answer：Okay, so I need to find the volume of a truncated right circular cone, also known as a frustum. The problem gives me the large base radius, which is 12 cm, the small base radius, which is 6 cm, and the height of the truncated cone, which is 10 cm. First, I remember that the volume of a frustum can be calculated using a specific formula, but I'm not exactly sure what it is. Maybe I can derive it or think about how it relates to the volumes of cones. I recall that a frustum is basically a cone with the top cut off, so maybe I can find the volume of the original large cone and subtract the volume of the smaller cone that was removed.To do that, I need to figure out the height of the original large cone before it was truncated. The problem gives me the height of the frustum as 10 cm, but I don't know the height of the entire large cone. Hmm, how can I find that?I remember that the two cones—the original large cone and the smaller cone that was removed—are similar. That means their dimensions are proportional. The radius of the large base is 12 cm, and the radius of the small base is 6 cm. So the ratio of the radii is 12:6, which simplifies to 2:1. This ratio should also apply to their heights.Let me denote the height of the smaller cone as ( x ). Then, the height of the original large cone would be ( x + 10 ) cm because the frustum's height is 10 cm. Since the ratio of the radii is 2:1, the ratio of the heights should also be 2:1. So, I can set up the proportion:[frac{x + 10}{x} = frac{12}{6} = 2]Wait, that doesn't seem right. Let me think again. If the large cone has a radius of 12 and the small cone has a radius of 6, then the ratio of the large cone's radius to the small cone's radius is 2:1. Therefore, the ratio of their heights should also be 2:1. So, the height of the large cone is twice the height of the small cone.Let me denote the height of the small cone as ( x ). Then, the height of the large cone would be ( 2x ). But the height of the frustum is the difference between the heights of the large cone and the small cone, which is given as 10 cm. So:[2x - x = 10 x = 10 text{ cm}]So, the height of the small cone is 10 cm, and the height of the large cone is ( 2x = 20 ) cm.Now that I have the heights of both cones, I can calculate their volumes. The formula for the volume of a cone is:[V = frac{1}{3}pi r^2 h]First, let's find the volume of the large cone:[V_{text{large}} = frac{1}{3}pi (12)^2 (20) = frac{1}{3}pi (144)(20) = frac{1}{3}pi (2880) = 960pi text{ cm}^3]Next, the volume of the small cone:[V_{text{small}} = frac{1}{3}pi (6)^2 (10) = frac{1}{3}pi (36)(10) = frac{1}{3}pi (360) = 120pi text{ cm}^3]To find the volume of the frustum, I subtract the volume of the small cone from the volume of the large cone:[V_{text{frustum}} = V_{text{large}} - V_{text{small}} = 960pi - 120pi = 840pi text{ cm}^3]Alternatively, I remember there's a direct formula for the volume of a frustum:[V = frac{1}{3}pi h (R^2 + Rr + r^2)]Where ( h ) is the height of the frustum, ( R ) is the radius of the large base, and ( r ) is the radius of the small base. Plugging in the values:[V = frac{1}{3}pi (10) (12^2 + 12 times 6 + 6^2) = frac{1}{3}pi (10) (144 + 72 + 36) = frac{1}{3}pi (10) (252) = frac{1}{3}pi (2520) = 840pi text{ cm}^3]Both methods give me the same result, so I'm confident that the volume of the truncated cone is ( 840pi ) cubic centimeters.

answer：Okay, so I'm trying to figure out which of the given approximate formulas for the diameter of a sphere is the most accurate. The formulas are:A: ( d approx sqrt[3]{frac{16}{9}V} )B: ( d approx sqrt[3]{2V} )C: ( d approx sqrt[3]{frac{300}{157}V} )D: ( d approx sqrt[3]{frac{21}{11}V} )I remember that the volume ( V ) of a sphere is given by the formula ( V = frac{4}{3}pi r^3 ), where ( r ) is the radius. Since the diameter ( d ) is twice the radius, ( d = 2r ), so ( r = frac{d}{2} ). Substituting this into the volume formula gives:( V = frac{4}{3}pi left(frac{d}{2}right)^3 )Simplifying that, ( left(frac{d}{2}right)^3 = frac{d^3}{8} ), so:( V = frac{4}{3}pi cdot frac{d^3}{8} = frac{pi d^3}{6} )So, rearranging this to solve for ( d ):( d^3 = frac{6V}{pi} )Taking the cube root of both sides:( d = sqrt[3]{frac{6V}{pi}} )So, the exact formula for the diameter is ( d = sqrt[3]{frac{6V}{pi}} ). The approximate formulas given in the options are of the form ( d approx sqrt[3]{frac{a}{b}V} ). So, comparing this to the exact formula, we can see that ( frac{a}{b} ) should be approximately ( frac{6}{pi} ).Given that ( pi approx 3.14159 ), let's compute ( frac{6}{pi} ):( frac{6}{3.14159} approx 1.90986 )So, the ideal constant ( frac{a}{b} ) should be approximately 1.90986.Now, let's compute ( frac{a}{b} ) for each option:A: ( frac{16}{9} approx 1.77778 )B: ( 2 ) (since ( frac{2}{1} = 2 ))C: ( frac{300}{157} approx 1.91083 )D: ( frac{21}{11} approx 1.90909 )So, let's list them:- A: ~1.77778- B: 2- C: ~1.91083- D: ~1.90909We want the value closest to 1.90986. Let's compute the differences:- A: |1.90986 - 1.77778| ≈ 0.13208- B: |1.90986 - 2| ≈ 0.09014- C: |1.90986 - 1.91083| ≈ 0.00097- D: |1.90986 - 1.90909| ≈ 0.00077So, the differences are:- A: ~0.13208- B: ~0.09014- C: ~0.00097- D: ~0.00077Clearly, both C and D are very close to the ideal value, with D being slightly closer than C. Therefore, option D is the most accurate approximation.Alternatively, another way to think about it is by expressing each formula in terms of an approximation for ( pi ). If we rearrange the exact formula ( d = sqrt[3]{frac{6V}{pi}} ) to solve for ( pi ), we get:( pi = frac{6V}{d^3} )But in the approximate formulas, ( d approx sqrt[3]{frac{a}{b}V} ), so substituting this into the equation for ( pi ):( pi approx frac{6V}{left(sqrt[3]{frac{a}{b}V}right)^3} = frac{6V}{frac{a}{b}V} = frac{6b}{a} )So, each formula implies an approximation for ( pi ) as ( frac{6b}{a} ). Let's compute this for each option:A: ( frac{6 times 9}{16} = frac{54}{16} = 3.375 )B: ( frac{6 times 1}{2} = frac{6}{2} = 3 )C: ( frac{6 times 157}{300} = frac{942}{300} = 3.14 )D: ( frac{6 times 11}{21} = frac{66}{21} approx 3.142857 )Now, comparing these approximations to the true value of ( pi approx 3.14159 ):- A: 3.375 (difference ≈ 0.23341)- B: 3 (difference ≈ 0.14159)- C: 3.14 (difference ≈ 0.00159)- D: ~3.142857 (difference ≈ 0.001267)So, again, both C and D are very close to the true value, with D being slightly closer. Therefore, option D provides the most accurate approximation for the diameter of a sphere given its volume.In summary, by either comparing the constants directly or by evaluating the implied approximation for ( pi ), option D is the most accurate.

answer：To find the modulus of the complex number ( z = frac{1 - 3i}{1 + i} ), I will first simplify the expression by eliminating the imaginary unit from the denominator. This can be done by multiplying both the numerator and the denominator by the conjugate of the denominator, which is ( 1 - i ).Multiplying the numerator and denominator by ( 1 - i ), we get:[z = frac{(1 - 3i)(1 - i)}{(1 + i)(1 - i)}]Next, I will expand both the numerator and the denominator:[text{Numerator: } (1 - 3i)(1 - i) = 1 cdot 1 + 1 cdot (-i) - 3i cdot 1 - 3i cdot (-i) = 1 - i - 3i + 3i^2]Since ( i^2 = -1 ), this simplifies to:[1 - 4i + 3(-1) = 1 - 4i - 3 = -2 - 4i]For the denominator:[(1 + i)(1 - i) = 1 cdot 1 + 1 cdot (-i) + i cdot 1 + i cdot (-i) = 1 - i + i - i^2]Again, using ( i^2 = -1 ), this becomes:[1 - i + i - (-1) = 1 + 1 = 2]Now, the simplified form of ( z ) is:[z = frac{-2 - 4i}{2} = -1 - 2i]To find the modulus of ( z ), I use the formula for the modulus of a complex number ( a + bi ), which is ( sqrt{a^2 + b^2} ):[|z| = sqrt{(-1)^2 + (-2)^2} = sqrt{1 + 4} = sqrt{5}]Therefore, the modulus of the complex number ( z ) is ( sqrt{5} ).

answer：Okay, so I have this problem about a rectangular region XYZW. Inside this rectangle, there are four squares placed in a specific way. Three of them are smaller squares with equal side lengths, and they cover the left side of the rectangle. Then there's one larger square on the right side that doesn't overlap with the smaller ones. The area of each smaller square is given as 4 square inches, and I need to find the area of the entire rectangle.Alright, let me break this down. First, since each smaller square has an area of 4 square inches, I can find their side lengths. The area of a square is side length squared, so if the area is 4, then the side length must be the square root of 4, which is 2 inches. So each small square has sides of 2 inches.Now, there are three of these small squares on the left side of the rectangle. I need to figure out how they are arranged. The problem says they cover the left side, so I'm assuming they are placed either vertically or horizontally. But since they are squares, their arrangement might affect the dimensions of the rectangle.If the three small squares are placed vertically, each one would contribute to the height of the rectangle. Since each has a side length of 2 inches, three of them stacked vertically would make the height of the rectangle 2 * 3 = 6 inches. Alternatively, if they are placed horizontally, the width contributed by them would be 2 * 3 = 6 inches. But since they are on the left side, I think it's more likely that they are arranged vertically, contributing to the height of the rectangle.But wait, the problem mentions that the larger square is on the right side without overlapping interiors. So, if the three small squares are on the left, and the larger square is on the right, they must both fit within the same rectangle without overlapping. That suggests that the height of the rectangle must accommodate both the small squares and the larger square.Hmm, maybe I need to consider the arrangement more carefully. If the three small squares are placed side by side horizontally on the left, their combined width would be 3 * 2 = 6 inches, and their height would be 2 inches. Then, the larger square on the right would