3. Binomial Series
⚫ A binomial squared becomes
⚫ A binomial cubed becomes
( )
+ = + +
2 2 2
2
a b a ab b
( ) ( )( )
+ = + +
3 2
a b a b a b
( )( )
= + + +
2 2
2
a b a ab b
= + + + + +
2 2
3 3
2 2
2 2
a b a b ab
a b b
a
= + + +
3 2 2 3
3 3
a a b ab b
4. Binomial Series (cont.)
⚫ As you may recall from Algebra II, the coefficients
correspond to rows from Pascal’s Triangle
0 1
1 1 1
2 1 2 1
3 1 3 3 1
4 1 4 6 4 1
5 1 5 10 10 5 1
6 1 6 15 20 15 6 1
6. Binomial Series (cont.)
⚫ Example: Expand
a = 2x and b = 1; the exponents begin and end at 5
(a goes down while b goes up). Looking at row 5 on the
triangle, our coefficients are 1, 5, 10, 10, 5, 1, so we write
our expression as follows:
(Notice that the exponents apply to the entire term of the
binomial, not just the variable.)
( )
+
5
2 1
x
( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )
+ + + + +
5 4 3 2 2 3 4 5
5 10 1
1 1 1
2 2 2 1
2 1
5 2
0
x x x x x
= + + + + +
5 4 3 2
32 80 80 40 10 1
x x x x x
+ + + + +
5 4 3 2 2 3 4 5
5 10 10 5
a a a a a
b b b b b
7. Binomial Series (cont.)
⚫ Consider the binomial series :
If we multiply the coefficient of a term by a fraction
consisting of the exponent of a over the term number,
we get the coefficient of the next number.
( )
+
7
a b
= + + + + + + +
7 6 5 2 4 3 3 4 2 5 6 7
7 21 35 35 21 7
a a b a b a b a b a b ab b
8
7
6
5
4
3
2
1
= =
exp. 7
coeff. 1 7,
term # 1
=
6
7 21,
2
5
21 =35, ...
3
8. Binomial Series (cont.)
⚫ Now let’s see what happens to if we don’t
simplify the fractions as we calculate them:
( )
+
8
a b
1
2
3
4
5
8
a
7
8
1
a b
6 2
8 7
1 2
a b
5 3
8 7 6
1 2 3
a b
4 4
8 7 6 5
1 2 3 4
a b
Do you see the pattern?
What is it?
9. Binomial Series (cont.)
⚫ The coefficients of a binomial series can be written as
factorials. For example, let’s look at the coefficient for
the fourth term:
=
8 7 6 8 7 6
1 2 3 1 2 3
=
8 7 6 5!
1 2 3 5!
=
8!
3! 5!
10. Binomial Series (cont.)
⚫ Looking back at the original expression:
Notice how the numbers in the coefficient expression
are found elsewhere in the expression.
⚫ 8 is the value of the exponent to which (a + b) is
raised.
⚫ 5 is the value of a’s exponent and 3 is the value of b’s.
⚫ The exponent of b is always one less than the term
number (4).
( )
+ = + +
5 3
8 !
... ...
! !
5
3
8
a b a b
11. Binomial Theorem
⚫ The formula for the term containing br of (a + b)n,
therefore, is
or nCr
⚫ Example: Find the term containing y6 of
( )
−
−
!
! !
n r r
n
a b
r n r
n
r
=
( )
10
8
x y
−
12. Binomial Theorem (cont.)
⚫ The formula for the term containing br of (a + b)n,
therefore, is
or nCr
⚫ Example: Find the term containing y6 of
( )
−
−
!
! !
n r r
n
a b
r n r
n
r
=
( )
10
8
x y
−
( )( ) ( ) ( )
6
10 6 4 6
10
8 210 262144
6
x y x y
−
− =
4 6
55,050,240x y
=
14. Binomial Theorem (cont.)
⚫ Example: Find the term in which contains f 9.
Since n = 15, n – r = 15 – 9 = 6. Therefore the term is
(When dealing with negative terms such as f, recall that
even exponents will produce positive terms and odd
exponents will produce negative terms.)
( )
15
e f
−
( )
9
6 6 9
15
5005
6
e f e f
− = −
15. Binomial Theorem (cont.)
⚫ Similarly, the kth term of binomial expansion of
is found by realizing that the exponent of b will be k – 1,
which gives us the formula:
(replace r with k – 1)
( )
n
a b
+
( )
1 1
1
n k k
n
a b
k
− − −
−
17. Binomial Theorem (cont.)
⚫ Example: Find the 4th term of
n = 12, k = 4, which means that k – 1 = 3
( )
12
2c d
−
( ) ( ) ( )( )
9 3 9 3
12
2 220 512
3
c d c d
− = −
9 3
112,640c d
= −
18. Binomial Theorem
⚫ Your calculator can also find the coefficient:
4th term of
n = 12, k – 1 = 3, n – (k –1) = 9
2l9¢b5312,3·
( )
12
2c d
−
19. Binomial Theorem (cont.)
⚫ Desmos can also find the coefficient using a function
called nCr(n, r):
4th term of
n = 12, k – 1 = 3, n – (k –1) = 9
( )
12
2c d
−
( ) ( )
9 3
12
2
3
c d
−