answer：Okay, so I have this problem about four rectangles arranged in a specific way, and I need to find the area of triangle ABC. Let me try to visualize this setup first.Each rectangle has a length of 2 units and a height of 1 unit. They are arranged so that each rectangle shares exactly one side with another rectangle. That probably means they're connected edge-to-edge, forming some kind of shape. Maybe like a bigger rectangle or a more complex figure?Given the coordinates: A is at (2, 0), B is at (0, 1), and C is at (2, 1). Also, A and B are vertices from different rectangles, and C is where two rectangles meet. Hmm, so C is a corner where two rectangles come together.Let me plot these points mentally. Point A is at (2, 0), which is on the x-axis, two units to the right. Point B is at (0, 1), which is on the y-axis, one unit up. Point C is at (2, 1), which is two units to the right and one unit up. So, triangle ABC has vertices at these three points.I think the best way to find the area of triangle ABC is to use the coordinate points. There's a formula for the area of a triangle given three vertices: Area = (1/2) |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|Let me plug in the coordinates into this formula.So, x₁ = 2, y₁ = 0 (point A)x₂ = 0, y₂ = 1 (point B)x₃ = 2, y₃ = 1 (point C)Plugging into the formula:Area = (1/2) |2*(1 - 1) + 0*(1 - 0) + 2*(0 - 1)|Calculating each term:First term: 2*(1 - 1) = 2*0 = 0Second term: 0*(1 - 0) = 0*1 = 0Third term: 2*(0 - 1) = 2*(-1) = -2Adding them up: 0 + 0 - 2 = -2Taking the absolute value: |-2| = 2Then multiplying by 1/2: (1/2)*2 = 1So, the area is 1 unit². That seems straightforward.Wait, let me double-check. Maybe I can use another method to confirm. How about using the base and height?Looking at points A(2,0), C(2,1), and B(0,1). If I consider AC as one side, the length from A to C is vertical, from (2,0) to (2,1), so that's 1 unit. Then, the base could be AC with length 1.But then, what's the height? The height would be the horizontal distance from point B to the line AC. Since AC is a vertical line at x=2, the horizontal distance from B(0,1) to x=2 is |0 - 2| = 2 units.Wait, but that would make the area (1/2)*base*height = (1/2)*1*2 = 1. Same result. Okay, that confirms it.Alternatively, I can plot the points on a graph. A is at (2,0), which is on the x-axis. C is at (2,1), directly above A. B is at (0,1), which is on the y-axis, one unit up. So, triangle ABC is a right triangle? Let me see.From A(2,0) to C(2,1) is vertical, and from C(2,1) to B(0,1) is horizontal. So, the triangle has a right angle at C. Therefore, the legs are AC and CB.Length of AC is 1 unit, and length of CB is from (2,1) to (0,1), which is 2 units. So, area is (1/2)*1*2 = 1. Yep, same answer.Wait, but hold on, is the triangle a right triangle? Because point B is at (0,1), and point C is at (2,1). So, CB is a horizontal line from (2,1) to (0,1), which is 2 units. AC is vertical from (2,0) to (2,1), which is 1 unit. So, yes, the triangle is a right triangle with legs 1 and 2, so area is 1.Alternatively, using vectors or determinants, but I think the methods I've used are sufficient.So, all methods point to the area being 1. Therefore, the answer should be D) 1.**Final Answer**boxed{D}

answer：First, I need to determine how much money was earned from recycling newspapers. Since the family collected 20 kilograms of newspapers and they receive 1.50 for every 5 kilograms, I can calculate the earnings from newspapers by dividing 20 by 5 and then multiplying by 1.50.Next, I'll subtract the earnings from newspapers from the total amount received to find out how much was earned from recycling cans. This will help me set up an equation to solve for the number of cans.Finally, I'll solve the equation to find the total number of cans collected by the family.

answer：First, I need to determine the cost of one T-shirt. Since each T-shirt costs 7 more than a pair of socks and a pair of socks costs 6, the cost of one T-shirt is:[6 + 7 = 13 text{ dollars}]Next, I calculate the total cost for one member. Each member needs 2 pairs of socks and 2 T-shirts. The cost for one pair of socks is 6, and the cost for one T-shirt is 13. Therefore, the total cost for one member is:[2 times (6 + 13) = 2 times 19 = 38 text{ dollars}]Finally, I calculate the number of members in the league by dividing the total cost by the cost per member:[frac{3510}{38} = 92.3684]Since the number of members cannot be a fraction, I round to the nearest whole number:[text{Nearest whole number of members} = 92]Thus, the number of members in the Velocity Volleyball League is 92.

answer：Okay, so I have this function f(x) which is a sum of absolute value expressions minus 24. The function is given by:f(x) = |7x - 3a + 8| + |5x + 4a - 6| + |x - a - 8| - 24And I need to find the value of the parameter 'a' that will make the minimum value of this function as small as possible. Hmm, okay. Let me think about how to approach this.First, I remember that absolute value functions have V-shaped graphs, and their minimums occur at the points where the expression inside the absolute value is zero. So, for each term in the function, I can find the critical points where the expression inside the absolute value equals zero. These points are important because the function f(x) will change its behavior around these points.So, let's find these critical points for each absolute value term.1. For the first term, |7x - 3a + 8|, the critical point is where 7x - 3a + 8 = 0. Solving for x, we get x = (3a - 8)/7.2. For the second term, |5x + 4a - 6|, the critical point is where 5x + 4a - 6 = 0. Solving for x, we get x = (-4a + 6)/5.3. For the third term, |x - a - 8|, the critical point is where x - a - 8 = 0. Solving for x, we get x = a + 8.So, these are the three critical points where each absolute value term changes its behavior. Now, since f(x) is a sum of these absolute value terms, the