Finding a closed form formula for a recursive sequence.

Question

$a_0 = 1, a_1 = 3$ and

$a_n = 2a_{n-1} + a_{n-2}$

How do I obtain a closed form for this sequence? We have learned about the method of characteristic roots, but I am unsure how to take that and turn this into a closed form.

Answer 1

Take $\lambda^2-2\lambda-1=0$ and find values of $C_1$ and $C_2$, for which $$a_n=C_1\lambda_1^n+C_2\lambda_2^n.$$ Use $a_0=1$ and $a_1=3$.

Good luck!

Answer 2

This recursion formula is the Pell-Lucas sequence, albeit with different initial conditions. It is in the family of generalized Fibonacci sequences, such as $f_n=af_{n-1}+bf_{n-2}$ with arbitrary $f_0$ and $f_1$. Using the methods described here, it is relatively straightforward to show that

$$ \begin{align}f_n &=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{af_0}{2} \frac{\alpha^n+\beta^n}{\alpha+\beta}\\ &=\frac{\left( {{f}_{1}}-{{f}_{0}}\beta \right)\,{{\alpha }^{n}}-\left( {{f}_{1}}-{{f}_{0}}\alpha\right)\,{{\beta }^{n}}}{\alpha -\beta}\\ &=f_1F_n+bf_0F_{n-1} \end{align}$$

where

$$ \alpha,\beta=(a\pm\sqrt{a^2+4b})/2\\ F_n=\frac{\alpha^n-\beta^n}{\alpha-\beta} $$

Specializing to the present case, we find that

$$ \alpha,\beta=\delta,\delta^*=1\pm\sqrt{2}\\ $$

where $\delta$ is the the so-called silver ratio and $\delta^*$ denotes the conjugate. This is consistent with the Pell-Lucas solution. As a side note, the initial conditions for Pell-Lucas are $f_0=f_1=2$ and the solution reduces to the simple Binet-type form $f_n=\delta^n+{\delta^*}^n$.

You have a number of choices as to how to express the solution.