Happy Tuesday, friends Worldwide from Dr. TJ Gunn in Houston Texas USA…

Here are the counting numbers (given by nn ) and their leading diagonal:

Lesson 2 Sequence 2

Here are the square numbers (given by n2n2) and their leading diagonal:

Lesson 2 Sequence 3

And here are the cube numbers (given by n3n3) and their leading diagonal:

Lesson 2 Sequence 4

And we could keep going!

Lesson 2 Sequence 5

Here’s a question:

What would be the diagonal for the sequence given by n2+nn2+n?
Would it just be the sum of the two individual diagonals? (That is, sum across the rows of the diagonals.)

Lesson 2 Sequence Question
Lesson 2 Sequence Answer

Let’s check. Here’s the sequence for n2+nn2+n and its difference table.

Yes … It’s leading diagonal is the sum of the diagonals n2+nn2+n for and nn!

This example suggests a strategy for finding formulas for sequences from the standard leading diagonals.

If you are given a sequence, find the leading diagonal of its difference table.

If you recognize that leading diagonal as a combination of standard leading diagonals, then you will have a candidate formula for the sequence!

This is, of course, assuming we trust patterns and choose to believe everything will work magically. But we can always check any answer we obtain from this method by plugging in numbers. So we need a third step:

Check to see if your formula actually works.

EXAMPLE 5: Find a formula for the sequence: 1  6  15  28  45  66  … (under the assumption we can trust patterns!)

Answer: Here’s its difference table:

Lesson 2 Practice 5
Lesson 2 Practice 4b

Is its leading diagonal a combination of any standard ones?

Notice that we have zeros in the fourth position onwards. This tells us we won’t be using n3n3.

Also notice we have “4” in the third position and only n2n2 has an entry in the third position, the number 2. It is clear we will need two of those twos.

Lesson 2 Practice 5c

We need a 5 in the second position and right now we have double 3 to give 6. This shows we need -1 of the nn diagonal.

The “5” and the “4” are all set, and we see that we don’t need any of the 1 diagonal.

2n2–n2n2–n does the trick and we can check that is does indeed produce the sequence 1, 6, 15, 28, 45, 66 … . So we know we are right!

PRACTICE 6: Use this method to find a formula for the sequence 2, 2, 4, 8, 14, 22, 32,….

Comment: Just so you know… The answer is n2–3n+4n2–3n+4. But see if you can get this answer from looking at the leading diagonal.

PRACTICE 7: Find a formula for the sequence 1 , 3, 15, 43, 93, 171, 283,….

PRACTICE 8: Find a formula for the sequence 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ….

Some Jargon:

Lesson 2 Practice 8

As we have seen, a sequence with constant first differences is called linear. For example, 2, 5, 8, 11, 14, … is linear.

Lesson 2 Practice 8b

A sequence with constant second differences (but not constant first differences) is called quadratic. For example, the sequence 2, 3, 6, 11, 18, 27… is quadratic. (We’ll explain this name as the course progresses.)

As we have seen, it seems that any quadratic sequence can be written as a formula involving n2n2 and nn and 1. For example, 2, 3, 6, 11, 18, 27, … has formula n2–2n+3n2–2n+3.

A sequence with constant third differences (but not constant second differences) is called cubic. Our work suggests that any cubic sequence is given by a formula involving n3n3, n2n2, nn and 1.

A sequence with constant fourth differences is called quartic. One with constant fifth differences quintic. (And constant sixth differences? Seventh differences? What might the names for these be?)

Lesson 2 Practice 9

Question: What might we call a sequence that is already constant? Does it seem appropriate to call a sequence such as 3, 3, 3, 3, 3, 3, … a constant sequence?
PRACTICE 9: Let S(n)S(n) be the number of squares one can find in an n×nn×n grid of squares. For example, S(3)=14S(3)=14 because there are fourteen squares to be found in a 3×33×3 grid:

a) Find S(1),S(2),S(4)S(1),S(2),S(4), and S(5)S(5).
b) Use the difference method (trusting patterns) to find a general formula for S(n)S(n).

OPTIONAL CHALLENGE: What’s 12+22+32+42+⋯+100212+22+32+42+⋯+1002?

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