Seductive Recursive Sequences

It's easy to show using number base expansion that, in general, for a recursively defined sequence $a_n=xa_{n-1}+1$ , $a_n=\frac{x^n-1}{x-1}$ . In this case, $\alpha_n=2^n-1$ and $\beta_n=\frac{8^n-1}{7}$ . Because $\frac{2016}{672} = 3$ , the fraction becomes $\frac{\alpha_{3n}}{\beta_n}=\frac{2^{3n}-1}{\frac{8^n-1}{7}}=7\frac{8^n-1}{8^n-1}=7$ .

Literally my first post, so go easy on me. I'm also not much of a trained mathematician, but I've experience in doing contest questions.

$\alpha_{N} = \sum_{k = 0}^{N - 1} 2^{k} = 2^{N} - 1$

$\beta_{J} = \sum_{k = 0}^{J - 1} 8^{k} = 8^{J} - 1 = 2^{3 * J} - 1$ $$

$\implies \frac{\alpha_{N}}{\beta{J}} = \frac{7 * (2^{N} - 1)}{2^{3 * J} - 1}$ $$

For $N = 2016$ and $J = 672 \implies 3 * J = 2016 \implies$ $$

$\frac{\alpha_{2016}}{\beta_{672}} = \boxed{7}$ .