MATH SOLVE

4 months ago

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

Accepted Solution

A:

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

Recursively:

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,

[tex]a_2=2a_1[/tex]

[tex]a_3=2a_2[/tex]

[tex]a_4=2a_3[/tex]

and so in general,

[tex]a_n=2a_{n-1}[/tex]

Explicitly:

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

[tex]a_n=2a_{n-1}=2^2a_{n-2}=2^3a_{n-3}=\cdots=2^{n-2}a_2=2^{n-1}a_1[/tex]

(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

[tex]a_n=2^{n-1}\cdot8[/tex]

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

[tex]a_{15}=2^{15-1}\cdot8=2^{14}\cdot2^3=2^{17}=131,072[/tex]

Recursively:

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,

[tex]a_2=2a_1[/tex]

[tex]a_3=2a_2[/tex]

[tex]a_4=2a_3[/tex]

and so in general,

[tex]a_n=2a_{n-1}[/tex]

Explicitly:

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

[tex]a_n=2a_{n-1}=2^2a_{n-2}=2^3a_{n-3}=\cdots=2^{n-2}a_2=2^{n-1}a_1[/tex]

(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

[tex]a_n=2^{n-1}\cdot8[/tex]

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

[tex]a_{15}=2^{15-1}\cdot8=2^{14}\cdot2^3=2^{17}=131,072[/tex]