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

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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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