## Recommended Posts

I was arguing with my teacher today, about something more complicated than it needed to be, but i wanted to know about the more complicated version, and she did not want to tell me, so now i need to ask someone on hear! please help me...

Ok, we were working with patterns... and this was the pattern...

2,5,11,23,47,95,192,386

The Math for the pattern the easy way is...

`n2+1`

Also there is another pattern...

`n3 (when "3") doubles each time...`

How would i write that in a mathamatical formula? such as n2+1?

If you do not understand, i will try and explain more, if you need it, but please help, thank you!

##### Share on other sites

The 13 shouldn't be there. 23 I believe.

The best I can give you is n+[3(2x)],

Your x will go up once every time. Kind of difficult to explain, you would use words with that, I don't know the mathematical sentence for it though. Hope it helps. ##### Share on other sites

n3 (when "3") doubles each time...

Would be written:

n(3^n)

When you say "doubles each time" that implies an exponential function (3 to the power of n, in your example).

If n=1 then you get 1(3^1)=1(3)=3

If n=2 then you get 2(3^2)=2(9)=18

If n=3 then you get 3(3^3)=3(27)=

Hmmm... I don't know, I can't count that high. Is that what you were looking for?

##### Share on other sites

n3 (when "3") doubles each time...

Would be written:

n(3^n)

When you say "doubles each time" that implies an exponential function (3 to the power of n, in your example).

If n=1 then you get 1(3^1)=1(3)=3

If n=2 then you get 2(3^2)=2(9)=18

If n=3 then you get 3(3^3)=3(27)=

Hmmm... I don't know, I can't count that high. Is that what you were looking for?

well you cannot use the powers of 3, because you need to use odd numbers such as "6,12,18,ect..." to keep the pattern going...

and 3(27)=81 i belive...

##### Share on other sites

well you cannot use the powers of 3, because you need to use odd numbers such as "6,12,18,ect..." to keep the pattern going...

Ummm... those aren't odd numbers. I'm not sure what you want then.

##### Share on other sites

boltbait is almost right

I believe it is 2+(3^(n-1))

2,5,11,23,47,95,192,386

if n=1 --> 2+(3^0) <-> 2+0 = 2

n=2 --> 2+(3^1) <-> 2+3 = 5

n=3 --> 2+(3^2) <-> 2+9 = 11

etc.

I'm on the 9th grade too! ##### Share on other sites

well i do not want to use the -1 method... thats why i was wondering about this...

hm... lemme see, hears a better discription if its evan better? lol

n+3, but "3" needs to be "6" next and "9" after that... and so on... you know?

##### Share on other sites

Ah, yes. Now I understand what you want.

Yes, trickman is on to it. I would just use * instead of x for multiplication to be less confusing. Or, just leave it out.

n+3(n-1)

EDIT: trickman, you made a small error... remember, that any number raised to the power of 0 = 1.

##### Share on other sites

well i do not want to use the -1 method... thats why i was wondering about this...

hm... lemme see, hears a better discription if its evan better? lol

n+3, but "3" needs to be "6" next and "9" after that... and so on... you know?

Well I think the only way to do this is the n-1 method... ##### Share on other sites

Ah, yes. Now I understand what you want.

Yes, trickman is on to it. I would just use * instead of x for multiplication to be less confusing. Or, just leave it out.

n+3(n-1)

Actually it is n+3^(n-1). I corrected it. ##### Share on other sites

Like i said, i do not want to use the (-1) method, thats the easier way, i wanted to use the harder way like this...

n+3, but "3" needs to be "6" next and "9" after that... and so on... you know?

##### Share on other sites

boltbait is almost right

I believe it is 2+(3^(n-1))

2,5,11,23,47,95,192,386

if n=1 --> 2+(3^0) 2+0 = 2

Slight problem there. Remember 3^0 = 1

(Any number raised to the power of 0 = 1.)

##### Share on other sites

n+3, but "3" needs to be "6" next and "9" after that... and so on... you know?

In your first post you said that the 3 should "double" each time, but going from 6 to 9 is not doubling.

So then, you want this pattern?

n+3, where n=1

n+6, where n=2

n+9, where n=3

n+12, where n=4

n+15, where n=5

If so, your formula would be:

n+3n

or simply

4n

##### Share on other sites

Yep thats what i wanted! thanks BoltBait!!!

##### Share on other sites

• 1 year later...

You think that's hard.

Try graphing absolute inequality equations. I'm doing that in algebra II right now.

----

D                  E                  S                  T                  I                 N                  Y

##### Share on other sites

damn my dads been on the internet for a while and id of awnsered this easily i did this a few months ago...yes im in year 9 too lol and we did this now were on some of the final topics ##### Share on other sites http://www.xkcd.com ##### Share on other sites

Ooh, those are my favorite. :-)

The Doctor: There was a goblin, or a trickster, or a warrior... A nameless, terrible thing, soaked in the blood of a billion galaxies. The most feared being in all the cosmos. And nothing could stop it, or hold it, or reason with it. One day it would just drop out of the sky and tear down your world.
Amy: But how did it end up in there?
The Doctor: You know fairy tales. A good wizard tricked it.
River Song: I hate good wizards in fairy tales; they always turn out to be him.

##### Share on other sites

Ooh, those are my favorite. :-)

We talked about force the other day in Physical Science...and I thought of it.   