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## Algebra 1

### Course: Algebra 1 > Unit 9

Lesson 4: Constructing geometric sequences- Explicit & recursive formulas for geometric sequences
- Recursive formulas for geometric sequences
- Explicit formulas for geometric sequences
- Converting recursive & explicit forms of geometric sequences
- Converting recursive & explicit forms of geometric sequences
- Geometric sequences review

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# Converting recursive & explicit forms of geometric sequences

CCSS.Math:

Sal solves the following problem: The explicit formula of a geometric sequence is g(x)=9*8^(x-1). Find the recursive formula of the sequence. Created by Sal Khan.

## Want to join the conversation?

- At0:25, why does x have to be a positive integer?(10 votes)
- This is just the definition of the function. While you can still figure out the result of g(x) for non-positive integers, the function is not defined for these values. Imagine the function g(x) is used to determine how many grams of soap you would need to clean x plates. You could still determine the result of the function for fractional or negative or complex values of x, but these values wouldn't be within the domain of the problem, since you can't have "-3.4 plates".(28 votes)

- So then how do you convert a recursive formula (with the information of the first g(x) number) into an explicit formula?(3 votes)
- Good question!

Well, the key pieces of information in both the explicit and recursive formulas are**the first term of the sequence**and**the constant amount that you change the terms by, aka the common ratio**(notice: the name "common ratio" is specific to geometric sequences, the name that applies to arithmetic seq. is "common difference") .

For example, you have the recursive formula:`g(1)=`

**9**`g(x)=g(x-1)*(`

**8**)

9 is the first term of the sequence, and 8 is the common ratio.

An explicit formula is structured as:`g(x)=(1st term of seq.)*(common ratio)^(x-1)`

Substitute for the variables, and you get the explicit formula:`g(x)= 9*8^(x-1)`

And there you have it!

Feel free to ask any more questions if you would like me to clarify my answer. ^-^

*Edit: I forgot to mention that the first term of a sequence is called the initial value.(7 votes)

- What are some practical applications that we could use recursive formula for instead of explicit formula?(2 votes)
- One example is the Fibonacci sequence which has to define n1 and n2 to complete the sequence.(8 votes)

- Why would one use the recursive? The explicit is so much easier and straightforward.(3 votes)
- Sometimes the recursive formula does a better job conveying how the sequence behaves. Take the Fibonacci sequence:

a(1)=0

a(2)=1

a(n+1)=a(n-1)+a(n) for all n>1

To get the next term, add up the previous two terms. This becomes {0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55...}(3 votes)

- Where can I find the formal syntax/grammar rules for defining these functions/sets/sequences? Sal gives several versions, which I love, but some seem more math-like than others. I'm new to this and I suspect that this shows up a lot in math.(3 votes)
- IF you are still interested, search google or other math sites to help bring you up to speed here.(1 vote)

- when would you ever write or use a formula in its recursive form? I mean, by saying f(x) = f(x-1), you just add that -1 back by performing an operation on it, for e.g, *8, isn't it easier just to say that f(x) = f(x), which is already self evident?(1 vote)
- Yes, I believe what you are saying is correct. I believe Sal (and most every math teacher) is identifying the function. "Function" is used interchangeably with "f(x)." It's like saying: "The function is f(x-1)" or "f(x) is f(x-1)." Kinda like how we say "PIN number" which literally means "personal identification number number." but hearing the word "number" is actually helpful because it helps us recognize the term faster.(5 votes)

- On the practise, KA will not allow me to input
*to the power of n - 1*Is there any work around because I am very bad at converting it?!(2 votes)- 2^n-1

is the same as

2^n*2^-1

therefore equals

2^n*1/2(2 votes)

- Hey guys, I'm SUPER confused by these kind of problems:

{*f(1)=4*

{*f(n)=f(n−1)⋅(−0.5)***Find an explicit formula for***f(n).**f(n)*= ?

In these problems, when you're given the answer/hints, the format is change to:

"*f(n)=4⋅0.5^(n-1)*"**WITH NO EXPLANATION WHY**!! Can someone please help me understand how to solve these properly 🙏🏻(2 votes)- Hi. Okay. So lets start with the problem you gave. f(1) =4 and f (n)=f(n-1)*(-0.5). If someone were to ask you "what is f(50). So plug in f(50) for the formula. you would get f(50)=f(50-1)*(-0.5). Which when you work it out is f(50)=f(49)*(-0.5). Well then you need to know what f(49) is. So you would have to find f(49) So you plug it in. f(49)=f(49-1)*(-.05) which when you work it out is f(49)=f(48)*(-.05). So now you need to know what f(48) is. This is why it is called "recursive", the answer needs you to get the answer of the previous term to get your final answer. You would have to get 47, then 46, then 45 until you got all the way back to 1. f(1)=4. so to get f(2) you would plug in f(2)=f(2-1)*(-0.5) which when you work out gives you f(2)=f(1)*(-0.5). since you know f(1)=4 because it was given to you , you can plug in the numbers. f(2)=4*(-0.5) = f(2)=-2. then you could use f(2) to work out f(3) till you got to f(50). But if you notice to get f(2) you had to multiply f(1) * -0.5 one time. to get f(3) you had to multiply f(1) * -0.5 two times. to get f(4) you had to multiply f(1) * -0.5 three times. So now you should see a pattern that whatever n equals you have to multiply f(1) n-1 times -0.5 to get the answer. so like f(50) , that means you have to multiply f(1) * -0.5 49 times. Since f(1) is always 4 this means you can say 4 * -0.5 (whatever your n is , minus one, and you have to multiply times 0.5 this many times. So in mathematical terms this means 4 * (-0.5)^n-1 times. I know that is a mouthful, but it is about the best way I can explain it. Feel free to ask any questions for further explanation. :)(1 vote)

