Question

How would I write this as a geometric recursive and explicit formula? Algebra 2

Answer 1

Denote the sequence by $a_n$ with $n\ge1$.

Recursively:

The first number is $a_1=8$, and each successive number is twice the previous one. So $a_2=16$, $a_3=32$, $a_4=64$, ... .

In other words,

$a_2=2a_1$
$a_3=2a_2$
$a_4=2a_3$

and so in general,

$a_n=2a_{n-1}$

Explicitly:

We can arrive at this formula by, in a way, working backwards. If $a_n=2a_{n-1}$, that means $a_{n-1}=2a_{n-2}$, and so $a_n=2(2a_{n-2})=2^2a_{n-2}$, and so on. We would end up with

$a_n=2a_{n-1}=2^2a_{n-2}=2^3a_{n-3}=\cdots=2^{n-2}a_2=2^{n-1}a_1$

(You'll notice a pattern here: the exponent of the 2 and the index of the earlier term in the sequence add up to $n$. For example, $2^1a_{n-1}\implies 1+(n-1)=n$.)

or simply

$a_n=2^{n-1}\cdot8$

Then the 15th term is obtained immediately by evaluating this rule at $n=15$. We get

$a_{15}=2^{15-1}\cdot8=2^{14}\cdot2^3=2^{17}=131,072$