🧠:<think>Okay, so I have this problem about modeling the New Orleans Saints' win ratio over 30 seasons using a sinusoidal function. It's divided into two parts. Let me try to tackle them step by step.Starting with part 1: They give me the function ( W(n) = A sin(Bn + C) + D ). I need to find A and D. They also mention that the maximum win ratio observed was 0.875 and the minimum was 0.125. Hmm, okay. I remember that for sinusoidal functions, the amplitude A is related to the maximum and minimum values. Specifically, the amplitude is half the difference between the maximum and minimum values. So, let me write that down.The maximum value of ( W(n) ) is ( D + A ) because the sine function oscillates between -1 and 1, so when it's multiplied by A, it goes from -A to A, and then adding D shifts it up. Similarly, the minimum value is ( D - A ). So, given that the maximum is 0.875 and the minimum is 0.125, I can set up equations:( D + A = 0.875 )( D - A = 0.125 )Now, if I add these two equations together, the A terms will cancel out:( (D + A) + (D - A) = 0.875 + 0.125 )Simplifying:( 2D = 1.0 )So, ( D = 0.5 ). That makes sense because the average of the maximum and minimum should be the vertical shift D. Now, to find A, I can plug D back into one of the equations. Let's take the first one:( 0.5 + A = 0.875 )Subtracting 0.5 from both sides:( A = 0.375 )So, A is 0.375 and D is 0.5. That seems straightforward.Moving on to part 2: They mention that the win ratio has a periodic trend with a complete cycle every 5 seasons. So, the period of the sinusoidal function is 5. I know that the period of a sine function ( sin(Bn + C) ) is ( frac{2pi}{B} ). Therefore, setting this equal to 5:( frac{2pi}{B} = 5 )Solving for B:( B = frac{2pi}{5} )So, B is ( frac{2pi}{5} ). Got that.Now, they also mention that the win ratio in the first season, ( W(1) = 0.5 ). I need to find the phase shift C. Let me plug in n = 1 into the function:( W(1) = A sin(B cdot 1 + C) + D = 0.5 )We already know A is 0.375 and D is 0.5, so substituting those in:( 0.375 sinleft( frac{2pi}{5} cdot 1 + C right) + 0.5 = 0.5 )Subtracting 0.5 from both sides:( 0.375 sinleft( frac{2pi}{5} + C right) = 0 )Divide both sides by 0.375:( sinleft( frac{2pi}{5} + C right) = 0 )When does sine equal zero? At integer multiples of π. So,( frac{2pi}{5} + C = kpi ), where k is an integer.Solving for C:( C = kpi - frac{2pi}{5} )Now, since the sine function is periodic, we can choose the smallest k that gives us a C within the standard range, typically between 0 and ( 2pi ). Let's see.If k = 0: ( C = -frac{2pi}{5} ). That's negative, but we can add ( 2pi ) to get it into the standard range:( C = -frac{2pi}{5} + 2pi = frac{8pi}{5} )If k = 1: ( C = pi - frac{2pi}{5} = frac{5pi}{5} - frac{2pi}{5} = frac{3pi}{5} )If k = 2: ( C = 2pi - frac{2pi}{5} = frac{10pi}{5} - frac{2pi}{5} = frac{8pi}{5} ), which is the same as k=0.So, the possible values for C are ( frac{3pi}{5} ) and ( frac{8pi}{5} ). But since the sine function is periodic, both would result in the same graph shifted by a full period, but since we're dealing with a specific starting point, n=1, we need to see which one makes sense.Wait, let me think. If we take C = ( frac{3pi}{5} ), then the argument of the sine function at n=1 is ( frac{2pi}{5} + frac{3pi}{5} = pi ), which is 180 degrees, where sine is zero. Similarly, if C = ( frac{8pi}{5} ), then the argument is ( frac{2pi}{5} + frac{8pi}{5} = frac{10pi}{5} = 2pi ), which is also zero. So both satisfy the condition.But typically, phase shifts are given in the range [0, 2π), so both are valid. However, since the function is sinusoidal, the choice of C affects the starting point of the sine wave. If we take C = ( frac{3pi}{5} ), the sine function starts at ( frac{3pi}{5} ) when n=0, but since we're starting at n=1, it's a bit tricky.Wait, maybe it's better to think about the phase shift in terms of when the maximum or