M1410905

# M1410905 - (Chapter 9 Discrete Math 9.32 PART C EXPANDING...

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(Chapter 9: Discrete Math) 9.30 SECTION 9.5: THE BINOMIAL THEOREM PART A: INTRO How do we expand a + b ( ) n , where n is a nonnegative integer? a + b ( ) 0 = 1 a + b ( ) 1 = a + b a + b ( ) 2 = a 2 + 2 ab + b 2 a + b ( ) 3 = a + b ( ) a + b ( ) 2 Do first = a 3 + 3 a 2 b + 3 ab 2 + b 3 a + b ( ) 10 = YUK! Look for patterns Take a + b ( ) 3 . Let’s look at the variable parts of the terms. a + b ( ) 3 = a 3 Starts with a 3 + 3 a 2 b + 3 ab 2 + b 3 Ends with b 3 At each step, • the exponent on a by 1 • the exponent on b by 1 a 0 = 1, b 0 = 1, so they are “invisible.” Each term has degree 3.

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(Chapter 9: Discrete Math) 9.31 In general, a + b ( ) n = a n + + b n n + 1 ( ) terms ( n is a whole number) (# terms = power + 1) What about the coefficients? They are given by … PART B: PASCAL’S TRIANGLE The ingredients: “Tent” of “1”s Any other entry = sum of the two entries immediately above Row n gives the coefficients for a + b ( ) n . a + b ( ) 0 = 1 a + b ( ) 1 = a + b a + b ( ) 2 = a 2 + 2 ab + b 2 a + b ( ) 3 = a 3 + 3 a 2 b + 3 ab 2 + b 3 a + b ( ) 4 = a 4 + 4 a 3 b + 6 a 2 b 2 + 4 ab 3 + b 4 (See my website for some magical properties of Pascal’s triangle!)
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Unformatted text preview: (Chapter 9: Discrete Math) 9.32 PART C: EXPANDING POWERS OF GENERAL BINOMIALS Example Expand and simplify 3 x − y ( ) 3 using the Binomial Theorem. Solution We will use the template for a + b ⎡ ⎣ ⎤ ⎦ 3 . 3 x − y ( ) 3 = 3 x Sub a = 3 x + − y ( ) Sub b = − y ⎡ ⎣ ⎢ ⎢ ⎢ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ⎥ ⎥ ⎥ 3 = a + b [ ] 3 = a 3 + 3 a 2 b + 3 ab 2 + b 3 Sub back: a ← 3 x ( ) , b ← − y ( ) = 3 x ( ) 3 + 3 3 x ( ) 2 − y ( ) + 3 3 x ( ) − y ( ) 2 + − y ( ) 3 Simplify. First, do powers. = 27 x 3 + 3 9 x 2 ( ) − y ( ) + 3 3 x ( ) y 2 ( ) − y 3 = 27 x 3 − 27 x 2 y + 9 xy 2 − y 3...
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