- Pascal's Triangle
- Formula for the Binomial Coefficient (Numbers in Pascal's Triangle)
- Formula for the Binomial Coefficients (Pascal's Observation)
- Relationships Between Coefficients
- Natural Numbers, Again

Our interest here is with the Binomial Theorem. The Theorem deals with equations of the form:

[1.1]

That is the expansion of factors consisting of 2 items (in this case a and b) raised to a power (Other factors can be included, which we will mention later).

In studying these equations, we can note that the sum of the powers in each term is equal to the power on the left-hand side. We can write the coefficients in a table for further study:

Table 1: Pascal's Triangle

Pascal's Triangle | |||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|

k | |||||||||||

n | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |

10 | 1 | 10 | 45 | 120 | 210 | 252 | 210 | 120 | 45 | 10 | 1 |

9 | 1 | 9 | 36 | 84 | 126 | 126 | 84 | 36 | 9 | 1 | |

8 | 1 | 8 | 28 | 56 | 70 | 56 | 28 | 8 | 1 | ||

7 | 1 | 7 | 21 | 35 | 35 | 21 | 7 | 1 | |||

6 | 1 | 6 | 15 | 20 | 15 | 6 | 1 | ||||

5 | 1 | 5 | 10 | 10 | 5 | 1 | |||||

4 | 1 | 4 | 6 | 4 | 1 | ||||||

3 | 1 | 3 | 3 | 1 | |||||||

2 | 1 | 2 | 1 | ||||||||

1 | 1 | 1 | |||||||||

0 | 1 |

In the table, n is the power, k is the term number. We fill the table in from the values of the equations (The blank cells are zero).

We call the numbers in the triangle Binomial Coefficients and represent them by , where n refers to the n-position in the column and k refers to the row position. At the moment, this symbol is defined as the n, k position in the table. When n and k are positive integers (or zero), this symbol is, in fact, a combination, but it is always a Binomial Coefficient, and only sometimes a combination. It is read as "n over k". At the moment, however, we do not know what it is apart from the given definition.

A careful study of the table, reveals that any number is made up of the sum of the number above it and the one to the left. Using our symbol:

[1.2, the Addition Law]

Where n is the row number and k is the column number in the table above.

Of course, we haven't proved this yet.

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[1.3]

[1.4]

It is also evident that when k>n, and n and k are nonnegative integers, then the term is zero:

[1.5]

(This is why, when n and k are nonnegative integers, the Binomial Series seems to be finite, because later terms are zero)

Looking at term 2, we can note that each number is a sum of natural numbers. And

[1.6]

So,

[1.7]

I do not recognise the numbers in the term 3's.

[Actually the formula is: , revealing the Upper Summation Formula, which we will consider later]

Therefore, I collect more terms, and expand Pascal's Triangle thus:

k | ||||||||
---|---|---|---|---|---|---|---|---|

n |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |

0 | 1 | |||||||

1 |
1 | 1 | ||||||

2 |
1 | 2 | 1 | |||||

3 |
1 | 3 | 3 | 1 | ||||

4 |
1 | 4 | 6 | 4 | 1 | |||

5 |
1 | 5 | 10 | 10 | 5 | 1 | ||

6 |
1 | 6 | 15 | 20 | 15 | 6 | 1 | |

7 |
1 | 7 | 21 | 35 | 35 | 21 | 7 | 1 |

8 |
1 | 8 | 28 | 56 | 70 | 56 | 28 | 8 |

9 |
1 | 9 | 36 | 84 | 126 | 126 | 84 | 36 |

10 |
1 | 10 | 45 | 120 | 210 | 252 | 210 | 120 |

11 |
1 | 11 | 55 | 165 | 330 | 462 | 462 | 330 |

We can use the method of differences to find out what the formula is. For term 3, we create the difference table:

n | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
---|---|---|---|---|---|---|---|

Series | 0 | 1 | 4 | 10 | 20 | 35 | 56 |

Δ1 | 1 | 3 | 6 | 10 | 15 | 21 | |

Δ2 | 2 | 3 | 4 | 5 | 6 | ||

Δ3 | 1 | 1 | 1 | 1 |

Because the Δ3 terms are all constants, we know the equation is a cubic, and can be represented as:

ax

Through making simultaneous equations, we find that equation is:

[1.8]

However, this is the sum of n terms, and term 3 is a sum to n-2, which gives:

[1.9]

In this way, we can derive formulae for the terms of Pascal's Triangle, and hence the Binomial Coefficients.