minimum occurs. But since we only know the value at n=1, which is 0.5, which is the average value D, that suggests that at n=1, the sine function is at zero. So, the sine function crosses the midline at n=1. Depending on the phase shift, it could be going up or down.But without more information, both C = ( frac{3pi}{5} ) and C = ( frac{8pi}{5} ) are possible. However, since ( frac{8pi}{5} ) is equivalent to ( -frac{2pi}{5} ), which would mean the sine wave is shifted to the right by ( frac{2pi}{5} ), but in terms of the function's argument, it's equivalent to a negative shift.But in the context of the problem, since we're starting at n=1, and the period is 5, it's probably more straightforward to take the smallest positive phase shift, which would be ( frac{3pi}{5} ). Alternatively, maybe it's better to leave it as ( -frac{2pi}{5} ) because that would mean the function is shifted to the left, but since n starts at 1, it might not matter.Wait, actually, let's consider the general solution for C. Since sine is zero at integer multiples of π, we can write:( frac{2pi}{5} + C = kpi )So, ( C = kpi - frac{2pi}{5} )To find the principal value, we can choose k such that C is between 0 and 2π. Let's try k=1:( C = pi - frac{2pi}{5} = frac{5pi - 2pi}{5} = frac{3pi}{5} )Which is approximately 1.885 radians, which is less than 2π, so that's a valid value.If we take k=0:( C = -frac{2pi}{5} ), which is negative, but we can add 2π to get it into the range:( C = -frac{2pi}{5} + 2pi = frac{8pi}{5} ), which is approximately 5.026 radians, which is also valid.So, both are valid, but since the problem doesn't specify any additional constraints, we can choose either. However, in many cases, the phase shift is expressed as a positive value less than 2π, so ( frac{3pi}{5} ) might be the preferred answer.But wait, let me double-check. If I plug C = ( frac{3pi}{5} ) into the function, then at n=1:( W(1) = 0.375 sinleft( frac{2pi}{5} + frac{3pi}{5} right) + 0.5 = 0.375 sin(pi) + 0.5 = 0 + 0.5 = 0.5 ), which is correct.Similarly, if I plug C = ( frac{8pi}{5} ):( W(1) = 0.375 sinleft( frac{2pi}{5} + frac{8pi}{5} right) + 0.5 = 0.375 sin(2pi) + 0.5 = 0 + 0.5 = 0.5 ), which is also correct.So both values of C satisfy the condition. However, since the problem doesn't specify whether the function is increasing or decreasing at n=1, we can't determine the exact phase shift without more information. But since both are valid, perhaps we need to consider the general solution.But in the context of the problem, since it's a sinusoidal function modeling win ratios over seasons, and we're starting at n=1 with a win ratio of 0.5, which is the average, it's likely that the function is either at a peak, trough, or crossing the midline. Since 0.5 is the average, it's crossing the midline. The question is, is it rising or falling?But without more data points, we can't determine the direction. Therefore, both phase shifts are possible. However, in the absence of additional information, perhaps the simplest solution is to take C = ( frac{3pi}{5} ), as it's the smaller positive angle.Alternatively, if we consider that the function could be at a peak or trough at n=1, but since 0.5 is the average, it's not a peak or trough. So, it's crossing the midline. The direction (rising or falling) isn't specified, so both possibilities exist.But since the problem asks for the value of C, and both are valid, perhaps we need to express it in terms of k. However, since the problem likely expects a specific value, and given that in trigonometric functions, phase shifts are often given in the range [0, 2π), I think ( frac{3pi}{5} ) is the answer they're looking for.So, to recap:1. A = 0.375, D = 0.52. B = ( frac{2pi}{5} ), C = ( frac{3pi}{5} )Wait, but let me just confirm the calculation for B again. The period is 5, so ( frac{2pi}{B} = 5 ), so B = ( frac{2pi}{5} ). Yes, that's correct.And for C, since ( sin(pi) = 