💡Alright, so I have this probability problem here about a cube and two people selecting lines between the centers of the cube's faces. The question is asking for the probability that the two lines they draw are parallel but not coincident. Hmm, okay, let me try to break this down step by step.First, let me visualize the cube. A cube has six faces, and each face has a center point. So, there are six points in total. Person A is going to pick any two of these six points and draw a straight line between them. Similarly, Person B is going to do the same. I need to find the probability that these two lines are parallel but not on top of each other.Alright, so the first thing I should figure out is how many different lines each person can draw. Since each person is choosing two points out of six, the number of possible lines is the combination of 6 points taken 2 at a time. The formula for combinations is C(n, k) = n! / (k!(n - k)!), right? So, plugging in the numbers, C(6, 2) = 6! / (2! * 4!) = (6 * 5) / (2 * 1) = 15. So, each person has 15 possible lines they can draw.Now, since both Person A and Person B are independently selecting lines, the total number of possible pairs of lines they can draw is 15 * 15 = 225. That makes sense because for each of Person A's 15 lines, Person B can choose any of their 15 lines, so 15 multiplied by 15 gives all possible combinations.Next, I need to figure out how many of these 225 pairs result in parallel but non-coincident lines. To do this, I should probably analyze the structure of the cube and the lines connecting the centers of its faces.I remember that the centers of the cube's faces form an octahedron. An octahedron has eight triangular faces, twelve edges, and six vertices. Each vertex of the octahedron corresponds to the center of a cube's face. So, the lines connecting these centers are essentially the edges of the octahedron.Now, in an octahedron, each edge is connected to two vertices, and each vertex is connected to four others. But I'm more interested in the directions of these edges because parallel lines would have the same direction.Wait, in a cube, the centers of opposite faces are aligned along the cube's main axes. So, for example, the centers of the top and bottom faces are aligned along the vertical axis, the centers of the front and back faces are aligned along the depth axis, and the centers of the left and right faces are aligned along the horizontal axis.But when connecting centers of adjacent faces, the lines would be along the face diagonals or the space diagonals of the cube. Hmm, actually, no. The centers of adjacent faces are connected by edges of the octahedron, which are along the cube's edges.Wait, maybe I need to think differently. Let me consider the cube with coordinates to make it clearer. Let's assign coordinates to the centers of the cube's faces. Suppose the cube is axis-aligned with side length 2, so the centers of the faces are at (±1, 0, 0), (0, ±1, 0), and (0, 0, ±1). So, these are the six points.Now, if I connect these points, the lines can be along the coordinate axes or along the face diagonals. Let me see:1. Lines along the x-axis: connecting (1, 0, 0) and (-1, 0, 0)2. Lines along the y-axis: connecting (0, 1, 0) and (0, -1, 0)3. Lines along the z-axis: connecting (0, 0, 1) and (0, 0, -1)These are the three main axes, each with only one line connecting the two opposite centers.Then, there are lines connecting centers of adjacent faces. For example, connecting (1, 0, 0) to (0, 1, 0), or (1, 0, 0) to (0, -1, 0), etc. These lines are along the face diagonals.Wait, but in the octahedron, each edge is between two centers of adjacent faces. So, each edge is a line connecting two points where one coordinate is ±1 and the other two are 0. So, for example, (1,0,0) connected to (0,1,0), (0,-1,0), (0,0,1), and (0,0,-1). So, each of these points has four connections.But in terms of lines, each edge is unique. So, how many unique directions do these lines have?Well, in the octahedron, each edge is part of a triangle, but in terms of direction, each edge has a direction vector. For example, the line from (1,0,0) to (0,1,0) has a direction vector of (-1,1,0). Similarly, the line from (1,0,0) to (0,0,1) has a direction vector of (-1,0,1). So, each of these edges has a unique direction vector.Wait, but actually, some of these direction vectors might be scalar multiples of each other, meaning they are parallel. For example, the line from (1,0,0) to (0,1,0) has direction vector (-1,1,0), and the line from (-1,0,0) to (0,-1,0) also has direction vector (1,-1,0), which is just the negative, so they are parallel.Similarly, the line from (1,0,0) to (0,0,1) has direction vector (-1,0,1), and the line from (-1,0,0) to (0,0,-1) has direction vector (1,0,-1), which is again the negative, so they are parallel.So, in the octahedron, each pair of opposite edges are parallel. So, how many such pairs are there?Let me think. Each edge has an opposite edge that's parallel. Since the octahedron has 12 edges, and each pair of opposite edges counts as one pair of parallel lines, so there are 6 pairs of parallel edges.Wait, but in our problem, we're considering lines, not edges. So, each line is defined by two