have to fit into the remaining space of the rectangle.But the problem says the larger square is placed on the right side without overlapping interiors. So, if the three small squares are on the left, the larger square must be adjacent to them either vertically or horizontally. If the small squares are arranged vertically, their combined height would be 6 inches, and the larger square would have to match that height. If the small squares are arranged horizontally, their combined width would be 6 inches, and the larger square would have to match that width.Wait, the problem doesn't specify whether the larger square is placed vertically or horizontally relative to the small squares. Maybe I need to figure that out.Let me think. If the three small squares are arranged vertically on the left, each with a height of 2 inches, then the total height of the rectangle would be 6 inches. The larger square on the right would have to have the same height as the rectangle, which is 6 inches, so its side length would be 6 inches. Then, the area of the larger square would be 6^2 = 36 square inches.But wait, the problem says the larger square is placed on the right side without overlapping interiors. If the small squares are arranged vertically, the larger square would have to be placed next to them horizontally. So, the width of the rectangle would be the width of the three small squares plus the width of the larger square.But the small squares are arranged vertically, so their width is just 2 inches each. Wait, no, if they are arranged vertically, their combined height is 6 inches, but their width is still 2 inches each. So, the total width contributed by the three small squares would be 2 inches, and the larger square would have to be placed next to them, so the total width of the rectangle would be 2 + side length of the larger square.But the larger square's side length is 6 inches, as we thought earlier, so the total width would be 2 + 6 = 8 inches. Then, the height of the rectangle is 6 inches, so the area would be 8 * 6 = 48 square inches. But wait, that seems too big because the total area of the small squares is 3 * 4 = 12 square inches, and the larger square is 36 square inches, so total area would be 12 + 36 = 48 square inches. That seems consistent.But let me check if the larger square can be placed differently. If the three small squares are arranged horizontally on the left, each with a width of 2 inches, their combined width would be 6 inches, and their height would be 2 inches. Then, the larger square on the right would have to match the height of the rectangle, which is 2 inches, so its side length would be 2 inches, but that can't be because it's supposed to be larger than the small squares. So that arrangement doesn't make sense.Alternatively, if the larger square is placed vertically next to the three small squares arranged horizontally, then the height of the rectangle would be the same as the height of the larger square. If the larger square has a side length larger than 2 inches, say 's', then the height of the rectangle would be 's' inches. The width of the rectangle would be the combined width of the three small squares plus the width of the larger square. But the three small squares arranged horizontally have a width of 6 inches, and the larger square's width is 's' inches. So the total width would be 6 + s inches. The height is 's' inches, so the area would be (6 + s) * s.But we also know that the area of the larger square is s^2, and the total area of the rectangle is the sum of the areas of the four squares, which is 3*4 + s^2 = 12 + s^2. So, setting that equal to the area of the rectangle, we have (6 + s) * s = 12 + s^2. Simplifying, 6s + s^2 = 12 + s^2, which leads to 6s = 12, so s = 2 inches. But that contradicts the fact that the larger square is larger than the small squares. So this arrangement is not possible.Therefore, the only feasible arrangement is the three small squares arranged vertically on the left, each with a height of 2 inches, making the total height of the rectangle 6 inches. The larger square on the right must then have a height of 6 inches, so its side length is 6 inches, and its area is 36 square inches. The width of the rectangle is the width of the three small squares (2 inches) plus the width of the larger square (6 inches), making the total width 8 inches. Therefore, the area of the rectangle is 8 inches * 6 inches = 48 square inches.Wait, but earlier I thought the area would be 28 square inches. Did I make a mistake somewhere? Let me check.Wait, in the initial problem, it says that the three smaller squares cover the left side and the larger square covers the right side without overlapping interiors. So, maybe the arrangement is such that the three small squares are placed side by side horizontally on the left, each with a width of 2 inches, making the total width of the left side 6 inches. Then, the larger square is placed on the right side, but it has to fit within the