answer：Alright, so I've got this math problem here: Calculate (1 + √2)^2023 multiplied by (1 - √2)^2023. Hmm, that's a pretty big exponent, 2023. I wonder how I can tackle this without getting lost in the calculations. Let me think step by step.First off, I remember that when you have two expressions raised to the same power and you multiply them together, there's a property of exponents that might help. Specifically, (a * b)^n = a^n * b^n. So, maybe I can combine these two terms first before dealing with the exponent.Let me write that down: (1 + √2)^2023 * (1 - √2)^2023. If I apply that exponent rule, it becomes [(1 + √2)(1 - √2)]^2023. Okay, that seems manageable. Now, I need to simplify the expression inside the brackets.Hmm, (1 + √2)(1 - √2) looks familiar. Isn't that a difference of squares? Yeah, the formula is (a + b)(a - b) = a² - b². So, applying that here, I get 1² - (√2)². Let me compute that: 1 - (√2)^2. Wait, (√2)^2 is just 2, right? So, 1 - 2 equals -1.So now, the expression simplifies to (-1)^2023. Okay, so what's (-1) raised to the 2023rd power? I remember that when you raise -1 to an odd exponent, it stays negative, and when you raise it to an even exponent, it becomes positive. Since 2023 is an odd number, (-1)^2023 should be -1.Let me double-check that. If I have (-1)^1 = -1, (-1)^2 = 1, (-1)^3 = -1, and so on. Yeah, it alternates between -1 and 1 depending on whether the exponent is odd or even. Since 2023 is odd, it should indeed be -1.So, putting it all together: (1 + √2)^2023 * (1 - √2)^2023 simplifies to (-1)^2023, which is -1. That seems straightforward, but let me make sure I didn't skip any steps or make any mistakes.Wait, another way to think about this is to recognize that (1 + √2) and (1 - √2) are conjugates. Their product is always a rational number, which in this case is -1, as I found earlier. So, raising them both to the same power and then multiplying should just give (-1) raised to that power. Since 2023 is odd, it's -1.I think that's solid. I don't see any errors in my reasoning. It all checks out. So, the final answer should be -1.

answer：Okay, so I have this problem here about an isosceles triangle with one of the interior angles measuring 70 degrees, and I need to find the measure of the vertex angle. Hmm, let's see. First, I remember that in an isosceles triangle, two sides are equal, and the angles opposite those sides are also equal. So, there are two equal angles, which are called the base angles, and the third angle is the vertex angle.Now, the problem says one of the interior angles is 70 degrees. That could mean a couple of things. Either the 70-degree angle is one of the base angles, or it's the vertex angle. I think I need to consider both possibilities because the problem doesn't specify which one it is.Let me start by recalling that the sum of all interior angles in any triangle is 180 degrees. So, if I can figure out the measures of the other angles based on whether 70 degrees is a base angle or the vertex angle, I can find the measure of the vertex angle.First, let's assume that the 70-degree angle is one of the base angles. Since the triangle is isosceles, the other base angle would also be 70 degrees. So, if both base angles are 70 degrees, I can find the vertex angle by subtracting the sum of the base angles from 180 degrees.Let me write that down:If base angles = 70 degrees each,then vertex angle = 180 - (70 + 70) = 180 - 140 = 40 degrees.So, in this case, the vertex angle would be 40 degrees.But wait, what if the 70-degree angle is actually the vertex angle? Then, the two base angles would be equal, and I can find their measures by subtracting the vertex angle from 180 degrees and then dividing by two.Let me calculate that:If vertex angle = 70 degrees,then sum of base angles = 180 - 70 = 110 degrees.Since the two base angles are equal,each base angle = 110 / 2 = 55 degrees.So, in this scenario, the vertex angle is 70 degrees, and the base angles are 55 degrees each.Hmm, so depending on whether the given 70-degree angle is a base angle or the vertex angle, the measure of the vertex angle can be either 40 degrees or 70 degrees.But wait, the problem specifically asks for the measure of the vertex angle. So, does that mean I need to consider both possibilities and present both answers? Or is there a way to determine which one it is?I think the problem doesn't specify whether the 70-degree angle is a base angle or the vertex angle, so both cases are possible. Therefore, the measure of the vertex angle could be either 40 degrees or 70 degrees.Let me double-check my calculations to make sure I didn't make any mistakes.First case:- Base angles = 70 degrees each- Vertex angle = 180 - (70 + 70) = 40 degreesThat seems correct.Second case:- Vertex angle = 70 degrees- Sum of base angles = 180 - 70 = 110 degrees- Each base angle = 110 / 2 = 55 degreesThat also seems correct.So, both scenarios are valid, and the vertex angle can be either 40 degrees or 70 degrees depending on whether the given 70-degree angle is a base angle or the vertex angle.I think that's all there is to it. I don't see any other possibilities because in an isosceles triangle, only two angles can be equal, and the third one is different. So, the given angle has to be either one of the equal base angles or the unique vertex angle.Therefore, the measure of the vertex angle is either 40 degrees or 70 degrees.