overall function f(x) will have its minimum somewhere around these critical points. It might be at one of these points or somewhere in between, depending on how the terms interact.I think the strategy here is to find the value of 'a' such that the minimum of f(x) is minimized. To do this, perhaps I should express f(x) in terms of these critical points and then find the value of 'a' that minimizes the expression.Alternatively, maybe I can consider the function f(x) as a piecewise linear function and find where its slope changes from negative to positive, which would indicate the minimum point. The slope of f(x) changes at each critical point, so I need to analyze the slopes in each interval defined by these critical points.But before diving into that, maybe I can simplify the function f(x) by combining like terms or looking for patterns. Let me see:f(x) = |7x - 3a + 8| + |5x + 4a - 6| + |x - a - 8| - 24Hmm, I notice that each absolute value term has a coefficient for x: 7, 5, and 1. Maybe I can factor out the x terms or express the function in terms of x and 'a' separately.Alternatively, perhaps I can express f(x) as a function of x for a given 'a', find its minimum in terms of 'a', and then find the 'a' that minimizes this minimum.Yes, that seems like a good approach. So, for a fixed 'a', f(x) is a function of x, and I can find its minimum value. Then, I can find the 'a' that makes this minimum as small as possible.To find the minimum of f(x) for a fixed 'a', I can consider the critical points I found earlier: x1 = (3a - 8)/7, x2 = (-4a + 6)/5, and x3 = a + 8.I need to order these critical points on the x-axis because the behavior of f(x) changes at each of these points. Depending on the order, the function f(x) will have different expressions in different intervals.So, let's find the order of x1, x2, and x3.First, let's compute x1, x2, and x3 in terms of 'a':x1 = (3a - 8)/7x2 = (-4a + 6)/5x3 = a + 8Now, let's see how these compare. To find the order, I can set up inequalities between them.First, compare x1 and x2:(3a - 8)/7 vs. (-4a + 6)/5Multiply both sides by 35 to eliminate denominators:5(3a - 8) vs. 7(-4a + 6)15a - 40 vs. -28a + 42Bring all terms to one side:15a - 40 + 28a - 42 > 043a - 82 > 043a > 82a > 2So, if a > 2, then x1 > x2. If a < 2, then x1 < x2. If a = 2, then x1 = x2.Similarly, let's compare x1 and x3:(3a - 8)/7 vs. a + 8Multiply both sides by 7:3a - 8 vs. 7a + 56Bring all terms to one side:3a - 8 - 7a - 56 > 0-4a - 64 > 0-4a > 64a < -16So, if a < -16, then x1 > x3. If a > -16, then x1 < x3. If a = -16, then x1 = x3.Next, compare x2 and x3:(-4a + 6)/5 vs. a + 8Multiply both sides by 5:-4a + 6 vs. 5a + 40Bring all terms to one side:-4a + 6 - 5a - 40 > 0-9a - 34 > 0-9a > 34a < -34/9 ≈ -3.777...So, if a < -3.777..., then x2 > x3. If a > -3.777..., then x2 < x3. If a = -34/9, then x2 = x3.Okay, so now we have the order of x1, x2, x3 depending on the value of 'a'. This is getting a bit complicated, but maybe we can consider different cases based on the value of 'a'.Case 1: a < -16In this case, from above, x1 > x3 and x2 > x3 (since a < -16 < -34/9). So, the order is x3 < x2 < x1 or x3 < x1 < x2? Wait, let me check.Wait, when a < -16, x1 > x3 and x2 > x3. But how do x1 and x2 compare? Earlier, we saw that for a < 2, x1 < x2. Since a < -16 < 2, so x1 < x2. Therefore, the order is x3 < x1 < x2.Case 2: -16 ≤ a < -34/9In this range, x1 < x3 (since a ≥ -16) and x2 > x3 (since a < -34/9). Also, since a < 2, x1 < x2. So, the order is x1 < x3 < x2.Case 3: -34/9 ≤ a < 2Here, x1 < x3 (since a ≥ -16) and x2 < x3 (since a ≥ -34/9). Also, since a < 2, x1 < x2. So, the order is x1 < x2 < x3.Case 4: a ≥ 2In this case, x1 > x2 (since a ≥ 2) and x1 < x3 (since a ≥ -16). Also, x2 < x3 (since a ≥ -34/9). So, the order is x2 < x1 < x3.Okay, so we have four cases based on the value of 'a', each with a different order of the critical points. For each case, I can write the expression for f(x) in each interval and find where the minimum occurs.But this seems quite involved. Maybe there's a smarter way. I recall that the minimum of a sum of absolute values occurs at the median of the critical points. Is that applicable here?Wait, yes, in general, the minimum of the sum of absolute deviations is achieved at the median of the points. So, perhaps the minimum of f(x) occurs at the median of x1, x2, x3.But in our case, the coefficients of x in each absolute value term are different: 7, 5, and 1. So, it's not exactly the same as the median of points, but rather a weighted median.Hmm, I think the minimum occurs at a point where the sum of the coefficients of x on one side equals the sum on the other side. Let me recall.Yes, for a function like f(x) = |x - a| + |x - b| + |x - c|, the minimum occurs at the median of a, b, c. But when the coefficients are different, like f(x) = k1|x - a| + k2|x - b| + k3|x - c|, the minimum occurs where the cumulative weight reaches half of the total weight.So, in our case, the coefficients are 7, 5, and 1. The total weight is 7 + 5 + 1 = 13. So, the minimum occurs at the point where the cumulative weight from the left reaches 6.5.So, let's order the critical points x1, x2, x3 and find the point where the cumulative weight from the left is 6.5.But the critical points are x1, x2, x3, and their order depends on 'a' as we saw earlier. So, depending on the case, the order changes, and hence the point where the cumulative weight reaches 6.5 changes.This seems complicated, but perhaps manageable.Let me try to formalize this.Given the critical points x1, x2, x3 with weights 7, 5, 1 respectively, the minimum occurs at the point where the sum of weights to the left is equal to the sum of weights to the right, or as close as possible.So, total weight is 13. Half of that is 6.5. So, we need to find the point where the cumulative weight from the left is 6.5.So, depending on the order of x1, x2, x3, the cumulative weight will be built by adding the