The Theorem

 Expand a power of a binomial using Pascal’s triangle or notation.  Find a specific term of a binomial expansion. The binomial theorem is used to raise a binomial (a + b) to relatively large powers. To better understand the theorem consider the following powers of (a+b): a b1  a b a b2  a2  2abb2 a b3  a3 3a2b 3ab2 b3 a b4  a4  4a3b 6a2b2  4ab3 b4 a b5  a5 5a4b 10a3b2 10a2b3 5ab4 b5 Note the following patterns for the expansion of ab n •1. There are n+1 terms, the first anda n last b n •2. The exponents of a decrease and exponents of b increase •3. The sum of the exponents of a and b in each term is n Using these patterns the expansion of 8 looks like a b a  b8  a8  ?a7b  ?a6b2  ?a5b3  ?a4b4  and the problem now comes down to finding the value of each coefficient. This can be done using Pascal’s triangle. a b0 1 a b1 1 1 a b2 1 2 1 a b3 1 3 3 1 a b4 1 4 6 4 1 a b5 1 5 10 10 5 1 Pascal’s Triangle The Binomial Theorem Using Pascal’s Triangle Example • Expand (u  v)4. Solution: We have (a + b)n, where a = u, b = v, and n = 4. We use the 5th row of Pascal’s Triangle: 1 4 6 4 1 Then we have:

()()u v4 1 u 4  4()()()()() u 3  v 1  6 u 2  v 2 4 u 1()()  v 3 1  v 4 u4 4 u 3 v  6 u 2 v 2  4 uv 3  v 4 Another• Expand Example (x  3y)4. • a = x, b = 3y, and n = 4. We use the 5th row of Pascal’s triangle: 1 4 6 4 1 • Then we have

1()x4  4 ()(3) x 3  y 1  6 ()(3) x 2  y 2  4 ()(3) x  y 3 1 (3)  y 4 x4 12 x 3 y  54 x 2 y 2  108 xy 3  81 y 4 Although Pascal’s triangle can be used to expand a binomial, as the value of the exponent gets larger, it becomes more and more tedious to use this method. The binomial theorem is used for these larger expansions. Before proceeding to the theorem we need some additional notation. The of the first n natural numbers is denoted n! and is called n factorial. n!123...n 1n

and 0! = 1 5!=(1)(2)(3)(4)(5) = 120 The : let n and r be nonnegative integersr  n with The binomial coefficient is denoted by and C or n nr   is defined by r 

n C    n! nr    r  r!n  r! Evaluate the expression:

5! 868 2!3!    335 n  n  In general      r  n  r The Binomial Theorem Using Factorial Notation Use binomial theorem to find (3a  2b)6 1 6 15 20

6 6 6 5 1 6 4 2 6 3 3  (3a)   (3a) (2b)   (3a) (2b)   (3a) (2b)  0 1  2 3 6 1 15 6 2 4 6 1 5 6 6  (3a) (2b)   (3a) (2b)    (2b) 4 5 6

 729a6  2916a5b  4860a4b2  4320a3b3  2160a2b4  576ab5  64b6 Example Finding a Specific Term • Finding the (r + 1)st Term n The (r + 1)st term of (a + b) is n n r r ab. r Example: Find the 7th term in the expansion (x2  2y)11. First, we note that 7 = 6 + 1. Thus, k = 6, a = x2, b = 2y, and n = 11. Then the 7th term of the expansion is

11 6 5 11 26611! 2 10 6  x 2 y or  x  2 y , or 29,568 x y 6 6!5!