answer：Okay, so I have this triangle ABC, and the sides opposite angles A, B, and C are a, b, and c respectively. I know that a is 3, and there's this equation given: 1 + (tan A)/(tan B) = (2c)/b. I need to find the maximum value of b + c. Hmm, okay, let me try to figure this out step by step.First, I remember that in triangles, the Law of Sines relates the sides and angles. It says that a/sin A = b/sin B = c/sin C = 2R, where R is the radius of the circumscribed circle. Maybe that can help me here. Also, the Law of Cosines relates the sides with the cosine of an angle: a² = b² + c² - 2bc cos A. Since I know a is 3, maybe I can use that later.Looking at the given equation: 1 + (tan A)/(tan B) = (2c)/b. I need to manipulate this equation to find a relationship between the sides or angles. Let me rewrite tan A and tan B in terms of sine and cosine. So, tan A = sin A / cos A and tan B = sin B / cos B. Therefore, (tan A)/(tan B) = (sin A / cos A) / (sin B / cos B) = (sin A cos B) / (sin B cos A). So substituting back into the equation: 1 + (sin A cos B)/(sin B cos A) = (2c)/b. Let me write that as 1 + (sin A cos B)/(sin B cos A) = (2c)/b.Hmm, maybe I can combine the terms on the left side. Let me find a common denominator. The first term is 1, which can be written as (sin B cos A)/(sin B cos A). So adding that to (sin A cos B)/(sin B cos A) gives [sin B cos A + sin A cos B] / (sin B cos A). Wait, the numerator looks familiar. Isn't sin B cos A + sin A cos B equal to sin(A + B)? Because sin(A + B) = sin A cos B + cos A sin B. Yes, that's right. So the numerator is sin(A + B). So now, the left side becomes sin(A + B)/(sin B cos A). And the equation is sin(A + B)/(sin B cos A) = (2c)/b. But in a triangle, A + B + C = π, so A + B = π - C. Therefore, sin(A + B) = sin(π - C) = sin C. So substituting that in, the equation becomes sin C / (sin B cos A) = (2c)/b.Now, from the Law of Sines, I know that c = 2R sin C and b = 2R sin B. So c / b = sin C / sin B. Therefore, 2c / b = 2 sin C / sin B. So substituting back into the equation: sin C / (sin B cos A) = 2 sin C / sin B. Hmm, let's write that out: [sin C / (sin B cos A)] = [2 sin C / sin B]. I can cancel sin C from both sides since sin C isn't zero in a triangle (angles are between 0 and π, so sine is positive). Similarly, sin B isn't zero. So cancelling sin C and sin B, we get 1 / cos A = 2. So 1 / cos A = 2 implies that cos A = 1/2. Therefore, angle A is π/3 or 60 degrees because cos(π/3) = 1/2.Okay, so angle A is 60 degrees. Now, I can use the Law of Cosines to relate the sides. The Law of Cosines says a² = b² + c² - 2bc cos A. Since a is 3 and cos A is 1/2, plugging in, we get 3² = b² + c² - 2bc*(1/2). Simplifying, that's 9 = b² + c² - bc.So, 9 = b² + c² - bc. Now, I need to find the maximum value of b + c. Let me denote S = b + c. I want to maximize S.I know that (b + c)² = b² + 2bc + c². From the equation above, I have b² + c² = 9 + bc. So substituting into the square of S, we get S² = (9 + bc) + 2bc = 9 + 3bc. So S² = 9 + 3bc.Therefore, S = sqrt(9 + 3bc). To maximize S, I need to maximize bc. So, what's the maximum value of bc given that 9 = b² + c² - bc?Hmm, this is a constrained optimization problem. I need to maximize bc given that b² + c² - bc = 9. Let me think about how to approach this.I can use the method of Lagrange multipliers, but maybe there's a simpler way using inequalities. Let me recall that for any real numbers, b² + c² ≥ 2bc. So, from the equation, 9 = b² + c² - bc ≥ 2bc - bc = bc. So, 9 ≥ bc. Therefore, bc ≤ 9.So the maximum value of bc is 9. Therefore, substituting back into S² = 9 + 3bc, the maximum S² is 9 + 3*9 = 9 + 27 = 36. Therefore, S = sqrt(36) = 6.Wait, but when does equality hold? For bc to be 9, we need b² + c² - bc = 9 and bc = 9. Let me check if that's possible.If bc = 9, then from the equation 9 = b² + c² - 9, so b² + c² = 18. Also, since bc = 9, we can write b² + c² = (b + c)² - 2bc = S² - 18. But we have b² + c² = 18, so S² - 18 = 18, which gives S² = 36, so S = 6. But also, for bc to be maximum, we need b = c because of the AM-GM inequality, which states that for positive real numbers, the product is maximized when the numbers are equal, given a fixed sum. Wait, but in this case, we have a fixed value of b² + c² - bc. Hmm, maybe it's still when b = c.Let me test if b = c. If b = c, then from the equation 9 = b² + b² - b*b = 2b² - b² = b². So b² = 9, so b = 3, and c = 3. Therefore, b + c = 6. So that's consistent.Therefore, the maximum value of b + c is 6, achieved when b = c = 3.Wait, but let me double-check. If b = c = 3, then angle A is 60 degrees, and sides b and c are both 3. So, triangle ABC would be an equilateral triangle? Because all sides are equal? Wait, no, because a is 3, and b and c are 3, so actually, it is an equilateral triangle. All sides are 3, so all angles are 60 degrees. That makes sense.But in the given equation, 1 + tan A / tan B = 2c / b. If all angles are 60 degrees, tan A = tan B = tan 60 = sqrt(3). So tan A / tan B = 1. Therefore, 1 + 1 = 2, and 2c / b = 2*3 / 3 = 2. So 2 = 2, which holds true. So that checks out.Therefore, everything seems consistent. So the maximum value of b + c is 6.**Final Answer**The maximum value of ( b + c ) is boxed{6}.

