# Find the missing term in these sequences

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 oeis.org/… – Derek Jennings Nov 23 '10 at 16:24
@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. – Ross Millikan Nov 23 '10 at 16:33
@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. – Derek Jennings Nov 23 '10 at 16:52
What an absurd problem. Any number whatsoever would be a valid answer without a more specific description of restrictions. – Andrés E. Caicedo Nov 23 '10 at 17:46
It is absurd indeed; it pisses me off all the more that such things frequently crop up in "intelligence tests"... – J. M. Nov 24 '10 at 1:18

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. – Derek Jennings Nov 23 '10 at 19:32
This is certainly true. It just helps us to remember that math may present multiple answers. =] – Tyler Clark Nov 23 '10 at 23:47

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

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Good insight. I think this one is better. – Ross Millikan Nov 23 '10 at 16:23
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. – Ross Millikan Nov 23 '10 at 16:22
I don't like either. I wonder what kind of test it is.. – gerry Nov 23 '10 at 16:25
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. – Quixotic Nov 23 '10 at 18:15

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.

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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.

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