0 ) and ( sin(2pi) = 0 ), both satisfy the condition. But since we're starting at n=1, and the function is crossing the midline, the phase shift could be such that it's either starting to rise or fall. But without knowing the next value, we can't determine the direction. So, perhaps both are acceptable, but I think the answer expects ( frac{3pi}{5} ).Alternatively, maybe I made a mistake in assuming the phase shift. Let me think again.The general form is ( sin(Bn + C) ). So, the phase shift is ( -C/B ). So, if we have ( Bn + C = B(n + C/B) ), so the phase shift is ( -C/B ). So, if we set ( -C/B = 0 ), then C=0, but that's not our case.Wait, no. The phase shift is the horizontal shift. So, if the function is ( sin(B(n - h)) ), then h is the phase shift. So, comparing to ( sin(Bn + C) ), it's ( sin(B(n + C/B)) ), so the phase shift is ( -C/B ).So, in our case, the phase shift is ( -C/B ). So, if we want the function to cross the midline at n=1, then the phase shift should be such that when n=1, the argument is a multiple of π.But perhaps this is overcomplicating. Since we've already found that C can be ( frac{3pi}{5} ) or ( frac{8pi}{5} ), and both satisfy the condition, I think either is acceptable. But since the problem is asking for the value of C, and not all possible values, perhaps we need to express it as ( frac{3pi}{5} ) because it's the principal value.Alternatively, maybe the answer expects C to be in terms of negative angle, but I think positive angles are more standard.So, I think I'll go with C = ( frac{3pi}{5} ).Wait, but let me check the calculation again. If I plug C = ( frac{3pi}{5} ), then at n=1:( Bn + C = frac{2pi}{5} + frac{3pi}{5} = pi ), so ( sin(pi) = 0 ), which gives W(1) = 0.5, correct.If I take C = ( frac{8pi}{5} ), then:( Bn + C = frac{2pi}{5} + frac{8pi}{5} = 2pi ), so ( sin(2pi) = 0 ), which also gives W(1) = 0.5.So, both are correct. But since the problem is about a cycle every 5 seasons, starting at n=1, and we're given that W(1)=0.5, which is the average, it's likely that the function is either at a peak, trough, or crossing the midline. Since it's crossing the midline, the phase shift could be such that it's either starting to rise or fall. But without more data, we can't determine the direction. Therefore, both C = ( frac{3pi}{5} ) and C = ( frac{8pi}{5} ) are valid.But since the problem is asking for the value of C, and not all possible values, perhaps we need to express it as ( frac{3pi}{5} ) because it's the principal value. Alternatively, if we consider the phase shift as a negative angle, it would be ( -frac{2pi}{5} ), but that's equivalent to ( frac{8pi}{5} ).Hmm, this is a bit confusing. Maybe I should express both possibilities, but I think the answer expects a single value. Given that, I'll go with ( frac{3pi}{5} ) as the phase shift.So, to summarize:1. A = 0.375, D = 0.52. B = ( frac{2pi}{5} ), C = ( frac{3pi}{5} )I think that's it.</think>

🧠:<think>Alright, so I'm trying to solve these two problems about a graphic novelist who uses fractal geometry in their work. Let me take it step by step.Starting with the first problem: Calculate the total area covered by all 100 pages of the novel. Each page is a square, and the side length decreases by a factor of 1/2 each time. The first page is a square with side length 1 unit, so its area is 1x1=1 square unit. The next page has a side length of 1/2, so its area is (1/2)^2 = 1/4. The third page would then be (1/4)^2 = 1/16, and so on.Hmm, this seems like a geometric series where each term is (1/4) times the previous term. The first term, a, is 1, and the common ratio, r, is 1/4. Since there are 100 pages, we need the sum of the first 100 terms of this series.I remember the formula for the sum of an infinite geometric series is S = a / (1 - r), but this is for an infinite number of terms. However, since 100 is a large number, maybe the sum is close to the infinite