Blaise
Pascal (1623-1662) is associated with the triangle of numbers which
bears his name, although it is known as Tartaglio's Triangle in Italy,
and was known at least 700 years before Pascal by Indian, Chinese, and
other mathematicians, perhaps a long time before that too.
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:
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.
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Formula
for the Binomial Coefficients (Numbers in Pascal's Triangle)
By studying the table, we note that
[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:
ax3+bx2+cx+d
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.
Formula
for the Binomial Coefficients (Pascal's Observation)
[3.3]
The table below was called the arithmetic triangle. 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
How the table is created
The k values relate to the row, and the n values relate to the start of
a diagonal. These numbers correspond to Pascal's Triangle.
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.
Pascal's Observation
For
each successive number on the diagonal, the lower relates to the upper
as horizontal distance to the left of the lower to the beginning of the
diagonal relates to the horizontal distance to the right of the upper
to the beginning of the diagonal, both inclusive.
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.
Modern Formulation
Using the notation for our regular Pascal's Triangle: [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]
Finding
the Values of the Coefficients (Numbers in Pascal's Triangle)
From Pascal's
Triangle, we have noted that the coefficient of term 0 is 1,
and the coefficient of term 2 is n. We have calculated terms 2 and 3,
and might have an idea what the general formula for the numbers might
be, but would need to do some series algebra to collect more
information.
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).
Relationships
Between Coefficients
Table 2 is reproduced from above, and is the Arithmetic Triangle. The
rows are made up of sums from the row above.
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
Upper Summation
Each row relates to the respective n over k terms. For
instance, in row 2, the coefficient, 10, is the 5 over 2 term in Pascal's Triangle.
It is a sum of part of the upper row, so: [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]
Parallel Summation
In a
similar fashion as before, we can note that the sums of numbers in
Pascal's Triangle are the numbers just above the last number in the
diagonal.
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
For instance, the highlighted yellow rows have their sum as red square above highlighted in red. [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.
Natural
Numbers, Again
The equation below shows the sum of the natural numbers as the sum of
Binomial Coefficients: [3.1.1]
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