0
$\begingroup$

I recall learning in school how to convert arithmetic and geometric sequence formulas between recursive and explicit, but I don't remember learning a systematic method to approach it. For example, let $a_n=2a_{n-1}+2$ where $a_0=5$. How would I convert this to an explicit form defined in terms of $n$? Beyond this specific example, is there a generalized method to do so?

$\endgroup$
1
$\begingroup$

There is a systematic way of solving something like $a_n=2a_{n-1}+2a_{n-2}$, which is called a linear recurrence relation. For example this pdf shows you how to do it.

But for your specific example you have a constant $2$, so it's not a linear recurrence. However you can rewrite it as $a_n+2=2(a_{n-1}+2),a_0+2=7$ where it's obvious that $a_n+2=7\cdot 2^n$ so $a_n=7\cdot 2^n-2$.

For a completely general recurrence there isn't really a systematic way to convert it to a formula, but you can always try tricks like the above (linear recurrence, substitution) to see if they work.

$\endgroup$
1
$\begingroup$

It's equivalent to $\left(~\mbox{with}\ a_{0} = 5~\right)$: \begin{align} a_{n} + 2 & = 2\left(a_{n - 1} + 2\right) = 2^{2}\left(a_{n - 2} + 2\right) = 2^{3}\left(a_{n - 3} + 2\right) = \cdots = 2^{n}\left(a_{0} + 2\right) = 7 \times 2^{n} \\[5mm] \implies & \bbox[15px,border:1px solid navy]{a_{n} = 7 \times 2^{n} - 2} \end{align}

$\endgroup$
1
$\begingroup$

Leu us consider the general case of $$a_n=\alpha\, a_{n-1}+\beta$$ Let first $a_n=b_n+\gamma$ which makes $$b_n+\gamma=\alpha\,(b_{n-1}+\gamma)+\beta\implies b_n=\alpha\,b_{n-1}+(\alpha\gamma+\beta-\gamma)$$ To come back to something simple, let (assuming $\alpha\neq 1$) $$\alpha\gamma+\beta-\gamma=0 \implies \gamma=\frac \beta {1-\alpha}$$ So, we are left with $$ b_n=\alpha\,b_{n-1} \implies b_n=c \,\alpha^{n-1}\implies a_n=c\, \alpha^{n-1}+\frac \beta {1-\alpha}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.