weights in the order of the points.So, for example, in Case 1: x3 < x1 < x2Weights: x3 has weight 1, x1 has weight 7, x2 has weight 5.Cumulative weights:From left: x3 (1), then x1 (1+7=8), then x2 (8+5=13)We need cumulative weight 6.5. So, between x3 and x1, since 1 < 6.5 < 8.So, the minimum occurs somewhere between x3 and x1.But wait, in this case, the function f(x) is decreasing before x3, then increasing after x3? Wait, no, the function is piecewise linear, and the slope changes at each critical point.Wait, maybe I need to think about the slope of f(x) in each interval.The slope of f(x) is the sum of the derivatives of each absolute value term. The derivative of |kx + c| is k*sign(kx + c). So, for each term:- The derivative of |7x - 3a + 8| is 7*sign(7x - 3a + 8)- The derivative of |5x + 4a - 6| is 5*sign(5x + 4a - 6)- The derivative of |x - a - 8| is 1*sign(x - a - 8)So, the slope of f(x) is 7*sign(7x - 3a + 8) + 5*sign(5x + 4a - 6) + 1*sign(x - a - 8)The minimum occurs where the slope changes from negative to positive, i.e., where the total slope transitions from negative to positive.So, in each interval between critical points, the slope is constant, and we can find where it changes sign.Therefore, to find the minimum, we need to find the interval where the slope changes from negative to positive.Given that, perhaps I can compute the slope in each interval and find where it crosses zero.But since the critical points depend on 'a', this is going to be a bit involved.Alternatively, maybe I can set the derivative equal to zero and solve for x in terms of 'a', but since the derivative is piecewise constant, it's not straightforward.Wait, perhaps another approach. Since the minimum occurs where the weighted sum of the signs changes from negative to positive, maybe I can set up an equation where the total weight to the left equals the total weight to the right.Given the weights 7, 5, 1, the total weight is 13. So, the point where the cumulative weight from the left is 6.5 is the point where the minimum occurs.So, depending on the order of x1, x2, x3, we can determine where this point is.Let me try to formalize this.Suppose the critical points are ordered as x_A < x_B < x_C with weights w_A, w_B, w_C.We need to find the point where the cumulative weight from the left is 6.5.So, if w_A < 6.5, then the point is between x_A and x_B, such that w_A + t*w_B = 6.5, where t is the fraction from x_A to x_B.Similarly, if w_A + w_B < 6.5, then the point is between x_B and x_C.But in our case, the weights are 7, 5, 1, so depending on the order, the cumulative weights will be different.Wait, actually, the weights are associated with each critical point, but the critical points themselves are functions of 'a'. So, the order of the critical points affects how the cumulative weights add up.This is getting quite complex. Maybe I can instead consider that the minimum occurs at a specific critical point, and then find 'a' such that this point minimizes f(x).Alternatively, perhaps I can express f(x) in terms of 'a' and then find the minimum over 'a'.Wait, let me try to express f(x) as a function of 'a' for a given x.But that might not be straightforward.Alternatively, maybe I can consider that the minimum of f(x) occurs at one of the critical points, so I can evaluate f(x) at each critical point and then find the 'a' that minimizes the minimum value.So, let's compute f(x) at each critical point x1, x2, x3.First, at x = x1 = (3a - 8)/7:f(x1) = |7x1 - 3a + 8| + |5x1 + 4a - 6| + |x1 - a - 8| - 24But since x1 is the critical point for the first term, |7x1 - 3a + 8| = 0.So, f(x1) = 0 + |5x1 + 4a - 6| + |x1 - a - 8| - 24Now, let's compute 5x1 + 4a - 6:5x1 + 4a - 6 = 5*(3a - 8)/7 + 4a - 6 = (15a - 40)/7 + 4a - 6Convert 4a to 28a/7 and -6 to -42/7:= (15a - 40 + 28a - 42)/7 = (43a - 82)/7Similarly, compute x1 - a - 8:x1 - a - 8 = (3a - 8)/7 - a - 8 = (3a - 8 - 7a - 56)/7 = (-4a - 64)/7So, f(x1) = |(43a - 82)/7| + |(-4a - 64)/7| - 24Simplify:= (|43a - 82| + | -4a - 64|)/7 - 24Similarly, we can compute f(x2) and f(x3), but this might take a while. Maybe focusing on f(x1) is enough for now.So, f(x1) = (|43a - 82| + | -4a - 64|)/7 - 24We can write this as:f(x1) = (|43a - 82| + |4a + 64|)/7 - 24Now, to minimize f(x1), we need to minimize the numerator |43a - 82| + |4a + 64|.This is a function of 'a', and we can find its minimum.Let me denote g(a) = |43a - 82| + |4a + 64|We need to find the value of 'a' that minimizes g(a).To minimize g(a), we can consider the points where the expressions inside the absolute values change sign, i.e., where 43a - 82 = 0 and 4a + 64 = 0.Solving 43a - 82 = 0 gives a = 82/43 ≈ 1.9065Solving 4a + 64 = 0 gives a = -16So, the critical points for g(a) are at a = -16 and a = 82/43.We can analyze g(a) in the intervals determined by these points: a < -16, -16 ≤ a ≤ 82/43, and a > 82/43.Let's compute g(a) in each interval.1. For a < -16:Both 43a - 82 and 4a + 64 are negative.So, g(a) = -(43a - 82) - (4a + 64) = -43a + 82 -4a -64 = -47a + 18This is a linear function with slope -47, which is decreasing as 'a' increases. So, the minimum in this interval occurs at a = -16.Compute g(-16):g(-16) = |43*(-16) - 82| + |4*(-16) + 64| = | -688 -82 | + | -64 + 64 | = | -770 | + | 0 | = 770 + 0 = 7702. For -16 ≤ a ≤ 82/43:In this interval, 4a + 64 ≥ 0 (since a ≥ -16), but 43a - 82 ≤ 0 (since a ≤ 82/43).So, g(a) = -(43a - 82) + (4a + 64) = -43a + 82 + 4a + 64 = -39a + 146This is a linear function with slope -39, which is decreasing as 'a' increases. So, the minimum in this interval occurs at a = 82/43.Compute g(82/43):g(82/43) = |43*(82/43) - 82| + |4*(82/43) + 64| = |82 - 82| + |328/43 + 64| = 0 + |7.6279 + 64| = 71.6279But let's compute it exactly:4*(82/43) = 328/43 = 7.6279...64 = 2752/43So, 328/43 + 2752/43 = 3080/43 = 71.6279...So, g(82/43) = 71.6279...3. For a > 82/43:Both 43a - 82 and 4a + 64 are positive.So, g(a) = (43a - 82) + (4a + 64) = 47a - 18This is a linear