points, and two lines are parallel if their direction vectors are scalar multiples.So, in the cube's face centers, how many unique directions do the lines have?We have the three main axes: x, y, z. Each has only one line connecting the two opposite centers.Then, for the face diagonals, we have lines like (1,0,0) to (0,1,0), which is a diagonal on the xy-plane. Similarly, (1,0,0) to (0,0,1) is a diagonal on the xz-plane, and (0,1,0) to (0,0,1) is a diagonal on the yz-plane.Each of these face diagonals has an opposite counterpart. For example, (1,0,0) to (0,1,0) is parallel to (-1,0,0) to (0,-1,0). Similarly, (1,0,0) to (0,0,1) is parallel to (-1,0,0) to (0,0,-1), and (0,1,0) to (0,0,1) is parallel to (0,-1,0) to (0,0,-1).So, for each face diagonal direction, there are two lines that are parallel. Therefore, for each of these three face diagonal directions, we have two lines each, making a total of 6 lines that are in three different parallel directions.Additionally, we have the three main axes, each with one line. So, in total, the lines can be categorized into six groups: three groups of two parallel lines (the face diagonals) and three groups of one line (the main axes).Wait, but in terms of counting the number of parallel line pairs, we need to consider how many pairs of lines are parallel but not coincident.So, for each of the three face diagonal directions, there are two lines. The number of ways to choose two lines from these two is C(2,2) = 1 for each direction. Since there are three such directions, that gives 3 pairs.Similarly, for the main axes, each direction only has one line, so there are no pairs of lines in those directions.But wait, in the octahedron, each edge is part of a pair of parallel edges. So, for each of the 12 edges, there is one other edge that is parallel to it. So, the total number of parallel line pairs is 12, because each of the 12 edges has one parallel counterpart, but since each pair is counted twice, we have 12 / 2 = 6 pairs.Wait, now I'm getting confused. Let me clarify.In the octahedron, there are 12 edges. Each edge has exactly one other edge that is parallel to it. So, the number of unique pairs of parallel edges is 12 / 2 = 6. So, there are 6 pairs of parallel lines.But in our problem, we're considering lines, not edges. So, each line is defined by two points, and two lines are parallel if their direction vectors are scalar multiples.So, how many unique directions do these lines have? For the main axes, we have three directions: x, y, z. Each direction has only one line.For the face diagonals, we have three different directions: xy, xz, yz. Each direction has two lines, one in each "direction" (positive and negative).So, in total, we have 3 (main axes) + 3 (face diagonals) = 6 unique directions.But for the face diagonals, each direction has two lines. So, for each of these three directions, there are two lines. Therefore, the number of pairs of parallel lines is 3 directions * C(2,2) = 3 pairs.Wait, but earlier I thought it was 6 pairs because of the octahedron's edges. Hmm.Let me think differently. Each line can be paired with another line that's parallel. So, for each line, how many lines are parallel to it?For the main axes lines, each has only one line in their direction, so they don't have any parallel lines except themselves, which we don't count because we need non-coincident lines.For the face diagonal lines, each has exactly one other line that's parallel to it. So, for each of the six face diagonal lines, there is one other line parallel to it.But wait, actually, in the octahedron, each edge is part of a pair of parallel edges. So, for each edge, there is exactly one other edge that's parallel. Therefore, the total number of parallel line pairs is 12 edges / 2 = 6 pairs.But in our case, the lines are defined by two points, so each line is an edge of the octahedron. Therefore, the number of parallel line pairs is 6.Wait, but in the problem, we have two people selecting lines. So, the total number of possible pairs is 15 * 15 = 225. The number of favorable outcomes is the number of pairs where the two lines are parallel but not coincident.So, if there are 6 pairs of parallel lines, each pair consisting of two distinct lines, then the number of favorable outcomes is 6 * 2 = 12? Wait, no.Wait, actually, for each pair of parallel lines, how many ordered pairs does that correspond to?If there are 6 pairs of parallel lines, each pair consists of two lines. So, for each such pair, the number of ordered pairs where Person A picks one and Person B picks the other is 2 (since A could pick the first and B the second, or vice versa). So, for each of the 6 pairs, there are 2 ordered pairs, so total favorable outcomes would be 6 * 2 = 12.Alternatively, if we think of unordered pairs, the number of unordered pairs of parallel lines is 6, each contributing 2 ordered pairs, so 12 ordered pairs.Therefore, the number of favorable outcomes is 12.Hence, the probability is 12 / 225 = 4 / 75.Wait, but let me double-check. So, total number of lines is 15. Number of parallel line pairs is 6, each contributing 2 ordered pairs, so 12. Therefore, probability is 12 / 225 = 4 / 75.Yes, that seems right.So, the answer should be D: 4/75.