Table 2 (Arithmetic Triangle)

n | ||||||||||

0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
||

k | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
---|---|---|---|---|---|---|---|---|---|---|

1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | ||

2 | 1 | 3 | 6 | 10 | 15 | 21 | 28 | |||

3 | 1 | 4 | 10 | 20 | 35 | 56 | ||||

4 | 1 | 5 | 15 | 35 | 70 | |||||

5 | 1 | 6 | 21 | 56 | ||||||

6 | 1 | 7 | 28 | |||||||

7 | 1 | 8 | ||||||||

8 | 1 |

We make this table by writing down a series of 1's, which is the series of constants.

Next we simply add these, so the numbers in the row below, 1, 2, ... are the sum of the first row, up to the given position. So, 4 in row 2, is the sum of the 1's up to 4's position. These are, of course, the natural numbers.

The next row, 1, 3, 6.. is composed of the sums of the previous row, and each cell is a part sum of the natural numbers.

In a similar manner each row is the part sums of the previous row.

Our familiar Binomial Coefficients appear on the diagonals. So the first 1 represents the Binomial Expansion of power zero (n=0). The next diagonal, 1, 1 is the expansion when n=1. And the last one is the expansion when n=8.

So any diagonal represent the Binomial Coefficients of successive n's.

For instance, so 35 (the lower) relates to 21 (the upper) as 5 relates to 3.

What this does is to give us a way of relating successive Binomial Coefficients.

[1.3.1]

Where n is the power and k is the term number.

That is, we can find a formula to find the next term in an expansion by rearranging 1.3.1:

[1.3.2]

Now we note Pascal's observation, we can calculate the values quite easily in a recursive manner.

[1.3.3]

The formula seem to fly out of the equation, and we can easily generalise:

[1.3.4]

The factorials become very clear. So we can confidently say:

And, because substituting (n-k) for k gives the same result, we can say:

[1.3.5, Symmetry Law]

Symmetry is evident by observing that in Pascal's Triangle, the kth coefficient is equal to the (n-k)th coefficient.

Of course, n is a nonnegative integer (non-negative).

As mentioned before, the n and k values relate to Pascal's Triangle: in Table 2, the k's relate to the rows, but the n's relate to the diagonals.

Table 2 (Arithmetic Triangle) | ||||||||||

n | ||||||||||

0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
||

k | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |
---|---|---|---|---|---|---|---|---|---|---|

1 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | ||

2 | 1 | 3 | 6 | 10 | 15 | 21 | 28 | |||

3 | 1 | 4 | 10 | 20 | 35 | 56 | ||||

4 | 1 | 5 | 15 | 35 | 70 | |||||

5 | 1 | 6 | 21 | 56 | ||||||

6 | 1 | 7 | 28 | |||||||

7 | 1 | 8 | ||||||||

8 | 1 |

[2.1.1]

Keeping with the coefficient, 10, we note that 5 over 2 is the sum of the first 4 terms from 1 to (n-1), n=5.

In general, we find:

[2.1.2]

And by writing n+1 for n:

[2.1.3]

Pascal's Triangle |
||||||||

k |
||||||||

n |
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |

12 |
1 | 12 | 66 | 220 | 495 | 792 | 924 | 792 |

11 |
1 | 11 | 55 | 165 | 330 | 462 | 462 | 330 |

10 |
1 | 10 | 45 | 120 | 210 | 252 | 210 | 120 |

9 |
1 | 9 | 36 | 84 | 126 | 126 | 84 | 36 |

8 |
1 | 8 | 28 | 56 | 70 | 56 | 28 | 8 |

7 |
1 | 7 | 21 | 35 | 35 | 21 | 7 | 1 |

6 |
1 | 6 | 15 | 20 | 15 | 6 | 1 | |

5 |
1 | 5 | 10 | 10 | 5 | 1 | ||

4 |
1 | 4 | 6 | 4 | 1 | |||

3 |
1 | 3 | 3 | 1 | ||||

2 |
1 | 2 | 1 | |||||

1 |
1 | 1 | ||||||

0 |
1 |

[2.2.1]

The general formula is:

[2.2.2, Parallel Summation]

This is the sum of Binomial Coefficients which increase in value in a parallel manner, so the distance between the upper and lower indices is constant.

[3.1.1]

From 2.1.3, we know that the right-hand sum is:

[3.1.2]

Giving us the sum of the first n natural numbers.

Here the basic expression used above is simply:

[3.1.3]

However, it isn't that hard to figure out:

[3.1.4]

And we more or less just write down:

[3.1.5]

Giving us the sum of the squares of the natural numbers:

[3.1.6]

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