function with slope 47, which is increasing as 'a' increases. So, the minimum in this interval occurs at a = 82/43.Compute g(82/43):Same as above, which is 71.6279...So, putting it all together, the minimum of g(a) occurs at a = 82/43, where g(a) ≈ 71.6279.Therefore, the minimum value of f(x1) is:f(x1) = g(a)/7 - 24 ≈ 71.6279/7 - 24 ≈ 10.2325 - 24 ≈ -13.7675But wait, we need to ensure that this is indeed the minimum of f(x). Because we evaluated f(x) at x1, but the actual minimum might occur at a different critical point or somewhere else.However, since we found that the minimum of g(a) occurs at a = 82/43, and f(x1) is expressed in terms of g(a), it suggests that setting a = 82/43 minimizes f(x1). But we need to verify if this is indeed the global minimum.Alternatively, perhaps we should consider the minimum of f(x) over all x and 'a', but that might be more complex.Wait, another approach: Since f(x) is a sum of absolute values minus 24, the minimum of f(x) will be achieved when the sum of the absolute values is minimized. So, the minimum of f(x) is equal to the minimum of the sum of absolute values minus 24.Therefore, to minimize f(x), we need to minimize the sum S(x) = |7x - 3a + 8| + |5x + 4a - 6| + |x - a - 8|.So, the problem reduces to minimizing S(x) over x and 'a', and then subtracting 24.But since we need to find 'a' such that the minimum of f(x) is as small as possible, we can focus on minimizing S(x).Given that S(x) is a sum of absolute values, its minimum occurs at a weighted median of the critical points, as I thought earlier.But since the weights are different, the weighted median approach applies.The weighted median is the point where the total weight on either side is as equal as possible.Given the weights 7, 5, 1, the total weight is 13. So, the weighted median is the point where the cumulative weight reaches 6.5.So, depending on the order of the critical points, the weighted median will be at a specific critical point or somewhere between two.But since the critical points depend on 'a', we need to find 'a' such that the weighted median is achieved, minimizing S(x).This is getting quite involved, but perhaps we can set up equations to find 'a' such that the weighted median condition is satisfied.Alternatively, since we found that the minimum of g(a) occurs at a = 82/43, and this corresponds to the point where the derivative of g(a) changes sign, perhaps this is indeed the optimal 'a'.But let me check.Wait, when a = 82/43, let's compute the critical points:x1 = (3a - 8)/7 = (3*(82/43) - 8)/7 = (246/43 - 344/43)/7 = (-98/43)/7 = -14/43 ≈ -0.3256x2 = (-4a + 6)/5 = (-4*(82/43) + 6)/5 = (-328/43 + 258/43)/5 = (-70/43)/5 = -14/43 ≈ -0.3256x3 = a + 8 = 82/43 + 8 = 82/43 + 344/43 = 426/43 ≈ 9.9065So, at a = 82/43, x1 = x2 ≈ -0.3256, and x3 ≈ 9.9065.So, the critical points are x1 = x2 ≈ -0.3256 and x3 ≈ 9.9065.So, the order is x1 = x2 < x3.Therefore, the function S(x) has critical points at x ≈ -0.3256 and x ≈ 9.9065.Now, let's compute the slope of S(x) in different intervals.1. For x < -0.3256:All terms inside the absolute values are negative.So, S(x) = -(7x - 3a + 8) - (5x + 4a - 6) - (x - a - 8) = -7x + 3a - 8 -5x -4a +6 -x +a +8 = (-7x -5x -x) + (3a -4a +a) + (-8 +6 +8) = -13x + 0a +6So, slope = -132. For -0.3256 < x < 9.9065:Here, 7x - 3a + 8 > 0 (since x > x1), 5x + 4a -6 > 0 (since x > x2), but x - a -8 < 0 (since x < x3).So, S(x) = (7x - 3a +8) + (5x +4a -6) - (x -a -8) = 7x -3a +8 +5x +4a -6 -x +a +8 = (7x +5x -x) + (-3a +4a +a) + (8 -6 +8) = 11x + 2a +10So, slope = 113. For x > 9.9065:All terms inside the absolute values are positive.So, S(x) = (7x -3a +8) + (5x +4a -6) + (x -a -8) = 7x -3a +8 +5x +4a -6 +x -a -8 = (7x +5x +x) + (-3a +4a -a) + (8 -6 -8) = 13x +0a -6So, slope = 13Therefore, the slope of S(x) changes from -13 to 11 at x ≈ -0.3256, and then from 11 to 13 at x ≈ 9.9065.So, the function S(x) is decreasing for x < -0.3256, then increasing for x > -0.3256, but with a smaller slope until x ≈ 9.9065, and then a larger slope.Wait, but the slope changes from -13 to 11 at x ≈ -0.3256, which is a change from negative to positive, indicating a minimum at x ≈ -0.3256.But wait, the slope was -13 before x ≈ -0.3256 and 11 after. So, the function is decreasing before x ≈ -0.3256 and increasing after, meaning the minimum occurs at x ≈ -0.3256.But wait, that's the same as x1 and x2, which are equal at a = 82/43.So, at a = 82/43, the minimum of S(x) occurs at x ≈ -0.3256, which is the point where both x1 and x2 coincide.Therefore, the minimum of S(x) is achieved at x = x1 = x2, and the minimum value is:S(x1) = |7x1 -3a +8| + |5x1 +4a -6| + |x1 -a -8|But since x1 is the critical point for the first term, |7x1 -3a +8| = 0.Similarly, since x1 = x2, |5x1 +4a -6| = 0.So, S(x1) = 0 + 0 + |x1 -a -8| = |x1 -a -8|But x1 = (3a -8)/7, so:|x1 -a -8| = |(3a -8)/7 -a -8| = |(3a -8 -7a -56)/7| = |(-4a -64)/7| = | -4(a +16)/7 | = (4|a +16|)/7But at a = 82/43, let's compute this:a = 82/43 ≈ 1.9065So, a +16 ≈ 17.9065Thus, |a +16| = 17.9065Therefore, S(x1) = (4 * 17.9065)/7 ≈ 71.626/7 ≈ 10.232So, S(x1) ≈ 10.232Therefore, f(x1) = S(x1) -24 ≈ 10.232 -24 ≈ -13.768But wait, can f(x) be negative? Let me check.Yes, because f(x) is the sum of absolute values minus 24. So, if the sum of absolute values is less than 24, f(x) will be negative.But is this the minimum value? Or can it be made even smaller?Wait, but we found that at a = 82/43, the minimum of S(x) is approximately 10.232, so f(x) ≈ -13.768.Is this the smallest possible minimum? Or is there a way to make it even smaller?Wait, perhaps if we choose 'a' such that the minimum of S(x) is even smaller, then f(x) would be even smaller (more negative). But we found that the minimum of g(a) = |43a -82| + |4a +64| occurs at a = 82/43, giving g(a) ≈ 71.6279, which corresponds to S(x1) ≈ 10.232.But let's see if we can get S(x) smaller than 10.232 by choosing a different 'a'.Wait, perhaps not, because we