answer：Okay, so I have these two trigonometry problems to solve. Let me take them one at a time. Starting with problem (1): Simplify the expression:[frac{tan(3pi - alpha) cos(2pi - alpha) sin(-alpha + frac{3pi}{2})}{cos(-alpha - pi) sin(-pi + alpha) cos(alpha + frac{5pi}{2})}]Hmm, this looks a bit intimidating with all those angles, but I remember that trigonometric functions have periodicity and certain identities that can help simplify them. Let me recall the periodicity and the identities for tangent, sine, and cosine.First, I know that:- The period of tangent is (pi), so (tan(theta + pi) = tan theta).- The period of sine and cosine is (2pi), so (sin(theta + 2pi) = sin theta) and (cos(theta + 2pi) = cos theta).- Also, there are co-function identities and phase shift identities, like (sin(theta + frac{pi}{2}) = cos theta) and (cos(theta + frac{pi}{2}) = -sin theta).- Additionally, odd and even properties: (sin(-theta) = -sin theta) and (cos(-theta) = cos theta).Let me try to simplify each term in the numerator and the denominator step by step.Starting with the numerator:1. (tan(3pi - alpha)): - Since the period of tangent is (pi), (tan(3pi - alpha) = tan(pi - alpha)). - I remember that (tan(pi - alpha) = -tan alpha). So this simplifies to (-tan alpha).2. (cos(2pi - alpha)): - The period of cosine is (2pi), so (cos(2pi - alpha) = cos(-alpha)). - Cosine is an even function, so (cos(-alpha) = cos alpha).3. (sin(-alpha + frac{3pi}{2})): - Let me rewrite this as (sin(frac{3pi}{2} - alpha)). - Using the identity (sin(frac{3pi}{2} - alpha) = -cos alpha). Wait, is that right? Let me think. - Alternatively, I can use the sine of a difference: (sin(A - B) = sin A cos B - cos A sin B). - So, (sin(frac{3pi}{2} - alpha) = sin frac{3pi}{2} cos alpha - cos frac{3pi}{2} sin alpha). - (sin frac{3pi}{2} = -1) and (cos frac{3pi}{2} = 0). - Therefore, this becomes (-1 cdot cos alpha - 0 cdot sin alpha = -cos alpha). - So, (sin(-alpha + frac{3pi}{2}) = -cos alpha).Putting it all together, the numerator becomes:[-tan alpha cdot cos alpha cdot (-cos alpha)]Let me compute this step by step:First, multiply (-tan alpha) and (cos alpha):[-tan alpha cdot cos alpha = -frac{sin alpha}{cos alpha} cdot cos alpha = -sin alpha]Then, multiply this result by (-cos alpha):[-sin alpha cdot (-cos alpha) = sin alpha cos alpha]So, the numerator simplifies to (sin alpha cos alpha).Now, moving on to the denominator:1. (cos(-alpha - pi)): - Cosine is even, so (cos(-alpha - pi) = cos(alpha + pi)). - Using the identity (cos(alpha + pi) = -cos alpha).2. (sin(-pi + alpha)): - Sine is odd, so (sin(-pi + alpha) = -sin(pi - alpha)). - Using the identity (sin(pi - alpha) = sin alpha), so this becomes (-sin alpha).3. (cos(alpha + frac{5pi}{2})): - Let me subtract (2pi) to bring it within the first rotation: (frac{5pi}{2} - 2pi = frac{pi}{2}). - So, (cos(alpha + frac{5pi}{2}) = cos(alpha + frac{pi}{2} + 2pi) = cos(alpha + frac{pi}{2})). - Using the identity (cos(alpha + frac{pi}{2}) = -sin alpha).Putting it all together, the denominator becomes:[-cos alpha cdot (-sin alpha) cdot (-sin alpha)]Let me compute this step by step:First, multiply (-cos alpha) and (-sin alpha):[-cos alpha cdot (-sin alpha) = cos alpha sin alpha]Then, multiply this result by (-sin alpha):[cos alpha sin alpha cdot (-sin alpha) = -cos alpha sin^2 alpha]So, the denominator simplifies to (-cos alpha sin^2 alpha).Now, putting the numerator and denominator together:[frac{sin alpha cos alpha}{- cos alpha sin^2 alpha}]Let me simplify this fraction. First, I can cancel out (cos alpha) from numerator and denominator:[frac{sin alpha}{- sin^2 alpha} = frac{1}{- sin alpha} = -frac{1}{sin alpha}]So, the simplified expression is (-csc alpha). But since the question asks to simplify, and (-csc alpha) is a simplified form, I think that's the answer.Wait, but let me double-check my steps because sometimes signs can be tricky.Starting with the numerator:- (tan(3pi - alpha) = tan(pi - alpha) = -tan alpha)- (cos(2pi - alpha) = cos alpha)- (sin(-alpha + frac{3pi}{2}) = sin(frac{3pi}{2} - alpha) = -cos alpha)So numerator: (-tan alpha cdot cos alpha cdot (-cos alpha))Which is: (-tan alpha