Expand the expression (1-2x)2.

Another one Princess — let's solve it. Question 27: The term independent of (x) in the expansion of (1-2x)^2 is Step 1: Expand the expression (1-2x)^2. (1-2x)^2 = (1)^2 - 2(1)(2x) + (2x)^2 = 1 - 4x + 4x^2 Step 2: Identify the term independent of x. The term independent of x is the constant term. In the expansion 1 - 4x + 4x^2, the constant term is 1. The term independent of x is 1. Since 1 is not among options a) 15, b) -15, c) 32, d) -32, the correct option is e) None of the above. Question 28: In the expansion of (a-b)^n, the sum of the coefficients is Step 1: To find the sum of coefficients of a polynomial, substitute 1 for each variable. For the expansion of (a-b)^n, substitute a=1 and b=1. Sum of coefficients = (1-1)^n = 0^n. Step 2: Evaluate 0^n. If n > 0, then 0^n = 0. In binomial expansion, n is typically a positive integer. The sum of the coefficients is 0. This matches option a) 0. Question 29: The middle term in the expansion of (1+x)^10 is Step 1: Determine the number of terms in the expansion. For (a+b)^n, there are n+1 terms. Here n=10, so there are 10+1=11 terms. Step 2: Find the position of the middle term. Since there are 11 terms (an odd number), there is one middle term. Its position is (n+1+1)/(2) = (11+1)/(2) = (12)/(2) = 6. So, it's the 6-th term. Step 3: Use the general term formula T_k+1 = nk a^n-k b^k. For the 6-th term, k+1=6 k=5. For (1+x)^10, a=1, b=x, n=10. T_6 = 105 (1)^10-5 (x)^5 T_6 = 105 (1)^5 (x)^5 T_6 = 105 x^5 The question asks for "the middle term", and the options are powers of x. This implies we should identify the variable part of the middle term. The middle term is 105 x^5. The variable part is x^5. This matches option a) x^5. Question 30: The coefficient of the term containing x^3 in the expansion of (1-2x)^-4 is Step 1: Use the generalized binomial theorem for (1+y)^n = 1 + ny + (n(n-1))/(2!)y^2 + (n(n-1)(n-2))/(3!)y^3 + Here, y = -2x and n = -4. We need the term containing x^3, which corresponds to the term with y^3. The coefficient of y^3 is (n(n-1)(n-2))/(3!). Step 2: Substitute n=-4 and y=-2x into the y^3 term. The term is ((-4)(-4-1)(-4-2))/(3!) (-2x)^3. = ((-4)(-5)(-6))/(3 × 2 × 1) (-2)^3 x^3 = (-120)/(6) (-8) x^3 = (-20) (-8) x^3 = 160x^3 Step 3: Identify the coefficient of x^3. The coefficient of x^3 is 160. The coefficient of the term containing x^3 is 160. Since 160 is not among options a) -4, b) -8, c) 8, d) 4, the correct option is e) None of the above. Question 31: What is the term independent of x in the expansion of (1+x)^7? Step 1: Use the general term formula T_k+1 = nk a^n-k b^k. For (1+x)^7, a=1, b=x, n=7. The term independent of x is the term where the power of x is 0. So, k=0. Step 2: Substitute k=0 into the general term formula. T_0+1 = 70 (1)^7-0 (x)^0 T_1 = 70 (1)^7 (1) Step 3: Calculate the value. 70 = 1 T_1 = 1 · 1 · 1 = 1 The term independent of x is 1. This matches option a) 1. Question 32: The coefficient of the second term in the expansion of (3-2x)^5 is Step 1: Use the general term formula T_k+1 = nk a^n-k b^k. For (3-2x)^5, a=3, b=-2x, n=5. We need the second term, so k+1=2 k=1. Step 2: Substitute k=1 into the general term formula. T_1+1 = 51 (3)^5-1 (-2x)^1 T_2 = 51 (3)^4 (-2x) Step 3: Calculate the values. 51 = 5 3^4 = 81 T_2 = 5 · 81 · (-2x) T_2 = 405 · (-2x) T_2 = -810x Step 4: Identify the coefficient of the second term. The coefficient of the second term is -810. The coefficient of the second term is -810. Since -810 is not among options a) -120, b) 240, c) -240, d) 120, the correct option is e) None of the above. Question 33: The sum of the coefficients in the expansion of (a+b)^n is Step 1: To find the sum of coefficients of a polynomial, substitute 1 for each variable. For the expansion of (a+b)^n, substitute a=1 and b=1. Sum of coefficients = (1+1)^n = 2^n. The sum of the coefficients is 2^n. This matches option c) 2^n. Send me the next one 📸