sum? But the problem says I can use the formula for the sum of an infinite geometric series, so maybe it's okay to approximate or treat it as infinite.Wait, actually, the formula for the sum of the first n terms is S_n = a(1 - r^n)/(1 - r). So, if I use that, I can plug in n=100, a=1, r=1/4.Calculating that: S_100 = 1*(1 - (1/4)^100)/(1 - 1/4). Simplifying the denominator: 1 - 1/4 = 3/4. So, S_100 = (1 - (1/4)^100)/(3/4) = (4/3)*(1 - (1/4)^100).Now, (1/4)^100 is an extremely small number, practically negligible. So, for all intents and purposes, the total area is approximately 4/3. But since the problem says to use the formula for the infinite series, maybe I can just write it as 4/3 without worrying about the (1/4)^100 term. Although, technically, it's a finite sum, but with 100 terms, it's so close to 4/3 that the difference is insignificant.Moving on to the second problem: The graphic novelist subdivides each page into smaller squares. The first page is divided into 4 squares, the second into 16, the third into 64, and so on. I need to find the total number of smaller squares across all 100 pages.Looking at the pattern: 4, 16, 64,... This is a geometric sequence where each term is multiplied by 4 each time. So, the first term a is 4, and the common ratio r is 4.Again, we can use the sum formula for a geometric series. Since we have 100 terms, we'll use the finite sum formula: S_n = a*(r^n - 1)/(r - 1). Plugging in the values: S_100 = 4*(4^100 - 1)/(4 - 1) = 4*(4^100 - 1)/3.Simplifying that: (4/3)*(4^100 - 1). That's a massive number, but I think that's the correct expression. Alternatively, since 4^100 is 4 raised to the 100th power, which is astronomically large, but mathematically, that's the sum.Wait, let me double-check the pattern. The first page is 4 squares, which is 4^1. The second is 16, which is 4^2. The third is 64, which is 4^3. So yes, each term is 4^n where n is the page number. So the total is the sum from n=1 to n=100 of 4^n.Which is indeed a geometric series with a=4, r=4, so the sum is (4*(4^100 - 1))/(4 - 1) = (4/3)*(4^100 - 1). That seems right.So, summarizing:1. Total area is approximately 4/3, considering the infinite sum.2. Total number of squares is (4/3)*(4^100 - 1).I think that's it. Although the numbers are huge, especially the second one, but that's just how exponential growth works.Final Answer1. The total area covered by all 100 pages is boxed{dfrac{4}{3}} square units.2. The total number of smaller squares used across all 100 pages is boxed{dfrac{4}{3} left(4^{100} - 1right)}.</think>

🧠:<think>Okay, so Sarah is analyzing the viewership of two Australian TV shows, "Neighbours" and "Home and Away," over 10 weeks. I need to figure out the total viewership for each show and then compare them to see which one is more popular over this period.Starting with the first part, the viewership for "Neighbours" is given by the function N(i) = 5000 + 200i² for each week i, where i ranges from 1 to n. Since n is 10, we need to calculate the total viewership over 10 weeks. To find the total viewership, I think I need to sum N(i) from i=1 to i=10. That means I have to compute the sum of 5000 + 200i² for each week and then add them all together. Let me write that down:Total viewership for "Neighbours" = Σ (from i=1 to 10) [5000 + 200i²]I can split this sum into two separate sums because addition is linear. So, it becomes:Σ 5000 + Σ 200i²Calculating the first sum, Σ 5000 from i=1 to 10. Since 5000 is a constant, this is just 5000 multiplied by the number of terms, which is 10. So, 5000 * 10 = 50,000.Now, the second sum is Σ 200i² from i=1 to 10. I can factor out the 200, so it becomes 200 * Σ i² from i=1 to 10.I remember that the sum of squares formula is Σ i² = n(n + 1)(2n + 1)/6. Plugging n=10 into this formula:Σ i² = 10*11*21/6Let me compute that step by step. First, 10*11 is 110. Then, 110*21 is 2310. Dividing that by 6: 2310 / 6 = 385.So, Σ i² from 1 to 10 is 385. Therefore, the second sum is 200 * 385 = 77,000.Adding both sums together: 50,000 + 77,000 = 127,000.So, the total viewership for "Neighbours" over 10 weeks is 127,000.Now, moving on to "Home and Away." The viewership for this show is given by H(i) = 4000(1.05)^i for each week i. Again, we need the total viewership over 10 weeks, so we have to compute the sum from i=1 to 10 of 4000(1.05)^i.This looks like a geometric series because each term is a constant multiple of the previous term. The general form of a geometric series is Σ ar^(i-1) from i=1 to n, but in this case, our exponent is i, not i-1. So, let me adjust for that.Let me write the sum as:Σ (from i=1 to 10) 4000(1.05)^iThis can be rewritten as 4000 * Σ (1.05)^i from i=1 to 10.I know the formula for the sum of a geometric series starting from i=0 is S = a*(r^(n+1) - 1)/(r - 1). But since our series starts at i=1, we can adjust it by subtracting the term at i=0.So, Σ (1.05)^i from i=1 to 10 = [Σ (1.05)^i from i=0 to 10] - (1.05)^0Which is [ (1.05)^(11) - 1 ] / (1.05 - 1) - 1Calculating that step by step:First, compute (1.05)^11. Let me calculate that. 1.05 to the power of 11. I can use logarithms or just compute step by step.But maybe it's easier to use the formula:Sum = a*(r^n - 1)/(r - 1) when starting from i=0.But since we're starting from i=1, it's Sum = a*(r^n - 1)/(r - 1) - a, where a is the first term. Wait, no. Let me think.Wait, actually, if we have the sum from i=1 to n of r^i, it's equal to r*(r^n - 1)/(r - 1). Let me verify that.Yes, because:Sum from i=1 to n of r^i = r + r² + ... + r^n = r*(1 + r + r² + ... + r^(n-1)) = r*( (r^n - 1)/(r - 1) )So, that's correct.Therefore, in our case, r = 1.05, n = 10.So, Sum = 1.05*( (1.05)^10 - 1 ) / (1.05 - 1 )Compute that:First, compute (1.05)^10. Let me calculate that. 1.05^10 is approximately... I remember that 1.05^10 is roughly 1.62889.So, (1.05)^10 ≈ 1.62889Then, (1.05)^10 - 1 ≈ 0.62889Multiply by 1.05: 1.05 * 0.62889 ≈ 0.6603345Divide by (1.05 - 1) which is 0.05:0.6603345 / 0.05 ≈ 13.20669So, the sum Σ (1.05)^i from i=1 to 10 ≈ 13.20669Therefore, the total viewership for "Home and Away" is 4000 multiplied by this sum:4000 * 13.20669 ≈ 4000 * 13.20669Calculating that:4000 * 13 = 52,0004000 * 0.20669 ≈ 4000 * 0.2 = 800, and 4000 * 0.00669 ≈ 26.76So, total ≈ 800 + 26.76 = 826.76Adding to 52,000: 52,000 + 826.76 ≈ 52,826.76So, approximately 52,826.76 viewers.Wait, that seems low compared to "Neighbours" which had 127,000. But let me double-check my calculations because 52,826 seems low for a 10-week period with exponential growth.Wait, perhaps I made a mistake in calculating the sum.Let me recalculate the sum Σ (1.05)^i from i=1 to 10.Using the formula:Sum = r*(r^n - 1)/(r - 1)r = 1.05, n = 10.So, Sum = 1.05*(1.05^10 - 1)/(0.05)We have 1.05^10 ≈ 1.62889So, 1.05*(1.62889 - 1) = 1.05*(0.62889) ≈ 0.6603345Divide by 0.05: 0.6603345 / 0.05 = 13.20669So, that part is correct.Therefore, 4000 * 13.20669 ≈ 52,826.76Wait, but 4000 per week, even with growth, over 10 weeks, shouldn't it be higher? Let me check the formula again.Wait, maybe I misapplied the formula. Let me think again.The formula for the sum of a geometric series starting at i=1 is:Sum = a*(r^n - 1)/(r - 1) where a is the first term.But in our case, the first term when i=1 is 4000*(1.05)^1 = 4200.Wait, no, actually, the formula is for the sum of r^i, not multiplied by a constant. So, in our case, the sum is 4000*(1.05 + (1.05)^2 + ... + (1.05)^10)Which is 4000*(Sum from i=1 to 10 of (1.05)^i )Which is 4000*( [ (1.05)*( (1.05)^10 - 1 ) ] / (1.05 - 1 ) )Which is what I did earlier, so that's correct.So, 4000 * 13.20669 ≈ 52,826.76Wait, but 13.20669 is the sum of (1.05)^i from i=1 to 10, so multiplying by 4000 gives the total viewership.But let me compute it more accurately.First, compute (1.05)^10 precisely.Using a calculator:1.05^1 = 1.051.05^2 = 