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What would be the best way in finding a general term $a_n, n>3$ for the recursive sequence?

$$a_n = \dfrac{6a_{n-1}^2a_{n-3} -8 a_{n-1}a_{n-2}^2}{a_{n-2}a_{n-3}}$$

where $a_1 = 1 ; a_2 = 2 ; a_3 = 24$ ;

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  • $\begingroup$ @Krish He wants general term in terms of $n$ $\endgroup$ – Dheeraj Kumar Dec 30 '14 at 12:15
  • $\begingroup$ such a term can not ever be found $\endgroup$ – Dr. Sonnhard Graubner Dec 30 '14 at 12:30
  • $\begingroup$ @Dr.SonnhardGraubner see my answer, such term could be found $\endgroup$ – RE60K Dec 30 '14 at 12:35
  • $\begingroup$ yes indeed veryfy your formula $\endgroup$ – Dr. Sonnhard Graubner Dec 30 '14 at 13:03
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Given the recursive sequence:

$$a_n = \dfrac{6a_{n-1}^2a_{n-3} -8 a_{n-1}a_{n-2}^2}{a_{n-2}a_{n-3}}\tag{E}$$

$(\bf E)$ can be rearranged to: $$\frac{a_n}{a_{n-1}}=6\frac{a_{n-1}}{a_{n-2}}-8\frac{a_{n-2}}{a_{n-3}}\tag{1}$$ Seeing a pattern in $(1)$ let a new term $t_n$ be: $$t_n=\frac{a_n}{a_{n-1}}\tag{2}$$ Now $t_n$ follows the recursive sequence: $$t_n=6t_{n-1}-8t_{n-2}\tag{3}$$ The characteristic equation of $t_n$ is: $$r^2=6r-8\iff r^2-6r+8=0\implies r=2,4\tag{4}$$ So $t_n$ is given by the relation: $$t_n=a2^n+b4^n\tag{5}$$ Also we know that $t_2=2,t_3=12$. So after substituting: $$\begin{align}4a+16b=2\quad\mid&\quad8a+64b=12\tag{6}\\ a=-0.5\quad\mid&\quad b=0.25\tag{7}\end{align}$$ Putting back values of $a$ and $b$: $$t_n=4^{n-1}-2^{n-1}\tag{8}$$ Now from $(2)$ and $(8)$: $$a_n=(4^{n-1}-2^{n-1})a_{n-1}\tag{9}$$ Following this pattern: $$\begin{align}a_n&=(4^{n-1}-2^{n-1})(4^{n-2}-2^{n-2})...(4^{2-1}-2^{2-1})\underbrace{a_1}_{a_1=1}\tag{10}\\a_n&=\left(\prod_{k=1}^{n-1}(4^{k}-2^{k})\right)\tag{11;$n\ge2$}\\\end{align}$$

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  • $\begingroup$ Very nice solution. Thank you. +1 $\endgroup$ – Bumblebee Dec 30 '14 at 12:34
  • $\begingroup$ @Nilan is it whole correct, so you find any errors? $\endgroup$ – RE60K Dec 30 '14 at 12:36
  • $\begingroup$ I did not understand, haw did you get the last line? $\endgroup$ – Bumblebee Dec 30 '14 at 12:37
  • $\begingroup$ @Nilan will edit $\endgroup$ – RE60K Dec 30 '14 at 12:37
  • $\begingroup$ can you prove this by induction or so? $\endgroup$ – Dr. Sonnhard Graubner Dec 30 '14 at 13:02

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