cdot cos alpha cdot (-cos alpha) = (-tan alpha) cdot (-cos^2 alpha) = tan alpha cos^2 alpha)Wait, hold on, that contradicts my earlier step. Let me recast this.Wait, no, I think I made a mistake in the multiplication order.Wait, (-tan alpha cdot cos alpha cdot (-cos alpha)) is equal to:First, (-tan alpha cdot cos alpha = -sin alpha), as I did before.Then, (-sin alpha cdot (-cos alpha) = sin alpha cos alpha). So that was correct.Denominator:- (cos(-alpha - pi) = cos(alpha + pi) = -cos alpha)- (sin(-pi + alpha) = -sin(pi - alpha) = -sin alpha)- (cos(alpha + frac{5pi}{2}) = cos(alpha + frac{pi}{2}) = -sin alpha)So denominator: (-cos alpha cdot (-sin alpha) cdot (-sin alpha))Compute step by step:First, (-cos alpha cdot (-sin alpha) = cos alpha sin alpha)Then, (cos alpha sin alpha cdot (-sin alpha) = -cos alpha sin^2 alpha)So denominator is (-cos alpha sin^2 alpha)Therefore, the entire expression is:[frac{sin alpha cos alpha}{- cos alpha sin^2 alpha} = frac{1}{- sin alpha} = -csc alpha]So, yes, that seems consistent. So the simplified expression is (-csc alpha), which can also be written as (-frac{1}{sin alpha}).Wait, but in the initial problem statement, the user wrote:"The original expression (= dfrac {-tan alphacdot cos alphacdot (-cos alpha)}{-cos alpha cdot (-sin alpha )cdot (-sin alpha )}=- dfrac {1}{sin alpha })."So, according to the user, the simplified expression is (- dfrac {1}{sin alpha }), which is the same as (-csc alpha). So that's consistent with my result.Okay, so problem (1) is simplified to (-csc alpha).Moving on to problem (2):Given (tan alpha = dfrac{1}{4}), find the value of:[frac{1}{2cos^2 alpha - 3 sin alpha cos alpha}]Hmm, okay. So I need to compute this expression given that (tan alpha = frac{1}{4}).First, let me recall that (tan alpha = frac{sin alpha}{cos alpha}), so if (tan alpha = frac{1}{4}), then (sin alpha = frac{1}{sqrt{1 + 16}} = frac{1}{sqrt{17}}) and (cos alpha = frac{4}{sqrt{17}}), assuming (alpha) is in the first quadrant where sine and cosine are positive. But since the problem doesn't specify, I might need to consider the general case, but perhaps the expression is positive regardless.Alternatively, I can express everything in terms of (tan alpha). Let me see.The expression is:[frac{1}{2cos^2 alpha - 3 sin alpha cos alpha}]I can factor out (cos alpha) in the denominator:[frac{1}{cos alpha (2 cos alpha - 3 sin alpha)}]Alternatively, I can divide numerator and denominator by (cos^2 alpha) to express everything in terms of (tan alpha). Let me try that.Divide numerator and denominator by (cos^2 alpha):Numerator becomes (frac{1}{cos^2 alpha}), which is (1 + tan^2 alpha).Denominator becomes (2 - 3 tan alpha).Wait, let me write it step by step.Starting with the denominator:[2cos^2 alpha - 3 sin alpha cos alpha]Divide numerator and denominator by (cos^2 alpha):Numerator: (1) divided by (cos^2 alpha) is (1/cos^2 alpha = sec^2 alpha = 1 + tan^2 alpha).Denominator: (2cos^2 alpha - 3 sin alpha cos alpha) divided by (cos^2 alpha) is:[2 - 3 frac{sin alpha}{cos alpha} = 2 - 3 tan alpha]Therefore, the entire expression becomes:[frac{1 + tan^2 alpha}{2 - 3 tan alpha}]Since (tan alpha = frac{1}{4}), substitute this value:First, compute (1 + tan^2 alpha):[1 + left(frac{1}{4}right)^2 = 1 + frac{1}{16} = frac{16}{16} + frac{1}{16} = frac{17}{16}]Next, compute (2 - 3 tan alpha):[2 - 3 cdot frac{1}{4} = 2 - frac{3}{4} = frac{8}{4} - frac{3}{4} = frac{5}{4}]Therefore, the expression becomes:[frac{frac{17}{16}}{frac{5}{4}} = frac{17}{16} cdot frac{4}{5} = frac{17 cdot 4}{16 cdot 5} = frac{68}{80} = frac{17}{20}]So, the value is (frac{17}{20}).Wait, let me check my steps again to make sure I didn't make a mistake.Starting with the expression:[frac{1}{2cos^2 alpha - 3 sin alpha cos alpha}]Expressed in terms of (tan alpha):[frac{1}{cos alpha (2 cos alpha - 3 sin alpha)} = frac{sec alpha}{2 cos alpha - 3 sin alpha}]But that seems more complicated. Alternatively, dividing numerator and denominator by (cos^2 alpha) is a better approach.So, numerator becomes (1/cos^2 alpha = 1 + tan^2 alpha), and denominator becomes (2 - 3 tan alpha), which is correct.Substituting (tan alpha = 