Binomial Theorem

lessonbunny.com

Answer Key & Explanations

Binomial Theorem · Grade 12 · Worksheet 3

1. Expand (4x - 3y)^7 = ? Answer: 16384x^7 - 86016x^6y + 193536x^5y^2 - 241920x^4y^3 + 181440x^3y^4 - 81648x^2y^5 + 20412xy^6 - 2187y^7 Solution: Use the binomial theorem: (a + b)^n = Σ C(n,k) * a^(n-k) * b^k Here a = 4x, b = -3y, n = 7 The binomial coefficients for n=7 are: 1, 7, 21, 35, 35, 21, 7, 1 k=0: C(7,0) * (4x)^7 * (-3y)^0 = 1 * 16384x^7 * 1 = 16384x^7 k=1: C(7,1) * (4x)^6 * (-3y)^1 = 7 * 4096x^6 * (-3y) = -86016x^6y k=2: C(7,2)…
   Full step-by-step solution

   Step 1: Use the binomial theorem: (a + b)^n = Σ C(n,k) * a^(n-k) * b^k Step 2: Here a = 4x, b = -3y, n = 7 Step 3: The binomial coefficients for n=7 are: 1, 7, 21, 35, 35, 21, 7, 1 Step 4: Expand each term: k=0: C(7,0) * (4x)^7 * (-3y)^0 = 1 * 16384x^7 * 1 = 16384x^7 k=1: C(7,1) * (4x)^6 * (-3y)^1 = 7 * 4096x^6 * (-3y) = -86016x^6y k=2: C(7,2) * (4x)^5 * (-3y)^2 = 21 * 1024x^5 * 9y^2 = 193536x^5y^2 k=3: C(7,3) * (4x)^4 * (-3y)^3 = 35 * 256x^4 * (-27y^3) = -241920x^4y^3 k=4: C(7,4) * (4x)^3 * (-3y)^4 = 35 * 64x^3 * 81y^4 = 181440x^3y^4 k=5: C(7,5) * (4x)^2 * (-3y)^5 = 21 * 16x^2 * (-243y^5) = -81648x^2y^5 k=6: C(7,6) * (4x)^1 * (-3y)^6 = 7 * 4x * 729y^6 = 20412xy^6 k=7: C(7,7) * (4x)^0 * (-3y)^7 = 1 * 1 * (-2187y^7) = -2187y^7 Step 5: Combine all terms: 16384x^7 - 86016x^6y + 193536x^5y^2 - 241920x^4y^3 + 181440x^3y^4 - 81648x^2y^5 + 20412xy^6 - 2187y^7

2. Mason, a computer science student, is designing a new encryption algorithm based on polynomial expansions. For one part of the algorithm, he needs to expand the expression (2x^2 - 3)^7 using the binomial theorem. What is the coefficient of the x^8 term in the expansion of (2x^2 - 3)^7? Answer: -15120 Solution: Write the general term for (a + b)^n: T_{k+1} = C(n, k) * a^(n-k) * b^k. Here n = 7, a = 2x^2, b = -3. The general term is: T_{k+1} = C(7, k) * (2x^2)^(7-k) * (-3)^k.
   Full step-by-step solution

   Step 1: Write the general term for (a + b)^n: T_{k+1} = C(n, k) * a^(n-k) * b^k. Step 2: Here n = 7, a = 2x^2, b = -3. Step 3: The general term is: T_{k+1} = C(7, k) * (2x^2)^(7-k) * (-3)^k. Step 4: The exponent of x in this term is 2(7 - k) = 14 - 2k. Step 5: For x^8, set 14 - 2k = 8 => 2k = 6 => k = 3. Step 6: Substitute k = 3 into the term: T_4 = C(7, 3) * (2x^2)^(4) * (-3)^3. Step 7: Compute each part: C(7, 3) = 35 (2x^2)^4 = 16 * x^8 (-3)^3 = -27 Step 8: Multiply: 35 * 16 * (-27) = 560 * (-27) = -15120. Step 9: The coefficient of x^8 is -15120. The answer is -15120.

3. Expand (2x - 3)^4 = ? Answer: 16x^4 - 96x^3 + 216x^2 - 216x + 81 Solution: We want to expand (2x - 3)^4. Use the binomial theorem. (a + b)^n = sum_{k=0}^n [C(n,k) * a^(n-k) * b^k] Here, a = 2x, b = -3, n = 4.
   Full step-by-step solution