💡Okay, so I have this equation to solve: (x + 11)/(x - 4) = (x - 3)/(x + 6). Hmm, it's a proportion, right? So I think I can use cross-multiplication to solve for x. Let me try that.First, I'll cross-multiply the numerators and denominators. That means I'll multiply (x + 11) by (x + 6) and set it equal to (x - 3) multiplied by (x - 4). So, writing that out, I get:(x + 11)(x + 6) = (x - 3)(x - 4)Alright, now I need to expand both sides of this equation. Let's start with the left side. Multiplying (x + 11) by (x + 6):x * x = x²x * 6 = 6x11 * x = 11x11 * 6 = 66So, adding all those up, the left side becomes x² + 6x + 11x + 66. Combining like terms, 6x + 11x is 17x, so the left side simplifies to x² + 17x + 66.Now, let's do the same for the right side: (x - 3)(x - 4).x * x = x²x * (-4) = -4x(-3) * x = -3x(-3) * (-4) = 12Adding those up, the right side becomes x² - 4x - 3x + 12. Combining like terms, -4x - 3x is -7x, so the right side simplifies to x² - 7x + 12.Now, putting it all together, the equation is:x² + 17x + 66 = x² - 7x + 12Hmm, I notice that both sides have an x² term. Maybe I can subtract x² from both sides to simplify. Let's try that:x² + 17x + 66 - x² = x² - 7x + 12 - x²Simplifying both sides, the x² terms cancel out:17x + 66 = -7x + 12Okay, now I have a linear equation. I need to solve for x. Let's get all the x terms on one side and the constants on the other. I'll add 7x to both sides to move the x terms to the left:17x + 7x + 66 = -7x + 7x + 12That simplifies to:24x + 66 = 12Now, I'll subtract 66 from both sides to isolate the term with x:24x + 66 - 66 = 12 - 66Which simplifies to:24x = -54Alright, now I can solve for x by dividing both sides by 24:x = -54 / 24Hmm, this fraction can be simplified. Both 54 and 24 are divisible by 6. Let's divide numerator and denominator by 6:-54 ÷ 6 = -924 ÷ 6 = 4So, x = -9/4.Wait a minute, I should check if this solution is valid in the original equation. Sometimes when you multiply both sides, you might introduce extraneous solutions or maybe divide by zero somewhere. Let me plug x = -9/4 back into the original equation to verify.First, let's compute the denominators:For the left side: x - 4 = (-9/4) - 4 = (-9/4) - (16/4) = (-25/4). That's not zero, so it's fine.For the right side: x + 6 = (-9/4) + 6 = (-9/4) + (24/4) = (15/4). Also not zero, so we're good.Now, let's compute both sides:Left side: (x + 11)/(x - 4) = [(-9/4) + 11]/[(-25/4)] = [( -9/4 + 44/4 )]/(-25/4) = (35/4)/(-25/4) = (35/4) * (-4/25) = -140/100 = -1.4Right side: (x - 3)/(x + 6) = [(-9/4) - 3]/[(15/4)] = [(-9/4 - 12/4)]/(15/4) = (-21/4)/(15/4) = (-21/4) * (4/15) = -84/60 = -1.4Both sides equal -1.4, so the solution checks out. Therefore, x = -9/4 is the correct solution.I think that's it. I didn't make any mistakes in my calculations, and the solution satisfies the original equation. So, x = -9/4 is the answer.

