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What is the next term of the following sequences ?

  • $3,8,13,24,41, \ldots$ Options: $\{ 70,75,80,85 \}$.
  • $4,23,60,121, \ldots$ Options: $\{212,221,241,242 \}$.

This two question is taken from a test paper of mine, I am no able to figure out the correct answer.

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    Remember your good friend: OEIS http://oeis.org/search?q=4,23,60,121&language=english&go=Search2010-11-23
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    @Derek Jennings: Unfortunately, for the other one it finds 98, not one of the options. Maybe another argument that it isn't such a good problem.2010-11-23
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    @Ross Yes, that's so. I think these sorts of problems are a bit suspect as they can lead some students to believe that there is only one solution. Anyway, whether or not the OEIS actually turns out to be of help in a particular case, it's certainly worth a look.2010-11-23
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    What an absurd problem. Any number whatsoever would be a valid answer without a more specific description of restrictions.2010-11-23
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    It is absurd indeed; it pisses me off all the more that such things frequently crop up in "intelligence tests"...2010-11-24
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    @Debanjan: Would you describe as precisely as possible the circumstances when this *test paper* was given, and what kind of entrance it gives or refuses access to? (I read your comment on gerry's answer.)2011-05-03

5 Answers 5

1

I am adding lots of answers here, sorry. I think since they are giving different information, they are best put separately.

If you have not had experience with this, you may be interested in how people derive formulas for these problems. Difference calculus is a nice method to use. Kelley and Peterson's textbook [1] is a very good resource for this (I tend to recommend this on a lot of different questions on here.)

Also, you can take a look at the notes I took in my difference equations course for this type of problem [2].

If you have any additional questions, I'd be happy to try to help you. Feel free to email me, my contact information is on my website ( http://www.tylerclark12.com/blog/?page_id=11 ). Good luck with your exam!

[1] Kelley, W. & Peterson, A. (2001). Difference Equations: An Introduction with Applications (2nd Ed.). San Diego, CA: Academic Press.

[2] http://www.tylerclark12.com/blog/wp-content/uploads/2010/10/Math_542_9-24.pdf

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For part (2), try adding $4$ to the numbers.

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    Good insight. I think this one is better.2010-11-23
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For the second one, take a look at the Online Encyclopedia of Integer Sequences[1]. Sequence #A030653 is what you are wanting.

(You can look at [1] for more extensive information including sources, but I will tell a bit about it here.)

Let $a(n)$ be the n-th number of the sequence for $n \in \{1,2,3...\}$.

Then we have that $a(n) = n^{3} + 3n^{2} + 3n - 3$.

Examples:

$a(1) = 1 + 3(1) + 3(1) - 3 = 4$

$a(2) = 8 + 3(4) + 3(2) -3 = 8 + 12 + 6 - 3 = 23$

...

For the next number in the sequence, we need to calculate $a(5)$. Using the formula above, we get,

$a(5) = 125 + 3(25) + 3(5) - 3 = 125 + 75 + 15 -3 = 212$.

So 212 is the next number in the sequence

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    For the second sequence $a(n)=(3n^4-22n^3+129n^2-126n+48)/8$ and so $a(5)=221.$ I just couldn't resist :-) But maybe someone will find this instructive.2010-11-23
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    This is certainly true. It just helps us to remember that math may present multiple answers. =]2010-11-23
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  1. 3+8+2=13,8+13+3=24,13+24+4=41,24+41+5=70
  2. I don't know

PS: you should use the tag "homework"

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    I was typing up the same. But I don't think this is a very good problem.2010-11-23
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    I don't like either. I wonder what kind of test it is..2010-11-23
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    This is not my homework, this comes from a national level entrance exam paper from it's quatitative aptitude and reasoning section.However,I don't know what they are trying to achieve by giving this sort of questions.2010-11-23
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To add to Gerry's answer...

For the first sequence, you can use a recurrence relation to find the answer.

We can say that,

$y(n+2) = y(n+1) + y(n) + n - 1$, $y(1) = 3$, and $y(2) = 8$.

Examples:

$y(3) = y(2) + y(1) + 3 - 1 = 8 + 3 + 3 - 1 = 13$

$y(4) = y(3) + y(2) + 4 - 1 = 13 + 8 + 4 - 1 = 24$

Now we are trying to find the next number in the sequence which is $y(6)$. So we have,

$y(6) = y(5) + y(4) + 6 - 1 = 41 + 24 + 6 - 1 = 70$.

Thus, 70 should be the next number in the sequence.

If I get a chance to find a formula for the sequence that does not involve recursion, I will post it here.