   We want to expand (2x - 3)^4. Step 1: Use the binomial theorem. The binomial theorem says: (a + b)^n = sum_{k=0}^n [C(n,k) * a^(n-k) * b^k] Here, a = 2x, b = -3, n = 4. Step 2: Write out each term for k = 0 to 4. For k = 0: C(4,0) * (2x)^4 * (-3)^0 = 1 * (16x^4) * 1 = 16x^4. For k = 1: C(4,1) * (2x)^3 * (-3)^1 = 4 * (8x^3) * (-3) = 4 * 8 * (-3) * x^3 = 32 * (-3) x^3 = -96x^3. For k = 2: C(4,2) * (2x)^2 * (-3)^2 = 6 * (4x^2) * 9 = 6 * 4 * 9 * x^2 = 24 * 9 x^2 = 216x^2. For k = 3: C(4,3) * (2x)^1 * (-3)^3 = 4 * (2x) * (-27) = 4 * 2 * (-27) x = 8 * (-27) x = -216x. For k = 4: C(4,4) * (2x)^0 * (-3)^4 = 1 * 1 * 81 = 81. Step 3: Add all terms together. 16x^4 - 96x^3 + 216x^2 - 216x + 81. Final answer: 16x^4 - 96x^3 + 216x^2 - 216x + 81.

4. Expand (3x + 2y)^5 using the binomial theorem = ? Answer: 243x^5 + 810x^4y + 1080x^3y^2 + 720x^2y^3 + 240xy^4 + 32y^5 Solution: Recall the binomial theorem: (a + b)^n = Σ(k=0 to n) [C(n,k) * a^(n-k) * b^k] For (3x + 2y)^5, a = 3x, b = 2y, n = 5 Use binomial coefficients from Pascal's triangle row 5: 1, 5, 10, 10, 5, 1 Term 1: C(5,0) * (3x)^5 * (2y)^0 = 1 * 243x^5 * 1 = 243x^5 Term 2: C(5,1) * (3x)^4 * (2y)^1 = 5 * 81x^4…
   Full step-by-step solution

   Step 1: Recall the binomial theorem: (a + b)^n = Σ(k=0 to n) [C(n,k) * a^(n-k) * b^k] Step 2: For (3x + 2y)^5, a = 3x, b = 2y, n = 5 Step 3: Use binomial coefficients from Pascal's triangle row 5: 1, 5, 10, 10, 5, 1 Step 4: Expand each term: Term 1: C(5,0) * (3x)^5 * (2y)^0 = 1 * 243x^5 * 1 = 243x^5 Term 2: C(5,1) * (3x)^4 * (2y)^1 = 5 * 81x^4 * 2y = 810x^4y Term 3: C(5,2) * (3x)^3 * (2y)^2 = 10 * 27x^3 * 4y^2 = 1080x^3y^2 Term 4: C(5,3) * (3x)^2 * (2y)^3 = 10 * 9x^2 * 8y^3 = 720x^2y^3 Term 5: C(5,4) * (3x)^1 * (2y)^4 = 5 * 3x * 16y^4 = 240xy^4 Term 6: C(5,5) * (3x)^0 * (2y)^5 = 1 * 1 * 32y^5 = 32y^5 Step 5: Combine all terms: 243x^5 + 810x^4y + 1080x^3y^2 + 720x^2y^3 + 240xy^4 + 32y^5

5. Noah is designing a suspension bridge and needs to model the parabolic shape of the main cable. The cable's height above the deck, in meters, is given by the binomial expansion of (x + 1)^6, where x represents the horizontal distance from the left tower in units of 10 meters. Write the expanded polynomial that models the cable's height, and determine the coefficient of the x^4 term. Answer: 15 Solution: Use the binomial theorem: (x + 1)^6 = sum_{k=0}^{6} C(6, k) * x^(6-k) * 1^k. The term with x^4 occurs when 6 - k = 4, so k = 2. The coefficient is C(6, 2) = 6!
   Full step-by-step solution

   Step 1: Use the binomial theorem: (x + 1)^6 = sum_{k=0}^{6} C(6, k) * x^(6-k) * 1^k. Step 2: The term with x^4 occurs when 6 - k = 4, so k = 2. Step 3: The coefficient is C(6, 2) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 30 / 2 = 15. Step 4: The full expansion is: x^6 + 6x^5 + 15x^4 + 20x^3 + 15x^2 + 6x + 1. Step 5: Therefore, the coefficient of the x^4 term is 15. The answer is 15.

6. Expand (4x - 5)^5 = ? Answer: 1024x^5 - 6400x^4 + 16000x^3 - 20000x^2 + 12500x - 3125 Solution: Use the binomial theorem: (a + b)^n = Σ C(n,k) * a^(n-k) * b^k Here a = 4x, b = -5, n = 5 The binomial coefficients for n=5 are: 1, 5, 10, 10, 5, 1 k=0: C(5,0) * (4x)^5 * (-5)^0 = 1 * 1024x^5 * 1 = 1024x^5 k=1: C(5,1) * (4x)^4 * (-5)^1 = 5 * 256x^4 * (-5) = -6400x^4 k=2: C(5,2) * (4x)^3 * (-5)^2…
   Full step-by-step solution

