0
$\begingroup$

I recently came across this interesting problem, where the progression was neither arithmetic nor geometric. The problem asks for an explicit formula for the following recursive formula:

$a_{1} = 2$

$a_{n} = 5(a_{n-1}+2)$

So the first five terms are $2, 20, 110, 560, 2810$.

I tried to distribute the $5$ in the second equation to get $5a_{n-1} + 10$ and then tried to apply the formula for a geometric sequence, which is $a_{1}(r)^{n-1}$, but got only $2(5)^{n-1}$, which clearly doesn't work for the sequence.

I also tried using finite differences on the first five terms to see if it was just a polynomial rule, but that didn't work.

How would you go about doing a problem like this?

$\endgroup$
  • $\begingroup$ Hint: Put $b_n=a_n+5/2$. What is the relation between $b_n$ and $b_{n-1}$ ? $\endgroup$ – Kelenner Apr 1 '17 at 16:40
  • $\begingroup$ There is a 5:1 ratio! Wow, that actually helped a lot and I solved the problem! However, I don't really get how you came up with the second sequence in the first place. Where can I learn and practice this kind of problem? $\endgroup$ – Aditya R Apr 1 '17 at 17:04
0
$\begingroup$

First begin by solving, as to why should be evident latter,

$$L=5(L+2)$$

The solution is $L=-2.5$, which we call a particular solution. Then we may write,

$$-2.5=5(-2.5+2) \tag{1}$$

Recall that the what we want to solve is,

$$a_{n}=5(a_{n-1}+2) \tag{2}$$

Luckily for us by subtracting the first equation from the second we get,

$$a_{n}+2.5=5(a_{n-1}+2.5)$$

Let $b_n=a_{n}+2.5$. The recursion transforms to (a homogenous linear recurrence):

$$b_{n}=5b_{n-1}$$

Because we must multiply by five each step, the solution to this is clearly,

$$b_{n}=5^{n-1}b_1$$

But by definition $b_1=a_1+2.5$ and $a_{n}=b_{n}-2.5$. So we get,

$$a_{n}=5^{n-1}(a_1+2.5)-2.5$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Interesting. But how did you come up with the $L = 5(L+2)$ in the first place? And why did you subtract the first from the second? Sorry if this is a beginner question, if somebody could point me to some resources where I could learn this would be helpful. $\endgroup$ – Aditya R Apr 1 '17 at 16:58
  • $\begingroup$ Often the first step to solving linear non-homogenous recurrences is to find a "particular" solution to the recurrence . Without the initial condition $a_1=2$ one may checK that $a_{n}=-2.5$ satisfies the recurrence. I recommend checking this out www3.cs.stonybrook.edu/~rezaul/Fall-2012/CSE548/…. Linear homogenous recurrences a class of recurrences we are comfortable with, in general recurrences tend to be much more complicated, and perhaps hopeless math.stackexchange.com/questions/147075/… in solving. $\endgroup$ – Ahmed S. Attaalla Apr 1 '17 at 17:23
  • $\begingroup$ Thanks. I will look into that. $\endgroup$ – Aditya R Apr 1 '17 at 17:31
  • $\begingroup$ Also, I found this in an Algebra 2 course if that helps. $\endgroup$ – Aditya R Apr 1 '17 at 17:54

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.