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Splash Screen. Five-Minute Check (over Lesson 10-4) Then/Now New Vocabulary Example 1: Power of a Binomial Sum Example 2: Power of a Binomial Difference Key Concept: Formula for the Binomial Coefficients of ( a + b ) n Example 3: Find Binomial Coefficients
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Five-Minute Check (over Lesson 10-4) Then/Now New Vocabulary Example 1: Power of a Binomial Sum Example 2: Power of a Binomial Difference Key Concept: Formula for the Binomial Coefficients of (a + b)n Example 3: Find Binomial Coefficients Example 4: Binomials with Coefficients Other than 1 Example 5: Real-World Example: Use Binomial Coefficients Key Concept: Binomial Theorem Example 6: Expand a Binomial Using the Binomial Theorem Example 7: Write a Binomial Expansion Using Sigma Notation Lesson Menu
Prove that 3 + 4 + 5 + … + (n + 2) = is valid for all positive integers n. 5–Minute Check 1
A. 5–Minute Check 1
B. 5–Minute Check 1
C. 5–Minute Check 1
D. 5–Minute Check 1
Prove that 3 + 4 + 5 + … + (n + 2) = is valid for all positive integers n. A. B. C. D. 5–Minute Check 1
Which of the following is an expression that is divisible by 3 for all positive integers n? A.2n– 1 B.3n– 1 C.4n– 1 D.5n– 1 5–Minute Check 2
You represented infinite series using sigma notation. (Lesson 10-1) • Use Pascal’s Triangle to write binomial expansions. • Use the Binomial Theorem to write and find the coefficients of specified terms in binomial expansions. Then/Now
binomial coefficients • Pascal’s triangle • Binomial Theorem Vocabulary
Power of a Binomial Sum A. Use Pascal’s Triangle to expand (a + b)6. Step 1 Write the series for (a + b)6, omitting the coefficients. Because the power is 6, this series should have 6 + 1 or 7 terms. Use the pattern of increasing and decreasing exponents to complete the series. a6b0 + a5b1 + a4b2 + a3b3 + a2b4 + a1b5 + a0b6 Exponents of a decrease from 6 to 0. Exponents of b increase from 0 to 6. Example 1
1 5 10 10 5 1 5th row 1 6 15 20 15 6 1 6th row Power of a Binomial Sum Step 2 Use the numbers in the sixth row of Pascal’s triangle as the coefficients of the terms. To find these numbers, extend Pascal’s triangle to the 6th row by adding corresponding numbers in the previous row. (a + b)6 = 1a6b0 + 6a5b1 + 15a4b2+ 20a3b3 + 15a2b4 + 6a1b5 + 1a0b6 (a + b)6 = a6+ 6a5b + 15a4b2+ 20a3b3 + 15a2b4 + 6ab5 + b6 Simplify. Example 1
Power of a Binomial Sum Answer: a6+ 6a5b + 15a4b2+ 20a3b3 + 15a2b4 + 6ab5 + b6 Example 1
Power of a Binomial Sum B. Use Pascal’s Triangle to expand (3x + 2)5. Step 1 Write the series for (a + b)5, omitting the coefficients and replacing a with 3x and b with 2. The series has 5 + 1 or 6 terms. (3x)5(2)0 + (3x)4(2)1 + (3x)3(2)2 + (3x)2(2)3 + (3x)1(2)4 + (3x)0(2)5 Exponents of 3x decrease from 5 to 0. Exponents of 2 increase from 0 to 5. Example 1
Power of a Binomial Sum Step 2 The numbers in the 5th row of Pascal’s triangle are 1, 5, 10, 10, 5, and 1. Use these numbers as the coefficients of the terms in the series. Then simplify. (3x + 2)5 = 1(3x)5(2)0 + 5(3x)4(2)1 + 10(3x)3(2)2 + 10(3x)2(2)3 + 5(3x)1(2)4 + 1(3x)0(2)5 (3x + 2)5 = 243x5 + 810x4 + 1080x3 + 720x2 + 240x + 32 Answer: 243x5 + 810x4 + 1080x3 + 720x2 + 240x + 32 Example 1
Use Pascal’s Triangle to expand (2x + 3)6. A. 64x6 + 576x5 + 2160x4 + 4320x3 + 162x2+ 486x + 729 B. 64x6 + 96x5 + 144x4 + 216x3 + 324x2 + 486x + 729 C. 64x6 + 576x5 + 2160x4 + 4320x3 + 4860x2+ 2916x + 729 D. 32x5 + 48x4 + 72x3 + 108x2 + 162x + 243 Example 1
Power of a Binomial Difference Use Pascal’s Triangle to expand (x – 2y)6. Because (x – 2y)6 = [x + (–2y)]6 , write the series for (a + b)6 , replacing a with x and b with –2y. Use the numbers in the 6th row of Pascal’s triangle, 1, 6, 15, 20, 15, 6, and 1, as the binomial coefficients. Then simplify. (x – 2y)6 = 1x6(–2y)0+ 6x5(–2y)1 + 15x4(–2y)2 + 20x3(–2y)3 + 15x2(–2y)4 + 6x1(–2y)5 + 1x0(–2y)6 = x6 – 12x5y + 60x4y2 – 160x3y3 + 240x2y4 – 192xy5 + 64y6 Example 2
Power of a Binomial Difference Answer:x6 – 12x5y + 60x4y2 – 160x3y3 + 240x2y4 – 192xy5 + 64y6 Example 2
