Expand: (x +
y)5
We start with the
binomial product:
(x + y)(x +
y)(x + y)(x + y) (x +
y)
Multiply the first
two factors:
(x + y)(x +
y)(x + y)(x + y) (x +
y)
(x
2+ 2xy +
y2)
(x + y)(x + y) (x
+ y)
Now multiply the
second two factors:
(x
2+ 2xy +
y2)
(x2
+ 2xy +
y2
)(x +
y)
Distribute the
terms in the bold back trinomial
into the bold red
trinomial
I'll split this
across lines to make it easier to
follow.
(
x2
(x
2
+ 2xy +
y2
)
+
2xy(x
2
+ 2xy +
y2
)
+ y
2(x
2
+ 2xy +
y2
) )
(x +
y)
Multiply these
terms line at a time.
( x
4 +
2x
3
y +
x
2
y2
+ 2x
3
y +
4x
2
y2
+ 2xy
3+
x2
y2
+ 2xy
3+
y4)
(x +
y)
Now combine like
terms (shown in colors.)
( x
4 +
4x
3
y +
6x
2
y2
+
4xy3
+
y4
) (x +
y)
Now we need to distribute the
trailing binomial across the
first expression.
(
x4
+
4x3
y + 6x2
y2
+
4xy3
+
y4
) (x)
+
(
x
4
+
4x3
y + 6x2
y2
+
4xy3
+
y4
)(y)
(x
5
+
4x
4
y +
6x
3
y2
+
4x
2
y3
+
xy
4
) +
(x4
y+ 4x3
y2
+
6x2
y3
+
4xy
4
+
y5
)
Now combine like terms one
more time (shown in
color).
x5
+5x4
y + 10x3
y2
+
10x2
y3
+
5xy
4
+
y5
Imagine expanding (x +
y)10
.
I'd be hard pressed. Algebra
is fun, but this lies beyond my
patience for sure!!
There's a better way, using
the Binomial
Theorem.
Binomial
theorem:
|
(x +
y)n
=
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Recall the sigma
(Σ) notation was introduced
when we discussed
polynomials.
|
If we
assign
ak
=
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Using this
notation
a0
,
a1
...,
an
for the coefficients we
have:
(x +
y)n =
a0x
n+
a1x
n-1y
+ a2x
n-2y
2+
... + an-1xy
n-1+
any
n
This expansion is called the
binomial
expansion of (x +
y)n.
And is 
Recall the
factorial
function:
n! =
n
•
(n-1)
•
(n-2)
•.
...
•
(n-(n-1))
=
n
•
(n-1)
•
(n-2)
•...•
1
Example, 5! = 5 • 4
• 3
• 2
• 1 =
120
Now, back to the binomial
expansion.
The coefficients a0,
a1, a2,
and a3 are:
(I colored the common factors
that cancel in red.)
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defined as
1
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|
|
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|
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
|
|
|
|
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defined as
1
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Substituting these values in for a0, a1, etc., we get:
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Example:
(x + y)5
n =
5
a0 =
1
a1 =
n = 5
a2 =
5(5-1)/2 = 10
a3 =
5(5-1)(5-2)/6 = 5(4)(3)/6 =
10
a4 =
5! / (4!(5-4)!) = 5! / (4!1!) =
5(4)(3)(2)(1)/4(3)(2)(1)(1) =
5
a5 =
1
(x +
y)5
= x5
+ 5x4y
+ 10 x3y
2
+ 10 x2
y3
+ 5x3y
4
+ y5
And there's a
way that these factorial
calculations are unnecessary.
This technique
uses Pascal's
Triangle.
Let's build an
isosceles triangle starting
from the apex working down
toward the base.
Level zero (the
apex) will contain one
node.
Level one will
have two node.
Level two will
have three nodes.
Level three will
have four nodes.
Level four will
have five nodes.
Etc.
These nodes will
be equally spaced from one side
of the triangle to the other
side on each
level.
Like
so:
Next we write
numbers in the circles one row
at a time.
The fist and
last nodes of each row have the
value 1.
Each node is the
sum of the the numbers of the
two nodes directly above, to
its upper left and upper
right.
Row zero has
one node, the apex, its value
is defined as
1.
Row one has
only a first and a last node,
so those two nodes have value
1.
Row two has
three nodes. Its first and last
nodes have value 1. The center
node has a 1 to its
upper left and another 1
to its upper right. These
values add to 2,
so the middle
node is assigned the number
2.
Row three has
four nodes. The first and last
nodes are assigned 1.
The second node in this row
from left to right has a 1 to
its upper left and a 2 its
upper right. These values add
to 3, so the value of this node
is assigned 3. The third
node in this row has a 2 to its
upper left and a 1 to its upper
right. These values add to 3 so
the value of this node is
assigned
3.
The triangle
below fills in the values for
the rows up to the row 5. the
plus signs show which previous
values are added to get a nodes
value. As an exercise, fill in
the values for the remaining
row, row 6. (The answer is
given later in this
page.)