same height as the small squares. Since the small squares are 2 inches tall, the larger square must also be 2 inches tall, but that can't be because it's supposed to be larger. So that doesn't work.Alternatively, maybe the three small squares are arranged vertically on the left, each 2 inches tall, making the total height 6 inches. Then, the larger square is placed on the right, also 6 inches tall, but its width would be equal to the width of the small squares, which is 2 inches. Wait, but then the larger square would have a side length of 2 inches, which is the same as the small squares, not larger. That doesn't make sense either.Hmm, I'm getting confused. Maybe I need to visualize this differently. Perhaps the three small squares are arranged in a way that their combined height equals the side length of the larger square. So, if each small square is 2 inches tall, three of them stacked vertically would make a height of 6 inches. Then, the larger square on the right would have a side length equal to that height, which is 6 inches. So, the larger square would have an area of 36 square inches.Now, the width of the rectangle would be the width of the three small squares plus the width of the larger square. Since each small square is 2 inches wide, three of them side by side would make a width of 6 inches. The larger square is 6 inches wide as well, so the total width of the rectangle would be 6 + 6 = 12 inches. Wait, but that would make the rectangle 12 inches wide and 6 inches tall, giving an area of 72 square inches. But that seems too large, and the total area of the squares would be 3*4 + 36 = 48 square inches, which doesn't match 72.Wait, no, the arrangement might not be that the three small squares are side by side horizontally and the larger square is next to them. Maybe the three small squares are arranged vertically on the left, each 2 inches wide and 2 inches tall, making the total height 6 inches. Then, the larger square is placed on the right, spanning the entire height of 6 inches, so its side length is 6 inches, making its area 36 square inches. The width of the rectangle would be the width of the three small squares (2 inches) plus the width of the larger square (6 inches), totaling 8 inches. Therefore, the area of the rectangle is 8 inches * 6 inches = 48 square inches. But the total area of the squares is 3*4 + 36 = 48 square inches, which matches. So that seems consistent.But earlier, I thought the area was 28 square inches. Maybe I was mistaken. Let me recast the problem.Alternatively, perhaps the three small squares are arranged in a 3x1 column on the left, each 2x2 inches, making the total height 6 inches and width 2 inches. Then, the larger square on the right must fit into the remaining space. Since the rectangle's height is 6 inches, the larger square must also be 6 inches tall, so its side length is 6 inches, making its area 36 square inches. The width of the rectangle is then 2 inches (from the small squares) plus 6 inches (from the larger square), totaling 8 inches. So, the area is 8*6=48 square inches.But wait, the problem says "without overlapping interiors," which suggests that the larger square is placed adjacent to the small squares without overlapping. So, if the small squares are 2 inches wide and 6 inches tall, the larger square must be 6 inches tall and 6 inches wide, placed next to them, making the total width 2+6=8 inches. So, the rectangle is 8 inches wide and 6 inches tall, area 48.But in the initial problem, the user's thought process concluded 28 square inches. Maybe I'm missing something. Let me try another approach.Suppose the three small squares are arranged horizontally on the left, each 2x2 inches, so their combined width is 6 inches and height is 2 inches. Then, the larger square must be placed vertically on the right, spanning the entire height of the rectangle. The height of the rectangle is determined by the larger square's side length. Let's say the larger square has a side length of 's' inches. Then, the height of the rectangle is 's' inches, and the width is 6 + s inches. The area of the rectangle is (6 + s) * s.The total area of the squares is 3*4 + s^2 = 12 + s^2. So, setting (6 + s)*s = 12 + s^2, we get 6s + s^2 = 12 + s^2, which simplifies to 6s = 12, so s=2 inches. But that makes the larger square the same size as the small squares, which contradicts the problem statement. So this arrangement is invalid.Therefore, the only valid arrangement is the three small squares arranged vertically on the left, each 2x2 inches, making the rectangle's height 6 inches and width 2 inches. The larger square on the right must then be 6x6 inches, making the total width 2 + 6 = 8 inches. Thus, the area of the rectangle is 8*6=48 square inches.Wait, but the initial thought process concluded 28 square inches. Maybe I need to re-examine that.Wait, perhaps the three small squares are arranged in a way that their combined height is equal to the side length of the larger