1/4):Numerator: (1 + (1/4)^2 = 1 + 1/16 = 17/16).Denominator: (2 - 3*(1/4) = 2 - 3/4 = 5/4).So, the fraction is ((17/16) / (5/4) = (17/16)*(4/5) = (17*4)/(16*5) = 68/80 = 17/20). That seems correct.Alternatively, I can compute (sin alpha) and (cos alpha) directly.Given (tan alpha = 1/4), so we can imagine a right triangle where the opposite side is 1 and the adjacent side is 4, so the hypotenuse is (sqrt{1^2 + 4^2} = sqrt{17}).Therefore, (sin alpha = 1/sqrt{17}) and (cos alpha = 4/sqrt{17}).Compute the denominator:[2 cos^2 alpha - 3 sin alpha cos alpha = 2 left(frac{16}{17}right) - 3 left(frac{1}{sqrt{17}}right) left(frac{4}{sqrt{17}}right)]Simplify each term:First term: (2 * (16/17) = 32/17).Second term: (3 * (1/sqrt{17}) * (4/sqrt{17}) = 3 * (4/17) = 12/17).So, denominator: (32/17 - 12/17 = 20/17).Therefore, the entire expression is (1 / (20/17) = 17/20). So, same result.So, that confirms that the answer is indeed (17/20).Therefore, problem (2) is solved.**Final Answer**(1) (boxed{-csc alpha})(2) (boxed{dfrac{17}{20}})

answer：Alright, so I have this problem about a scientific research institute allocating funds as rewards to its researchers. There are ten places, from first to tenth, and each subsequent place gets more than half of the remaining funds by 10,000 Yuan. The goal is to find out the total amount of funds allocated for these rewards, expressed in ten-thousand Yuan units.Okay, let's break this down. First, I need to understand how the funds are being distributed. The first place gets more than half of the total funds by 10,000 Yuan. Then, the second place gets more than half of the remaining funds by another 10,000 Yuan, and this pattern continues until the tenth place, which gets the last of the funds.Hmm, so each recipient is getting a bit more than half of what's left after the previous recipient has taken their share. This seems like a recursive process where each step depends on the previous one. Maybe I can model this with some sort of sequence or series.Let me denote the total funds as ( T ). The first place gets more than half of ( T ) by 10,000 Yuan. So, mathematically, that would be ( frac{T}{2} + 10,000 ). But since we're dealing in ten-thousand Yuan units, I can simplify this by dividing everything by 10,000. So, let's redefine ( T ) as the total funds in ten-thousand Yuan. Then, the first place gets ( frac{T}{2} + 1 ).After the first place has taken their share, the remaining funds would be ( T - left( frac{T}{2} + 1 right) = frac{T}{2} - 1 ). Now, the second place gets more than half of this remaining amount by 1 ten-thousand Yuan. So, the second place gets ( frac{1}{2} left( frac{T}{2} - 1 right) + 1 ).Let me compute that: ( frac{1}{2} times frac{T}{2} = frac{T}{4} ), and ( frac{1}{2} times (-1) = -frac{1}{2} ). So, the second place gets ( frac{T}{4} - frac{1}{2} + 1 = frac{T}{4} + frac{1}{2} ).Now, the remaining funds after the second place would be ( left( frac{T}{2} - 1 right) - left( frac{T}{4} + frac{1}{2} right) = frac{T}{2} - 1 - frac{T}{4} - frac{1}{2} = frac{T}{4} - frac{3}{2} ).This pattern seems to be getting complicated quickly. Maybe there's a better way to approach this. Perhaps I can model the amount each place receives as a sequence where each term depends on the previous one.Let me denote ( a_n ) as the amount the nth place receives. According to the problem, each recipient gets more than half of the remaining funds by 10,000 Yuan. So, if I think recursively, the amount the nth place receives is half of what's left after the (n-1)th place has taken their share, plus 1.But wait, actually, it's more than half, so it's half of the remaining funds plus 1. So, ( a_n = frac{1}{2} times text{remaining funds after } (n-1)text{th place} + 1 ).This seems like a recursive relationship. Maybe I can express this in terms of the previous amount.Let me think about the remaining funds after each step. Let ( R_n ) be the remaining funds after the nth place has received their reward. Then, ( R_n = R_{n-1} - a_n ).But ( a_n = frac{1}{2} R_{n-1} + 1 ). So, substituting that into the equation for ( R_n ), we get:( R_n = R_{n-1} - left( frac{1}{2} R_{n-1} + 1 