Row four has
five nodes. The first and last
nodes are assigned 1. The
second node is 1 + 3, the third
node is 3 + 3, and the fourth
node is 3 + 1. This gives
values 4, 6, and
4.
Row five has
six nodes. The first and last
nodes are assigned 1. The
second node is 1 + 4, the third
is 4 + 6, the fourth is 6 + 4
and the fifth is 4 + 1. This
gives the values 5, 10, 10, and
5.
List the values
for row 6 as an exercise. The
answer you will find later in
this document.
Now the row
number is the exponent n
for (x +
y)n
.
(row 0, n=0)
(x + y)0
= 1
(row 1, n
= 1) (x + y)1
= x + y
(row 2, n =
2) (x + y)2
= x2 + 2xy
+
y2
(row 3, n
= 3) (x + y)3
= x3 + 3x
2y + 3xy
2 +
y3
(row 4, n =
4) (x + y)4
= x4 + 4x
3y + 6x
2y2 +
4xy
3 +
y4
So here's how
you do it. Let's do n =
5.
First write the
powers of x from 5
down to 1
from left to right leaving room
around each:
x5
x4
x3
x2
x
Now next to
these terms write the powers of
y from 1
up to 5
starting with the second term
like so:
x5
x4 y
x3
y2
x2
y3
xy4
y5
Now fill in the
coefficients from row number 5
of Pascal's
triangle.
1x5
5x4y
10
x3y2
10 x2
y3 5
xy4
1y5
Now place the
plus sign from the original
binomial which we raised
to
the
5th power between
terms and you're
done.
x5
+
5 x4 y
+
10 x3
y2 +
10 x2
y3 +
5 x y4
+
y5
Note, this is
the same result as using the
algebra at the start of this
page. I believe this method is
easier (and
quicker.)
As an
exercise, with the last row
you added, expand (x +
y)6.
You should
get:
x6 +
6x5y +
15x4y2 +
20x3y3 +
15x2y4 +
6xy5 +
y6
What about (x
– y)n
?
Note that x
– y = x +
(–
y)
The only change
is the odd
power of a negative
number is
a negative
number.
For example (x
– y)5
is (x +
(–y))5
x5
5
x4
(-y)1
10
x3
(-y)2
10
x2
(-y)3
5
x (-y)
4
(-y)
5
x5
+ 5
x4
(-y) + 10
x3
y2
+ 10
x2
(-y
3
) + 5 x
y
4
+
-y
5
x5
–
5 x4 y + 10
x3 y2
–
10 x2 y3
+ 5 x y4
–
y5
The negative
signs come from the matching
odd (red)
exponents.
How about
(–x –
y)n
?
Well recall the
power distribution rule:
(ab)n
= anb
n
So, (–x
– y)n
= ((–1)(x +
y))n
(and here we have a = -1
and b =
(x+y))
Using the power
distribution rule
((–1)(x
+ y))n
=
(–1)n
(x + y)n
We have the
usual (x+y)n
multiplied throughout by
a -1 if n is
odd.
Example:
(–x –
y)3
write this as
((-1)(x +
y))3
which is
(–1)3
(x +
y)3
which is
–(x +
y)3
Now (x +
y)3 we know from
before is x3 +
3x2y +
3xy2 +
y3
so the answer is
–(x3 +
3x2y +
3xy2 +
y3) =
–x3 –
3x2y –
3xy2 –
y3
Example:
(x +
2y)3
First write the
x terms:
x3
x2
x
Thenwrite the y
terms next to the x terms:
Note: the y term is
2y not
y
x3
x2(2y)
3x(2y)2
(2y)3
Then write in
the coefficients from Pascal's
triangle for n =
3
x3
3x2(2y)
3x(2y)2
(2y)3
Now evaluate the
powers of the y terms, and
simplify the
terms:
x3
6x2 y
3x4y2
8y3
Finally, since
both original terms, x and 2y,
are positive, place plus
signs
between these
final terms (and in the second
term multiply the 3 and 4 to
get 12.)
x3 +
6x2 y +
12xy2 +
8y3
Answer from
before, the
6th
row in Pascal's triangle
is:
1 6 15
20 15 6 1
Example:
(– x +
y)7
((–
x) +
y)7
Same steps as
before:
(-x)7
(-x)6
(-x)5
(-x)4
(-x)3
(-x)2
(-x)
(-x)7
(-x)6 y
(-x)5 y2
(-x)4
y3
(-x)3 y4
(-x)2
y5
(-x)y6
y7
We need row
seven from Pascal's triangle.
Just above is row six, so
we
can directly get
row seven: 1 7 21 35
35 21 7
1
Now write these
in place:
(-x)7
7(-x)6 y
21(-x)5
y2
35(-x)4
y3 35
(-x)3 y4
21(-x)2
y5
7(-x)y6
y7
Evaluate the
exponents for the x
terms:
–
x7 + 7x6y
–
21x5y2 +
35x4y3
–
35x3y4 +
21x2y5
– 7xy6
+ y7
The negative
signs came from the odd powers
of (-x).
