# The Binomial Theorem

```The Binomial Theorem
In these notes we prove the binomial theorem, which says that for any integer n ≥ 1,
n
(x + y) =
n
X
n
ℓ
X
xℓ y n−ℓ =
ℓ=0
ℓ+m
ℓ
ℓ m
xy
n
ℓ
where
=
n!
ℓ!(n−ℓ)!
(Bn )
ℓ,m≥0
ℓ+m=n
Here n! (read “n factorial”) means 1 &times; 2 &times; 3 &times; &middot; &middot; &middot; &times; n so that, for example,
n
1
=
n
n−1
=
n
2
=
n
n−2
=
n!
1!(n−1)!
n!
2!(n−2)!
=
1&times;2&times;3&times;&middot;&middot;&middot;&times;(n−1)&times;n
1&times;2&times;3&times;&middot;&middot;&middot;&times;(n−1)
=
1&times;2&times;3&times;&middot;&middot;&middot;&times;(n−2)&times;(n−1)&times;n
(1&times;2)&times;(1&times;2&times;3&times;&middot;&middot;&middot;&times;(n−2))
=
=
n
n(n−1)
2
n!
= 1. As special cases of the Binomial
By convention 0! = 1 so that n0 = nn = 0!n!
Theorem, we have
n = 1 (x + y)1 = x + y
n = 2 (x + y)2 = x2 + 2xy + y 2
n = 3 (x + y)3 = x3 + 3x2 y + 3xy 2 + y 3
The proof is by induction on n. First we check
Proof of the Binomial Theorem:
that, when n = 1,
n
X
ℓ n−ℓ
n!
ℓ!(n−ℓ)! x y
=
ℓ n−ℓ n!
n=1
ℓ!(n−ℓ)! x y
ℓ n−ℓ n!
n=1
ℓ!(n−ℓ)! x y
+
ℓ=0
ℓ=0
=
1! 0 1
0!1! x y
+
1! 1 0
1!0! x y
ℓ=1
=x+y
so that (Bn ) is correct for n = 1. To complete the proof we have to show that, for any
integer n ≥ 2, (Bn ) is a consequence of (Bn−1 ). So pick any integer n ≥ 2 and assume that
(x + y)n−1 =
n−1
X
n−1
ℓ
ℓ n−1−ℓ
xy
ℓ=0
is true. Now compute
n
(x + y) = (x + y)
n−1
(x + y) =
n−1
X
n−1
ℓ
x
ℓ+1 n−1−ℓ
y
ℓ=0
c Joel Feldman.
September 18, 2012
+
n−1
X
n−1
ℓ
xℓ y n−ℓ
ℓ=0
The Binomial Theorem
1
The second sum has the same powers of x and y, namely xℓ y n−ℓ , as appear in (Bn ). The
make the powers of x and y in the first sum, namely xℓ+1 y n−1−ℓ look more like those of
(Bn ), we make the change of summation variable from ℓ to ℓ̃ = ℓ + 1. The first sum
n−1
X
n−1
ℓ
n
ℓ+1 n−1−ℓ X
x y
=
ℓ=0
n−1 ℓ̃ n−ℓ̃
xy
ℓ̃−1
ℓ̃=1
As ℓ̃ is just a dummy summation variable, we may call it anything we like. In particular,
we may rename ℓ̃ back to ℓ. So we now have
(x + y)n =
n
X
n−1
ℓ−1
X
ℓ n−ℓ n−1
xy
+
ℓ=1
n−1
ℓ
xℓ y n−ℓ
ℓ=0
n−1 ℓ n−ℓ = ℓ−1 x y
ℓ=n
+
n−1
ℓ
xℓ y n−ℓ ℓ=0
+
n−1
X
n−1
ℓ−1
+
n−1
ℓ
xℓ y n−ℓ
ℓ=1
Recalling that n! = n (n − 1)! we have
n
ℓ
=
n!
ℓ!(n−ℓ)!
=
n(n−1)!
ℓ(ℓ−1)! (n−ℓ)!
=
n n−1
ℓ ℓ−1
n
ℓ
=
n!
ℓ!(n−ℓ)!
=
n(n−1)!
ℓ! (n−ℓ)(n−ℓ−1)!
=
n−1
n
n−ℓ
ℓ
So
(x + y)n =
n−1
n−1
xn +
n−1
0
yn +
n−1
X
n−1
ℓ−1
+
n−1
ℓ
ℓ n−ℓ
xy
ℓ=1
= xn + y n +
n−1
X
n
ℓ
ℓ
n
+
n−ℓ
n
ℓ n−ℓ
xy
ℓ=1
= xn + y n +
n−1
X
n
ℓ
xℓ y n−ℓ
ℓ=1
=
n
X
n
ℓ
xℓ y n−ℓ
ℓ=0
as desired.
c Joel Feldman.
September 18, 2012
The Binomial Theorem
2
```