right) = frac{1}{2} R_{n-1} - 1 ).So, ( R_n = frac{1}{2} R_{n-1} - 1 ).This is a recursive formula for the remaining funds after each place. Now, we can try to solve this recurrence relation.First, let's write down the recurrence:( R_n = frac{1}{2} R_{n-1} - 1 ).This is a linear recurrence relation. To solve it, we can find the homogeneous solution and a particular solution.The homogeneous equation is ( R_n - frac{1}{2} R_{n-1} = 0 ). The characteristic equation is ( r - frac{1}{2} = 0 ), so ( r = frac{1}{2} ). Therefore, the homogeneous solution is ( R_n^{(h)} = C left( frac{1}{2} right)^n ).Now, for the particular solution, since the nonhomogeneous term is a constant (-1), we can assume a constant particular solution ( R_n^{(p)} = A ).Substituting into the recurrence:( A = frac{1}{2} A - 1 ).Solving for A:( A - frac{1}{2} A = -1 )( frac{1}{2} A = -1 )( A = -2 ).So, the general solution is:( R_n = R_n^{(h)} + R_n^{(p)} = C left( frac{1}{2} right)^n - 2 ).Now, we need to find the constant C using the initial condition. What is the initial condition? At n=0, before any rewards are given, the remaining funds are the total funds T. So, ( R_0 = T ).Plugging n=0 into the general solution:( R_0 = C left( frac{1}{2} right)^0 - 2 = C - 2 = T ).Therefore, ( C = T + 2 ).So, the solution becomes:( R_n = (T + 2) left( frac{1}{2} right)^n - 2 ).Now, we need to find the amount each place receives. Recall that ( a_n = frac{1}{2} R_{n-1} + 1 ).Let's express ( a_n ) in terms of T:( a_n = frac{1}{2} R_{n-1} + 1 = frac{1}{2} left[ (T + 2) left( frac{1}{2} right)^{n-1} - 2 right] + 1 ).Simplify this:( a_n = frac{1}{2} (T + 2) left( frac{1}{2} right)^{n-1} - 1 + 1 ).The -1 and +1 cancel out, so:( a_n = frac{1}{2} (T + 2) left( frac{1}{2} right)^{n-1} ).Simplify further:( a_n = (T + 2) left( frac{1}{2} right)^n ).So, each recipient's reward is ( a_n = (T + 2) left( frac{1}{2} right)^n ).Now, we have ten recipients, so n goes from 1 to 10. The total funds T is the sum of all ( a_n ) from n=1 to n=10.So, ( T = sum_{n=1}^{10} a_n = sum_{n=1}^{10} (T + 2) left( frac{1}{2} right)^n ).Factor out ( (T + 2) ):( T = (T + 2) sum_{n=1}^{10} left( frac{1}{2} right)^n ).Now, compute the sum ( sum_{n=1}^{10} left( frac{1}{2} right)^n ). This is a finite geometric series with first term ( a = frac{1}{2} ), common ratio ( r = frac{1}{2} ), and number of terms ( n = 10 ).The sum of a finite geometric series is ( S_n = a frac{1 - r^n}{1 - r} ).Plugging in the values:( S_{10} = frac{1}{2} times frac{1 - left( frac{1}{2} right)^{10}}{1 - frac{1}{2}} = frac{1}{2} times frac{1 - frac{1}{1024}}{frac{1}{2}} = frac{1}{2} times 2 times left( 1 - frac{1}{1024} right) = 1 - frac{1}{1024} = frac{1023}{1024} ).So, ( T = (T + 2) times frac{1023}{1024} ).Now, solve for T:( T = frac{1023}{1024} T + frac{1023}{1024} times 2 ).Subtract ( frac{1023}{1024} T ) from both sides:( T - frac{1023}{1024} T = frac{1023}{1024} times 2 ).Simplify the left side:( frac{1}{1024} T = frac{2046}{1024} ).Multiply both sides by 1024:( T = 2046 ).So, the total funds allocated for the rewards are 2046 ten-thousand Yuan.Wait, let me double-check this. If T = 2046, then the first place gets ( frac{2046}{2} + 1 = 1023 + 1 = 1024 ) ten-thousand Yuan. Then, the remaining funds are ( 2046 - 1024 = 1022 ).The second place gets ( frac{1022}{2} + 1 = 511 + 1 = 512 ). Remaining funds: ( 1022 - 512 = 510 ).Third place: ( frac{510}{2} + 1 = 255 + 1 = 256 ). Remaining: ( 510 - 256 = 254 ).Fourth place: ( frac{254}{2} + 1 = 127 + 1 = 128 ). Remaining: ( 254 - 128 = 126 ).Fifth place: ( frac{126}{2} + 1 = 63 + 1 = 64 ). Remaining: ( 126 - 64 = 62 ).Sixth place: ( frac{62}{2} + 1 = 31 + 1 = 32 ). Remaining: ( 62 - 32 = 30 ).Seventh place: ( frac{30}{2} + 1 = 15 + 1 = 16 ). Remaining: ( 30 - 16 = 14 ).Eighth place: ( frac{14}{2} + 1 = 7 + 1 = 8 ). Remaining: ( 14 - 8 = 6 ).Ninth place: ( frac{6}{2} + 1 = 3 + 1 = 4 ). Remaining: ( 6 - 4 = 2 ).Tenth place: ( frac{2}{2} + 1 = 1 + 1 = 2 ). Remaining: ( 2 - 2 = 0 ).Adding up all the rewards: 1024 + 512 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 = 2046. Yep, that checks out.So, the total amount of funds allocated is 2046 ten-thousand Yuan.