minimized g(a), which is related to S(x1). But maybe the minimum of S(x) can be smaller if we choose a different 'a' such that the minimum occurs at a different critical point.Alternatively, perhaps the minimum of S(x) is indeed minimized when a = 82/43, giving the smallest possible S(x) ≈ 10.232, and thus f(x) ≈ -13.768.But let's verify this by considering another critical point.Let's compute f(x2) at a = 82/43.x2 = (-4a +6)/5 = (-4*(82/43) +6)/5 = (-328/43 +258/43)/5 = (-70/43)/5 = -14/43 ≈ -0.3256So, x2 = x1 ≈ -0.3256Therefore, f(x2) = f(x1) ≈ -13.768Similarly, compute f(x3):x3 = a +8 = 82/43 +8 ≈ 9.9065Compute S(x3):S(x3) = |7x3 -3a +8| + |5x3 +4a -6| + |x3 -a -8|But x3 = a +8, so:|7x3 -3a +8| = |7(a +8) -3a +8| = |7a +56 -3a +8| = |4a +64||5x3 +4a -6| = |5(a +8) +4a -6| = |5a +40 +4a -6| = |9a +34||x3 -a -8| = |(a +8) -a -8| = |0| = 0So, S(x3) = |4a +64| + |9a +34|At a = 82/43:|4*(82/43) +64| = |328/43 + 2752/43| = |3080/43| ≈ 71.6279|9*(82/43) +34| = |738/43 + 1462/43| = |2200/43| ≈ 51.1628So, S(x3) ≈ 71.6279 + 51.1628 ≈ 122.7907Thus, f(x3) = S(x3) -24 ≈ 122.7907 -24 ≈ 98.7907Which is much larger than f(x1) ≈ -13.768Therefore, the minimum occurs at x1 = x2 ≈ -0.3256, giving f(x) ≈ -13.768.But is this the smallest possible minimum? Or can we choose a different 'a' to get a smaller f(x)?Wait, let's consider another value of 'a'. Suppose we choose 'a' such that x1, x2, x3 coincide, i.e., all three critical points are the same. Then, the function S(x) would have a single critical point, and the minimum would be achieved there.But is that possible?Set x1 = x2 = x3:(3a -8)/7 = (-4a +6)/5 = a +8First, set (3a -8)/7 = (-4a +6)/5Cross-multiplying:5*(3a -8) = 7*(-4a +6)15a -40 = -28a +4215a +28a = 42 +4043a = 82a = 82/43 ≈ 1.9065Now, check if this equals x3:x3 = a +8 = 82/43 +8 ≈ 9.9065But x1 = x2 ≈ -0.3256 ≠ x3 ≈ 9.9065So, they don't coincide. Therefore, it's not possible to have all three critical points coincide.Alternatively, perhaps we can set two critical points to coincide and see if that gives a better minimum.But we already saw that at a = 82/43, x1 = x2, and the minimum is achieved there.Alternatively, perhaps choosing 'a' such that x2 = x3.Set (-4a +6)/5 = a +8Multiply both sides by 5:-4a +6 = 5a +40-4a -5a = 40 -6-9a = 34a = -34/9 ≈ -3.777...So, at a = -34/9, x2 = x3.Compute x1 at a = -34/9:x1 = (3a -8)/7 = (3*(-34/9) -8)/7 = (-102/9 -72/9)/7 = (-174/9)/7 = (-58/3)/7 = -58/21 ≈ -2.7619So, the critical points are x1 ≈ -2.7619, x2 = x3 ≈ (-4*(-34/9) +6)/5 = (136/9 +54/9)/5 = 190/9 /5 = 38/9 ≈ 4.2222So, the order is x1 < x2 = x3.Now, compute S(x) at x1:S(x1) = |7x1 -3a +8| + |5x1 +4a -6| + |x1 -a -8|But x1 is the critical point for the first term, so |7x1 -3a +8| = 0.Compute the other terms:5x1 +4a -6 = 5*(-58/21) +4*(-34/9) -6 = (-290/21) + (-136/9) -6Convert to common denominator, which is 63:= (-290*3)/63 + (-136*7)/63 + (-6*63)/63= (-870)/63 + (-952)/63 + (-378)/63= (-870 -952 -378)/63 = (-2200)/63 ≈ -34.9206So, |5x1 +4a -6| ≈ 34.9206Compute x1 -a -8:x1 -a -8 = (-58/21) - (-34/9) -8 = (-58/21) +34/9 -8Convert to common denominator 63:= (-58*3)/63 + (34*7)/63 - (8*63)/63= (-174)/63 + 238/63 - 504/63= (-174 +238 -504)/63 = (-440)/63 ≈ -7.0So, |x1 -a -8| ≈ 7.0Therefore, S(x1) ≈ 0 + 34.9206 +7.0 ≈ 41.9206Thus, f(x1) ≈ 41.9206 -24 ≈ 17.9206Which is larger than the previous minimum of ≈ -13.768.Similarly, compute S(x2) = S(x3):At x = x2 = x3 ≈4.2222Compute S(x2):|7x2 -3a +8| + |5x2 +4a -6| + |x2 -a -8|But x2 is the critical point for the second term, so |5x2 +4a -6| =0Compute the other terms:7x2 -3a +8 =7*(38/9) -3*(-34/9) +8 = 266/9 +102/9 +72/9 = (266 +102 +72)/9 =440/9 ≈48.8889|x2 -a -8| = |38/9 - (-34/9) -8| = |38/9 +34/9 -72/9| = |(38 +34 -72)/9| = |0/9| =0So, S(x2) = |440/9| +0 +0 ≈48.8889Thus, f(x2) ≈48.8889 -24 ≈24.8889Which is larger than the previous minimum.Therefore, at a = -34/9, the minimum of f(x) is ≈17.9206, which is larger than the minimum at a =82/43.Therefore, a =82/43 gives a smaller minimum.Similarly, let's check another case, say a = -16.At a = -16, compute x1, x2, x3:x1 = (3*(-16) -8)/7 = (-48 -8)/7 = -56/7 = -8x2 = (-4*(-16) +6)/5 = (64 +6)/5 =70/5 =14x3 = -16 +8 =-8So, x1 =x3 =-8, x2=14So, the critical points are x1=x3=-8 and x2=14.Compute S(x) at x1=x3=-8:S(-8) = |7*(-8) -3*(-16) +8| + |5*(-8) +4*(-16) -6| + |(-8) - (-16) -8|Compute each term:|7*(-8) -3*(-16) +8| = |-56 +48 +8| =|0| =0|5*(-8) +4*(-16) -6| = |-40 -64 -6| = |-110| =110|(-8) - (-16) -8| = |(-8 +16 -8)| =|0| =0So, S(-8)=0 +110 +0=110Thus, f(-8)=110 -24=86Which is much larger than the previous minimum.Therefore, a =82/43 still gives the smallest minimum.Another test: let's choose a =0.Compute x1, x2, x3:x1=(0 -8)/7≈-1.1429x2=(0 +6)/5=1.2x3=0 +8=8So, order: x1≈-1.1429 <x2=1.2 <x3=8Compute S(x1):|7x1 -0 +8| + |5x1 +0 -6| + |x1 -0 -8|=|7*(-1.1429) +8| + |5*(-1.1429) -6| + |(-1.1429) -8|≈| -8 +8 | + | -5.7145 -6 | + | -9.1429 |≈0 +11.7145 +9.1429≈20.8574Thus, f(x1)=20.8574 -24≈-3.1426Which is better than some previous values, but not as good as a=82/43.Similarly, compute S(x2):|7x2 -0 +8| + |5x2 +0 -6| + |x2 -0 -8|=|7*1.2 +8| + |5*1.2 -6| + |1.2 -8|=|8.4 +8| + |6 -6| + |-6.8|≈16.4 +0 +6.8≈23.2Thus, f(x2)=23.2 -24≈-0.8Which is worse than f(x1) at a=0.Compute S(x3):|7*8 -0 +8| + |5*8 +0 -6| + |8 -0 -8|=|56 +8| + |40 -6| + |0|≈64 +34 +0≈98Thus, f(x3)=98 -24=74So, at a=0, the minimum of f(x) is≈-3.1426, which is better than some cases but not as good as a=82/43.Therefore, from these tests, it seems that a=82/43 gives the smallest minimum of f(x).But let's try to formalize this.We found that the minimum of g(a) = |43a -82| + |4a +64| occurs at a=82/43, giving g(a)=71.6279.Therefore, S(x1)=g(a)/7≈10.232Thus, f(x1)=10.232 -24≈-13.768Since this is the smallest value we've found so far, and considering that g(a) is minimized at a=82/43, it's likely that this is indeed the optimal value.Therefore, the value of 'a' that minimizes the minimum value of f(x) is a=82/43.

