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Appendix 3

THEORETIC ARITHMETIC

Triangular numbers

Square numbers

The sum of consecutive cubes

Triangular numbers

A formula for the triangular numbers

Square numbers

The sum of consecutive cubes

IN THIS TOPIC we will look at numbers themselves, not just their symbols: 1, 2, 3, 4, and so on. By doing so we will see unexpected structures that are inherent in the natural numbers.

Here is a number in the form of a triangle:

Triangular number

This is number 10. (Count themexclamation) Since 10 can be pictured in this way, we call 10 a triangular number.

Now, how do we generate a triangular number?  We begin with 1:

Triangular number

We say that 1 is the first triangular number.  To form the next, we add 2:

Triangular number

So the next triangular number is 3.  The number we add to the previous triangular number is called the gnomon (NOH-mon). We added the gnomon 2 to 1.

To form the next triangular number, we add the gnomon 3:

Triangular number

It produces the next triangular number, 6.

To form the next, we add 4:

Triangular number

And so the first four triangular numbers are 1, 3, 6, 10.  Each one is the sum of consecutive numbers.

  1.
1 + 2 = 3.
1 + 2 + 3 = 6.
1 + 2 + 3 + 4 = 10.

Problem 1.   Write the first ten triangular numbers.

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Do the problem yourself first!

1, 3, 6, 10, 15, 21, 28, 36, 45, 55.

The difference between consecutive triangles increases by 1.

A formula for the triangular numbers

We will now show that a triangular number -- the sum of consecutive numbers -- is given by this algebraic formula:

½n(n + 1),

where n is the last number in the sum. (For example, n = 4 in the last sum above.)

To see that, look at this oblong number, in which the base is one more than the height:

Oblong number

An oblong number is the product of a number with its successor, which is the one after.  Algebraically, it has the form

n(n + 1).

In this one, the height n is 4; that is, this oblong number is 4 × 5 = 20.

But an oblong is composed of two equal triangles:

oblong numbers

Therefore each triangle is half of the oblong.  Each one is

½n(n + 1).

The sum of consecutive numbers is equal to half the product of the
last number in the sum with its successor
.

Example.   Find the sum of the first 50 numbers -- that is, find the 50th triangular number.

Solution.  In the formula, we will put n = 50.  Then n + 1 = 51. Therefore the sum is

½(50 × 51) = ½(2550) = 1275.

Problem 2.   What is the 200th triangular number?

½(200 × 201) = ½(40,200) = 20,100.

Square numbers

square numbers

Just as a triangular number is a number that can appear as a triangle, so a square number can take the form of a square. 25 is a square number. If we call 1 the first square number, then by adding which gnomons was 25 produced?

square numbers

To 1 we added 3 to produce 4.

To 4 we added 5 to produce 9.

To 9 we added 7 to produce 16.

To 16 we added 9 to produce 25.

The gnomons of the squares are the odd numbers.

Every square number is a sum of consecutive odd numbers.

  1.
1 + 3 = 4.
1 + 3 + 5 = 9.
1 + 3 + 5 + 7 = 16.
1 + 3 + 5 + 7 + 9 = 25.

Now, how are square numbers related to triangular numbers?

triangular numbers

Every square is composed of two consecutive triangles.

Triangles:  1  3  6  10  15  21  28 . . .

1 + 3 = 4.
3 + 6 = 9.
6 + 10 = 16.
10 + 15 = 25.
15 + 21 = 36.

The sum of consecutive cubes

When the same number is repeated as a factor three times -- as 4 × 4 × 4 -- we call the product the 3rd power of that base; that product is commonly called a cube. (This is analogous to the volume of the solid figure called a cube.)

Here is number 4:

Number 4

Upon repeatedly adding it four times --

Square of 4

-- we have the 2nd power, or the square, of 4.

Upon repeatedly adding that power four times --

Cube of 4

-- we have the 3rd power, or the cube, of 4.

It will be convenient for the moment to express the cube of a number with the exponent 3.

13 = 1.
23 = 8.
33 = 27.
43 = 64.

We come now to one of the most remarkable facts in the structure of the natural numbers:

The sum of n consecutive cubes is equal to the square
of the nth triangle.

13 + 23 + 33 + . . . + n3 = (1 + 2 + 3 + . . . + n)2.

To see that, we will begin here:

The difference between the squares of two consecutive triangular numbers
is a cube.

Triangles:  1  3  6  10  15  21  28

32 − 12 = 23.
62 − 32 = 33.
102 − 62 = 43.
152 − 102 = 53.

The base of each cube is the difference of the two triangles.

Look -- here is the cube of 4:

Cubes

From it, let us separate this rectangular array --

Cubes

-- and reposition it here:

Cubes

Then we have the square of side 10 --

Cubes

-- minus the square of side 6exclamation

The difference between the squares of those two consecutive triangular numbers is equal to a cube.

Therefore,

Cubes

The sum of those four cubes is equal to the square of the fourth triangle.

*

Alternatively, since every square number is the sum of consecutive odd numbers, so is the square of a triangular number.

12 = 1
(1 + 2)2 = 1 + 3 + 5 = 9
(1 + 2 + 3)2 = 1 + 3 + 5 + 7 + 9 + 11 = 36
(1 + 2 + 3 + 4)2 = 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 100

Therefore, the difference of those squares -- each cube -- will also be the sum of consecutive odd numbers, although not starting with 1.

1 = 1
8 = 3 + 5
27 = 7 + 9 + 11
64 = 13 + 15 + 17 + 19

Again, the sum of those four cubes is equal to the square of the fourth triangle.

Cubes

We proved that by mathematical induction
in Topic 27 of Precalculus.

Cubes

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