answer：Okay, so I have this problem here: "A fair coin is tossed 4 times. What is the probability of exactly two consecutive heads?" The answer choices are given from A to E, with fractions like 1/16, 1/8, 3/16, 1/4, and 3/8. I need to figure out which one is correct.First, I should understand what the question is asking. It's about tossing a coin four times and finding the probability that there are exactly two consecutive heads. That means in the four tosses, there should be one instance where two heads come up in a row, and the other two tosses shouldn't form another pair of consecutive heads. Also, I think it's important that there are exactly two heads in a row, not more. So, if I get three heads in a row, that wouldn't count because it's more than two consecutive heads.Alright, let's break it down. When dealing with probability, I know that the probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. So, I need to find both the total number of possible outcomes when tossing a coin four times and the number of favorable outcomes where exactly two consecutive heads occur.Starting with the total number of possible outcomes. Since each toss of a fair coin has two possible results—heads (H) or tails (T)—and the coin is tossed four times, the total number of possible outcomes should be 2^4, which is 16. So, there are 16 different sequences of heads and tails possible.Now, I need to find the number of favorable outcomes where there are exactly two consecutive heads. To do this, I can list all possible sequences of four coin tosses and then count how many of them have exactly two consecutive heads. But that might take a while. Maybe there's a smarter way to approach this.Alternatively, I can think about the different positions where the two consecutive heads can occur in the four tosses. Since we're dealing with four tosses, the two consecutive heads can start at position 1, 2, or 3. Let's consider each case:1. **Case 1: Consecutive heads start at position 1 (i.e., the first two tosses are heads).** - So, the first two tosses are H, H. Now, the third and fourth tosses should not form another pair of consecutive heads. That means the third toss can be either H or T, but if it's H, then the fourth toss must be T to avoid having three consecutive heads. Wait, actually, the problem says exactly two consecutive heads, so if the third toss is H, then we have two consecutive heads starting at position 2 as well, which would mean more than one instance of two consecutive heads. But the question is about exactly two consecutive heads, so I think that means only one instance of two consecutive heads in the entire sequence. So, if I have H, H, H, T, that would have two instances of two consecutive heads: positions 1-2 and 2-3. So, that wouldn't count because it's more than one instance. Therefore, after H, H, the third toss must be T to avoid having another consecutive head. Then, the fourth toss can be either H or T, but if it's H, does that create another instance of two consecutive heads? Let's see: H, H, T, H. In this case, the first two are H, H, and the last two are T, H. So, only one instance of two consecutive heads. Similarly, H, H, T, T would also have only one instance. So, from this case, we have two favorable outcomes: H, H, T, H and H, H, T, T.2. **Case 2: Consecutive heads start at position 2 (i.e., the second and third tosses are heads).** - So, the sequence would be something like X, H, H, Y, where X and Y are either H or T. But we need to ensure that there are no other instances of two consecutive heads. So, the first toss (X) can't be H because then we'd have H, H, H, Y, which would have two instances of two consecutive heads. Therefore, X must be T. Similarly, the fourth toss (Y) can't be H because then we'd have T, H, H, H, which would have two instances of two consecutive heads. Therefore, Y must be T. So, the only favorable outcome in this case is T, H, H, T.3. **Case 3: Consecutive heads start at position 3 (i.e., the third and fourth tosses are heads).** - This would be similar to Case 1. The sequence would be something like X, Y, H, H. To avoid having another instance of two consecutive heads, the second toss (Y) must be T. The first toss (X) can be either H or T, but if it's H, then we have H, T, H, H. Is that acceptable? Let's check: H, T, H, H. The first two tosses are H, T, which is not two consecutive heads, and the last two are H, H, which is exactly one instance. So, that's acceptable. Similarly, T, T, H, H is also acceptable. Therefore, from this case, we have two favorable outcomes: H, T, H, H and T, T, H, H.Now, let's tally up all the favorable outcomes we found:- From Case 1: H, H, T, H and H, H, T, T (2 outcomes)- From Case 2: T, H, H, T (1 outcome)- From Case 3: H, T, H, H and T, T, H, H (2 outcomes)So, in total, we have 2 + 1 + 2 = 5 favorable outcomes.Wait a minute, that's 5, but I thought earlier that the answer might be 4. Did I make a mistake? Let me double-check.Looking back at Case 1: H, H, T, H and H, H, T, T. That's correct, two outcomes.Case 2: T, H, H, T. That's one outcome.Case 3: H, T, H, H and T, T, H, H. That's two outcomes.So, 2 + 1 + 2 = 5. Hmm, but I thought the answer was 4. Maybe I'm overcounting somewhere.Wait, let's list all the favorable outcomes explicitly:1. H, H, T, H2. H, H, T, T3. T, H, H, T4. H, T, H, H5. T, T, H, HSo, that's five sequences. But let me check if any of these have more than one instance of two consecutive heads.1. H, H, T, H: Only one instance at the beginning.2. H, H, T, T: Only one instance at the beginning.3. T, H, H, T: Only one instance in the middle.4. H, T, H, H: Only one instance at the end.5. T, T, H, H: Only one instance at the end.So, all five are valid. But wait, the total number of possible