Pascal’s Triangle and the Binomial Theorem

The Binomial Theorem Strategy only: how do we expand these? 1. (x + 2)2 2. (2x + 3)2 3. (x – 3)3 4. (a + b)4

The Binomial Theorem Solutions 1. (x + 2)2 = x2 + 2(2)x + 22 = x2 + 4x + 4 2. (2x + 3)2 = (2x)2 + 2(3)(2x) + 32 = 4x2 + 12x + 9 3. (x – 3)3 = (x – 3)(x – 3)2 = (x – 3)(x2 – 2(3)x + 32) = (x – 3)(x2 – 6x + 9) = x(x2 – 6x + 9) – 3(x2 – 6x + 9) = x3 – 6x2 + 9x – 3x2 + 18x – 27 = x3 – 9x2 + 27x – 27 4. (a + b)4 = (a + b)2(a + b)2 = (a2 + 2ab + b2)(a2 + 2ab + b2) = a2(a2 + 2ab + b2) + 2ab(a2 + 2ab + b2) + b2(a2 + 2ab + b2) = a4 + 2a3b + a2b2 + 2a3b + 4a2b2 + 2ab3 + a2b2 + 2ab3 + b4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

Isn’t there an easier way? THAT is a LOT of work! Isn’t there an easier way?

The Binomial Theorem

The binomial theorem provides a useful method for raising any binomial to a nonnegative integral power. Consider the patterns formed by expanding (x + y)n. (x + y)0 = 1 1 term (x + y)1 = x + y 2 terms (x + y)2 = x2 + 2xy + y2 3 terms (x + y)3 = x3 + 3x2y + 3xy2 + y3 4 terms (x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 5 terms 6 terms (x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5 Notice that each expansion has n + 1 terms. Example: (x + y)10 will have 10 + 1, or 11 terms.

Consider the patterns formed by expanding (x + y)n. (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 (x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5 1. The exponents on x decrease from n to 0. The exponents on y increase from 0 to n. 2. Each term is of degree n. Example: The 5th term of (x + y)10 is a term with x6y4.”

The coefficients of the binomial expansion are called binomial coefficients. The coefficients have symmetry. (x + y)5 = x5 + 5x4y + 10x3y2 + 10x2y3 + 5xy4 + y5 1 The first and last coefficients are 1. The coefficients of the second and second to last terms are equal to n. Example: What are the last 2 terms of (x + y)10 ? Since n = 10, the last two terms are 10xy9 + 1y10. The coefficient of xn–ryr in the expansion of (x + y)n is written or nCr . So, the last two terms of (x + y)10 can be expressed as 10C9 xy9 + 10C10 y10 or as xy 9 + y10.

The triangular arrangement of numbers below is called Pascal’s Triangle. 0th row 1 1 1 1st row 1 + 2 = 3 1 2 1 2nd row 1 3 3 1 3rd row 6 + 4 = 10 1 4 6 4 1 4th row 1 5 10 10 5 1 5th row Each number in the interior of the triangle is the sum of the two numbers immediately above it. The numbers in the nth row of Pascal’s Triangle are the binomial coefficients for (x + y)n .

Introducing: Pascal’s Triangle Take a moment to copy the first 6 rows. What patterns do you see? Row 5 Row 6

Example: Use the fifth row of Pascal’s Triangle to generate the sixth row and find the binomial coefficients , , 6C4 and 6C2 . 5th row 1 5 10 10 5 1 6th row 1 6 15 20 15 6 1 6C0 6C1 6C2 6C3 6C4 6C5 6C6 = 6 = and 6C4 = 15 = 6C2. There is symmetry between binomial coefficients. nCr = nCn–r

Example: Use Pascal’s Triangle to expand (2a + b)4. 1 1 1st row 1 2 1 2nd row 1 3 3 1 3rd row 1 4 6 4 1 4th row 0th row 1 (2a + b)4 = 1(2a)4 + 4(2a)3b + 6(2a)2b2 + 4(2a)b3 + 1b4 = 1(16a4) + 4(8a3)b + 6(4a2b2) + 4(2a)b3 + b4 = 16a4 + 32a3b + 24a2b2 + 8ab3 + b4

The symbol n! (n factorial) denotes the product of the first n positive integers. 0! is defined to be 1. 1! = 1 4! = 4 • 3 • 2 • 1 = 24 6! = 6 • 5 • 4 • 3 • 2 • 1 = 720 n! = n(n – 1)(n – 2)  3 • 2 • 1 Formula for Binomial Coefficients For all nonnegative integers n and r, Example:

Example: Use the formula to calculate the binomial coefficients 10C5, 10C0, and .