square. So, each small square is 2 inches tall, three of them make 6 inches, so the larger square is 6 inches tall, making its area 36. Then, the width of the rectangle is 2 inches (from the small squares) plus 6 inches (from the larger square), totaling 8 inches. So, area is 8*6=48.But maybe the problem is that the three small squares are arranged such that their combined width equals the side length of the larger square. So, each small square is 2 inches wide, three of them make 6 inches, so the larger square is 6 inches wide, making its area 36. Then, the height of the rectangle is 2 inches (from the small squares) plus 6 inches (from the larger square), totaling 8 inches. So, area is 6*8=48.Wait, but that would mean the larger square is 6x6, placed vertically next to the three small squares arranged horizontally. But then the height of the rectangle would be 6 inches, and the width would be 6 inches (from the larger square) plus 2 inches (from the small squares), totaling 8 inches. So, area is 8*6=48.I think I'm going in circles here. Let me try to visualize it differently. Maybe the three small squares are placed in a way that their combined area plus the larger square's area equals the rectangle's area. But that's not necessarily the case because the squares might not cover the entire rectangle.Wait, no, the problem says that the three small squares cover the left side and the larger square covers the right side without overlapping interiors. So, the entire rectangle is covered by the four squares without overlapping. Therefore, the total area of the rectangle is the sum of the areas of the four squares, which is 3*4 + s^2, where s is the side length of the larger square.But earlier, when I assumed the larger square is 6x6, the total area would be 3*4 + 36 = 48, which would mean the rectangle's area is 48. But the initial thought process concluded 28, so maybe I'm missing something.Wait, perhaps the larger square is not 6x6. Let me think again. If the three small squares are arranged vertically, each 2x2, making the height 6 inches. Then, the larger square must fit into the remaining space on the right. If the rectangle's height is 6 inches, the larger square must be 6 inches tall, so its side length is 6 inches, making its area 36. The width of the rectangle is 2 (from the small squares) + 6 (from the larger square) = 8 inches. So, area is 8*6=48.Alternatively, if the three small squares are arranged horizontally, each 2x2, making the width 6 inches. Then, the larger square must fit into the remaining space vertically. The height of the rectangle would be the same as the larger square's side length. Let's say the larger square has side length 's'. Then, the height of the rectangle is 's' inches, and the width is 6 + s inches. The area of the rectangle is (6 + s)*s.The total area of the squares is 3*4 + s^2 = 12 + s^2. So, (6 + s)*s = 12 + s^2 => 6s + s^2 = 12 + s^2 => 6s = 12 => s=2. But that makes the larger square the same size as the small squares, which contradicts the problem statement. So this arrangement is invalid.Therefore, the only valid arrangement is the three small squares arranged vertically, making the rectangle's height 6 inches, and the larger square on the right with side length 6 inches, making the width 8 inches. So, the area is 48 square inches.Wait, but the initial thought process concluded 28. Maybe I'm overcomplicating it. Let me try to see where 28 comes from.If the three small squares are arranged such that their combined height is equal to the side length of the larger square, then each small square is 2 inches tall, three of them make 6 inches. So, the larger square is 6x6, area 36. Then, the width of the rectangle is 2 inches (from the small squares) plus 6 inches (from the larger square), totaling 8 inches. So, area is 8*6=48.Alternatively, maybe the larger square is placed such that its side length is equal to the combined width of the three small squares. Each small square is 2 inches wide, three of them make 6 inches. So, the larger square is 6x6, area 36. The height of the rectangle is 2 inches (from the small squares) plus 6 inches (from the larger square), totaling 8 inches. So, area is 6*8=48.Wait, maybe the problem is that the larger square is placed such that its side length is equal to the side length of the small squares multiplied by some factor. If the three small squares are arranged vertically, their total height is 6 inches, so the larger square must be 6 inches tall, making its area 36. Then, the width of the rectangle is 2 inches (small squares) plus 6 inches (larger square), totaling 8 inches. So, area is 8*6=48.I think I'm stuck on 48. Maybe the initial thought process was wrong. Let me check the initial thought process again.The initial thought process said:1. Each small square has side length 2 inches.2. Total area of small squares is 12.3. Larger square has side length twice that of small squares, so 4 inches, area 16.4. Total area is 12 + 16 = 28.But wait, if