answer：Alright, so I have this problem about an ellipse, and it's divided into three parts. Let me try to tackle each part step by step. I'm a bit nervous because ellipses can be tricky, but I'll give it a shot.Starting with part (I): I need to write down the equation of ellipse C and the coordinates of the foci. The given equation is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a > b > 0). The foci are (F_1) and (F_2), which are on the x-axis since the major axis is along the x-axis because (a > b).The problem states that the sum of distances from a point (P(1, frac{sqrt{3}}{2})) on the ellipse to the foci (F_1) and (F_2) is 4. I remember that one of the defining properties of an ellipse is that the sum of the distances from any point on the ellipse to the two foci is constant and equal to (2a). So, if that sum is 4, then (2a = 4), which means (a = 2).Now, since (P(1, frac{sqrt{3}}{2})) lies on the ellipse, plugging its coordinates into the ellipse equation should satisfy it. Let's do that:[frac{1^2}{2^2} + frac{left(frac{sqrt{3}}{2}right)^2}{b^2} = 1]Simplifying:[frac{1}{4} + frac{frac{3}{4}}{b^2} = 1]Let me solve for (b^2):Subtract (frac{1}{4}) from both sides:[frac{frac{3}{4}}{b^2} = frac{3}{4}]Wait, that can't be right. Let me check my steps again.Wait, the equation after plugging in P is:[frac{1}{4} + frac{frac{3}{4}}{b^2} = 1]So, subtract (frac{1}{4}):[frac{frac{3}{4}}{b^2} = frac{3}{4}]Multiply both sides by (b^2):[frac{3}{4} = frac{3}{4}b^2]Then, divide both sides by (frac{3}{4}):[1 = b^2]So, (b^2 = 1), which means (b = 1) since (b > 0).Therefore, the equation of the ellipse is:[frac{x^2}{4} + y^2 = 1]Now, the coordinates of the foci. For an ellipse, the distance from the center to each focus is (c), where (c^2 = a^2 - b^2). So, let's compute (c):[c^2 = 4 - 1 = 3 implies c = sqrt{3}]Since the major axis is along the x-axis, the foci are at ((pm sqrt{3}, 0)). So, (F_1 = (-sqrt{3}, 0)) and (F_2 = (sqrt{3}, 0)).Okay, part (I) seems done. The equation is (frac{x^2}{4} + y^2 = 1) and the foci are at ((pm sqrt{3}, 0)).Moving on to part (II): Let Q be a moving point on ellipse C, and we need to find the equation of the trajectory of the midpoint T of segment (F_1Q).Hmm, so T is the midpoint between (F_1) and Q. Let me denote Q as ((x_0, y_0)), which lies on the ellipse, so it satisfies (frac{x_0^2}{4} + y_0^2 = 1).The midpoint T will have coordinates:[x = frac{-sqrt{3} + x_0}{2}, quad y = frac{0 + y_0}{2}]So,[x = frac{x_0 - sqrt{3}}{2}, quad y = frac{y_0}{2}]I can solve these for (x_0) and (y_0):[x_0 = 2x + sqrt{3}, quad y_0 = 2y]Since Q lies on the ellipse, substituting (x_0) and (y_0) into the ellipse equation:[frac{(2x + sqrt{3})^2}{4} + (2y)^2 = 1]Let me simplify this:First, expand ((2x + sqrt{3})^2):[(2x + sqrt{3})^2 = 4x^2 + 4xsqrt{3} + 3]Divide by 4:[frac{4x^2 + 4xsqrt{3} + 3}{4} = x^2 + xsqrt{3} + frac{3}{4}]Then, ((2y)^2 = 4y^2).So, putting it all together:[x^2 + xsqrt{3} + frac{3}{4} + 4y^2 = 1]Subtract (frac{3}{4}) from both sides:[x^2 + xsqrt{3} + 4y^2 = frac{1}{4}]Hmm, this seems a bit messy. Maybe I can complete the square for the x-terms.Looking at (x^2 + xsqrt{3}), let's complete the square:Take the coefficient of x, which is (sqrt{3}), divide by 2: (frac{sqrt{3}}{2}), square it: (frac{3}{4}).So,[x^2 + xsqrt{3} = left(x + frac{sqrt{3}}{2}right)^2 - frac{3}{4}]Substitute back into the equation:[left(x + frac{sqrt{3}}{2}right)^2 - frac{3}{4} + 4y^2 = frac{1}{4}]Add (frac{3}{4}) to both sides:[left(x + frac{sqrt{3}}{2}right)^2 + 4y^2 = 1]Ah, that looks nicer. So, the equation of the trajectory of point T is:[left(x + frac{sqrt{3}}{2}right)^2 + 4y^2 = 1]That's an ellipse centered at (left(-frac{sqrt{3}}{2}, 0right)) with semi-major axis 1 along the x-axis and semi-minor axis (frac{1}{2}) along the y-axis.Okay, part (II) seems done.Now, part (III): Line l passes through the fixed point M(0, 2) and intersects ellipse C at two distinct points A and B. If angle AOB is acute (where O is the origin), find the range of values for the slope (k_0) of line l.Alright, so line l passes through M(0, 2). Let me denote the slope as (k), so the equation of line l is:[y = kx + 2]This line intersects the ellipse (frac{x^2}{4} + y^2 = 1). Let's find the points of intersection by substituting (y = kx + 2) into the ellipse equation.Substituting:[frac{x^2}{4} + (kx + 2)^2 = 1]Let me expand this:First, expand ((kx + 2)^2):[k^2x^2 + 4kx + 4]So, the equation becomes:[frac{x^2}{4} + k^2x^2 + 4kx + 4 = 1]Multiply through by 4 to eliminate the denominator:[x^2 + 4k^2x^2 + 16kx + 16 = 4]Combine like terms:[(1 + 4k^2)x^2 + 16kx + 12 = 0]So, we have a quadratic in x:[(1 + 4k^2)x^2 + 16kx + 12 = 0]For the line to intersect the ellipse at two distinct points, the discriminant must be positive. Let's compute the discriminant (Delta):[Delta = (16k)^2 - 4(1 + 4k^2)(12)]Calculate each part:First, ((16k)^2 = 256k^2)Then, (4(1 + 4k^2)(12) = 48(1 + 4k^2) = 48 + 192k^2)So,[Delta = 256k^2 - (48 + 192k^2) = 256k^2 - 48 - 192k^2 = 64k^2 - 48]For two distinct real roots, (Delta > 0):[64k^2 - 48 > 0 implies 64k^2 > 48 implies k^2 > frac{48}{64} = frac{3}{4}]So, (k^2 > frac{3}{4}), which means (|k| > frac{sqrt{3}}{2}). That's one condition.Now, the angle AOB is acute. O is the origin, so vectors OA and OB are from the origin to points A and B. The angle between OA and OB is acute, which means the dot product of vectors OA and OB is positive.Let me denote points A and B as (A(x_1, y_1)) and (B(x_2, y_2)). Then,[overrightarrow{OA} cdot overrightarrow{OB} = x_1x_2 + y_1y_2 > 0]So, I need to find the condition when (x_1x_2 + y_1y_2 > 0).First, let's find expressions for (x_1 + x_2) and (x_1x_2) from the quadratic equation.From the quadratic equation:[(1 + 4k^2)x^2 + 16kx + 12 = 0]Sum of roots:[x_1 + x_2 = -frac{16k}{1 + 4k^2}]Product of roots:[x_1x_2 = frac{12}{1 + 4k^2}]Now, (y_1 = kx_1 + 2) and (y_2 = kx_2 + 2). So,[y_1y_2 = (kx_1 + 2)(kx_2 + 2) = k^2x_1x_2 + 2k(x_1 + x_2) + 4]So, the dot product is:[x_1x_2 + y_1y_2 = x_1x_2 + k^2x_1x_2 + 2k(x_1 + x_2) + 4]Factor out (x_1x_2):[(1 + k^2)x_1x_2 + 2k(x_1 + x_2) + 4]Now, substitute the expressions from above:[(1 + k^2)left(frac{12}{1 + 4k^2}right) + 2kleft(-frac{16k}{1 + 4k^2}right) + 4]Let me compute each term:First term:[(1 + k^2)left(frac{12}{1 + 4k^2}right) = frac{12(1 + k^2)}{1 + 4k^2}]Second term:[2kleft(-frac{16k}{1 + 4k^2}right) = -frac{32k^2}{1 + 4k^2}]Third term is just 4, which can be written as (frac{4(1 + 4k^2)}{1 + 4k^2}) to have the same denominator.So, combining all terms:[frac{12(1 + k^2) - 32k^2 + 4(1 + 4k^2)}{1 + 4k^2}]Let me expand the numerator:First, expand each part:12(1 + k^2) = 12 + 12k^2-32k^2 remains as is.4(1 + 4k^2) = 4 + 16k^2So, adding all together:12 + 12k^2 - 32k^2 + 4 + 16k^2Combine like terms:Constants: 12 + 4 = 16k^2 terms: 12k^2 - 32k^2 + 16k^2 = (12 - 32 + 16)k^2 = (-4)k^2So, numerator is 16 - 4k^2.Thus, the dot product is:[frac{16 - 4k^2}{1 + 4k^2}]We need this to be greater than 0:[frac{16 - 4k^2}{1 + 4k^2} > 0]Since the denominator (1 + 4k^2) is always positive (as (k^2 geq 0)), the inequality reduces to:[16 - 4k^2 > 0 implies 16 > 4k^2 implies 4 > k^2 implies k^2 < 4 implies |k| < 2]So, combining this with the earlier condition that (|k| > frac{sqrt{3}}{2}), we have:[frac{sqrt{3}}{2} < |k| < 2]Therefore, the slope (k_0) must satisfy:[-2 < k_0 < -frac{sqrt{3}}{2} quad text{or} quad frac{sqrt{3}}{2} < k_0 < 2]So, the range of (k_0) is ((-2, -frac{sqrt{3}}{2}) cup (frac{sqrt{3}}{2}, 2)).Let me just double-check my steps to make sure I didn't make any mistakes.1. Found the equation of the ellipse correctly by using the given point and the sum of distances.2. For part (II), correctly found the midpoint coordinates and substituted into the ellipse equation, then simplified to get the trajectory equation.3. For part (III), set up the line equation, substituted into the ellipse, found the discriminant condition, then computed the dot product condition, leading to the range of k.Everything seems to check out. I think I did it right!**Final Answer**Ⅰ. The equation of ellipse C is (boxed{dfrac{x^2}{4} + y^2 = 1}) and the foci are (boxed{F_1(-sqrt{3}, 0)}) and (boxed{F_2(sqrt{3}, 0)}).Ⅱ. The equation of the trajectory of midpoint T is (boxed{left(x + dfrac{sqrt{3}}{2}right)^2 + 4y^2 = 1}).Ⅲ. The range of values for the slope (k_0) is (boxed{(-2, -dfrac{sqrt{3}}{2}) cup (dfrac{sqrt{3}}{2}, 2)}).

