finding a 'recursive formula' with first 4 #'s in sequence?

[katie2012]

Hello,

The question says "Which of the following is a recursive formula for the sequence whose first four terms are 4, 2, –4, and –22? "

And then it's multiple choice:
A) a^n = 3a[sub:1g6wpz8z]n – 1[/sub:1g6wpz8z] – 1
B) a^n = 3a[sub:1g6wpz8z]n – 1[/sub:1g6wpz8z] – 10
C) a^n = -3a[sub:1g6wpz8z]n-1[/sub:1g6wpz8z]+1
D) a^n = -3a[sub:1g6wpz8z]n-1[/sub:1g6wpz8z]-10

I'm not really sure what it is I should do to solve this because in the past I've just had to solve for the first x terms of the sequence, not find the formula itself. So... do I just plug in the first term (4) into each formula until it works out to 2 for the second term, -4 for the third, and -22 for the fourth? I already tried plugging it in but I got kind of confused...

Any help would be appreciated!

 

[Denis]

Re: finding a 'recursive formula' with first 4 #'s in sequen

 4   2   -4    -22   -76 .......
  -2  -6   -18  -54 ......

Next term is -76 : ya'll ok now?

 

[katie2012]

Re: finding a 'recursive formula' with first 4 #'s in sequen

 4   2   -4    -22   -76 .......
  -2  -6   -18  -54 ......

Next term is -76 : ya'll ok now?

Yeah I kind of figured it out on my own actually... but what I was looking for was what the actual formula was, not the next few numbers in the series...

Next time you try to help someone make sure you know what it is they actually need help with!

 

[Denis]

Re: finding a 'recursive formula' with first 4 #'s in sequen

Purpose of my post was to give you a hint (we normally don't do homework here):
the hint was showing the differences in the terms: -2 , -6 , -18 , -54 , ....
so you could see that a certain difference was 3 times the previous.

Let me know if you are still offended by that post: I will delete it.

 

[mmm4444bot]

Next time you try to help someone make sure you know what it is they actually need help with!

Are you trying to help Denis, with this statement?

Nobody here has yet been able to figure out what Denis actually needs help with. :wink:

 

[Denis]

Re:

Nobody here has yet been able to figure out what Denis actually needs help with. :wink:

Including myself!!!

 

[TchrWill]

Re: finding a 'recursive formula' with first 4 #'s in sequen

The question says "Which of the following is a recursive formula for the sequence whose first four terms are 4, 2, –4, and –22? "

And then it's multiple choice:
A) a^n = 3a[sub8xxflgd]n – 1[/sub8xxflgd] – 1
B) a^n = 3a[sub8xxflgd]n – 1[/sub8xxflgd] – 10
C) a^n = -3a[sub8xxflgd]n-1[/sub8xxflgd]+1
D) a^n = -3a[sub8xxflgd]n-1[/sub8xxflgd]-10

I'm not really sure what it is I should do to solve this because in the past I've just had to solve for the first x terms of the sequence, not find the formula itself. So... do I just plug in the first term (4) into each formula until it works out to 2 for the second term, -4 for the third, and -22 for the fourth? I already tried plugging it in but I got kind of confused...

-4, -2, 4, 22, 76, 238, 724, 2182, 6556, 19678, derives from a(n)=3*a(n-1)+10

Look in on http://oeis.org/search?q=4%2C+2%2C+-4%2 ... &go=Search

Are you sure you have your signs right?

 

[Mrspi]

Re: finding a 'recursive formula' with first 4 #'s in sequen

Hello,

The question says "Which of the following is a recursive formula for the sequence whose first four terms are 4, 2, –4, and –22? "

And then it's multiple choice:
A) a^n = 3a[sub:2hzlpntq]n – 1[/sub:2hzlpntq] – 1
B) a^n = 3a[sub:2hzlpntq]n – 1[/sub:2hzlpntq] – 10
C) a^n = -3a[sub:2hzlpntq]n-1[/sub:2hzlpntq]+1
D) a^n = -3a[sub:2hzlpntq]n-1[/sub:2hzlpntq]-10

I'm not really sure what it is I should do to solve this because in the past I've just had to solve for the first x terms of the sequence, not find the formula itself. So... do I just plug in the first term (4) into each formula until it works out to 2 for the second term, -4 for the third, and -22 for the fourth? I already tried plugging it in but I got kind of confused...

Any help would be appreciated!

Maybe I'm confused....do you really mean that the left side of each of these formulas is a[sup:2hzlpntq]n[/sup:2hzlpntq]? That's how a^n would be interpreted. Then, if a = 2, you'd have a[sup:2hzlpntq]2[/sup:2hzlpntq] = "one of the expressions on the right side" Somehow, that doesn't seem correct.

If you really meant that the left side was a[sub:2hzlpntq]n[/sub:2hzlpntq], which is the way such an exercise would NORMALLY look, and
you are given four possible answers, of which ONE is supposed to be correct, I think your method should work.

You know that the first term of the sequence is 4. That would be a[sub:2hzlpntq]1[/sub:2hzlpntq]

Let's try answer choice A. a[sub:2hzlpntq]1[/sub:2hzlpntq] = 4.
a[sub:2hzlpntq]n[/sub:2hzlpntq] = 3*a[sub:2hzlpntq]n - 1[/sub:2hzlpntq] - 1
So,
a[sub:2hzlpntq]2[/sub:2hzlpntq] = 3*a[sub:2hzlpntq]1[/sub:2hzlpntq] - 1
a[sub:2hzlpntq]2[/sub:2hzlpntq] = 3*4 - 1
a[sub:2hzlpntq]2[/sub:2hzlpntq] = 12 - 1
a[sub:2hzlpntq]2[/sub:2hzlpntq] = 11
That's not the second term we have, so we can discard answer choice A.

Do the same thing with the other possibilities, and I think you'll find which of those choices is correct.

 

[TchrWill]

Re: finding a 'recursive formula' with first 4 #'s in sequen

The question says "Which of the following is a recursive formula for the sequence whose first four terms are 4, 2, –4, and –22? "

And then it's multiple choice:
A) a^n = 3a[sub:ix4hq17e]n – 1[/sub:ix4hq17e] – 1
B) a^n = 3a[sub:ix4hq17e]n – 1[/sub:ix4hq17e] – 10
C) a^n = -3a[sub:ix4hq17e]n-1[/sub:ix4hq17e]+1
D) a^n = -3a[sub:ix4hq17e]n-1[/sub:ix4hq17e]-10

I'm not really sure what it is I should do to solve this because in the past I've just had to solve for the first x terms of the sequence, not find the formula itself. So... do I just plug in the first term (4) into each formula until it works out to 2 for the second term, -4 for the third, and -22 for the fourth? I already tried plugging it in but I got kind of confused...

Upon reexamining your problem statement, it becomes clear that it is simply the reversed sign of the sequence I mentioned earlier.

-4, -2, 4, 22, 76, 238, 724, 2182, 6556, 19678, which derives from a(n)=3a(n-1)+10 where -4 is the 0th term as shown.

n...0...1...2...3....4....5...
a..-4..-2...4..22..76..238...

Your sequence then becomes

n...0...1...2...3....4.....5...
a...4...2..-4.-22..-76..-238...

deriving from a(n) = 3a(n-1) - 10

Look in on http://oeis.org/search?q=4%2C+2%2C+-4%2 ... &go=Search again.