the larger square has side length twice that of the small squares, that would be 4 inches, making its area 16. Then, the rectangle's dimensions would be based on that.If the three small squares are arranged vertically, each 2 inches tall, making the height 6 inches. The larger square is 4x4 inches, so its height is 4 inches. But that doesn't match the height of the rectangle, which is 6 inches. So, that arrangement wouldn't work because the larger square would only cover part of the height.Alternatively, if the larger square is 4x4, then the height of the rectangle would be 4 inches, and the three small squares arranged vertically would have a total height of 6 inches, which exceeds the rectangle's height. That doesn't work either.So, the initial thought process might have made a mistake by assuming the larger square's side length is twice that of the small squares. Instead, the larger square's side length should be equal to the total height contributed by the three small squares, which is 6 inches, making its area 36. Therefore, the total area of the rectangle is 48 square inches.I think the initial thought process was incorrect in assuming the larger square's side length is twice that of the small squares. Instead, it should be equal to the total height contributed by the three small squares, leading to a larger area.

answer：Okay, so I have this problem where there are n positive numbers, x₁, x₂, ..., xₙ, and their sum is equal to 1. I need to find the smallest possible value of S, where S is the largest among the fractions x₁/(1+x₁), x₂/(1+x₁+x₂), ..., xₙ/(1+x₁+x₂+...+xₙ). Hmm, that sounds a bit complicated, but let me try to break it down.First, let me understand what each fraction represents. For each k from 1 to n, the fraction is xₖ divided by the sum of 1 and all the previous x's up to xₖ. So, for example, the first fraction is x₁/(1+x₁), the second is x₂/(1+x₁+x₂), and so on until the last one, which is xₙ/(1+x₁+x₂+...+xₙ). Since all x's are positive and their total sum is 1, each denominator is at least 1 and at most 2, right? Because 1 plus the sum of x's is 1 + 1 = 2.So, each fraction is less than or equal to 1, but I need to find the smallest possible maximum among these fractions. That is, I want to arrange the x's such that the largest of these fractions is as small as possible. This seems like an optimization problem where I need to minimize the maximum value.Let me think about how to approach this. Maybe I can use some inequality or optimization technique. Since all the denominators are cumulative sums starting from 1, each subsequent fraction has a larger denominator. So, the first fraction x₁/(1+x₁) is the largest because the denominator is the smallest, and each subsequent fraction is smaller because the denominator increases. But wait, is that always the case? Let me check with an example.Suppose n = 2. Then, x₁ + x₂ = 1. The fractions are x₁/(1+x₁) and x₂/(1+x₁+x₂). The second denominator is 2 because 1 + x₁ + x₂ = 1 + 1 = 2. So, the second fraction is x₂/2. The first fraction is x₁/(1+x₁). Let's say x₁ is 1/2, then x₂ is also 1/2. Then, the first fraction is (1/2)/(1 + 1/2) = (1/2)/(3/2) = 1/3, and the second fraction is (1/2)/2 = 1/4. So, the maximum is 1/3. If I make x₁ larger, say x₁ = 2/3, then x₂ = 1/3. The first fraction is (2/3)/(1 + 2/3) = (2/3)/(5/3) = 2/5, and the second fraction is (1/3)/2 = 1/6. So, the maximum is 2/5, which is larger than 1/3. If I make x₁ smaller, say x₁ = 1/3, then x₂ = 2/3. The first fraction is (1/3)/(4/3) = 1/4, and the second fraction is (2/3)/2 = 1/3. So, the maximum is 1/3 again. Hmm, interesting. So, in this case, the maximum is minimized when x₁ = x₂ = 1/2, giving the maximum S = 1/3.Wait, but in the second case, when x₁ = 1/3, the maximum was also 1/3. So, maybe the maximum is minimized when all the fractions are equal? Let me test that idea.If I want both fractions to be equal, then x₁/(1+x₁) = x₂/2. Since x₂ = 1 - x₁, substituting, we get x₁/(1+x₁) = (1 - x₁)/2. Let's solve for x₁:x₁/(1+x₁) = (1 - x₁)/2Cross-multiplying:2x₁ = (1 - x₁)(1 + x₁)2x₁ = 1 - x₁²Bring all terms to one side:x₁² + 2x₁ - 1 = 0Using quadratic formula:x₁ = [-2 ± sqrt(4 + 4)] / 2 = [-2 ± sqrt(8)] / 2 = [-2 ± 2√2]/2 = -1 ± √2Since x₁ must be positive, x₁ = -1 + √2 ≈ 0.4142. Then x₂ = 1 - x₁ ≈ 0.5858.Calculating the fractions:x₁/(1+x₁) ≈ 0.4142 / (1 + 0.4142) ≈ 0.4142 / 1.4142 ≈ 0.2929x₂/2 ≈ 0.5858 / 2 ≈ 0.2929So, both fractions are equal, and S = 0.2929, which is approximately 1 - 1/√2 ≈ 0.2929. Wait, that's interesting. So, in the case of n=2, the minimal S is 1 - 1/√2.But in my earlier example, when x₁ = x₂ = 1/2, the maximum was 1/3 ≈ 0.3333, which is larger than 0.2929. So, making the fractions equal gives a lower maximum. So, perhaps in general, the minimal S is achieved when all the fractions are equal?Let me test this for n=3.For n=3, x₁ + x₂ + x₃ = 1. The fractions are:x₁/(1+x₁), x₂/(1+x₁+x₂), x₃/(1+x₁+x₂+x₃) = x₃/2.If I want all three fractions equal, let's denote