- I was inspired by these formulas to make an explicit and recursive formula for the remaining debt after payment per term on a mortgage, with a fixed payment per term plan. It seems I've managed to make a recursive one that works:

n = term

i = interest

p = terms per interest period (i.e. 12 terms per year)

R = remaining debt

a = original amount of the debt

m = monthly payment

Recursive formula R(n):

R(1) = a

R(n) = R(n-1)*(i/p+1)–m

I've been trying for over an hour now though, to convert this into an explicit one, with no luck. It seems that for a formula on the given issue to work, it is necessary to refer back to the previous term, and that you can't make explicit formulas for any sequence, even if you can make a recursive one. It would make sense, as I can't really see the value of recursive formulas if explicit ones always work. Explicit ones are a lot simpler and faster. Would I be right to make these conclusions?

Also interested to hear any feedback on the recursive formula. Does it check out? Could it be simpler?(2 votes) - I'm pretty sure this is beyond the scope of this lesson, but if I have the following recursion:

$C_n=6n*C_{n-1}$, is there any way to solve the recursion and solve for C_n explicitly?(2 votes)- when given a recursive formula, you must always be given at least one term in order to solve for it.

Let's say you are given C_2 = 12

Substitute 2 for n and you will find that:

C_2 = 6(2) * C_1

You know C_2, so you can solve for C1:

12 = 12 * C_1, and C_1 = 1

Once you have a term,*then*you can write an explicit formula:

C_n = 6n(1 vote)

## Video transcript

So I have the function g
of x is equal to 9 times 8 to the x minus 1 power. And it's defined for
x being a positive-- or if x is a positive-- integer. If x is a positive integer. So we could say the
domain of this function, or all the valid inputs
here are positive integers. So 1, 2, 3, 4, 5,
on and on and on. So this is an explicitly
defined function. What I now want to do is to
write a recursive definition of this exact same function. That given an x, it'll give
the exact same outputs. So let's first just try
to understand the inputs and outputs here. So let's make a little table. Let's make a table here. And let's think
about what happens when we put in various x's
into this function definition. So the domain is
positive integers. So let's try a couple of them. 1, 2, 3, 4. And then see what the
corresponding g of x is. g of x. So when x is equal
to 1, g of x is 9 times 8 to the
1 minus 1 power, 9 times 8 to the 0
power, or 9 times 1. So g of x is going to be just 9. When x is 2, what's
going to happen? It'll be 9 times 8
to the 2 minus 1. So that's the same thing as
9 times 8 to the 1st power. And that's just going
to be 9 times 8. So that is 72. Actually let me just
write it that way. Let me write it
as just 9 times 8. 9 times 8. Then when x is equal to
3, what's going on here? Well this is going
to be 3 minus 1 is 2. So it's going to be 8 squared. So it's going to be
9 times 8 squared. So we could write that
as 9 times 8 times 8. I think you see a little
bit of a pattern forming. When x is 4, this is going to be
8 to the 4 minus 1 power, or 9 to the 3rd power. So that's 9 times
8 times 8 times 8. So this gives a
good clue about how we would define
this recursively. Notice, if our first term,
when x equals 1 is 9, every term after that is 8
times the preceding term. Is 8 times the preceding term. 8 times the preceding term. 8 times the preceding term. So let's define that as
a recursive function. So first define our base case. So we could say
g of x-- and I'll do this is a new color
because I'm overusing the red. I like the blue. g of x. Well we can define
our base case. It's going to be equal
to 9 if x is equal to 1. g of x equals 9 if x equals 1. So that took care of
that right over there. And then if it
equals anything else it equals the previous g of x. So if we're looking at--
let's go all the way down to x minus 1, and then an x. So if this entry right over
here is g of x minus 1, however many times
you multiply the 8s and we have a 9 in front,
so this is g of x minus 1. We know that g of x-- we know
that this one right over here is going to be the previous
entry, g of x minus 1. The previous entry times 8. So we could write
that right here. Times 8. So for any other
x other than 1, g of x is equal to the previous
entry-- so it's g of-- I'll do that in a blue color--
g of x minus 1 times 8. If x is greater than 1, or
x is integer greater than 1. Now let's verify that
this actually works. So let's draw
another table here. So once again, we're
going to have x and we're going to have g of x. But this time we're going to
use this recursive definition for g of x. And the reason
why it's recursive is it's referring to itself. In its own definition, it's
saying hey, g of x, well if x doesn't equal 1 it's going
to be g of x minus 1. It's using the function itself. But we'll see that it
actually does work out. So let's see... When x is equal to 1, so g
of 1-- well if x equals 1, it's equal to 9. It's equal to 9. So that was pretty
straightforward. What happens when x equals 2? Well when x equals 2, this
case doesn't apply anymore. We go down to this case. So when x is equal
to 2 it's going to be equivalent
to g of 2 minus 1. Let me write this down. It's going to be equivalent
to g of 2 minus 1 times 8, which is the same thing
as g of 1 times 8. And what's g of 1? Well g of 1 is right over here. g of 1 is 9. So this is going to
be equal to 9 times 8. Exactly what we got over here. And of course this was
equivalent to g of 2. So let me write this. This is g of 2. Let me scroll over
a little bit so I don't get all scrunched up. So now let's go to 3. Let's go to 3. And right now I'll
write g of 3 first. So g of 3 is equal
to-- we're going to this case-- it's equal
to g of 3 minus 1 times 8. So that's equal
to g of 2 times 8. Well what's g of 2? Well g of 2, we already
figured out is 9 times 8. So it's equal to 9 times 8--
that's g of 2-- times 8 again. And so you see we get
the exact same results. So this is the recursive
definition of this function.