🧠:<think>First, I need to determine how many people the vampire drains each week. I'll start by calculating the total number of people the werewolf eats over the 9 weeks.Since the werewolf eats 5 people each week, over 9 weeks it eats:5 people/week * 9 weeks = 45 peopleNext, I'll find out how many people are left for the vampire after the werewolf has eaten its share:Total village population - Werewolf's consumption = 72 people - 45 people = 27 peopleNow, I'll calculate how many people the vampire drains each week by dividing the remaining people by the number of weeks:27 people / 9 weeks = 3 people/weekTherefore, the vampire drains 3 people each week.</think>

🧠:<think>Alright, so I have this problem here: given positive numbers ( a ) and ( b ) satisfying ( 2a + b = 1 ), and the inequality ( 2sqrt{ab} - 4a^2 - b^2 leq t - frac{1}{2} ) always holds. I need to find the range of the real number ( t ). The options are A, B, C, D, which are various intervals involving ( sqrt{2}/2 ) and ( 1/2 ).First, let me try to understand the problem. We have two variables, ( a ) and ( b ), which are positive and satisfy the equation ( 2a + b = 1 ). So, ( a ) and ( b ) are related by this linear equation. The inequality given involves ( a ) and ( b ) in a more complex expression: ( 2sqrt{ab} - 4a^2 - b^2 leq t - frac{1}{2} ). This inequality must hold for all positive ( a ) and ( b ) satisfying ( 2a + b = 1 ). So, essentially, I need to find the minimum value of ( t ) such that this inequality is always true.Let me rewrite the inequality to make it clearer:( 2sqrt{ab} - 4a^2 - b^2 leq t - frac{1}{2} )If I rearrange this, I can write it as:( t geq 2sqrt{ab} - 4a^2 - b^2 + frac{1}{2} )So, ( t ) must be greater than or equal to this expression involving ( a ) and ( b ). Therefore, to find the range of ( t ), I need to find the maximum value of the right-hand side expression because ( t ) has to be at least as big as the maximum of that expression.Let me denote the expression on the right-hand side as a function ( f(a, b) ):( f(a, b) = 2sqrt{ab} - 4a^2 - b^2 + frac{1}{2} )So, my task reduces to finding the maximum value of ( f(a, b) ) given that ( 2a + b = 1 ) and ( a, b > 0 ). Once I find this maximum, that will be the minimum value of ( t ), and since ( t ) can be any real number greater than or equal to this maximum, the range of ( t ) will be from this maximum to infinity.Now, since ( 2a + b = 1 ), I can express ( b ) in terms of ( a ):( b = 1 - 2a )Since ( a ) and ( b ) are positive, ( 1 - 2a > 0 ) which implies ( a < frac{1}{2} ). So, ( a ) is in the interval ( (0, frac{1}{2}) ).Now, substituting ( b = 1 - 2a ) into ( f(a, b) ):( f(a) = 2sqrt{a(1 - 2a)} - 4a^2 - (1 - 2a)^2 + frac{1}{2} )Let me compute each term step by step.First, compute ( sqrt{ab} ):( sqrt{a(1 - 2a)} )Then, ( 2sqrt{ab} ) is:( 2sqrt{a(1 - 2a)} )Next, compute ( 4a^2 ):( 4a^2 )Then, compute ( b^2 ):( (1 - 2a)^2 = 1 - 4a + 4a^2 )So, ( -4a^2 - b^2 ) becomes:( -4a^2 - (1 - 4a + 4a^2) = -4a^2 -1 + 4a -4a^2 = -8a^2 + 4a -1 )Putting it all together, ( f(a) ) is:( 2sqrt{a(1 - 2a)} -8a^2 + 4a -1 + frac{1}{2} )Simplify the constants:( -1 + frac{1}{2} = -frac{1}{2} )So, ( f(a) = 2sqrt{a(1 - 2a)} -8a^2 + 4a - frac{1}{2} )This looks a bit complicated, but maybe