1.10251.05^3 ≈ 1.1576251.05^4 ≈ 1.215506251.05^5 ≈ 1.27628156251.05^6 ≈ 1.34009564061.05^7 ≈ 1.40710042261.05^8 ≈ 1.47745544371.05^9 ≈ 1.55132821591.05^10 ≈ 1.6288946267So, (1.05)^10 ≈ 1.6288946267Then, (1.05)^10 - 1 ≈ 0.6288946267Multiply by 1.05: 1.05 * 0.6288946267 ≈ 0.6603393579Divide by 0.05: 0.6603393579 / 0.05 ≈ 13.206787158So, the sum is approximately 13.206787158Multiply by 4000: 4000 * 13.206787158 ≈ 52,827.14863So, approximately 52,827.15 viewers.Wait, that still seems low. Let me check if I'm using the correct formula.Alternatively, maybe I should use the formula for the sum starting at i=0 and then subtract the i=0 term.Sum from i=1 to 10 of (1.05)^i = Sum from i=0 to 10 of (1.05)^i - (1.05)^0Sum from i=0 to 10 of (1.05)^i = (1.05)^11 - 1 / (1.05 - 1) ≈ (1.05)^11 - 1 / 0.05Compute (1.05)^11:1.05^11 = 1.05 * 1.6288946267 ≈ 1.7103393579So, (1.7103393579 - 1)/0.05 ≈ 0.7103393579 / 0.05 ≈ 14.206787158Subtract 1 (the i=0 term): 14.206787158 - 1 = 13.206787158Same result as before. So, the sum is indeed approximately 13.206787158Therefore, 4000 * 13.206787158 ≈ 52,827.15So, approximately 52,827 viewers for "Home and Away" over 10 weeks.Wait, but "Neighbours" had 127,000, which is more than double. That seems counterintuitive because "Home and Away" is growing exponentially, but maybe over 10 weeks, the quadratic growth of "Neighbours" is still higher.Wait, let me check the calculations again because 52,827 seems low for a show with exponential growth over 10 weeks.Wait, maybe I made a mistake in the formula. Let me think again.The viewership for "Home and Away" is H(i) = 4000*(1.05)^iSo, the total viewership is Σ H(i) from i=1 to 10 = 4000*(1.05 + 1.05² + ... + 1.05^10)Which is 4000*(Sum from i=1 to 10 of 1.05^i )Which is 4000*( [1.05*(1.05^10 - 1)] / (1.05 - 1) )Which is 4000*( [1.05*(1.6288946267 - 1)] / 0.05 )= 4000*( [1.05*0.6288946267] / 0.05 )= 4000*(0.6603393579 / 0.05 )= 4000*(13.206787158 )= 52,827.14863Yes, that's correct. So, approximately 52,827 viewers.Wait, but let me compute the actual sum manually to verify.Compute each week's viewership and sum them up:Week 1: 4000*1.05 = 4200Week 2: 4000*1.1025 = 4410Week 3: 4000*1.157625 ≈ 4630.5Week 4: 4000*1.21550625 ≈ 4862.025Week 5: 4000*1.2762815625 ≈ 5105.12625Week 6: 4000*1.3400956406 ≈ 5360.38256Week 7: 4000*1.4071004226 ≈ 5628.40169Week 8: 4000*1.4774554437 ≈ 5909.821775Week 9: 4000*1.5513282159 ≈ 6205.312864Week 10: 4000*1.6288946267 ≈ 6515.578507Now, let's sum these up:4200 + 4410 = 86108610 + 4630.5 = 13240.513240.5 + 4862.025 = 18102.52518102.525 + 5105.12625 ≈ 23207.6512523207.65125 + 5360.38256 ≈ 28568.0338128568.03381 + 5628.40169 ≈ 34196.435534196.4355 + 5909.821775 ≈ 40106.25727540106.257275 + 6205.312864 ≈ 46311.5701446311.57014 + 6515.578507 ≈ 52827.14865So, the manual sum gives approximately 52,827.15, which matches the earlier calculation. So, that's correct.Therefore, "Neighbours" has a total viewership of 127,000 over 10 weeks, while "Home and Away" has approximately 52,827.15 viewers over the same period.Comparing the two, 127,000 is more than double of 52,827, so "Neighbours" has a higher total viewership over the 10-week period.Wait, but just to make sure, let me re-express the total viewership for "Neighbours" again.Total viewership for "Neighbours" is Σ (5000 + 200i²) from i=1 to 10.Which is 10*5000 + 200*Σ i² from 1 to 10.Σ i² from 1 to 10 is 385, so 200*385=77,000.10*5000=50,000.Total=50,000 +77,000=127,000. Correct.So, yes, "Neighbours" has higher viewership over 10 weeks.</think>