them all equal to S.So,x₁/(1+x₁) = S => x₁ = S(1+x₁) => x₁ = S + Sx₁ => x₁(1 - S) = S => x₁ = S/(1 - S)Similarly, x₂/(1+x₁+x₂) = S. Let me denote y = 1 + x₁ + x₂. Then, x₂ = S y. But y = 1 + x₁ + x₂ = 1 + x₁ + S y. So, y = 1 + x₁ + S y => y(1 - S) = 1 + x₁. But x₁ = S/(1 - S), so:y(1 - S) = 1 + S/(1 - S) => y(1 - S) = (1 - S + S)/ (1 - S) )? Wait, let me compute:1 + S/(1 - S) = (1 - S + S)/(1 - S) = 1/(1 - S). So,y(1 - S) = 1/(1 - S) => y = 1/(1 - S)^2.But y = 1 + x₁ + x₂ = 1 + S/(1 - S) + S y.Wait, substituting y:1 + S/(1 - S) + S y = ySo,1 + S/(1 - S) = y - S y = y(1 - S)But we have y = 1/(1 - S)^2, so:1 + S/(1 - S) = (1 - S) * y = (1 - S) * [1/(1 - S)^2] = 1/(1 - S)But 1 + S/(1 - S) = [ (1 - S) + S ] / (1 - S ) = 1/(1 - S). So, that checks out.Now, moving on to the third fraction: x₃/(1 + x₁ + x₂ + x₃) = x₃/2 = S => x₃ = 2S.But since x₁ + x₂ + x₃ = 1, we have:x₁ + x₂ + x₃ = S/(1 - S) + S y + 2S = 1But y = 1/(1 - S)^2, so x₂ = S y = S/(1 - S)^2.So,x₁ + x₂ + x₃ = S/(1 - S) + S/(1 - S)^2 + 2S = 1Let me compute each term:First term: S/(1 - S)Second term: S/(1 - S)^2Third term: 2SSo, adding them together:S/(1 - S) + S/(1 - S)^2 + 2S = 1Let me factor out S:S [ 1/(1 - S) + 1/(1 - S)^2 + 2 ] = 1Let me compute the expression inside the brackets:Let me denote t = 1 - S, so S = 1 - t.Then,1/t + 1/t² + 2 = (1 + 1/t + 2t)/t²? Wait, maybe better to compute as is.Wait, let me compute:1/(1 - S) + 1/(1 - S)^2 + 2 = [ (1 - S) + 1 ] / (1 - S)^2 + 2 = (2 - S)/(1 - S)^2 + 2Wait, that might not be helpful. Alternatively, let me compute:1/(1 - S) + 1/(1 - S)^2 + 2 = [ (1 - S) + 1 ] / (1 - S)^2 + 2 = (2 - S)/(1 - S)^2 + 2Hmm, not sure. Maybe better to substitute t = 1 - S.Let t = 1 - S, so S = 1 - t.Then,1/(1 - S) = 1/t1/(1 - S)^2 = 1/t²2 remains 2.So, the expression becomes:1/t + 1/t² + 2 = (1 + t)/t² + 2Wait, 1/t + 1/t² = (t + 1)/t²So, total expression is (t + 1)/t² + 2 = (t + 1 + 2t²)/t²So, going back:S [ (t + 1 + 2t²)/t² ] = 1But S = 1 - t, so:(1 - t) [ (t + 1 + 2t²)/t² ] = 1Multiply both sides by t²:(1 - t)(t + 1 + 2t²) = t²Expand the left side:(1)(t + 1 + 2t²) - t(t + 1 + 2t²) = t + 1 + 2t² - t² - t - 2t³ = (t - t) + (1) + (2t² - t²) - 2t³ = 1 + t² - 2t³So,1 + t² - 2t³ = t²Subtract t² from both sides:1 - 2t³ = 0 => 2t³ = 1 => t³ = 1/2 => t = (1/2)^{1/3} = 2^{-1/3}So, t = 2^{-1/3}, which means S = 1 - t = 1 - 2^{-1/3}Therefore, in the case of n=3, the minimal S is 1 - 2^{-1/3}.Wait, that seems to follow a pattern. For n=2, S = 1 - 2^{-1/2}; for n=3, S = 1 - 2^{-1/3}. So, perhaps in general, for n, the minimal S is 1 - 2^{-1/n}.Let me test this hypothesis for n=1. If n=1, then x₁=1, and the fraction is 1/(1+1)=1/2. So, S=1/2, which is 1 - 2^{-1/1}=1 - 1/2=1/2. That checks out.So, it seems that the minimal S is 1 - 2^{-1/n} for each n.Now, how do we achieve this? For each k, the fraction xₖ / (1 + x₁ + ... +xₖ) = S.Let me denote yₖ = 1 + x₁ + ... +xₖ. Then, y₀=1, y₁=1+x₁, y₂=1+x₁+x₂, ..., yₙ=2.Then, xₖ = yₖ - yₖ₋₁.Given that xₖ / yₖ = S, so xₖ = S yₖ.But xₖ = yₖ - yₖ₋₁ = S yₖ => yₖ - yₖ₋₁ = S yₖ => yₖ₋₁ = yₖ (1 - S)So, yₖ = yₖ₋₁ / (1 - S)This is a geometric progression. Starting from y₀=1, y₁=1/(1 - S), y₂=1/(1 - S)^2, ..., yₙ=1/(1 - S)^n.But we know that yₙ=2, so:1/(1 - S)^n = 2 => (1 - S)^n = 1/2 => 1 - S = 2^{-1/n} => S = 1 - 2^{-1/n}Which confirms our earlier result.Therefore, the minimal S is 1 - 2^{-1/n}, and it is achieved when yₖ = 1/(1 - S)^k, so xₖ = yₖ - yₖ₋₁ = 2^{-k/n} - 2^{-(k-1)/n}.Wait, let me compute that:Since 1 - S = 2^{-1/n}, so yₖ = 1/(1 - S)^k = (2^{1/n})^k = 2^{k/n}Thus, yₖ = 2^{k/n}, so xₖ = yₖ - yₖ₋₁ = 2^{k/n} - 2^{(k-1)/n}.So, each xₖ is 2^{(k-1)/n} (2^{1/n} - 1).Therefore, the minimal S is 1 - 2^{-1/n}, and it is achieved when xₖ = 2^{k/n} - 2^{(k-1)/n} for each k from 1 to n.Let me verify this for n=2:x₁ = 2^{1/2} - 2^{0} = √2 - 1 ≈ 0.4142x₂ = 2^{2/2} - 2^{1/2} = 2 - √2 ≈ 0.5858Which matches our earlier result.Similarly, for n=3:x₁ = 2^{1/3} - 1 ≈ 1.26 - 1 = 0.26x₂ = 2^{2/3} - 2^{1/3} ≈ 1.5874 - 1.26 ≈ 0.3274x₃ = 2^{3/3} - 2^{2/3} = 2 - 1.5874 ≈ 0.4126Let's check if x₁ + x₂ + x₃ ≈ 0.26 + 0.3274 + 0.4126 ≈ 1, which is correct.Now, let's compute the fractions:x₁/(1+x₁) ≈ 0.26 / (1 + 0.26) ≈ 0.26 / 1.26 ≈ 0.2063x₂/(1+x₁+x₂) ≈ 0.3274 / (1 + 0.26 + 0.3274) ≈ 0.3274 / 1.5874 ≈ 0.2063x₃/2 ≈ 0.4126 / 2 ≈ 0.2063So, all fractions are equal to approximately 0.2063, which is 1 - 2^{-1/3} ≈ 1 - 0.7937 ≈ 0.2063. Perfect, that works.Therefore, the minimal S is indeed 1 - 2^{-1/n}, and the x's are given by xₖ = 2^{k/n} - 2^{(k-1)/n} for each k from 1 to n.I think this makes sense. By setting up the fractions to be equal, we ensure that none of them exceed the minimal possible maximum, and the geometric progression of the cumulative sums yₖ allows us to solve for the exact values of xₖ. This approach seems solid, and the examples for n=2 and n=3 confirm the result.So, in conclusion, the smallest possible value of S is 1 - 2^{-1/n}, and this is achieved when each xₖ is set to 2^{k/n} - 2^{(k-1)/n} for k from 1 to n.