Use Pascal’s Triangle to expand (3x – y)6. A. 729x6 – 243x5y + 81x4y2 – 27x3y3 + 9x2y4 – 3xy5+ y6 B. 729x6 – 1458x5y + 1215x4y2 – 540x3y3 + 135x2y4– 18xy5 + y6 C. 729x6 + 1458x5y + 1215x4y2 + 540x3y3 + 135x2y4+ 18xy5 + y6 D. 3x6 – 18x5y + 45x4y2 – 60x3y3 + 45x2y4 – 18xy5 + y6 Example 2
Subtract. Rewrite 10! as 10 • 9 • 8 • 7! and divide out common factorials. Find Binomial Coefficients Find the coefficient of the fourth term in the expansion of (a – b)10. To find the coefficient of the fourth term, evaluate nCrfor n = 10 and r = 4 – 1 or 3. Example 3
Simplify. Find Binomial Coefficients The coefficient of the fourth term in the expansion of (a – b)10 is 120. Answer:120 Example 3
Find the coefficient of the 5th term in the expansion of (a – b)8. A. 28 B. 56 C. 70 D. 280 Example 3
Subtract. Binomials with Coefficients Other than 1 Find the coefficient of the x7y term in the expansion of (3x – 4y)8. For (3x – 4y)8 to have the form (a + b)n, let a = 3x and b = –4y. The coefficient of the term containing an – r brin the expansion of (a + b)nis given by nCr. So, to find the binomial coefficient of the term containing a7b in the expansion of (a + b)8, evaluate nCrfor n = 8 and r = 1. Example 4
Rewrite 8! as 8 • 7! and divide out common factorials. Simplify. Binomials with Coefficients Other than 1 Thus, the binomial coefficient of the a7b term in (a + b)8 is 8. Substitute 3x for a and –4y for b to find the coefficient of the x7y term in the original binomial expansion. 8a7b = 8(3x)7(–4y) a = 3x and b = –4y = –69,984x7y Simplify. Therefore, the coefficient of the x7y term in the expansion of (3x – 4y)8 is –69,984. Example 4
Binomials with Coefficients Other than 1 Answer: –69,984 Example 4
Find the coefficient of the x2y4 term in the binomial expansion of (2x – 3y)6. A. –4320 B. –2916 C. 324 D. 4860 Example 4
A success in this situation is landing on green, so p = and q = 1 – or . Use Binomial Coefficients GAMES During a player’s turn in a certain board game, players must spin a spinner. The four possible colors the spinner can land on are green, blue, red, or yellow. If the probability for all four colors is equal, what is the probability of landing on green 5 times out of 10 spins? Example 5
p = , q = , n = 10, and x = 5 Use a calculator. Use Binomial Coefficients Each spin represents a trial, so n = 10. You want to find the probability that the player lands on green 5 times out of those 10 trials, so let x = 5. To find this probability, find the value of the term nCx px qn – xin the expansion of (p + q)n. Example 5
Use Binomial Coefficients So, the probability of landing on green 5 times out of 10 spins is about 5.84%. Answer:about 5.84% Example 5
A. B. C. D. COIN TOSS A fair coin is flipped 10 times. Find the probability of getting exactly 3 tails. Example 5
Expand a Binomial Using the Binomial Theorem A. Use the Binomial Theorem to expand (2t + 3u)3. Apply the Binomial Theorem to expand (a + b)3, where a = 2t and b = 3u. (2t + 3u)3= 3C0(2t)3(3u)0 + 3C1 (2t)2(3u)1 + 3C2 (2t)1(3u)2 + 3C3(2t)0(3u)3 = 1(8t3)(1) + 3(4t2)(3u) + 3(2t )( 9u2) + 1(1)(27u3) = 8t3 + 36t2u + 54tu2 + 27u3 Answer:8t3 + 36t2u + 54tu2 + 27u3 Example 6
Expand a Binomial Using the Binomial Theorem B. Use the Binomial Theorem to expand (a – 2b2)4. Apply the Binomial Theorem to expand (a + b)4, where a = a and b = –2b2. (a – 2b2)4= 4C0 a4(–2b2)0 + 4C1 a3(–2b2)1 + 4C2 a2(–2b2)2 + 4C3 a1(–2b2)3 + 4C4 a0(–2b2)4 = 1a4(1) + 4a3(–2b2) + 6a2(4b4) + 4a(–8b6) + 1(1)(16b8) = a4 – 8a3b2 + 24a2b4 – 32ab6 + 16b8 Answer:a4 – 8a3b2 + 24a2b4 – 32ab6 + 16b8 Example 6
Use the Binomial Theorem to expand (3t2– 2u)4. A. 81t8 – 216t6u + 216t4u2 – 96t2u3 + 16u4 B. 81t8 – 54t6u + 36t4u2 – 24t2u3 + 16u4 C. 81t8 + 216t6u + 216t4u2 + 96t2u3 + 16u4 D. 81t4 – 54t3u + 36t2u2 – 24tu3 + 16u4 Example 6
Answer: Write a Binomial Expansion Using Sigma Notation Represent the expansion of (3x – 5y)17 using sigma notation. Apply the Binomial Theorem to represent the expansion of (a + b)17using sigma notation, where a = 3x and b = –5y. Example 7
A. B. C. D. Represent the expansion of (2y – 6z)18 using sigma notation. Example 7