🧠:<think>Okay, so I have this problem where Professor Alex is using a Markov Chain to model voter behavior between three political parties: A, B, and C. The transition matrix P is given, and I need to solve two sub-problems. Let me tackle them one by one.Sub-problem 1: Determine the steady-state distribution of voter preferences.Alright, the steady-state distribution is a vector π = (π_A, π_B, π_C) such that πP = π and the sum of π_A + π_B + π_C = 1. So, I need to find the stationary distribution of this Markov chain.First, I remember that for a steady-state distribution, the vector π must satisfy the equation πP = π. That means each component of π is equal to the sum of the products of π and the corresponding column of P.So, writing out the equations:1. π_A = π_A * 0.6 + π_B * 0.3 + π_C * 0.12. π_B = π_A * 0.2 + π_B * 0.5 + π_C * 0.33. π_C = π_A * 0.2 + π_B * 0.2 + π_C * 0.6And also, π_A + π_B + π_C = 1.Hmm, so that's three equations, but since the sum is 1, maybe I can reduce it to two equations.Let me rewrite the equations:From equation 1:π_A = 0.6π_A + 0.3π_B + 0.1π_CSubtract 0.6π_A from both sides:0.4π_A = 0.3π_B + 0.1π_CLet me write this as:0.4π_A - 0.3π_B - 0.1π_C = 0 ...(1)From equation 2:π_B = 0.2π_A + 0.5π_B + 0.3π_CSubtract 0.5π_B from both sides:0.5π_B = 0.2π_A + 0.3π_CSo:-0.2π_A + 0.5π_B - 0.3π_C = 0 ...(2)Equation 3:π_C = 0.2π_A + 0.2π_B + 0.6π_CSubtract 0.6π_C:0.4π_C = 0.2π_A + 0.2π_BDivide both sides by 0.2:2π_C = π_A + π_BSo:π_A + π_B - 2π_C = 0 ...(3)Now, I have three equations:(1): 0.4π_A - 0.3π_B - 0.1π_C = 0(2): -0.2π_A + 0.5π_B - 0.3π_C = 0(3): π_A + π_B - 2π_C = 0And the constraint:(4): π_A + π_B + π_C = 1Hmm, maybe I can express π_C from equation (3) in terms of π_A and π_B.From equation (3):π_C = (π_A + π_B)/2So, let's substitute π_C into equations (1) and (2).Substituting into equation (1):0.4π_A - 0.3π_B - 0.1*(π_A + π_B)/2 = 0Let me compute 0.1*(π_A + π_B)/2 = 0.05π_A + 0.05π_BSo, equation (1) becomes:0.4π_A - 0.3π_B - 0.05π_A - 0.05π_B = 0Combine like terms:(0.4 - 0.05)π_A + (-0.3 - 0.05)π_B = 00.35π_A - 0.35π_B = 0Divide both sides by 0.35:π_A - π_B = 0 => π_A = π_BInteresting, so π_A equals π_B.Now, substitute π_A = π_B into equation (2):-0.2π_A + 0.5π_B - 0.3π_C = 0But since π_A = π_B, let's write:-0.2π_A + 0.5π_A - 0.3π_C = 0Combine terms:( -0.2 + 0.5 )π_A - 0.3π_C = 00.3π_A - 0.3π_C = 0Divide both sides by 0.3:π_A - π_C = 0 => π_A = π_CWait, so π_A = π_B and π_A = π_C? That would mean π_A = π_B = π_C.But let's check if that's consistent with equation (3):From equation (3): π_A + π_B - 2π_C = 0If π_A = π_B = π_C, then:π_A + π_A - 2π_A = 0 => 0 = 0, which is true.So, all three are equal. Therefore, π_A = π_B = π_C.But wait, let's check if that's consistent with equation (1):From equation (1): 0.4π_A - 0.3π_B - 0.1π_C = 0If π_A = π_B = π_C, then:0.4π_A - 0.3π_A - 0.1π_A = 0(0.4 - 0.3 - 0.1)π_A = 0 => 0 = 0, which is true.Similarly, equation (2):-0.2π_A + 0.5π_B - 0.3π_C = 0Again, substituting π_A = π_B = π_C:-0.2π_A + 0.5π_A - 0.3π_A = 0(-0.2 + 0.5 - 0.3)π_A = 0 => 0 = 0, which is true.So, all equations are satisfied if π_A = π_B = π_C.But wait, if all are equal, then each is 1/3, since π_A + π_B + π_C = 1.So, π_A = π_B = π_C = 1/3.But hold on, let me verify this because sometimes when solving linear systems, especially with Markov chains, the solution might not be as straightforward.Alternatively, maybe I made a mistake in the substitution.Wait, let's see:From equation (3): π_A + π_B = 2π_CIf π_A = π_B, then 2π_A = 2π_C => π_A = π_C.So, yes, π_A = π_B = π_C.Therefore, each is 1/3.But let me cross-verify this with the transition matrix.In steady state, the distribution should satisfy πP = π.Let me compute πP where π = [1/3, 1/3, 1/3].Compute each component:First component: (1/3)(0.6) + (1/3)(0.3) + (1/3)(0.1) = (0.6 + 0.3 + 0.1)/3 = 1.0/3 = 1/3Second component: (1/3)(0.2) + (1/3)(0.5) + (1/3)(0.3) = (0.2 + 0.5 + 0.3)/3 = 1.0/3 = 1/3Third component: (1/3)(0.2) + (1/3)(0.2) + (1/3)(0.6) = (0.2 + 0.2 + 0.6)/3 = 1.0/3 = 1/3So, yes, it does satisfy πP = π.Therefore, the steady-state distribution is π = (1/3, 1/3, 1/3).Wait, but that seems a bit counterintuitive because the transition matrix doesn't have equal rows. Let me check if the chain is regular.Looking at the transition matrix P:- From Party A, 60% stay, 20% go to B, 20% go to C.