Question 1120309: 3,9,12,21,33,54 what are the next two numbers in the sequence
Found 3 solutions by solver91311, MathTherapy, greenestamps:
Answer by solver91311(24713)   (Show Source):
Answer by MathTherapy(10552)  (Show Source):
You can put this solution on YOUR website!
3,9,12,21,33,54 what are the next two numbers in the sequence
Close to but NOT a Fibonacci series. Nonetheless, the 1st 2 terms are 3 and 9, and every term after those is the sum of the 2 preceding terms.
Therefore, the 7th is 33 + 54 = 87, and the 8th is 54 + 87 = 141. That's it!!
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
The solution by the other tutor shows you a way that you can always find AN answer to any problem like this. Given a finite sequence of n numbers, you can always find a polynomial of degree (n-1) that produces that sequence.
HOWEVER....
In fact ALL problems like this are invalid, unless they SPECIFY that the sequence is supposed to be defined by a polynomial of least possible degree.
The whole truth is that you could put ANY two numbers next and it would be a valid sequence.
So, in general, any problem like this that asks for the next number(s) in a sequence, without specifying the type of sequence, is not a valid problem.
Supposing that the sequence is in fact produced by a polynomial of degree 5, if you are only asked to find the next number(s), then there is no need to find the polynomial, by the process described by the other tutor. You can find subsequent numbers in the sequence without finding the polynomial itself, using the method of finite differences.
With the method of finite differences, you make a diagram showing the terms of the sequence, the differences between successive terms, the differences between those differences, and so on, until you find a row of constant differences. If the n-th row of differences is constant, then the sequence can be produced by a polynomial of degree n.
For the sequence in this problem, the diagram looks like this:
3 9 12 21 33 54
6 3 9 12 21
-3 6 3 9
9 -3 6
-12 9
21
We didn't find a row of constant differences; the 5th row of differences contains the single number 21.
At this point, we can CHOOSE to make the sequence defined by a polynomial of degree 5 by adding more "21"s to the 5th row of difference; then working back up the array of numbers we can find subsequent numbers in the original sequence.
Since the problem asks for the next two numbers in the sequence, we add "21" two more times:
3 9 12 21 33 54 120 318
6 3 9 12 21 66 198
-3 6 3 9 45 132
9 -3 6 36 87
-12 9 30 51
21 21 21
So if the given sequence is produced by a polynomial of degree 5, then the next two numbers in the sequence are 120 and 318.
That would be the result you get if you use the other tutor's process to find the polynomial.
Note, however, that there are an infinite number of polynomials that will produce the given sequence of numbers. The polynomial is unique if it degree 5; but it can be any degree higher than 5.
To create another sequence that uses a polynomial of higher degree to produce the given sequence of 6 numbers, we can make the next number in the 5th row of differences any number we want, then use the new number in the 6th row of differences as the common difference, leading to a sequence produced by a polynomial of degree 6.
It might look like this....
3 9 12 21 33 54 108 234
6 3 9 12 21 54 126
-3 6 3 9 33 72
9 -3 6 24 39
-12 9 18 15
21 9 -3
-12 -12
Now the next two numbers in the sequence -- now produced by a polynomial of degree 6 -- are 108 and 234.
That sequence was created by making the second number in the row of 5th differences "9". Clearly choosing any different number there would result in a different sequence, produce by a different polynomial of degree 6.
And we could go on forever; for example not getting a row of constant difference until the 10th row, meaning the sequence could be produced by a polynomial of degree 10.
In conclusion....
(1) Most importantly, NO problem like this is a valid problem unless it specifies what kind of pattern (e.g., a polynomial) the sequence is.
(2) You can ALWAYS find AN answer that produces a given finite sequence of n numbers by assuming the sequence is produced by a polynomial of degree (n-1); if you are only asked to find subsequent numbers in the sequence, you do not need to determine the polynomial itself.
(3) You can produce different subsequent numbers in the sequence assuming that it is produced by a polynomial of degree HIGHER than (n-1); and there are an infinite number of such sequences.
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Addendum....
Tutor @maththerapy has observed that the sequence is a Lucas sequence (like Fibonacci but starting with different first two numbers).
But you have to ignore his "that's it" proclamation at the end of his response.
His answer is another POSSIBLE answer that makes good mathematical sense; but it is no better than any of the infinite number of possible answers that can be obtained assuming the sequence is produced by a polynomial.
And, as I stated early in my original response, any next two numbers will make a valid sequence.
If I answered that the next two numbers were "pi to the 5/3 power" and "the square root of 34902.2", then that would be every bit as good an answer as any other.
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