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## Integrated math 1

### Unit 12: 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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# Explicit & recursive formulas for geometric sequences

Sal finds an explicit formula of a geometric sequence given the first few terms of the sequences. Then he explores equivalent forms the explicit formula and finds the corresponding recursive formula.

## Want to join the conversation?

- On the practice, how do you make "n-1" into one exponent because when I try to type it all into one exponent it wont work. :((2 votes)
- On a side note: If you got a negative constant ratio, don't forget to wrap it as well.

Like this: a * (-8)^(n-1)

Or otherwise your answer might not get accepted.(2 votes)

- so if the sequence was 3,6,12 would the equation be g(22) = 3 x 2^21. The final solution should be g(22)= 3 x 2097152 which is g(22) = 6291456?(6 votes)
- This is a question,in general,How do you know when to use an Explicit or Recursive equation to solve a problem?(2 votes)
- Both equations require that you know the first term and the common ratio. Since you need the same information for both, ultimately it comes down to which formula best suits your needs.

The recursive formula requires that you know the term directly before the term you are looking to find. Therefore, if you are looking for a term that is within close proximity (ie find the 4th given the 1st) the recursive will probably be easiest and require less work.

The explicit formula will allow you to find any term without having to work through each consecutive term. This makes it particularly useful for finding terms that are a good way down the sequence (ie. find the 108th term).(2 votes)

- For one of the practice problems (Practice: Explicit formulas for geometric sequences) it says:

Haruka and Mustafa were asked to find the explicit formula for 4, 12, 36, 108

Haruka said g(n)= 4*3^n

Mustafa said g(n)= 4*4^n-1

the answer was that both of them were incorrect but I do not understand why that is the case. Could you explain how neither Mustafa or Haruka is correct?(4 votes)- "n" represents the term

Start with Haruka:

If n=1: g(1)=4*3^1 = 4*3 = 12

The first term is suppose to be 4, not 12.

Now for Mustafa:

If n=1: g(1) = 4*4^(1-1) = 4*4^0 = 4^1 = 4

so first term is ok.

If n=2: g(2) = 4*4^(2-1) = 4*4 = 16

The 2nd term is suppose to be 12, not 16.

The correct version would be: g(n)= 4*3^(n-1)

Hope this helps.(2 votes)

- shouldn't the 1/2 be in parenthesis? should read (1/2)^(n-1)?(2 votes)
- how would I solve a Geometric Series like S4/S8 = 1/17 with 3 being the first term?(3 votes)
- If the first term is 3, then the 4th term is 3r³ and the 8th is 3r⁷ for common ratio r. So

S4/S8=1/17

(3r³)/(3r⁷)=1/17

1/r⁴=1/17

r⁴=17

r=±⁴√17

So you have your first term and common ratio.(2 votes)

- What exactly is a recursive function?(2 votes)
- The recursive formula is heavier, the better is the explicit one, isn't it?(2 votes)

- Why is one half to the negative one equals 2?(3 votes)
- Negative exponents are changed to positive exponents by using the reciprocal of the base. The reciprocal of 1/2 = 2/1 = 2

Thus, (1/2)^(-1) = 2

You can find the lessons about negative exponents at: https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-numbers-operations/cc-8th-pos-neg-exponents/v/negative-exponents(2 votes)

- g(1)=168*1/2^(1-1)=168*1/2^0=168 ? 1/2^0 equals 1? I couldn't understand, please help.(3 votes)
- Anything to the 0 power is 1 except 0. If you have (1/2)^2=1/2*1/2 = 1/4. Multiply both sides by 2 to get (1/2)^1 = 1/2, do it again to get (1/2)^0 = 1.(2 votes)

- At2:55, how does Sal rewrite the g(n) equation from 168 x 1^(n-1)/2 to 168/2^n-1 ?(2 votes)
- EDIT

I neglected to mention that all of 1/2 is being brought to the n-1 power, so it's not (1^(n-1))/2

END OF EDIT

(1/2)^(n-1) =

(1^(n-1))/(2^(n-1)) =

1/(2^(n-1))

so then 168 * (1/2)^(n-1) =

168 * 1/(2^(n-1)) =

168/(2^(n-1))

I hope you were able to follow that, but if not let me know. Some hints I can offer, 1 to any power is still 1, and a fraction a/b to a power makes the power effect the numerator and denominator, so (a/b)^x = (a^x)/(b^x).

