $b_n=\frac1{a_n}$. Then $b_1=1$, $b_2=\frac19$, and $b_{n+2}=6b_{n+1}-9b_n$ for every $n\geqslant1$. This affine recursion of order $2$ has a double root at $__$ hence $b_n=$ $(An+B)\cdot$ $__$ ${}^n$ for every $n$. Identifying $A$ and $B$ yields finally $a_n=$ $______$ for every $n\geqslant1$.

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After seeing this hint I can see why you would try this. I suppose the key is that the right hand side is of the form $$\frac{ab}{k_1a + k_2b}$$ which doesn't lend itself to simplification. However, if you consider its reciprocal you have $$\frac{k_1a+k_2b}{ab} = \frac{k_1a}{ab} + \frac{k_2b}{ab} = k_1\frac{1}{b} + k_2\frac{1}{a}$$ which suggests the aforementioned substitution. –  Michael Albanese Apr 28 '13 at 13:11

As in Did's answer, but simpler to handle: $$ b_{n + 2} = 6 b_{n + 1} - 9 b_n \qquad b_1 = 1, b_2 = 1/9 $$ Define $B(z) = \sum_{n \ge 0} b_{n + 1} z^n$. By the properties of generating functions: $$ \frac{B(z) - b_1 - b_2 z}{z^2} = 6 \frac{B(z) - b_1}{z} - 9 B(z) $$ From here: $$ B(z) = \frac{9 - 53 z}{9 - 54 z + 81 z^2} = \frac{53}{27} \frac{1}{1 - 3 z} - \frac{26}{27} \frac{1}{(1 - 3 z)^2} $$ By the expansion (a form of the binomial theorem): $$ (1 - u)^{-m} = \sum_{n \ge 0} \binom{n + m - 1}{m - 1} z^n $$ we get: $$ b_{n + 1} = \frac{53}{27} 3^n - \frac{26}{27} \binom{n + 1}{1} 3^n = (27 - 26 n) \cdot 3^{n - 3} $$ so that finally: $$ a_n = \frac{1}{(53 - 26 n) \cdot 3^{n - 4}} $$

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