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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You're on a roll — here are the solutions to question 2. 2. a) Find the binomial expansion of sqrt(1-8x), |x| < (1)/(8), in ascending powers of x up to and including the term in x^3, simplifying each term. Step 1: Rewrite the expression in the form (1+y)^n. sqrt(1-8x) = (1-8x)^(1)/(2) Here, n = (1)/(2) and y = -8x. Step 2: Apply the binomial expansion formula (1+y)^n = 1 + ny + (n(n-1))/(2!)y^2 + (n(n-1)(n-2))/(3!)y^3 + . (1-8x)^(1)/(2) = 1 + ((1)/(2))(-8x) + (1)/(2)((1)/(2)-1)2!(-8x)^2 + (1)/(2)((1)/(2)-1)((1)/(2)-2)3!(-8x)^3 + Step 3: Simplify each term. = 1 + ((1)/(2))(-8x) + (1)/(2)(-(1)/(2))2(64x^2) + (1)/(2)(-(1)/(2))(-(3)/(2))6(-512x^3) + = 1 - 4x + (-1)/(4)2(64x^2) + (3)/(8)6(-512x^3) + = 1 - 4x - (1)/(8)(64x^2) + (3)/(48)(-512x^3) + = 1 - 4x - 8x^2 - 32x^3 + The binomial expansion is 1 - 4x - 8x^2 - 32x^3. b) Show that, when x = (1)/(100), the exact value of sqrt(1-8x) is sqrt(23)5. Step 1: Substitute x = (1)/(100) into the expression sqrt(1-8x). sqrt(1 - 8((1)/(100))) Step 2: Simplify the expression. = sqrt(1 - (8)/(100)) = sqrt(1 - (2)/(25)) = sqrt((25)/(25) - (2)/(25)) = sqrt((23)/(25)) = sqrt(23)sqrt(25) = sqrt(23)5 This shows that the exact value of sqrt(1-8x) is sqrt(23)5. c) Substitute x = (1)/(100) into the expansion in part (a) and hence obtain an approximation to sqrt(23). Give your answer to 5 decimal places. Step 1: Substitute x = (1)/(100) = 0.01 into the expansion from part (a): 1 - 4x - 8x^2 - 32x^3. sqrt(1-8x) ≈ 1 - 4(0.01) - 8(0.01)^2 - 32(0.01)^3 Step 2: Calculate the value. = 1 - 0.04 - 8(0.0001) - 32(0.000001) = 1 - 0.04 - 0.0008 - 0.000032 = 0.96 - 0.0008 - 0.000032 = 0.9592 - 0.000032 = 0.959168 Step 3: Use the result from part (b) to find an approximation for sqrt(23). We know sqrt(23)5 ≈ 0.959168. sqrt(23) ≈ 5 × 0.959168 sqrt(23) ≈ 4.79584 Step 4: Round the answer to 5 decimal places. sqrt(23) ≈ 4.79584 The approximation to sqrt(23) is 4.79584. d) Given that (2+x)^5 + (2-x)^5 = A + Bx^2 + Cx^4, find the constants A, B and C. Step 1: Expand (2+x)^5 using the binomial theorem (a+b)^n = _k=0^n nk a^n-k b^k. (2+x)^5 = 502^5x^0 + 512^4x^1 + 522^3x^2 + 532^2x^3 + 542^1x^4 + 552^0x^5 = 1(32)(1) + 5(16)x + 10(8)x^2 + 10(4)x^3 + 5(2)x^4 + 1(1)x^5 = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5 Step 2: Expand (2-x)^5. This is similar to (2+x)^5, but with alternating signs for odd powers of x. (2-x)^5 = 502^5(-x)^0 + 512^4(-x)^1 + 522^3(-x)^2 + 532^2(-x)^3 + 542^1(-x)^4 + 552^0(-x)^5 = 1(32)(1) + 5(16)(-x) + 10(8)x^2 + 10(4)(-x^3) + 5(2)x^4 + 1(1)(-x^5) = 32 - 80x + 80x^2 - 40x^3 + 10x^4 - x^5 Step 3: Add the two expansions. (2+x)^5 + (2-x)^5 = (32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5) + (32 - 80x + 80x^2 - 40x^3 + 10x^4 - x^5) Combine like terms: = (32+32) + (80x-80x) + (80x^2+80x^2) + (40x^3-40x^3) + (10x^4+10x^4) + (x^5-x^5) = 64 + 0x + 160x^2 + 0x^3 + 20x^4 + 0x^5 = 64 + 160x^2 + 20x^4 Step 4: Compare this result with A + Bx^2 + Cx^4. By comparing the coefficients: A = 64 B = 160 C = 20 The constants are A=64, B=160, C=20. Send me the next one 📸