💡Okay, so I have this problem where I need to find the value of ( k ) such that the tetrahedron formed by the points ( P_1(1,2,3) ), ( P_2(2,4,1) ), ( P_3(1,k,5) ), and ( P_4(4,k+1,3) ) has a volume of 1. Hmm, I remember that the volume of a tetrahedron can be found using the scalar triple product of vectors. Let me try to recall the exact formula.I think the formula for the volume ( V ) is given by:[V = frac{1}{6} | vec{AB} cdot (vec{AC} times vec{AD}) |]Where ( vec{AB} ), ( vec{AC} ), and ( vec{AD} ) are vectors from point ( A ) to points ( B ), ( C ), and ( D ) respectively. So, in this case, I can choose ( P_1 ) as point ( A ), and then ( P_2 ), ( P_3 ), and ( P_4 ) as points ( B ), ( C ), and ( D ).First, I need to find the vectors ( vec{P_1P_2} ), ( vec{P_1P_3} ), and ( vec{P_1P_4} ).Calculating each vector:- ( vec{P_1P_2} = P_2 - P_1 = (2-1, 4-2, 1-3) = (1, 2, -2) )- ( vec{P_1P_3} = P_3 - P_1 = (1-1, k-2, 5-3) = (0, k-2, 2) )- ( vec{P_1P_4} = P_4 - P_1 = (4-1, (k+1)-2, 3-3) = (3, k-1, 0) )Okay, so now I have the three vectors:- ( vec{AB} = (1, 2, -2) )- ( vec{AC} = (0, k-2, 2) )- ( vec{AD} = (3, k-1, 0) )Next, I need to compute the cross product ( vec{AC} times vec{AD} ). Let me recall that the cross product of two vectors ( vec{u} = (u_1, u_2, u_3) ) and ( vec{v} = (v_1, v_2, v_3) ) is given by:[vec{u} times vec{v} = (u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1)]So, applying this to ( vec{AC} ) and ( vec{AD} ):Let me denote ( vec{AC} = (0, k-2, 2) ) as ( (u_1, u_2, u_3) ) and ( vec{AD} = (3, k-1, 0) ) as ( (v_1, v_2, v_3) ).Calculating each component of the cross product:- First component: ( u_2v_3 - u_3v_2 = (k-2)(0) - (2)(k-1) = 0 - 2(k - 1) = -2k + 2 )- Second component: ( u_3v_1 - u_1v_3 = (2)(3) - (0)(0) = 6 - 0 = 6 )- Third component: ( u_1v_2 - u_2v_1 = (0)(k-1) - (k-2)(3) = 0 - 3(k - 2) = -3k + 6 )So, the cross product ( vec{AC} times vec{AD} ) is:[(-2k + 2, 6, -3k + 6)]Now, I need to compute the dot product of ( vec{AB} ) with this cross product. Let me denote ( vec{AB} = (1, 2, -2) ) as ( (a, b, c) ) and the cross product as ( (d, e, f) ).The dot product is:[a cdot d + b cdot e + c cdot f]Substituting the values:[1 cdot (-2k + 2) + 2 cdot 6 + (-2) cdot (-3k + 6)]Let me compute each term:- First term: ( 1 cdot (-2k + 2) = -2k + 2 )- Second term: ( 2 cdot 6 = 12 )- Third term: ( -2 cdot (-3k + 6) = 6k - 12 )Adding all these together:[(-2k + 2) + 12 + (6k - 12) = (-2k + 6k) + (2 + 12 - 12) = 4k + 2]So, the scalar triple product is ( 4k + 2 ). The volume is one-sixth of the absolute value of this:[V = frac{1}{6} |4k + 2|]We are given that the volume ( V ) is 1, so:[frac{1}{6} |4k + 2| = 1]Multiplying both sides by 6:[|4k + 2| = 6]This absolute value equation leads to two possible equations:1. ( 4k + 2 = 6 )2. ( 4k + 2 = -6 )Let me solve each equation for ( k ).First equation:[4k + 2 = 6 4k = 6 - 2 4k = 4 k = 1]Second equation:[4k + 2 = -6 4k = -6 - 2 4k = -8 k = -2]So, the possible values for ( k ) are 1 and -2. Let me just verify if these make sense.If ( k = 1 ), then ( P_3 = (1, 1, 5) ) and ( P_4 = (4, 2, 3) ). Plugging back into the volume formula, it should give 1. Similarly, if ( k = -2 ), then ( P_3 = (1, -2, 5) ) and ( P_4 = (4, -1, 3) ). I think both should satisfy the volume condition.Wait, just to be thorough, maybe I should plug these values back into the scalar triple product to ensure the volume is indeed 1.For ( k = 1 ):Scalar triple product: ( 4(1) + 2 = 6 ). So, volume is ( frac{1}{6} times 6 = 1 ). Correct.For ( k = -2 ):Scalar triple product: ( 4(-2) + 2 = -8 + 2 = -6 ). Taking absolute value, it's 6. Volume is ( frac{1}{6} times 6 = 1 ). Correct.So, both values are valid. Therefore, ( k ) can be either 1 or -2.**Final Answer**The possible values of ( k ) are boxed{1} and boxed{-2}.

