# Precalculus Name: Notes on the Binomial Theorem 10/4/13 I

```Precalculus
Notes on the Binomial Theorem
Name: _____________________________________________________
10/4/13
I. Binomial Coefficients
A. Split students into groups and have each group expand a binomial with different degrees. Have one
group member from each group put the final result on the board in order.
( x  y )0  1
( x  y )1  x  y
( x  y ) 2  x 2  2 xy  y 2
( x  y )3  x 3  3 x 2 y  3xy 2  y 3
( x  y ) 4  x 4  4 x 3 y  6 x 2 y 2  4 xy 3  y 4
( x  y )5  x5  5 x 4 y  10 x3 y 2  10 x 2 y 3  5 xy 4  y 5
( x  y )6  x 6  6 x 5 y  15 x 4 y 2  20 x 3 y 3  15 x 2 y 4  6 xy 5  y 6
B. What observations can you make about the expansions of (𝑥 + 𝑦)𝑛 ?
- There are n + 1 terms in each expansion.
- x and y have symmetrical roles: As the powers of x decrease by one, the posers of y increase by
one.
- The sum of the powers of each term is n
- The first term is 𝑥 𝑛 and the last term is 𝑦 𝑛 and the coefficient of each is one.
- The coefficients are symmetrical.
II. Finding the coefficients
A. The hardest part of a binomial expansion is finding the coefficients of the interior terms.
B. To find the binomial coefficients we use the Binomial Theorem.
C. The Binomial Theorem:
In the expansion of ( x  y)n ,
( x  y ) n  x n  nx n 1 y  ...  n Cm x n  m y m  ...nxy n 1  y n
n
n
( x  y )n     x nk y k
k 0  k 
Where the coefficient of x nm y m is given by:
n
Cm 
n!
(n  m)!m !
D. Example: Find the following binomial coefficients and state what position they would be in the expansion.
1. 8 C2
2. 10 C3
3. 7 C0
4. 6 C6
Third
Fourth
First
5.
6.
7.
7
C3
Fourth
7
Fifth
C4
12
Seventh
C1
Second
8.
12
C11
Twelfth
III. Pascal’s Triangle
A. Every row has the number 1 at the beginning and end.
B. Each number is obtained by adding the two numbers
immediately above it.
C. These numbers are the coefficients of the binomial
expansions.
D. The nth row gives the coefficients of ( x  y)n .
E. Use Pascal’s Triangle to find the coefficients of ( x  y )8
1, 8, 28, 56, 70, 56, 28, 8, 1
IV. Binomial Expansions
A. Write the expansion of ( x  1)3 :
( x  1)3  x3  3x 2 (1)  3x(1)2  1(1)3
( x  1)3  x3  3x 2  3x  1
B. Write the expansion of ( x  2) 4 :
( x  2)4  x 4  4 x3 (2)  6 x 2 (2) 2  4 x(2)3  (2) 4
( x  2)4  x 4  8 x3  24 x 2  32 x  16
C. Write the expansion of (9  2 x)5 :
(9  2 x)5  (9)5  5(9)4 (2 y)  10(9)3 (2 y) 2  10(9) 2 (2 y)3  5(9)(2 y) 4  (2 y)5
(9  2 y)5  59,049  65,610 y  29,160 y 2  6, 480 y 3  720 y 4  32 y 5
V. Finding a specific term in a Binomial Expansion
A. Find the sixth term in the expansion of (3a  2b)12 :
- Let 𝑥 = 3𝑎 𝑎𝑛𝑑 𝑦 = 2𝑏 and in the sixth term the exponents would be: 𝑥 7 𝑦 5
- To find the coefficient given by Pascal’s triangle, recall that we can use the Binomial Theorem:
-
n
Cm  12 C5 
12!
 792
7!5!
- Therefore, the sixth term of the expansion is:
12
C5 x 7 y 5  792(3a)7 (2b)5  55, 427,328a 7b5
B. Find the coefficient of the given term in the expansion of the binomial:
Binomial: ( x  3)9
Term: ax 6
This is the fourth term of the expansion so, use:
n
Cm x n  m y m  9 C3 x 6 (3)3  84 x 6 (27)  2268 x 6
Binomial:
(2 x  y)14
Term:
ax9 y 5
This is the sixth term in the expansion, so use:
n
Cm x n  m y m  14 C5 (2 x)9 y 5  2002(512) x 9 y 5  1, 025, 024 x 9 y 5
```