ADVANCED
A Topic from
Combinatorial
Analysis
There's more to
Pascal's triangle than meets
the eye. There are times when
we have a collection of n items
and we're interested in how
many different unique choices
we have for a unique subset of
the items. Recall that subsets
are the same if they contain
the same elements (order has no
meaning in a
set.)
Pascal's
triangle can answer this
question for us.
For example say
we have 5 tennis players and we
want to know
how many ways we
can choose from these players a
doubles team.
We can use
pascal's triangle. We go to row
5 and choose the third number
from the left, which is 10. We
can reason this out as well by
asking, “How many ways
can we choose the first
player?” The answer is 5.
Once we've chosen that player,
how many ways can we choose the
second. That answer is 4. So we
have 5 * 4 = 20 ways. But for
each of these pairs there's a
duplicate, so we must divide by
2. 20/2 = 10. There are 10 ways
to choose a doubles tennis team
from 5 players. Ok, let's
tabulate our solution, i.e.,
let's write it
out.
Let the players
be labeled a, b, c, d, and
e.
|
player
|
a
|
b
|
c
|
d
|
e
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|
a
|
|
ab
|
ac
|
ad
|
ae
|
|
b
|
ba
|
|
bc
|
bd
|
be
|
|
c
|
ca
|
cb
|
|
cd
|
ce
|
|
d
|
da
|
db
|
dc
|
|
de
|
|
e
|
ea
|
eb
|
ec
|
ed
|
|
These are all
the possible combinations of
two players out of 5 players.
Each player in a row can be
paired with the other four
players in the columns. I've
color coded the duplicates. Ten
combinations in the table above
the gray diagonal replicate the
other ten combinations below
the gray diagonal, so we can
eliminate either ten
pairs.
Now back to
Pascal's
triangle.
Each row
represents a collection of
n
items, 0, 1, 2, 3, 4 and
5. Each diagonal from the
left represent how many
items, k,
at a time are being
selected, 0, 1, 2, 3, 4,
and. Each k has its own
colored
diagonal.
There is only
one way to choose 0 items from
n items, including n = 0 items,
and that is, no choice is made.
This choice is the left edge of
the triangle
along the left
diagonal where k = 0. And each
value on this diagonal is
1.
Now look at the
diagonal for k = 1. We
are asking the question, how
many ways can n items be
selected 1 at a time. Well the
answer is n, for all n, except
n = 0. For n = 0, you cannot
choose 1 item, the question
becomes meaningless. As you
follow this diagonal you see
that the value in each row on
this diagonal is the value for
n.
Continuing, for
the diagonal k = 2, we
are asking how many ways can 2
items out of n items be
selected (ignoring the order of
selection as usual)? For n = 1,
there is no entry: you can't
choose 2 items from a
collection that only has 1
item. For n = 2, in that
diagonal the value is 1; you
can only choose 2 out of 2
items 1 way. For n = 3, you can
choose the first item 3 ways,
then the second from the
remaining 2 items 2 ways which
makes 3 * 2 = 6 ways, but we
have a duplicate for each
choice so we eliminate the
duplicates, 6/2 = 3, which is
the entry in this diagonal for
n = 3. From the triangle, for n
= 4 there are 6 ways and for n
= 5 there are 10 ways. (Reason
these two out as an exercise;
answers are
below.)
For the diagonal
k = 3, we are asking how
many ways can 3 items be chosen
out of n items (ignoring the
order of selection)? For n = 0,
1, and 2 this is impossible.
But for n = 3, 4, and 5 we get
1, 4, and 10 respectively. For
n = 5 we can choose the first
item 5 ways, this leaves 4
items from which we can choose
the second item 4 ways, this
leaves 3 items. Out of these 3
items we can choose item 3, 3
ways. So we have 5 * 4 * 3 = 60
ways. Divide by 3 * 2 = 6 and
get 60/6 = 10 ways. Six? Not
two? Each triplet has 5 others
that are the same (3 * 2 * 1) =
6. So we need to divide the
total number of triplets by 6
to get the number of unique
triplets.
Example, for
3 items a, b and c the possible
triplets are:
abc, acb,
bac, bca, cab,
cba
We don't care
about order so all six are the
same. This is why me must
divide by 6.
For diagonal
k = 4, on our triangle
we have n = 4 and 5 that are
meaningful. For n = 4 we get 1,
there is only 1 way to choose 4
items out of 4 items. For n = 5
we have 5 ways to choose the
first value, 4 ways to choose
the next, and so forth to get
5*4*3*2*1 = 120. As argued
previously this means we have 4
* 3 *2 * 1 = 24 copies of each
group of 4 items. So the number
of unique groups = 120/24 =
5.
Exercise
answers:
for k = 2 and n
= 4
There are 4 ways
to choose the first item. Once
that item is removed that
leaves 3 to choose from, so
there are 3 ways to choose the
second item. That give s 4 * 3
= 12. Each pair has a duplicate
so we must divide by 2, 12/2 =
6.
Likewise,
for n = 5, there
are 5 ways to choose the first
one and then 4 ways to choose
the second. That gives 5 * 4 =
20. Each pair has a duplicate
so we have 20/2 =
10.
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