💡First, I need to find the coefficient of the ( x^2 ) term in the expansion of the sum ( 1 + (1+x) + (1+x)^2 + ldots + (1+x)^9 ).I recall that the binomial expansion of ( (1+x)^k ) is given by:[(1+x)^k = sum_{n=0}^{k} binom{k}{n} x^n]where ( binom{k}{n} ) is the binomial coefficient.To find the coefficient of ( x^2 ), I need to consider the ( x^2 ) term from each expansion ( (1+x)^k ) where ( k ) ranges from 2 to 9. For each ( k ), the coefficient of ( x^2 ) is ( binom{k}{2} ).So, the coefficients to sum are:[binom{2}{2}, binom{3}{2}, ldots, binom{9}{2}]Using the Hockey-Stick Identity, which states:[sum_{k=r}^{n} binom{k}{r} = binom{n+1}{r+1}]I can simplify the sum:[binom{2}{2} + binom{3}{2} + ldots + binom{9}{2} = binom{10}{3}]Calculating ( binom{10}{3} ):[binom{10}{3} = frac{10!}{3!(10-3)!} = frac{10 times 9 times 8}{3 times 2 times 1} = 120]Therefore, the coefficient of the ( x^2 ) term in the given expression is 120.

💡Okay, so I have this problem about arithmetic and geometric sequences. Let me try to break it down step by step.First, part (1) says: Prove that 0 < c < d, and deduce the value of c from b₂ < a₃.Alright, so we have an arithmetic sequence {a_n} with first term c and common difference d. So, the terms are a₁ = c, a₂ = c + d, a₃ = c + 2d, and so on.Then, we have a geometric sequence {b_n} with first term d and common ratio c. So, the terms are b₁ = d, b₂ = d * c, b₃ = d * c², etc.We are given that a₁ < b₁ < a₂ < b₂ < a₃. Let me write down these inequalities:1. a₁ < b₁: c < d2. b₁ < a₂: d < c + d3. a₂ < b₂: c + d < d * c4. b₂ < a₃: d * c < c + 2dSo, from the first inequality, c < d. That gives us 0 < c < d because if c were negative, d would have to be greater than c, but since we have a geometric sequence with common ratio c, if c is negative, the terms might alternate or behave unpredictably. But since all terms are integers, maybe c is positive? Wait, the problem says c and d are integers, but doesn't specify they are positive. Hmm.But let's see. From the second inequality, d < c + d. If we subtract d from both sides, we get 0 < c. So, c is positive. So, 0 < c < d.Okay, that's the first part done: 0 < c < d.Now, we need to deduce the value of c from b₂ < a₃. So, let's look at the fourth inequality: d * c < c + 2d.Let me rewrite that: d * c - c < 2d. Factor out c: c(d - 1) < 2d.Since c and d are integers and 0 < c < d, let's see possible small integer values for c and d.Let me consider c = 1. Then, the inequality becomes 1*(d - 1) < 2d ⇒ d - 1 < 2d ⇒ -1 < d, which is always true since d > c = 1. But let's check the third inequality: a₂ < b₂ ⇒ c + d < d * c. If c = 1, then 1 + d < d * 1 ⇒ 1 + d < d ⇒ 1 < 0, which is false. So c cannot be 1.Next, c = 2. Let's check the third inequality: a₂ < b₂ ⇒ 2 + d < d * 2 ⇒ 2 + d < 2d ⇒ 2 < d. Since c = 2 and 0 < c < d, d must be at least 3.Now, check the fourth inequality: b₂ < a₃ ⇒ 2d < 2 + 2d ⇒ 2d < 2 + 2d ⇒ 0 < 2, which is always true. So, c = 2 works.Wait, but let me check