EC3070 FINANCIAL DERIVATIVES
THE BINOMIAL THEOREM
Pascal’s Triangle and the Binomial Expansion.
To prove the binomial
theorem, it is necessary to invoke some fundamental principles of combinato
rial calculus. We shall develop the necessary results after derving the general
expression for a binomial expansion by a process of induction.
Consider the following binomial expansions:
(
p
+
q
)
0
= 1
,
(
p
+
q
)
1
=
p
+
q,
(
p
+
q
)
2
=
p
2
+ 2
pq
+
q
2
,
(
p
+
q
)
3
=
p
3
+ 3
p
2
q
+ 3
pq
2
+
q
3
,
(
p
+
q
)
4
=
p
4
+ 4
p
3
q
+ 6
p
2
q
2
+ 4
pq
3
+
q
4
,
(
p
+
q
)
5
=
p
5
+ 5
p
4
q
+ 10
p
3
q
2
+ 10
p
2
q
3
+ 5
pq
4
+
q
5
.
The generic expansion is in the form of
(
p
+
q
)
n
=
p
n
+
np
n
−
1
q
+
n
(
n
−
1)
2
p
n
−
2
q
2
+
n
(
n
−
1)(
n
−
2)
3!
p
n
−
3
q
3
+
· · ·
+
n
(
n
−
1)
· · ·
(
n
−
r
+ 1)
r
!
p
n
−
r
q
r
+
· · ·
+
n
(
n
−
1)(
n
−
2)
3!
p
3
q
n
−
3
+
n
(
n
−
1)
2
p
2
q
n
−
2
+
npq
n
−
1
+
p
n
.
In a tidier notation, this becomes
(
p
+
q
)
n
=
n
X
x
=0
n
!
(
n
−
x
)!
x
!
p
x
q
n
−
x
.
We can find the coe
ﬃ
cient of the binomial expansions of successive degrees
by the simple device known as Pascal’s triangle:
1
1
1
1
2
1
1
3
3
1
1
4
6
4
1
1
5
10
10
5
1
1
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THE BINOMIAL THEOREM
The numbers in each row but the first are obtained by adding two adjacent
numbers in the row above. The rule is true even for the units that border the
triangle if we suppose that there are some invisible zeros extending indefinitely
on either side of each row.
Instead of relying merely upon observation to establish the formula for the
binomial expansion, we should prefer to derive the formula by algebraic meth
ods. Before we do so, we must rea
ﬃ
rm some notions concerning permutations
and combinations that are essential to a proper derivation.
Permutations.
Let us consider a set of three letters
{
a, b, c
}
and let us find
the number of ways in which they can be can arranged in a distinct order. We
may pick any one of the three to put in the first position. Either of the two
remaining letters may be placed in the second position. The third position must
be filled by the unused letter.
With three ways of filling the first place, two
of filling the second and only one way of filling the third, there are altogether
3
×
2
×
1 = 6 di
ff
erent arrangements. These arrangements or permutations are
abc,
cab,
bca,
cba,
bac
acb.
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 Spring '12
 D.S.G.Pollock
 Probability theory, ob jects

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