answer：Alright, so I've got this problem here about arithmetic and geometric sequences, and I need to find their general formulas and then figure out the sum of another sequence derived from them. Let me try to break this down step by step.First, let's parse the problem:We have an arithmetic sequence {a_n} with a common difference d = 1. So, that means each term increases by 1 from the previous one. The general form of an arithmetic sequence is a_n = a_1 + (n - 1)d, right? Since d is 1, that simplifies to a_n = a_1 + (n - 1). So, a_n = a_1 + n - 1, or a_n = n + (a_1 - 1). Hmm, okay.Then, we have a geometric sequence {b_n} with a common ratio q = 2. The general form of a geometric sequence is b_n = b_1 * q^(n - 1). Since q is 2, that becomes b_n = b_1 * 2^(n - 1). Got that.Now, the problem says that 1 is the geometric mean of a_1 and b_1. The geometric mean of two numbers is the square root of their product. So, sqrt(a_1 * b_1) = 1. Squaring both sides, we get a_1 * b_1 = 1. Okay, that's one equation.Next, we're given vectors a and b, where vector a is (a_1, a_2) and vector b is (b_1, b_2). The dot product of these vectors is 5. So, a · b = a_1*b_1 + a_2*b_2 = 5.Let me write down what I know:1. a_1 * b_1 = 1 (from the geometric mean)2. a · b = a_1*b_1 + a_2*b_2 = 5Since {a_n} is arithmetic with d=1, a_2 = a_1 + 1. Similarly, since {b_n} is geometric with q=2, b_2 = b_1 * 2.So, substituting these into the dot product equation:a_1*b_1 + (a_1 + 1)*(b_1*2) = 5But from the first equation, a_1*b_1 = 1. So, substitute that in:1 + (a_1 + 1)*(2*b_1) = 5Simplify:1 + 2*b_1*(a_1 + 1) = 5Subtract 1:2*b_1*(a_1 + 1) = 4Divide both sides by 2:b_1*(a_1 + 1) = 2But we know from the first equation that a_1*b_1 = 1. So, let's denote a_1 as 'a' and b_1 as 'b' for simplicity.So, we have:1. a*b = 12. b*(a + 1) = 2From equation 1: b = 1/aSubstitute into equation 2:(1/a)*(a + 1) = 2Simplify:(a + 1)/a = 2Multiply both sides by a:a + 1 = 2aSubtract a:1 = aSo, a_1 = 1. Then, from equation 1, b_1 = 1/a_1 = 1/1 = 1.So, a_1 = 1 and b_1 = 1.Now, let's find the general formulas for {a_n} and {b_n}.For the arithmetic sequence:a_n = a_1 + (n - 1)*d = 1 + (n - 1)*1 = nSo, a_n = n.For the geometric sequence:b_n = b_1 * q^(n - 1) = 1 * 2^(n - 1) = 2^(n - 1)So, b_n = 2^(n - 1)Alright, that takes care of part (1). Now, moving on to part (2):We need to define c_n = 2^{a_n} * log_2(b_n), and find the sum of the first n terms, denoted as T_n.First, let's express c_n in terms of n.We know a_n = n, so 2^{a_n} = 2^n.And b_n = 2^{n - 1}, so log_2(b_n) = log_2(2^{n - 1}) = n - 1.Therefore, c_n = 2^n * (n - 1) = (n - 1)*2^n.So, c_n = (n - 1)*2^n.Now, we need to find T_n = c_1 + c_2 + ... + c_n.Let's write out the first few terms to see the pattern:c_1 = (1 - 1)*2^1 = 0*2 = 0c_2 = (2 - 1)*2^2 = 1*4 = 4c_3 = (3 - 1)*2^3 = 2*8 = 16c_4 = (4 - 1)*2^4 = 3*16 = 48And so on.So, T_n = 0 + 4 + 16 + 48 + ... + (n - 1)*2^nHmm, this looks like a series where each term is (k - 1)*2^k for k from 1 to n.But since c_1 is 0, we can start the summation from k=2 to n, but it's easier to include k=1 as 0.So, T_n = sum_{k=1}^n (k - 1)*2^kLet me denote S = sum_{k=1}^n (k - 1)*2^kWe can write this as S = sum_{k=1}^n (k*2^k - 2^k) = sum_{k=1}^n k*2^k - sum_{k=1}^n 2^kSo, S = [sum_{k=1}^n k*2^k] - [sum_{k=1}^n 2^k]We can compute these two sums separately.First, let's compute sum_{k=1}^n 2^k. That's a geometric series.Sum_{k=1}^n 2^k = 2*(2^n - 1)/(2 - 1) = 2*(2^n - 1) = 2^{n+1} - 2Okay, that's straightforward.Now, the more challenging part is sum_{k=1}^n k*2^k.I remember there's a formula for sum_{k=1}^n k*r^k, which is r*(1 - (n+1)*r^n + n*r^{n+1}) / (1 - r)^2In our case, r = 2.So, plugging in:sum_{k=1}^n k*2^k = 2*(1 - (n+1)*2^n + n*2^{n+1}) / (1 - 2)^2Simplify denominator: (1 - 2)^2 = 1So, numerator: 2*(1 - (n+1)*2^n + n*2^{n+1})Let me compute this:First, expand the numerator:2*1 - 2*(n+1)*2^n + 2*n*2^{n+1}Simplify each term:2 - 2*(n+1)*2^n + 2*n*2^{n+1}Note that 2^{n+1} = 2*2^n, so 2*n*2^{n+1} = 2*n*2*2^n = 4n*2^nSimilarly, 2*(n+1)*2^n = 2n*2^n + 2*2^n = (2n + 2)*2^nSo, putting it all together:2 - (2n + 2)*2^n + 4n*2^nCombine like terms:2 + [ - (2n + 2) + 4n ] *2^nSimplify inside the brackets:-2n - 2 + 4n = 2n - 2So, the expression becomes:2 + (2n - 2)*2^nFactor out 2:2 + 2*(n - 1)*2^nWhich is:2 + (n - 1)*2^{n+1}So, sum_{k=1}^n k*2^k = 2 + (n - 1)*2^{n+1}Therefore, going back to S:S = [sum_{k=1}^n k*2^k] - [sum_{k=1}^n 2^k] = [2 + (n - 1)*2^{n+1}] - [2^{n+1} - 2]Simplify:2 + (n - 1)*2^{n+1} - 2^{n+1} + 2Combine like terms:2 + 2 + (n - 1)*2^{n+1} - 2^{n+1}Which is:4 + (n - 2)*2^{n+1}So, T_n = 4 + (n - 2)*2^{n+1}Wait, let me double-check that.Wait, when we subtract [2^{n+1} - 2], it's minus 2^{n+1} plus 2.So, 2 + (n - 1)*2^{n+1} - 2^{n+1} + 2Which is 4 + (n - 2)*2^{n+1}Yes, that seems correct.So, T_n = 4 + (n - 2)*2^{n+1}Alternatively, we can factor out 2^{n+1}:T_n = (n - 2)*2^{n+1} + 4But it's probably fine as is.Let me test this formula with n=1, 2, 3 to see if it matches.For n=1:T_1 = c_1 = 0Plugging into the formula:4 + (1 - 2)*2^{2} = 4 + (-1)*4 = 4 - 4 = 0. Correct.For n=2:T_2 = c_1 + c_2 = 0 + 4 = 4Formula:4 + (2 - 2)*2^{3} = 4 + 0*8 = 4. Correct.For n=3:T_3 = 0 + 4 + 16 = 20Formula:4 + (3 - 2)*2^{4} = 4 + 1*16 = 20. Correct.For n=4:T_4 = 0 + 4 + 16 + 48 = 68Formula:4 + (4 - 2)*2^{5} = 4 + 2*32 = 4 + 64 = 68. Correct.Okay, seems like the formula works.So, summarizing:(1) The general term for the arithmetic sequence is a_n = n, and for the geometric sequence is b_n = 2^{n - 1}.(2) The sum T_n of the first n terms of {c_n} is T_n = 4 + (n - 2)*2^{n+1}**Final Answer**(1) The general formula for the arithmetic sequence is boxed{a_n = n} and for the geometric sequence is boxed{b_n = 2^{n-1}}.(2) The sum of the first n terms of the sequence {c_n} is boxed{T_n = 4 + (n - 2) cdot 2^{n+1}}.