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?
$\endgroup$ 21 Answer
$\begingroup$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,
$$a_{n}=5^{n-1}(a_1+2.5)-2.5$$
$\endgroup$ 4