And again, if anything doesn't make sense let me know.(2 votes)

## Video transcript

- [Voiceover] So, this table here where you're given a bunch of Ns, N equals one, two, three, four, and we get the corresponding G of N. And one way to think about
it is that this function, G, defines a sequence where N
is the term of the sequence. So for example, we could
say this is the same thing as the sequence where
the first term is 168, second term is 84, third term is 42, and fourth term is 21,
and we keep going on, and on, and on. Now, let's think about what
type of a sequence this is. If we think of it as starting at 168, and how do we go from 168 to 84? Well, one way, you could
say we subtract at 84, but another way to think about it is you multiply it by one half. So, times one half. And then to go from 84 to 42, you multiply by one half again. Times one half. And to go from 42 to 21, you
multiply by one half again. So, this right over here
is a geometric series. We're starting at a term
and every successive term is the previous term
times, it's often called the common ratio, times one half. So, how can we write G
of N, how can we define this explicitly in terms of N? And I encourage you to pause the video and think about how to do that. So, construct a, so,
if I say G of N equals, think of a function
definition that describes what we've just seen here starting at 168, and then multiplying
by one half every time you add a new term. Well, one way to think
about it is we start at 168, and then we're gonna multiply by one half, we're gonna multiply by one
half a certain number of times. So, we could view the exponent
as the number of times we multiply by one half. And how many times are we
gonna multiply by one half? The first term, we multiply
by one half zero times. The second term, we multiply
by one half one time. Third term, we multiply
by one half two times. Fourth term, we multiply
by one half three times. So, the figure, it seems
like whatever term we're on, we're multiplying by one half,
that term minus one times. And you can see that this works. If N is equal to one, you're going to have one minus one, that's just gonna be zero. One half to the zero's just one. So, you're just gonna get a 168. If N is two, well, two minus one, you're gonna multiply
by one half one time, which you see right over here, N is three, you're gonna multiply by one half twice. Three minus two is, or,
three minus one is two. You're gonna multiply by one half twice, and you see that right over there. So, this feels like a really
nice explicit definition for this geometric series. And you can think of it in other ways, you could write this
as G of N is equal to, let's see, one way you could write it, as, you could write it as 168,
and I'm just algebraically manipulating it over
two to the N minus one. Another way you could think about it is, well, let's use our exponent
properties a little bit, we could say G of N is
equal to, let's see, one half to the N minus
one, that's the same thing as one half, let me write this. It's equal to 168. Lemme do this in a different color. So, this part right over
here is the same thing as one half to the N. So, times one half to
the N, times one half to the negative one. One half to the negative one. Well, one half to the negative one is just two, is just two, so, this is times two. So, we could rewrite this whole thing as 168 times two is what? 336? 336, did I do that right? 160 times two would be 320, plus 16, two times eight, so yeah, 336. And then times one half to the N. Times one half to the N. So, these are equivalent statements. This one makes a little
bit more intuitive sense, it kinda jumps out at you,
you're starting at 168 and you're multiplying by one half. Whatever term you are minus one times. But this is algebraically
equivalent to this, to our original one. But, can we also define
G of N recursively? And I encourage you to pause
the video and try to do that. In a lot of ways, the recursive definition is a little bit more straight
forward, so let's do that. G, well, I'll make the
recursive function a different, well, I got, I'll stick
with G of N since it's on this table right over here. G of N is equal to, and so, let's see, if we're going to, when N equals one, if N is equal to one,
we're starting at 168. 168, and if N is greater than one and a whole number, so, if N, so, we're, this is gonna be defined
over all positive integers, and whole number, what are we gonna do? Well, we're gonna take
one half and multiply it times the previous term. So, it's gonna be one half
times G of N minus one. And you can verify that this works. If N is equal to one, we
just go right over here, it's gonna be 168. G of two is gonna be
one half times G of one, which is, of course, 168. so, 168 times one half is 84. G of three is gonna be
one half times G of two, which it is, G of three is
one half times G of two. So, this is how we would define, this is the explicit
definition of this sequence, this is a recursive function
to define this sequence.