Technique in Sequence: How Can I Use Formulas to Find the Next Terms?

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The discussion focuses on finding formulas for various numerical sequences. Participants suggest methods such as calculating differences and ratios between terms to identify patterns. One user successfully derives formulas for some sequences, while others express frustration with their understanding and seek further clarification. Techniques like the Lagrange interpolating polynomial and adjusting exponents are discussed as potential solutions. Overall, the thread emphasizes collaborative problem-solving in sequence analysis.
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Can anyone help me how to get the formula fast?

Example:
1) 2, 4, 8, 16 n=1
Answer: 2^n

Problem:
1) -2, 2, 22, 122... n=0
2) 8, 19, 41, 85, 173, 349... n=1
3) -1/3, 1/2, -3/4... n=1
 
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I am really inexperienced at sequences, especially when it comes to finding formulas; so please excuse my help as it could be a very unconventional way to approach the problems.

1) I don't know

2) I noticed that the difference between each term started with 11, and doubled each time so I assumed the formula would be in the form of y=a.2^{x+b}+c and after some guessing, I let a=11 because the first difference was 11... edit: broke a rule.
3) The sequence is being multiplied by \frac{-3}{2} since \frac{S_2}{S_1}=\frac{S_3}{S_2}=\frac{-3}{2} but I'm unsure where to go from there :blushing:
 
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sarah22 said:
Can anyone help me how to get the formula fast?

Problem:
1) -2, 2, 22, 122...
2) 8, 19, 41, 85, 173, 349...
3) -1/3, 1/2, -3/4...

Hi sarah22! :smile:

In problems like this, the methods that usually work are to write the numbers on one line, and either the differences or the ratios between successive numbers on their next line …

then do it again and again until you get a line that's easy (like 2 2 2 2… or 1 3 5 7 …) …

in at least one of your three problems, you'll have to use both differences and ratios (on different lines, of course) :wink:
 
edit: broke a rule.
 
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Mentallic, please do not give complete answers!
 
sarah22 said:
Can anyone help me how to get the formula fast?

Problem:
1) -2, 2, 22, 122...
2) 8, 19, 41, 85, 173, 349...
3) -1/3, 1/2, -3/4...

If the formula is a polynomial, try the Lagrange interpolating polynomial.
 
Ok I got number 2 and some in 3. I'm sorry I didn't put where the n starts. I'll edit my first post.

2) (11)(2^n) -3
3) (-1/3)(-3/2)^n if n = 0. but it starts with 1 so how can I do it?

I still don't get the other number. OMG. I'm doom.

@Mentallic
Ok I got some in number 3.

@tiny-tim
Yeah I'm doing that but still I don't get number 1 and 3.

@bpet
Can you teach how is it done?

Thanks everyone for helping me. :wink:
 
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Hi sarah22! :wink:

Try 1) … write out the differences and you get … ? :smile:
 
Ok here is it. I still don't get it. Oh no! I don't want to fail.

0 -(2) ?(n) n=0
4 -(2) 4(n) n=1
24 -(2) 12(n) n=2
144-(2) 48(n) n=3
 
  • #10
sarah22 said:
0 -(2) ?(n) n=0
4 -(2) 4(n) n=1
24 -(2) 12(n) n=2
144-(2) 48(n) n=3

eugh! you're adding! why are you adding? :cry:

take the differences :smile:
 
  • #11
sarah22 said:
3) (-1/3)(-3/2)^n if n = 0. but it starts with 1 so how can I do it?
You want to make the first term to the power of 0, (-1/3)(-3/2)^0

then the next term to the power of 1, (-1/3)(-3/2)^1

but for the first term n=1 and for the second, n=2.
Fiddle with the exponent a bit, I'm sure you'll get it :wink:

For #3, when you take the differences between the terms, notice the rate at which the numbers are rising. Is it by a multiple, a ratio, an exponential..?
 
  • #12
tiny-tim said:
eugh! you're adding! why are you adding? :cry:

take the differences :smile:

Ok, I got a technique from my friend. I'll just subtract the first term from the next then see how is it made.

-2, 2, 22, 122... n=0

2-2, 22-2, 122-22
0, 20, 100So on each term it is increase 5 times the previous.

(5^n)
1, 5, 25, 125

so to make the formula correct. You need to minus it by 3. that's it.

My professor didn't teach most of this technique. I really hate her!

Mentallic said:
You want to make the first term to the power of 0, (-1/3)(-3/2)^0

then the next term to the power of 1, (-1/3)(-3/2)^1

but for the first term n=1 and for the second, n=2.
Fiddle with the exponent a bit, I'm sure you'll get it :wink:

Yeah, I'm a bit crazy the time I posted it. lol. You just need to subtract n in 1.

(-1/3)(-3/2)^n-1

Mentallic said:
For #3, when you take the differences between the terms, notice the rate at which the numbers are rising. Is it by a multiple, a ratio, an exponential..?

I dunno. Is it exponential because I raise it by n?
 
  • #13
sarah22 said:
Ok, I got a technique from my friend. I'll just subtract the first term from the next then see how is it made.

-2, 2, 22, 122... n=0

2-2, 22-2, 122-22
0, 20, 100

did you do this yourself, or just copy it? either way you got it wrong :redface:

it's 2 - (-2) = 4, not 0 …

and how did you get
So on each term it is increase 5 times the previous.
from 0, 20, 100 ?
so to make the formula correct. You need to minus it by 3. that's it.

uhhh? where does the 3 come from? and what do you mean anyway?
My professor didn't teach most of this technique. I really hate her!

tush, sarah22 … that's no way to talk about your nice teacher! :rolleyes:

anyway, show us the formula, just to make sure you've got it now. :smile:
 
  • #14
sarah22 said:
Ok, I got a technique from my friend. I'll just subtract the first term from the next then see how is it made.

-2, 2, 22, 122... n=0

2-2, 22-2, 122-22
0, 20, 100


So on each term it is increase 5 times the previous.
So even though you got the difference between the first and second term wrong, you found the rate of increase correctly. Handy :smile:


sarah22 said:
(5^n)
1, 5, 25, 125

so to make the formula correct. You need to minus it by 3. that's it.
that's it? nearly. Test your formula S_n=5^n-3 and you'll see what you're missing.



sarah22 said:
Yeah, I'm a bit crazy the time I posted it. lol. You just need to subtract n in 1.

(-1/3)(-3/2)^n-1



I dunno. Is it exponential because I raise it by n?
Yes :smile:
 

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