Technique in Sequence.?

1. Mar 8, 2009

sarah22

Can anyone help me how to get the formula fast?

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

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

Last edited: Mar 8, 2009
2. Mar 8, 2009

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

Last edited: Mar 8, 2009
3. Mar 8, 2009

tiny-tim

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)

4. Mar 8, 2009

Mentallic

edit: broke a rule.

Last edited: Mar 8, 2009
5. Mar 8, 2009

tiny-tim

Mentallic, please do not give complete answers!!

6. Mar 8, 2009

bpet

If the formula is a polynomial, try the Lagrange interpolating polynomial.

7. Mar 8, 2009

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.

Last edited: Mar 8, 2009
8. Mar 8, 2009

tiny-tim

Hi sarah22!

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

9. Mar 8, 2009

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

10. Mar 8, 2009

tiny-tim

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

take the differences

11. Mar 9, 2009

Mentallic

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..?

12. Mar 9, 2009

sarah22

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.

(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!

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?

13. Mar 9, 2009

tiny-tim

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
from 0, 20, 100 ?
uhhh? where does the 3 come from? and what do you mean anyway?
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.

14. Mar 10, 2009

Mentallic

So even though you got the difference between the first and second term wrong, you found the rate of increase correctly. Handy

that's it? nearly. Test your formula $$S_n=5^n-3$$ and you'll see what you're missing.

Yes