This preview shows pages 1–2. Sign up to view the full content.
This theorem applies to the expansion of the expression (x+y)
n
.
Oddly enough, this expansion is tied to the counting (in
particular combinations), that we have been doing.
Consider multiplying this expression out long hand:
(x+y)(x+y).
....(x+y)
Now, each term from this expansion is a product of terms,
where each term is chosen from each binomial. One such term
is x
n
, which is created by choosing the x in each binomial.
Another term is x
n1
y. This can be created by choosing the y in
the first binomial and the x in all of the following binomials.
But, you could also get that same term by picking x from the
first binomial, y from the second, and x from the rest. In fact,
you can get that term in exactly n ways.
Thus, the expansion of
(x+y)
n
starts out as x
n
+ nx
n1
y. It seems
tedious to work this entire expansion out in this manner. We
know that each term is going to be of the form x
k
y
nk
, for
integers k that range from 0 to n, inclusive.
So now, the question becomes, can we get a general formula for
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '09

Click to edit the document details