if c can be 3. If c = 3, then from the third inequality: a₂ < b₂ ⇒ 3 + d < 3d ⇒ 3 < 2d ⇒ d > 1.5. Since d is integer and d > c = 3, d ≥ 4.Now, check the fourth inequality: b₂ < a₃ ⇒ 3d < 3 + 2d ⇒ 3d - 2d < 3 ⇒ d < 3. But d ≥ 4, which contradicts. So c cannot be 3.Similarly, c cannot be greater than 2 because it would lead to a contradiction in the fourth inequality. So c must be 2.Okay, so part (1) is done: 0 < c < d and c = 2.Now, part (2): If the sequence {a_n} has a total of 3n terms, with the sum of the first n terms being A, the sum of the next n terms being B, and the sum of the last n terms being C, find the value of (B² - AC)/(A - C)².Alright, let's recall that the sum of an arithmetic sequence is S_k = (k/2)(2a₁ + (k - 1)d).Given that {a_n} is arithmetic with a₁ = c = 2 and common difference d.So, the sum of the first n terms, A = S_n = (n/2)(2*2 + (n - 1)d) = (n/2)(4 + (n - 1)d).Similarly, the sum of the first 2n terms, S_{2n} = (2n/2)(4 + (2n - 1)d) = n(4 + (2n - 1)d).Then, the sum of the next n terms, B = S_{2n} - S_n = n(4 + (2n - 1)d) - (n/2)(4 + (n - 1)d).Let me compute that:B = n(4 + 2n d - d) - (n/2)(4 + n d - d)= n(4 + 2n d - d) - (n/2)(4 + n d - d)= 4n + 2n² d - n d - 2n - (n² d)/2 + (n d)/2= (4n - 2n) + (2n² d - (n² d)/2) + (-n d + (n d)/2)= 2n + (3n² d)/2 - (n d)/2Similarly, the sum of the first 3n terms, S_{3n} = (3n/2)(4 + (3n - 1)d).Then, the sum of the last n terms, C = S_{3n} - S_{2n} = (3n/2)(4 + (3n - 1)d) - n(4 + (2n - 1)d).Compute C:C = (3n/2)(4 + 3n d - d) - n(4 + 2n d - d)= (3n/2)(4 + 3n d - d) - 4n - 2n² d + n d= (6n + (9n² d)/2 - (3n d)/2) - 4n - 2n² d + n d= (6n - 4n) + (9n² d)/2 - 2n² d + (-3n d/2 + n d)= 2n + (5n² d)/2 + (-n d)/2Now, we have expressions for A, B, and C.Let me write them again:A = (n/2)(4 + (n - 1)d) = 2n + (n(n - 1)d)/2B = 2n + (3n² d)/2 - (n d)/2C = 2n + (5n² d)/2 - (n d)/2Wait, maybe there's a pattern here. Let me see:A = 2n + (n² d - n d)/2B = 2n + (3n² d - n d)/2C = 2n + (5n² d - n d)/2So, A, B, C are linear in n with coefficients involving d.Now, let's compute B² - AC and (A - C)².First, compute A - C:A - C = [2n + (n² d - n d)/2] - [2n + (5n² d - n d)/2] = (n² d - n d)/2 - (5n² d - n d)/2 = (-4n² d)/2 = -2n² dSo, (A - C)² = ( -2n² d )² = 4n⁴ d²Now, compute B² - AC.First, compute B²:B = 2n + (3n² d - n d)/2 = 2n + (3n² d)/2 - (n d)/2So, B² = [2n + (3n² d)/2 - (n d)/2]^2This will be a bit messy, but let's see.Alternatively, maybe there's a smarter way. Since A, B, C are linear in n, perhaps B is the average of A and C? Wait, let's check:From A, B, C expressions:A = 2n + (n² d - n d)/2C = 2n + (5n² d - n d)/2So, (A + C)/2 = [2n + (n² d - n d)/2 + 2n + (5n² d - n d)/2]/2 = [4n + (6n² d - 2n d)/2]/2 = [4n + 3n² d - n d]/2 = 2n + (3n² d - n d)/2 = BSo, B is the average of A and C. Therefore, B = (A + C)/2.Thus, 2B = A + C ⇒ A + C = 2B.Therefore, B² - AC can be expressed in terms of A and C.Let me compute B² - AC:Since B = (A + C)/2, then B² = (A + C)² / 4.So, B² - AC = (A + C)² / 