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

[sarah22]

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

 

[Mentallic]

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

 

[tiny-tim]

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!

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)

 

[Mentallic]

edit: broke a rule.

 

[tiny-tim]

Mentallic, please do not give complete answers!

 

[bpet]

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.

 

[sarah22]

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.

 

[tiny-tim]

Hi sarah22!

Try 1) … write out the differences and you get … ?

 

[sarah22]

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

 

[tiny-tim]

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?

take the differences ​

 

[Mentallic]

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

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..?

 

[sarah22]

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

take the differences ​

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!

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

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

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?

 

[tiny-tim]

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

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!

anyway, show us the formula, just to make sure you've got it now. ​

 

[Mentallic]

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

(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.

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