1. Math 143 Section 8.5 Binomial Theorem

2. (a + b) 2 =a 2 + 2ab + b 2 (a + b) 3 =a 3 + 3a 2 b + 3ab 2 + b 3 (a + b) 4 =a 4 + 4a 3 b + 6a 2 b 2 + 4ab 3 + b 4 (a + b) 5 =a 5 + 5a 4 b + 10a 3 b 2 + 10a 2 b 3 + 5ab 4 + b 5 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 (a + b) 1 =a + b

3. 5! = For a non-negative number, n, 5  4  3  2  1 = 120 9! = 9  8  7  6  5  4  3  2  1 = 362,880 n! = (n)(n – 1)(n – 2)   ... 2  1 1! = 1 0! = 1

4. We can use Pascal’s triangle to expand binomials, but it becomes large and cumbersome when the powers of the binomial are large. Therefore, the coefficients in a binomial expansion are often given in terms of factorials. For nonnegative integers n and r, with n  r, the expression is called a binomial coefficient and is defined by nrnr nrnr = n! r! (n – r)! 6262 3030 = 6! 2! · 4! = 6 · 5 · 4! 2! · 4! = 30 2 = 15 = 3! 0! · 3! = 1

5. For any positive integer, n (a + b) n = a n + a n-1 b + a n-2 b 2 + a n-3 b 3 +... + b n n0n0 n1n1 n2n2 n3n3n Expand: (x + 2) 5 5050 5151 5252 5353 5 5454 x 5 + x 4 (2) 1 + x 3 (2) 2 + x 2 (2) 3 + x(2) 4 + (2) 5 = = (1)(x 5 ) + 5(x 4 )(2) + 10(x 3 )(4) + 10(x 2 )(8) + 5(x)(16) + (1)(32) = x 5 + 10x 4 + 40x 3 + 80x 2 + 80x + 32

6. Expand: (3x – 2y) 4 4040 4141 4242 43434 (3x) 4 + (3x) 3 (-2y) 1 + (3x) 2 (-2y) 2 + (3x) 1 (-2y) 3 + (-2y) 4 = = (1)(81x 4 ) + 4(27x 3 )(-2y) + 6(9x 2 )(4y 2 ) + 4(3x)(-8y 3 ) + (1)(16y 4 ) = 81x 4 – 216x 3 y + 216x 2 y 2 – 96xy 3 + 16y 4 Find the first three terms of the following expansion (x 2 – 1) 20 20 0 20 1 20 2 (x 2 ) 20 + (x 2 ) 19 (-1) 1 + (x 2 ) 18 (-1) 2 +... = = x 40 – 20x 38 + 190x 36 –... = (1)(x 40 ) + 20(x 38 )(-1) + 190(x 36 )(1) +...

7. Find the 8 th term of the expansion of (2x – 3) 12 12 7 (2x) 5 (-3) 7 8 th term = = 792(32x 5 )(-2187) = - 55,427,328 x 5 In the expansion of (x + 2y) 15, find the term containing y 8 the term containing y 8 = 15 8 (x) 7 (2y) 8 = 6435(x 7 )(256y 8 ) = 1,647,360 x 7 y 8

8. f(x) = x 5 find and simplify f(x + h) – f(x) h f(x + h) – f(x) h = (x + h) 5 – x 5 h = (x 5 + 5x 4 h + 10x 3 h 2 + 10x 2 h 3 + 5xh 4 + h 5 ) – x 5 h = 5x 4 + 10x 3 h + 10x 2 h 2 + 5xh 3 + h 4 = 5x 4 h + 10x 3 h 2 + 10x 2 h 3 + 5xh 4 + h 5 h