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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?

Aditya R
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  • Hint: Put $b_n=a_n+5/2$. What is the relation between $b_n$ and $b_{n-1}$ ? – Kelenner Apr 01 '17 at 16:40
  • 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? – Aditya R Apr 01 '17 at 17:04

1 Answers1

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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,

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

Ahmed S. Attaalla
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  • 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. – Aditya R Apr 01 '17 at 16:58
  • 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 http://www3.cs.stonybrook.edu/~rezaul/Fall-2012/CSE548/CSE548-lectures-8-9.pdf. Linear homogenous recurrences a class of recurrences we are comfortable with, in general recurrences tend to be much more complicated, and perhaps hopeless http://math.stackexchange.com/questions/147075/why-is-solving-non-linear-recurrence-relations-hopeless in solving. – Ahmed S. Attaalla Apr 01 '17 at 17:23
  • Thanks. I will look into that. – Aditya R Apr 01 '17 at 17:31
  • Also, I found this in an Algebra 2 course if that helps. – Aditya R Apr 01 '17 at 17:54