What number is next in this sequence?

Question

Just wanted to give away points. Whoever gets it right gets 10 pts.

What number is next in this sequence?

1, 6, 15, 44, 109

Answer 1

Okay -- quick answer?  226.

Explanation?...

Let's look at the differences between consecutive terms:

6 - 1 = 5

15 - 6 = 9

44 - 15 = 29

109 - 44 = 65

Now -- these differences are not all the same.  Let's explore further.  Let's try looking at _second_ differences...

9 - 5 = 4

29 - 9 = 20

65 - 29 = 36

These differences are not the same either.  _Third_ differences?

20 - 4 = 16

36 - 20 = 16

Oooh!  These differences are the same!  Which means the next second difference should be 16 more than 36 -- namely, 52.  Which means that, if we go back to the first differences, the next first difference should be 52 more than 65 -- a.k.a. 117.  Which means that the next term in the original sequence should be 117 more than 109 -- which gives us our 226.

Now -- let's explore some even _deeper_ mathematics!  An arithmetic sequence (with equal first differences) can be described by a linear equation.  If it takes going down to the _second_ difference before you get a constant sequence, then the original sequence is governed by a _quadratic_ equation.  And if you have to go all the way to the _third_ differences, then you'd need a _cubic_ equation to generate the sequence.

So -- let's try a generic cubic equation...

t(n) = an^3 + bn^2 +cn + d

The initial term would be t(0), and it would have to equal a value (in this case) of "1."  (The next term will have to equal "6," the term after that "15," etc.)

t(0) = a(0)^3 + b(0)^2 + c(0) + d = 1

0 + 0 + 0 + d = 1

d = 1

t(1) = a(1)^3 + b(1)^2 + c(1) + d = 6

a + b + c + d = 6

a + b + c + 1 = 6  (We know that d = 1)

a + b + c = 5

t(2) = a(2)^3 + b(2)^2 + c(2) + d = 15

8a + 4b + 2c + 1 = 15

8a + 4b + 2c = 14

t(3) = a(3)^3 + b(3)^2 + c(3) + d = 44

27a + 9b + 3c + 1 = 44

27a + 9b + 3c = 43

Okay -- we now have three equations with three unknowns.  We can solve this system -- using elimination, using matrices, using any number of tools.  However you do it, you get the following values . . .

a = 8/3

b = -6

c = 25/3

And d, of course, was equal to 1.

Using these values, you can now write a specific function for your sequence:

t(n) = (8/3)n^3 - 6n^2 + (25/3)n + 1

This will generate any number of terms for you -- you just have to plug in an index number.

t(0) = 1

t(1) = 6

t(2) = 15

t(3) = 44

t(4) = 109

t(5) = 226

t(6) = 441

t(7) = 680

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t(10) = 2151

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t(50) = 318751

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t(100) = 2607501

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Just notice, please that t(100) actually means the _101st_ term of the sequence, not the 100th.  (It's the Year Zero problem in a different guise.)  The first term of the sequence is t(0), not t(1), so your index is "one off" from your normal count...

:-)

Answer 2

I cant explain it, but I get the feeling its 256 =/

Im no doubt wrong