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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Q7 a) To write down the binomial expansion of (1-y)^6, we use the binomial theorem (a+b)^n = _k=0^n nk a^n-k b^k. Here, a=1, b=-y, and n=6. Step 1: Write out the binomial expansion terms. (1-y)^6 = 60(1)^6(-y)^0 + 61(1)^5(-y)^1 + 62(1)^4(-y)^2 + 63(1)^3(-y)^3 + 64(1)^2(-y)^4 + 65(1)^1(-y)^5 + 66(1)^0(-y)^6 Step 2: Calculate the binomial coefficients and simplify each term. 60 = 1 61 = 6 62 = (6 × 5)/(2 × 1) = 15 63 = (6 × 5 × 4)/(3 × 2 × 1) = 20 64 = 15 65 = 6 66 = 1 Substitute these values: (1-y)^6 = 1(1)(1) + 6(1)(-y) + 15(1)(y^2) + 20(1)(-y^3) + 15(1)(y^4) + 6(1)(-y^5) + 1(1)(y^6) (1-y)^6 = 1 - 6y + 15y^2 - 20y^3 + 15y^4 - 6y^5 + y^6 The binomial expansion is 1 - 6y + 15y^2 - 20y^3 + 15y^4 - 6y^5 + y^6. Q7 b) To deduce the expansion of (1 - x + x^2)^6 in ascending powers of x as far as the term in x^5, we substitute y = (x - x^2) into the expansion from part (a). Step 1: Substitute y = (x - x^2) into the expansion of (1-y)^6. (1 - (x - x^2))^6 = (1 - x + x^2)^6 We need to expand each term of 1 - 6y + 15y^2 - 20y^3 + 15y^4 - 6y^5 + y^6 up to x^5. Step 2: Expand powers of y = (x - x^2) up to x^5. y = x - x^2 y^2 = (x - x^2)^2 = x^2 - 2x^3 + x^4 y^3 = (x - x^2)^3 = x^3 - 3x^2(x^2) + 3x(x^2)^2 - (x^2)^3 = x^3 - 3x^4 + 3x^5 - x^6 y^4 = (x - x^2)^4 = x^4(1-x)^4 = x^4(1 - 4x + ) = x^4 - 4x^5 + y^5 = (x - x^2)^5 = x^5(1-x)^5 = x^5(1 - ) = x^5 - Terms y^6 and higher will start with x^6 or higher, so they are not needed for terms up to x^5. Step 3: Substitute these expansions into the binomial expansion from part (a) and collect terms. (1 - x + x^2)^6 = 1 - 6(x - x^2) + 15(x^2 - 2x^3 + x^4) - 20(x^3 - 3x^4 + 3x^5) + 15(x^4 - 4x^5) - 6(x^5) + = 1 - 6x + 6x^2 + 15x^2 - 30x^3 + 15x^4 20x^3 + 60x^4 - 60x^5 + 15x^4 - 60x^5 6x^5 Step 4: Collect coefficients for each power of x. Constant term: 1 Coefficient of x: -6 Coefficient of x^2: 6 + 15 = 21 Coefficient of x^3: -30 - 20 = -50 Coefficient of x^4: 15 + 60 + 15 = 90 Coefficient of x^5: -60 - 60 - 6 = -126 The expansion of (1 - x + x^2)^6 up to the term in x^5 is: 1 - 6x + 21x^2 - 50x^3 + 90x^4 - 126x^5 Q7 c) To evaluate 0.91^6 correct to two decimal places by taking x=0.1 in (b). Step 1: Verify that x=0.1 makes (1 - x + x^2)^6 equal to 0.91^6. Substitute x=0.1 into (1 - x + x^2): 1 - 0.1 + (0.1)^2 = 1 - 0.1 + 0.01 = 0.9 + 0.01 = 0.91 So, (1 - x + x^2)^6 = (0.91)^6. Step 2: Substitute x=0.1 into the expansion obtained in part (b). (0.91)^6 ≈ 1 - 6(0.1) + 21(0.1)^2 - 50(0.1)^3 + 90(0.1)^4 - 126(0.1)^5 = 1 - 0.6 + 21(0.01) - 50(0.001) + 90(0.0001) - 126(0.00001) = 1 - 0.6 + 0.21 - 0.05 + 0.009 - 0.00126 Step 3: Calculate the sum. 1 - 0.6 = 0.4 0.4 + 0.21 = 0.61 0.61 - 0.05 = 0.56 0.56 + 0.009 = 0.569 0.569 - 0.00126 = 0.56774 Step 4: Round the result to two decimal places. 0.56774 ≈ 0.57 The value of 0.91^6 correct to two decimal places is 0.57. Drop the next question.