This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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The general term in the binomial expansion of (a+b)^n is given by the formula: T_k+1 = nk a^n-k b^k where nk = (n!)/(k!(n-k)!). Problem 14: Obtain the fourth term in the expansion of (x-y)^9. Here, a=x, b=-y, n=9. For the fourth term, k+1=4, so k=3. Step 1: Apply the general term formula. T_4 = 93 (x)^9-3 (-y)^3 Step 2: Calculate the binomial coefficient. 93 = (9!)/(3!(9-3)!) = (9!)/(3!6!) = (9 × 8 × 7)/(3 × 2 × 1) = 3 × 4 × 7 = 84 Step 3: Substitute the values and simplify. T_4 = 84 x^6 (-y^3) T_4 = -84x^6y^3 The correct option is B. The fourth term is B. -84x^6y^3. Problem 15: Obtain the fourth term in the expansion of (x^2 + 3)^6. Here, a=x^2, b=3, n=6. For the fourth term, k+1=4, so k=3. Step 1: Apply the general term formula. T_4 = 63 (x^2)^6-3 (3)^3 Step 2: Calculate the binomial coefficient. 63 = (6!)/(3!(6-3)!) = (6!)/(3!3!) = (6 × 5 × 4)/(3 × 2 × 1) = 20 Step 3: Substitute the values and simplify. T_4 = 20 (x^2)^3 (3^3) T_4 = 20 (x^6) (27) T_4 = 540x^6 The correct option is B. The fourth term is B. 540x^6. Problem 16: Obtain the sixth term in the expansion of (3x - 2y)^9. Here, a=3x, b=-2y, n=9. For the sixth term, k+1=6, so k=5. Step 1: Apply the general term formula. T_6 = 95 (3x)^9-5 (-2y)^5 Step 2: Calculate the binomial coefficient. 95 = (9!)/(5!(9-5)!) = (9!)/(5!4!) = (9 × 8 × 7 × 6)/(4 × 3 × 2 × 1) = 9 × 2 × 7 = 126 Step 3: Substitute the values and simplify. T_6 = 126 (3x)^4 (-2y)^5 T_6 = 126 (3^4 x^4) ((-2)^5 y^5) T_6 = 126 (81 x^4) (-32 y^5) T_6 = 126 × 81 × (-32) x^4 y^5 T_6 = 10206 × (-32) x^4 y^5 T_6 = -326592 x^4 y^5 The correct option is D. The sixth term is D. -326592x^4y^5. Problem 17: Obtain the second term in the expansion of (2x - 1)^2. Here, a=2x, b=-1, n=2. For the second term, k+1=2, so k=1. Step 1: Apply the general term formula. T_2 = 21 (2x)^2-1 (-1)^1 Step 2: Calculate the binomial coefficient. 21 = (2!)/(1!(2-1)!) = (2!)/(1!1!) = 2 Step 3: Substitute the values and simplify. T_2 = 2 (2x)^1 (-1)^1 T_2 = 2 (2x) (-1) T_2 = -4x The correct option is C. The second term is C. -4x. Problem 18: What is the coefficient of x^3 in (2 + 3x)^4? Here, a=2, b=3x, n=4. We need to find the term where the power of x is 3. The general term is T_k+1 = nk a^n-k b^k. Step 1: Write the general term for this expansion. T_k+1 = 4k (2)^4-k (3x)^k T_k+1 = 4k (2)^4-k (3^k x^k) Step 2: Identify the value of k for x^3. For the term containing x^3, we must have x^k = x^3, so k=3. Step 3: Substitute k=3 into the general term formula. T_3+1 = T_4 = 43 (2)^4-3 (3^3 x^3) T_4 = 43 (2)^1 (3^3) x^3 Step 4: Calculate the binomial coefficient and simplify. 43 = (4!)/(3!(4-3)!) = (4!)/(3!1!) = 4 T_4 = 4 × 2 × 27 x^3 T_4 = 8 × 27 x^3 T_4 = 216 x^3 The coefficient of x^3 is 216. The correct option is D. The coefficient of x^3 is D. 216.