   Step 1: Use the binomial theorem: (a + b)^n = Σ C(n,k) * a^(n-k) * b^k Step 2: Here a = 4x, b = -5, n = 5 Step 3: The binomial coefficients for n=5 are: 1, 5, 10, 10, 5, 1 Step 4: Expand each term: k=0: C(5,0) * (4x)^5 * (-5)^0 = 1 * 1024x^5 * 1 = 1024x^5 k=1: C(5,1) * (4x)^4 * (-5)^1 = 5 * 256x^4 * (-5) = -6400x^4 k=2: C(5,2) * (4x)^3 * (-5)^2 = 10 * 64x^3 * 25 = 16000x^3 k=3: C(5,3) * (4x)^2 * (-5)^3 = 10 * 16x^2 * (-125) = -20000x^2 k=4: C(5,4) * (4x)^1 * (-5)^4 = 5 * 4x * 625 = 12500x k=5: C(5,5) * (4x)^0 * (-5)^5 = 1 * 1 * (-3125) = -3125 Step 5: Combine all terms: 1024x^5 - 6400x^4 + 16000x^3 - 20000x^2 + 12500x - 3125

7. Hana is a botanist studying the spread of a rare fern species. She models the number of new fronds produced by a single plant over time using the expansion of (2x + 3y)^6, where x represents weeks and y represents the number of spores. Hana needs to find the coefficient of the term in the expansion that contains x^4 y^2 to understand the growth pattern at a specific stage. Using the binomial theorem, what is the coefficient of the x^4 y^2 term in the expansion of (2x + 3y)^6? Answer: 2160 Solution: Use the binomial theorem: (a + b)^6 = sum from k=0 to 6 of C(6, k) * a^(6-k) * b^k. Here, a = 2x and b = 3y. The term containing x^4 y^2 means the exponent of x is 4 and y is 2.
   Full step-by-step solution

   Step 1: Use the binomial theorem: (a + b)^6 = sum from k=0 to 6 of C(6, k) * a^(6-k) * b^k. Step 2: Here, a = 2x and b = 3y. The term containing x^4 y^2 means the exponent of x is 4 and y is 2. Step 3: The general term is C(6, k) * (2x)^(6-k) * (3y)^k. For x^4, we need 6-k = 4, so k = 2. Step 4: For k = 2, the term is C(6, 2) * (2x)^(6-2) * (3y)^2 = C(6, 2) * (2x)^4 * (3y)^2. Step 5: Calculate each part: C(6, 2) = 6!/(2! * 4!) = (6 * 5)/(2 * 1) = 15. Step 6: (2x)^4 = 2^4 * x^4 = 16 * x^4. Step 7: (3y)^2 = 3^2 * y^2 = 9 * y^2. Step 8: Multiply all parts: 15 * 16 * 9 * x^4 * y^2 = (15 * 16) = 240, then 240 * 9 = 2160. Step 9: The coefficient of the x^4 y^2 term is 2160. The answer is 2160.

8. Expand (4x - 3y)^5 = ? Answer: 1024x^5 - 3840x^4y + 5760x^3y^2 - 4320x^2y^3 + 1620xy^4 - 243y^5 Solution: Use the binomial theorem: (a + b)^n = Σ C(n,k) * a^(n-k) * b^k Here a = 4x, b = -3y, n = 5 The binomial coefficients for n=5 are: 1, 5, 10, 10, 5, 1 k=0: C(5,0) * (4x)^5 * (-3y)^0 = 1 * 1024x^5 * 1 = 1024x^5 k=1: C(5,1) * (4x)^4 * (-3y)^1 = 5 * 256x^4 * (-3y) = -3840x^4y k=2: C(5,2) * (4x)^3 *…
   Full step-by-step solution

   Step 1: Use the binomial theorem: (a + b)^n = Σ C(n,k) * a^(n-k) * b^k Step 2: Here a = 4x, b = -3y, n = 5 Step 3: The binomial coefficients for n=5 are: 1, 5, 10, 10, 5, 1 Step 4: Expand each term: k=0: C(5,0) * (4x)^5 * (-3y)^0 = 1 * 1024x^5 * 1 = 1024x^5 k=1: C(5,1) * (4x)^4 * (-3y)^1 = 5 * 256x^4 * (-3y) = -3840x^4y k=2: C(5,2) * (4x)^3 * (-3y)^2 = 10 * 64x^3 * 9y^2 = 5760x^3y^2 k=3: C(5,3) * (4x)^2 * (-3y)^3 = 10 * 16x^2 * (-27y^3) = -4320x^2y^3 k=4: C(5,4) * (4x)^1 * (-3y)^4 = 5 * 4x * 81y^4 = 1620xy^4 k=5: C(5,5) * (4x)^0 * (-3y)^5 = 1 * 1 * (-243y^5) = -243y^5 Step 5: Combine all terms: 1024x^5 - 3840x^4y + 5760x^3y^2 - 4320x^2y^3 + 1620xy^4 - 243y^5