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9.5 The Binomial Theorem

9.5 The Binomial Theorem. Let’s look at the expansion of (x + y) n. (x + y) 0 = 1. (x + y) 1 = x + y. (x + y) 2 = x 2 +2xy + y 2. (x + y) 3 = x 3 + 3x 2 y + 3xy 2 + y 3. (x + y) 4 = x 4 + 4x 3 y + 6x 2 y 2 + 4xy 3 + y 4. Expanding a binomial using Pascal’s Triangle.

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9.5 The Binomial Theorem

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  1. 9.5 The Binomial Theorem Let’s look at the expansion of (x + y)n (x + y)0 = 1 (x + y)1 = x + y (x + y)2 = x2 +2xy + y2 (x + y)3 = x3 + 3x2y + 3xy2 + y3 (x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4

  2. Expanding a binomial using Pascal’s Triangle Pascal’s Triangle • 1 • 1 • 1 2 1 1 3 3 1 1 4 6 4 1 Write the next row. 1 5 10 10 5 1 1 6 15 20 15 6 1

  3. Expand (x + 3)4 From Pascal’s triangle write down the 4th row. 1 4 6 4 1 These numbers are the same numbers that are the coefficients of the binomial expansion. The expansion of (a + b)4 is: 1a4b0 + 4a3b1 + 6a2b2 + 4a1b3 + 1a0b4 Notice that the exponents always add up to 4 with the a’s going in descending order and the b’s in ascending order. Now substitute x in for a and 3 in for b.

  4. 1a4b0 + 4a3b1 + 6a2b2 + 4a1b3 + 1a0b4 x4 + 4x3(3)1 + 6x2(3)2 + 4x(3)3 + 34 This simplifies to x4 + 12x3 + 54x2 + 108x + 81 Expand (x – 2y)4 This time substitute x in for a and -2y for b. Use ( ). x4 + 4x3(-2y)1 + 6x2(-2y)2 + 4x(-2y)3 + (-2y)4 The final answer is: x4 – 8x3y + 24x2y2 – 32xy3 + 16y4

  5. The Binomial Theorem In the expansion of (x + y)n (x + y)n = xn + nxn-1y + … + nCmxn-mym + … +nxyn-1 + yn the coefficient of xn-mym is given by

  6. Find the following binomial coefficients. 8C2 = 28 10C3 = 120 35 7C3 = 35 7C4 =

  7. Find the 6th term in the expansion of (3a + 2b)12 Using the Binomial Theorem, let x = 3a and y = 2b and note that in the 6th term, the exponent of y is m = 5 and the exponent of x is n – m = 12 – 5 = 7. Consequently, the 6th term of the expansion is: = 55,427,328 a7b5

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