outcomes is 16, so the probability would be 5/16, which is not one of the answer choices. The closest is 3/16 or 1/4.Hmm, maybe I made a mistake in identifying the favorable outcomes. Let me think again.Wait, perhaps I'm misunderstanding the question. It says "exactly two consecutive heads." Does that mean exactly two heads in a row, and the other two tosses are tails? Or does it mean that in the entire sequence, there is exactly one instance of two consecutive heads, and the rest can be anything except forming another two consecutive heads?I think it's the latter. So, my initial approach was correct, but maybe I'm missing something.Alternatively, maybe the answer is 4/16 = 1/4, which is option D. But according to my count, it's 5/16, which isn't an option. So, perhaps I'm overcounting.Wait, let me list all possible sequences with exactly two consecutive heads:1. H, H, T, T2. T, H, H, T3. T, T, H, H4. H, T, H, H5. H, H, T, HWait, that's five. But maybe H, H, T, H is actually having two instances of two consecutive heads? Let's see: H, H, T, H. The first two are H, H, and then the last two are T, H. So, only one instance of two consecutive heads. So, it's valid.Similarly, H, T, H, H: H, T, H, H. Only the last two are H, H. So, one instance.T, H, H, T: Only the middle two are H, H.H, H, T, T: Only the first two are H, H.T, T, H, H: Only the last two are H, H.So, all five are valid. But since the answer choices don't include 5/16, maybe I'm misinterpreting the question.Wait, perhaps the question is asking for exactly two heads in total, with exactly two consecutive. So, not more than two heads, and exactly two of them are consecutive.In that case, the total number of heads is two, and they are consecutive.So, let's approach it that way.Total number of sequences with exactly two heads: C(4,2) = 6.Out of these, how many have the two heads consecutive?The number of ways to have two consecutive heads in four tosses is 3: positions 1-2, 2-3, 3-4.So, for each of these, the other two tosses must be tails.So, the favorable outcomes would be:1. H, H, T, T2. T, H, H, T3. T, T, H, HSo, three outcomes.Therefore, the probability is 3/16, which is option C.Wait, but earlier I thought it was five outcomes, but that was when considering sequences with any number of heads, as long as there's exactly one instance of two consecutive heads. But if the question is specifically about exactly two heads in total, with exactly two consecutive, then it's three outcomes.So, I need to clarify the question.The question says: "What is the probability of exactly two consecutive heads?"It doesn't specify the total number of heads, just that there are exactly two consecutive heads. So, that could mean that in the entire sequence, there is exactly one instance of two consecutive heads, and the rest can be anything except forming another two consecutive heads.But in that case, the number of favorable outcomes is five, as I listed earlier, leading to 5/16, which isn't an option. So, perhaps the intended interpretation is that there are exactly two heads in total, and they are consecutive.In that case, the answer would be 3/16, which is option C.Alternatively, maybe the question is asking for sequences where the maximum number of consecutive heads is two, meaning that there are no three or four consecutive heads, but at least two consecutive heads.But that interpretation would include more sequences, so the probability would be higher.Given the answer choices, and the fact that 3/16 is an option, I think the intended interpretation is that there are exactly two heads in total, and they are consecutive.Therefore, the favorable outcomes are:1. H, H, T, T2. T, H, H, T3. T, T, H, HSo, three outcomes.Hence, the probability is 3/16, which is option C.But wait, earlier I thought it was five outcomes, but that was under a different interpretation. So, I need to make sure which interpretation is correct.Looking back at the question: "What is the probability of exactly two consecutive heads?"It doesn't specify the total number of heads, just that there are exactly two consecutive heads. So, that could mean that in the entire sequence, there is exactly one instance of two consecutive heads, and the rest can be anything except forming another two consecutive heads.But in that case, as I listed earlier, there are five such sequences:1. H, H, T, H2. H, H, T, T3. T, H, H, T4. H, T, H, H5. T, T, H, HSo, five outcomes, leading to a probability of 5/16, which isn't an option. Therefore, perhaps the intended interpretation is that there are exactly two heads in total, and they are consecutive.In that case, the answer is 3/16, which is option C.Alternatively, maybe the question is asking for sequences where the number of consecutive heads is exactly two, meaning that there are no more than two consecutive heads, and at least two consecutive heads.In that case, the favorable outcomes would include sequences with exactly two consecutive heads, but not more. So, that would include sequences like H, H, T, T; T, H, H, T; T, T, H, H; H, H, T, H; H, T, H, H; and T, H, H, H (but wait, T, H, H, H has three consecutive heads, which is more than two, so it shouldn't be included). Similarly, H, H, H, T has three consecutive heads, so it's excluded.So, the sequences are:1. H, H, T, T2. T, H, H, T3. T, T, H, H4. H, H, T, H5. H, T, H, HSo, five outcomes again, leading to 5/16, which isn't an option.Given that, perhaps the intended answer is 3/16, assuming that the question is asking for exactly two heads in total, which are consecutive.Therefore, the answer is C, 3/16.But I'm still a bit confused because the wording isn't entirely clear. However, given the answer choices, and the fact that 3/16 is an option, I think that's the intended answer.