4 - AC = (A² + 2AC + C²)/4 - AC = (A² + 2AC + C² - 4AC)/4 = (A² - 2AC + C²)/4 = (A - C)² / 4.Therefore, (B² - AC)/(A - C)² = ( (A - C)² / 4 ) / (A - C)² = 1/4.So, the value is 1/4.Wait, that seems too straightforward. Did I make a mistake?Let me verify:Given that B is the average of A and C, so B = (A + C)/2.Then, B² - AC = (A + C)² /4 - AC = (A² + 2AC + C²)/4 - AC = (A² - 2AC + C²)/4 = (A - C)² /4.Thus, (B² - AC)/(A - C)² = 1/4.Yes, that seems correct.So, part (2) answer is 1/4.Now, part (3): If the sums of the first n, first 2n, and first 3n terms of the sequence {b_n} are D, G, and H respectively, express H in terms of D and G, i.e., H = f(D, G), without involving the letter d.Alright, {b_n} is a geometric sequence with first term d and common ratio c = 2.So, the sum of the first k terms is S_k = d*(2^k - 1)/(2 - 1) = d*(2^k - 1).Therefore:D = sum of first n terms = d*(2^n - 1)G = sum of first 2n terms = d*(2^{2n} - 1)H = sum of first 3n terms = d*(2^{3n} - 1)We need to express H in terms of D and G without involving d.Let me see:From D = d*(2^n - 1), we can write d = D / (2^n - 1)Similarly, G = d*(2^{2n} - 1) = d*( (2^n)^2 - 1 ) = d*(2^n - 1)(2^n + 1)So, G = D*(2^n + 1)Thus, 2^n + 1 = G / DSimilarly, H = d*(2^{3n} - 1) = d*( (2^n)^3 - 1 ) = d*(2^n - 1)(2^{2n} + 2^n + 1)But 2^n - 1 = D / d, so H = d*(D / d)*(2^{2n} + 2^n + 1) = D*(2^{2n} + 2^n + 1)But we need to express 2^{2n} and 2^n in terms of D and G.We have 2^n + 1 = G / D ⇒ 2^n = (G / D) - 1Then, 2^{2n} = (2^n)^2 = ( (G / D) - 1 )²So, 2^{2n} + 2^n + 1 = ( (G/D - 1)^2 ) + (G/D - 1) + 1Let me compute that:= (G²/D² - 2G/D + 1) + (G/D - 1) + 1= G²/D² - 2G/D + 1 + G/D - 1 + 1= G²/D² - G/D + 1Therefore, H = D*(G²/D² - G/D + 1) = (G²/D) - G + DSo, H = (G²)/D - G + DAlternatively, H = (G² - GD + D²)/D = (G² - GD + D²)/DBut the problem says to express H in terms of D and G without involving d, so H = (G²)/D + D - G.Yes, that seems correct.Let me check:H = D*(2^{3n} - 1) = D*( (2^n)^3 - 1 ) = D*(2^n - 1)(2^{2n} + 2^n + 1)We have 2^n - 1 = D / d, but we expressed 2^n = (G/D) - 1, so 2^{2n} = (G/D - 1)^2.Thus, H = D*( (G/D - 1)^2 + (G/D - 1) + 1 )= D*( (G²/D² - 2G/D + 1) + (G/D - 1) + 1 )= D*( G²/D² - 2G/D + 1 + G/D - 1 + 1 )= D*( G²/D² - G/D + 1 )= G²/D - G + DYes, that's correct.So, part (3) answer is H = (G²)/D + D - G.

💡First, I start with the initial length of the thread, which is 12 cm.Next, I add two-thirds of the initial length to the thread. Calculating two-thirds of 12 cm gives me 8 cm. Adding this to the initial length results in a new length of 20 cm.Then, I multiply this new length by three-quarters. Three-quarters of 20 cm is 15 cm.After that, I divide the current length by one and a half. Dividing 15 cm by 1.5 gives me 10 cm.Finally, I remove one-fourth of this length. One-fourth of 10 cm is 2.5 cm. Subtracting this from 10 cm leaves me with 7.5 cm.To convert the remaining length to inches, I divide 7.5 cm by 2.54 cm per inch, resulting in approximately 2.95 inches.