answer：Okay, so I have this problem where I need to find ( b^2 ) for an ellipse given that its foci coincide with the foci of a hyperbola. The ellipse is given by the equation ( frac{x^2}{25} + frac{y^2}{b^2} = 1 ), and the hyperbola is given by ( frac{x^2}{196} - frac{y^2}{121} = frac{1}{49} ). Hmm, let me try to figure this out step by step.First, I remember that for both ellipses and hyperbolas, the foci are located at a distance ( c ) from the center along the major axis. For an ellipse, the relationship between ( a ), ( b ), and ( c ) is ( c^2 = a^2 - b^2 ), where ( a ) is the semi-major axis and ( b ) is the semi-minor axis. For a hyperbola, the relationship is ( c^2 = a^2 + b^2 ), where ( a ) is the distance from the center to the vertices, and ( b ) is related to the conjugate axis.So, my plan is to first find the foci of the hyperbola, then set those equal to the foci of the ellipse and solve for ( b^2 ).Starting with the hyperbola equation: ( frac{x^2}{196} - frac{y^2}{121} = frac{1}{49} ). I notice that this isn't in the standard form of a hyperbola, which is ( frac{(x - h)^2}{a^2} - frac{(y - k)^2}{b^2} = 1 ). To get it into standard form, I should divide both sides by ( frac{1}{49} ) to make the right-hand side equal to 1.Let me do that:Multiply both sides by 49:( 49 times left( frac{x^2}{196} - frac{y^2}{121} right) = 49 times frac{1}{49} )Simplify:( frac{49x^2}{196} - frac{49y^2}{121} = 1 )Simplify the fractions:( frac{x^2}{4} - frac{y^2}{frac{121}{49}} = 1 )So, now the hyperbola is in standard form:( frac{x^2}{4} - frac{y^2}{frac{121}{49}} = 1 )From this, I can see that ( a^2 = 4 ) and ( b^2 = frac{121}{49} ). Therefore, ( a = 2 ) and ( b = frac{11}{7} ).Now, to find the foci of the hyperbola, I need to calculate ( c ) using the formula ( c^2 = a^2 + b^2 ).Plugging in the values:( c^2 = 4 + frac{121}{49} )Let me compute that. First, convert 4 into a fraction with denominator 49:( 4 = frac{196}{49} )So,( c^2 = frac{196}{49} + frac{121}{49} = frac{317}{49} )Therefore, ( c = sqrt{frac{317}{49}} = frac{sqrt{317}}{7} )So, the foci of the hyperbola are at ( (pm frac{sqrt{317}}{7}, 0) ).Now, moving on to the ellipse. The ellipse equation is ( frac{x^2}{25} + frac{y^2}{b^2} = 1 ). Since the foci of the ellipse coincide with those of the hyperbola, the value of ( c ) for the ellipse is the same as for the hyperbola, which is ( frac{sqrt{317}}{7} ).For an ellipse, the relationship between ( a ), ( b ), and ( c ) is ( c^2 = a^2 - b^2 ). Here, ( a^2 = 25 ) because the denominator under ( x^2 ) is 25.So, plugging into the formula:( left( frac{sqrt{317}}{7} right)^2 = 25 - b^2 )Calculating the left side:( frac{317}{49} = 25 - b^2 )Now, solve for ( b^2 ):( b^2 = 25 - frac{317}{49} )Convert 25 into a fraction with denominator 49:( 25 = frac{1225}{49} )So,( b^2 = frac{1225}{49} - frac{317}{49} = frac{1225 - 317}{49} )Calculate ( 1225 - 317 ):Let me do this subtraction step by step:1225 minus 300 is 925.925 minus 17 is 908.So,( b^2 = frac{908}{49} )Hmm, let me check if this fraction can be simplified. 908 divided by 49. Let's see, 49 times 18 is 882, and 908 minus 882 is 26. So, 908/49 is equal to 18 and 26/49. Since 26 and 49 have no common factors besides 1, the fraction is already in its simplest form.Therefore, ( b^2 = frac{908}{49} ).Wait, let me double-check my steps to make sure I didn't make a mistake.1. I converted the hyperbola equation to standard form correctly by multiplying both sides by 49.2. Calculated ( a^2 = 4 ) and ( b^2 = 121/49 ).3. Then, ( c^2 = a^2 + b^2 = 4 + 121/49 = (196 + 121)/49 = 317/49 ). So, ( c = sqrt{317}/7 ). That seems right.4. For the ellipse, ( a^2 = 25 ), so ( c^2 = 25 - b^2 ).5. Plugged in ( c^2 = 317/49 ), so ( 317/49 = 25 - b^2 ).6. Converted 25 to 1225/49, subtracted 317/49 to get 908/49. That seems correct.I think all my steps are accurate. So, ( b^2 = 908/49 ). I can write this as a mixed number if needed, but since the question just asks for ( b^2 ), the fraction is fine.Just to make sure, let me compute 908 divided by 49:49 × 18 = 882908 - 882 = 26So, 908/49 = 18 + 26/49 = 18 26/49. But as an improper fraction, it's 908/49.Yes, that's correct.So, I think I've got the right answer here. The key was to first convert the hyperbola equation to standard form, find its foci, then use that same distance for the ellipse's foci to solve for ( b^2 ).**Final Answer**The value of ( b^2 ) is boxed{dfrac{908}{49}}.