Section 8 5 Binomial Theorem a b1 a

Section 8. 5 Binomial Theorem

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

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 ³ r, the expression n r is called a binomial coefficient and is defined by n r 6 2 3 0 = n! r! (n – r)! = 6! 2! · 4! = 6 2!· 5· · 4!4! = 3! 0! · 3! = 1 = 30 2 = 15

For any positive integer, n (a + b)n = n n-1 n n-2 2 n n-3 3 n a + a b +. . . + bn 0 1 2 3 n Expand: (x + 2)5 = 5 5 5 4 1 5 x + x (2) + 0 1 2 x 3(2)2 5 2 3 5 + x (2) + 4 3 x(2)4 5 + 5 (2)5 = (1)(x 5) + 5(x 4)(2) + 10(x 3)(4) + 10(x 2)(8) + 5(x)(16) + (1)(32) = x 5 + 10 x 4 + 40 x 3 + 80 x 2 + 80 x + 32

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

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

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

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