Binomial Theorem Example: Use the Binomial Theorem to expand (x4 + 2)3.

Although the Binomial Theorem is stated for a binomial which is a sum of terms, it can also be used to expand a difference of terms. Simply rewrite (x + y) n as (x + (– y)) n and apply the theorem to this sum. Example: Use the Binomial Theorem to expand (3x – 4)4.

Example: Use the Binomial Theorem to write the first three terms in the expansion of (2a + b)12 .

Example: Find the eighth term in the expansion of (x + y)13 . Think of the first term of the expansion as x13y 0 . The power of y is 1 less than the number of the term in the expansion. The eighth term is 13C7 x 6 y7. Therefore, the eighth term of (x + y)13 is 1716 x 6 y7.

The Binomial Theorem Use Pascal’s Triangle to expand (a + b)5. Use the row that has 5 as its second number. The exponents for a begin with 5 and decrease.      1a5b0 + 5a4b1 + 10a3b2 + 10a2b3 + 5a1b4 + 1a0b5 The exponents for b begin with 0 and increase. In its simplest form, the expansion is a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5. Row 5

The Binomial Theorem Use Pascal’s Triangle to expand (x – 3)4. First write the pattern for raising a binomial to the fourth power. 1 4 6 4 1   Coefficients from Pascal’s Triangle. (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4 Since (x – 3)4 = (x + (–3))4, substitute x for a and –3 for b. (x + (–3))4 = x4 + 4x3(–3) + 6x2(–3)2 + 4x(–3)3 + (–3)4 = x4 – 12x3 + 54x2 – 108x + 81 The expansion of (x – 3)4 is x4 – 12x3 + 54x2 – 108x + 81.

The Binomial Theorem Use the Binomial Theorem to expand (x – y)9. Write the pattern for raising a binomial to the ninth power. (a + b)9 = 9C0a9 + 9C1a8b + 9C2a7b2 + 9C3a6b3 + 9C4a5b4 + 9C5a4b5 + 9C6a3b6 + 9C7a2b7 + 9C8ab8 + 9C9b9 Substitute x for a and –y for b. Evaluate each combination. (x – y)9 = 9C0x9 + 9C1x8(–y) + 9C2x7(–y)2 + 9C3x6(–y)3 + 9C4x5(–y)4 + 9C5x4(–y)5 + 9C6x3(–y)6 + 9C7x2(–y)7 + 9C8x(–y)8 + 9C9(–y)9 = x9 – 9x8y + 36x7y2 – 84x6y3 + 126x5y4 – 126x4y5 + 84x3y6 – 36x2y7 + 9xy8 – y9 The expansion of (x – y)9 is x9 – 9x8y + 36x7y2 – 84x6y3 + 126x5y4 – 126x4y5 + 84x3y6 – 36x2y7 + 9xy8 – y9.

Let’s Try Some Expand the following a) (x-y5)3 b) (3x-2y)4

Let’s Try Some Expand the following (x-y5)3

Let’s Try Some Expand the following (3x-2y)4

Let’s Try Some Expand the following (3x-2y)4

How does this relate to probability? You can use the Binomial Theorem to solve probability problems. If an event has a probability of success p and a probability of failure q, each term in the expansion of (p + q)n represents a probability. Example: 10C2 * p8 q2 represents the probability of 8 successes in 10 tries

The Binomial Theorem Brianna makes about 90% of the shots on goal she attempts. Find the probability that Bri makes exactly 7 out of 12 consecutive goals. Since you want 7 successes (and 5 failures), use the term p7q5. This term has the coefficient 12C5. Probability (7 out of 10) = 12C5 p7q5 = • (0.9)7(0.1)5 The probability p of success = 90%, or 0.9. 12! 5! •7! = 0.0037881114 Simplify. Bri has about a 0.4% chance of making exactly 7 out of 12 consecutive goals.