# Lecture 8 Lecture 8, binomial theorem

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For unlimited access to Class Notes, a Class+ subscription is required. MATH 135 Winter 2009
Lecture VIII Notes
Binomial Theorem
Recall that we are trying to come up with a way of expanding (a+b)nwithout actually having to
expand it for each value of nin which we are interested. This is similar to wanting to come up with
“closed form” expressions for things like 12+ 22+· · · +n2.
Last time we introduced the notation n
r=n!
r!(nr)! and did a few calculations.
Binomial Theorem (Theorem 4.34)
If aand bare any numbers and nP, then
(a+b)n=n
0an+n
1an1b+· · · +n
ranrbr+· · · +n
n1abn1+n
nbn
Alternatively, we can write (a+b)n=
n
X
r=0 n
ranrbr.
We will prove this and do some calculations, but need to do look at a couple of preliminary re-
sults ﬁrst.
Proposition 4.33
If nand rare integers with 0 rn, then n
ris an integer.
Rationale
We will not formally prove this. However, last time we looked at n
ras the number of ways of
choosing robjects from among nobjects. Since this number of ways is an integer, then n
rshould
be an integer.
Proposition 4.32
If nand rare integers with 1 rn, then n+ 1
r=n
r+n
r1.
Aside
It is quite likely that you have seen this Proposition before in the following picture:
1
1 1
121
1331
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
What’s this called? Can you see how this relates to Proposition 4.32?
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