Finding the closed form ...

Question

I'm currently working through my text book, but it doesn't seem to cover this concept very well. :)

Find the closed form of $a_n=3a_{n-1}$ with initial condition $a_0=2$.

Answer 1

This is a geometric sequence. The terms of your sequence will look like this:

$$a_0 = 2$$

$$a_1 = 2\times3 = 6$$

$$a_2 = 2\times3^2 = 18$$

$$a_n = 2\times3^n$$

You can see from $a_n = 3a_{n-1}$ that each term is multiplied by $3$, so our ratio is $3$. Our initial term is $2$.

Answer 2

HINT: You have:

$$\begin{align*} a_1&=3a_0=2\cdot3\\ a_2&=3a_1=3\cdot3a_0=3^2a_0=2\cdot3^2\\ a_3&=3a_2=3\cdot3^2a_0=3^3a_0=2\cdot3^3\\ a_4&=3a_3=3\cdot3^3a_0=3^4a_0=2\cdot3^4 \end{align*}$$

Given this, what do you think $a_n$ is for arbitrary $n$?