- From Party B, 30% go to A, 50% stay, 20% go to C.- From Party C, 10% go to A, 30% go to B, 60% stay.Is this chain irreducible? Yes, because you can get from any state to any other state in one step. For example, A can go to B or C, B can go to A or C, C can go to A or B. So, it's irreducible.Is it aperiodic? The period of a state is the greatest common divisor of the lengths of all possible loops that start and end at that state. For each state, since there is a self-loop (e.g., A can stay at A with probability 0.6), the period is 1. So, it's aperiodic.Therefore, the Markov chain is ergodic, meaning it has a unique stationary distribution, which we found as (1/3, 1/3, 1/3). So, that must be correct.Sub-problem 2: Calculate the expected distribution after two election cycles.Given the initial distribution π_initial = (0.4, 0.3, 0.3), we need to compute π_initial * P^2.Alternatively, since matrix multiplication is associative, we can compute it step by step.First, compute π_initial * P to get the distribution after one cycle, then multiply that result by P again to get after two cycles.Let me compute π1 = π_initial * P.Compute each component:π1_A = 0.4*0.6 + 0.3*0.3 + 0.3*0.1Compute:0.4*0.6 = 0.240.3*0.3 = 0.090.3*0.1 = 0.03Sum: 0.24 + 0.09 + 0.03 = 0.36Similarly, π1_B = 0.4*0.2 + 0.3*0.5 + 0.3*0.3Compute:0.4*0.2 = 0.080.3*0.5 = 0.150.3*0.3 = 0.09Sum: 0.08 + 0.15 + 0.09 = 0.32π1_C = 0.4*0.2 + 0.3*0.2 + 0.3*0.6Compute:0.4*0.2 = 0.080.3*0.2 = 0.060.3*0.6 = 0.18Sum: 0.08 + 0.06 + 0.18 = 0.32So, after one cycle, π1 = (0.36, 0.32, 0.32)Now, compute π2 = π1 * P.Compute each component:π2_A = 0.36*0.6 + 0.32*0.3 + 0.32*0.1Compute:0.36*0.6 = 0.2160.32*0.3 = 0.0960.32*0.1 = 0.032Sum: 0.216 + 0.096 + 0.032 = 0.344π2_B = 0.36*0.2 + 0.32*0.5 + 0.32*0.3Compute:0.36*0.2 = 0.0720.32*0.5 = 0.160.32*0.3 = 0.096Sum: 0.072 + 0.16 + 0.096 = 0.328π2_C = 0.36*0.2 + 0.32*0.2 + 0.32*0.6Compute:0.36*0.2 = 0.0720.32*0.2 = 0.0640.32*0.6 = 0.192Sum: 0.072 + 0.064 + 0.192 = 0.328So, after two cycles, π2 = (0.344, 0.328, 0.328)Wait, let me verify these calculations step by step to make sure I didn't make any arithmetic errors.First, π1:π1_A: 0.4*0.6 = 0.24, 0.3*0.3 = 0.09, 0.3*0.1 = 0.03. Total 0.36. Correct.π1_B: 0.4*0.2 = 0.08, 0.3*0.5 = 0.15, 0.3*0.3 = 0.09. Total 0.32. Correct.π1_C: 0.4*0.2 = 0.08, 0.3*0.2 = 0.06, 0.3*0.6 = 0.18. Total 0.32. Correct.Then π2:π2_A: 0.36*0.6 = 0.216, 0.32*0.3 = 0.096, 0.32*0.1 = 0.032. Sum: 0.216 + 0.096 = 0.312 + 0.032 = 0.344. Correct.π2_B: 0.36*0.2 = 0.072, 0.32*0.5 = 0.16, 0.32*0.3 = 0.096. Sum: 0.072 + 0.16 = 0.232 + 0.096 = 0.328. Correct.π2_C: 0.36*0.2 = 0.072, 0.32*0.2 = 0.064, 0.32*0.6 = 0.192. Sum: 0.072 + 0.064 = 0.136 + 0.192 = 0.328. Correct.So, yes, the calculations seem correct.Alternatively, I could have computed P squared and then multiplied by π_initial, but since the numbers are manageable, step-by-step multiplication is fine.So, the expected distribution after two election cycles is approximately (0.344, 0.328, 0.328).Alternatively, to express these as fractions:0.344 is 344/1000 = 86/250 = 43/125 ≈ 0.3440.328 is 328/1000 = 82/250 = 41/125 ≈ 0.328But since the question doesn't specify, decimal form is probably fine.Wait, let me check 0.344 + 0.328 + 0.328 = 1.0, which is correct.So, that seems consistent.Summary of Thoughts:For Sub-problem 1, I set up the equations for the steady-state distribution, realized that all components are equal due to the symmetry in the equations, and confirmed that each is 1/3. For Sub-problem 2, I calculated the distribution after one cycle and then used that result to compute the distribution after two cycles, ensuring each step was accurate.Final AnswerSub-problem 1: The steady-state distribution is (boxed{left( frac{1}{3}, frac{1}{3}, frac{1}{3} right)}).Sub-problem 2: The expected distribution after two election cycles is (boxed{(0.344, 0.328, 0.328)}).</think>