1. Lecture 29 Section 6.7 Thu, Mar 3, 2005
The Binomial Theorem
Lecture 29
Section 6.7
Thu, Mar 3, 2005

2. (a + b)n = k = 0..n C(n, k)an – kbk.
The Binomial Theorem
Theorem: Given any numbers a and b and any nonnegative integer n,
(a + b)n = k = 0..n C(n, k)an – kbk.
Proof: Use induction on n.
Basic step: n = 0.
(a + b)0 = 1.
k=0..0 C(0, k)a0 – kbk = C(0, 0)a0b0 = 1.
Therefore, the statement is true when n = 0.

3. Proof, continued Inductive step
Suppose the statement is true for some n  0.
Then

4. Proof, continued

5. Proof, continued Thus, the statement is true for all n  0.
Therefore, the statement is true for n + 1.
Thus, the statement is true for all n  0.

6. 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.

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

8. Example: Calculating 1.0016 Compute 1.0016. 1.0016 = (1 + 0.001)6
= 1 + 6(0.001) + 15(0.001)2 + 20(0.001)3
+ 15(0.001)4 + 6(0.001)5 + (0.001)6 = =

9. Example: Approximating (1+x)n
Theorem: For small values of x,
(1 + x)n  1 + nx.
(1 + x)n  1 + nx + C(n, 2)x2
= 1 + nx + n(n – 1)/2 x2.
(1 + x)n  1 + nx + C(n, 2)x2 + C(n, 3)x3.
= 1 + nx + n(n – 1)/2 x2
+ n(n – 1)(n – 2)/6 x3.
And so on.

10. Example For example, (1 + x)10  1 + 10x + 45x2 + 120x3
when x is small.
Compute the value of (1 + x)10 and the approximation when x = .01.

11. Newton’s Generalization of the Binomial Theorem
Theorem: Given any numbers a, b, and n,
(a + b)n = k=0.. C(n, k)an – kbk.
where
C(n, k) = [n(n – 1)…(n – k + 1)]/k!
Note that n need not be an integer nor positive.

12. Example Expand (a + b)-1 in a series, showing the first 5 terms.
Compute the first 5 coefficients
C(-1, 0) = 1.
C(-1, 1) = -1.
C(-1, 2) = (-1)(-2)/2! = 1.
C(-1, 3) = (-1)(-2)(-3)/3! = -1.
C(-1, 4) = (-1)(-2)(-3)(-4)/4! = 1.

13. Example, continued
Therefore,

14. Example Expand (a + b)5/2 in a series, showing the first 5 terms.
Compute the first 5 coefficients
C(5/2, 0) = 1.
C(5/2, 1) = 5/2.
C(5/2, 2) = (5/2)(3/2)/2! = 15/8.
C(5/2, 3) = (5/2)(3/2)(1/2)/3! = 5/16.
C(5/2, 4) = (5/2)(3/2)(1/2)(-1/2)/4! = -5/128.

15. Example, continued
Therefore,

16. Example Approximate (1.2)5/2. Let a = 1 and b = 0.2 = 1/5.
Then b/a = 1/5.

17. The Multinomial Theorem
Theorem: In the expansion of
(a1 + … + ak)n,
the coefficient of a1n1a2n2…aknk is
n!/(n1!n2!…nk!)
where n = n1 + n2 + … + nk.

18. 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.

19. 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.
Find (a + b + c)4.