# 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. –  Andres 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
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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. –  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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