1. Lecture 34 Section 6.7 Wed, Mar 28, 2007
The Binomial Theorem
Lecture 34
Section 6.7
Wed, Mar 28, 2007

2. The Binomial Theorem
Theorem: Given any numbers a and b and any nonnegative integer n,

3. The Binomial Theorem Proof: Use induction on n.
Base case: Let n = 0. Then
(a + b)0 = 1 and
Therefore, the statement is true when n = 0.

4. Proof, continued Inductive step
Suppose the statement is true when n = k for some k  0.
Then

5. Proof, continued

6. Proof, continued Thus, the statement is true for all n  0.
Therefore, the statement is true when n = k + 1.
Thus, the statement is true for all n  0.

7. Example: Binomial Theorem
Expand (a + b)8.
C(8, 0) = C(8, 8) = 1.
C(8, 1) = C(8, 7) = 8.
C(8, 2) = C(8, 6) = 28.
C(8, 3) = C(8, 5) = 56.
C(8, 4) = 70.

8. Example: Binomial Theorem
Therefore,
(a + b)8 = a8 + 8a7b + 28a6b2 + 56a5b3
+ 70a4b4 + 56a3b5 + 28a2b6 + 8ab7 + b8.

9. Example: Calculating 1.018 Compute 1.018 on a calculator.
What do you see?

10. Example: Calculating 1.018 Compute 1.018 on a calculator.
What do you see?
1.018 =

11. Example: Calculating 1.016 1.018 = (1 + 0.01)8
= 1 + 8(0.01) + 28(0.01)2 + 56(0.01)3
+ 70(0.01)4 + 56(0.01) (0.01)6 + 8(0.01)7 + (0.01)8 = =

12. Example: Approximating (1+x)n
Theorem: For small values of x,
and so on.

13. Example For example, (1 + x)8  1 + 8x + 28x2 when x is small.
Compute the value of (1 + x)8 and the approximation when x = .03.
Do it again for x = -.03.

14. Expanding Trinomials
Expand (a + b + c)3.

15. Expanding Trinomials Expand (a + b + c)3.
(a + b + c)3 = ((a + b) + c)3
= (a + b)3 + 3(a + b)2c
+ 3(a + b)c2 + c3,
= (a3 + 3a2b + 3ab2 + b3)
+ 3(a2 + 2ab + b2)c
+ 3(a + b)c2
+ c3.

16. Expanding Trinomials
= a3 + 3a2b + 3ab2 + b3 + 3a2c + 6abc + 3b2c + 3ac2 + 3bc2 + c3.
What is the pattern?

17. Expanding Trinomials (a + b + c)3 = (a3 + b3 + c3)
+ 3(a2b + a2c + ab2 + b2c + ac2 + bc2)
+ 6abc.

18. The Multinomial Theorem
Theorem: In the expansion of
(a1 + … + ak)n,
the coefficient of a1n1a2n2…aknk is

19. Example: The Multinomial Theorem
Expand (a + b + c + d)3.
The terms are
a3, b3, c3, d3, with coefficient 3!/3! = 1.
a2b, a2c, a2d, ab2, b2c, b2d, ac2, bc2, c2d, ad2, bd2, cd2, with coefficient 3!/(1!2!) = 3.
abc, abd, acd, bcd, with coefficient 3!/(1!1!1!) = 6.

20. Example: The Multinomial Theorem
Therefore,
(a + b + c + d)3 = a3 + b3 + c3 + d3 + 3a2b
+ 3a2c + 3a2d + 3ab2 + 3b2c + 3b2d
+ 3ac2 + 3bc2 + 3c2d + 3ad2 + 3bd2
+ 3cd2 + 6abc + 6abd + 6acd + 6bcd.

21. Example: The Multinomial Theorem
Find (a + 2b + 1)4.

22. Actuary Exam Problem If we expand the expression (a + 2b + 3c)4,
what will be the sum of the coefficients?