Math: How Do I Find The Pattern's Rule?

Factual Questions
1

I can’t recall how to find the general rule when given the following set of numbers:
0, 3, 10, 21, 36…

Or, in a table:
for n=1, the result = 0
for n=2, the result = 3
for n=3, the result = 10
for n=4, the result = 21
for n=5, the result = 36

First, I found the first difference between all consecutive values which varies as:
for n=1, result = n/a
for n=2, result = 3
for n=3, result = 7
for n=4, result = 11
for n=5, result = 15

Then, I found the second difference (delta between the first differences), and we find a constant of k=4

Because it took two steps to get a constant delta, we know the general pattern is an^2 + bn + c…but I can’t solve it! Does anyone recall how to do this? One example in a geometry book suggests to notice that the results follow the following pattern: 13, 25, 37, 49… so the pattern is something like n(n^2 + c)…but I still can’t get it to work.

Last, what kind of series or progression is this?

  • Jinx
2

2n[sup]2[/sup] - 3n + 1

3

OK, that’s helpful. Now, can you explain how you found this solution? We’d like to learn a little if it’s not asking too much! - Jinx

4

You already have your answer:
(01), (13), (25), (37), (4*9), ((n-1)(2n-1))

5

I guess…but really it seems like, once you get to this step, you’re mentally asking yourself the difference between “n” and each factor. How would you explain the general technique to a high schooler…lucky guessing until it works? - Jinx

6

Oops! The previous heading was typed in haste. Yes, I see it works for all terms in the sequence. Thanks for your help… - Jinx

7

First off, this sort of sequence would generally be referred to as a quadratic sequence.

As far as a general method for solving it, once you know that it follows the pattern s(n) = a(n^2) + bn + c, you can use any three elements of the sequence (such as the first three) to solve for a, b, and c, just like any other time you’re solving a system of three equations in three unknowns. To run through one such solution quickly:

for n = 1:
(1) a + b + c = 0

for n = 2:
(2) 4a + 2b + c = 3

Subtracting (1) from (2):
(3) 3a + b = 3

for n = 3:
(4) 9a + 3b + c = 10

Subtracting (1) from (4):
(5) 8a + 2b = 10

Multiplying (4) by a factor of 2:
(6) 6a + 2b = 6

Subtracting (6) from (5):
(7) 2a = 4, which is easily solved as a = 2

Substituting the new value of a into (6):
(8) 12 + 2b = 6, which is solved as b = -3

Substituting the values of a and b into (1):
(9) 2 - 3 + c = 0, which is solved as c = 1

Therefore the general equation is 2(n^2) -3n + 1

Hope that helps.

8

There’s another explanation of the series, which gives the same results as earlier posters. It’s every other member of the series of triangular numbers:
0, [1], 3, [6], 10, [15], 21, [28], 36, [45], …

The differences between members of this series are just the natural numbers:
1, 2, 3, 4, 5, 6, 7, 8, 9, …

Note that if you add these in pairs, you get the differences in the original series:
1+2=3, 3+4=7, 5+6=11, 7+8=15, …

The usual formula for the triangular numbers is:
n(n-1)/2

If you add any two consecutive members of the series, you get a square, e.g.
28+36 = 64 = 8 squared

9

It looks like the OP’s question is a problem in “the calculus of finite differences.” For a good short introduction, see Chapter 2 of Martin Gardner’s Colossal Book of Mathematics.

10

Minor problem with my solution… the explanation for equation (6) should read:

Multiplying (3) by a factor of 2:

Makes a bit more sense that way. :smiley: (Do I get any marks taken off for that mistake? :smack: )

11

Plug in n=1 and n=2, and n=3 to get linear equations for a, b, and c

Nothing special. Just the evaluations of a polynomial at the sequence of positive integers.

12

Another solution: look it up.