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

Accepted Solution

Denote the sequence by [tex]a_n[/tex] with [tex]n\ge1[/tex].


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

In other words,


and so in general,



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


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

or simply


Then the 15th term is obtained immediately by evaluating this rule at [tex]n=15[/tex]. We get