I can simplify it further or find its maximum by taking the derivative.Alternatively, perhaps there's a smarter substitution or inequality I can use to find the maximum without calculus.Let me think about the expression ( 2sqrt{ab} -4a^2 -b^2 ). Maybe I can express this in terms of ( ab ) and use some inequality.Given that ( 2a + b = 1 ), I can express ( ab ) in terms of ( a ) or ( b ). Let me express ( ab ) as:( ab = a(1 - 2a) = a - 2a^2 )So, ( ab = a - 2a^2 ). Let me denote ( x = a ), so ( ab = x - 2x^2 ). Then, ( sqrt{ab} = sqrt{x - 2x^2} ).So, ( 2sqrt{ab} = 2sqrt{x - 2x^2} ).Also, ( 4a^2 + b^2 = 4x^2 + (1 - 2x)^2 = 4x^2 + 1 - 4x + 4x^2 = 8x^2 -4x +1 ).Wait, but in the original expression, it's ( -4a^2 -b^2 ). So, that would be ( - (4a^2 + b^2) = - (8x^2 -4x +1) = -8x^2 +4x -1 ).So, putting it all together, the expression ( 2sqrt{ab} -4a^2 -b^2 ) is:( 2sqrt{x - 2x^2} -8x^2 +4x -1 )Which is the same as ( f(a) ) above.So, perhaps I can consider ( f(x) = 2sqrt{x - 2x^2} -8x^2 +4x -1 ), and find its maximum for ( x in (0, 1/2) ).Alternatively, maybe I can use substitution to make it easier. Let me set ( y = sqrt{ab} ). Since ( ab = x - 2x^2 ), ( y = sqrt{x - 2x^2} ).But I'm not sure if that helps directly.Alternatively, perhaps I can use the Cauchy-Schwarz inequality or AM-GM inequality to bound the expression.Wait, let's see. The expression ( 2sqrt{ab} -4a^2 -b^2 ) can be rewritten as ( 2sqrt{ab} - (4a^2 + b^2) ).Given that ( 2a + b =1 ), maybe I can relate ( 4a^2 + b^2 ) to something.Let me compute ( (2a + b)^2 = 1^2 =1 ).Expanding ( (2a + b)^2 = 4a^2 +4ab + b^2 =1 ).So, ( 4a^2 + b^2 =1 -4ab ).Therefore, ( 4a^2 + b^2 =1 -4ab ).So, substituting back into the expression:( 2sqrt{ab} - (4a^2 + b^2) = 2sqrt{ab} - (1 -4ab) = 2sqrt{ab} -1 +4ab ).So, the expression simplifies to:( 2sqrt{ab} +4ab -1 ).So, the original inequality becomes:( 2sqrt{ab} +4ab -1 leq t - frac{1}{2} )Which simplifies to:( 2sqrt{ab} +4ab - frac{1}{2} leq t )So, ( t ) must be greater than or equal to ( 2sqrt{ab} +4ab - frac{1}{2} ).Therefore, to find the range of ( t ), I need to find the maximum value of ( 2sqrt{ab} +4ab - frac{1}{2} ) given that ( 2a + b =1 ) and ( a, b >0 ).Let me denote ( f(ab) = 2sqrt{ab} +4ab - frac{1}{2} ). So, I need to find the maximum of ( f(ab) ) where ( ab ) is subject to ( 2a + b =1 ).But ( ab ) is not independent; it's related to ( a ) and ( b ). So, perhaps I can express ( ab ) in terms of a single variable.From ( 2a + b =1 ), we have ( b =1 -2a ). So, ( ab = a(1 -2a) = a -2a^2 ).Let me denote ( x = a ), so ( ab = x -2x^2 ). Therefore, ( f(ab) = 2sqrt{x -2x^2} +4(x -2x^2) - frac{1}{2} ).Simplify this:( f(x) = 2sqrt{x -2x^2} +4x -8x^2 - frac{1}{2} )This is similar to what I had earlier. Now, to find the maximum of ( f(x) ) for ( x in (0, 1/2) ).Alternatively, perhaps I can consider ( ab ) as a variable, say ( y = ab ). Then, since ( 2a + b =1 ), we can find the maximum possible value of ( y ).Using AM-GM inequality: For positive numbers, ( 2a + b geq 2sqrt{2ab} ). Wait, let me check.Wait, AM-GM for two variables: ( frac{2a + b}{2} geq sqrt{2ab} ). So, ( frac{1}{2} geq sqrt{2ab} ), so ( sqrt{2ab} leq frac{1}{2} ), so ( 2ab leq frac{1}{4} ), so ( ab leq frac{1}{8} ).Therefore, the maximum value of ( ab ) is ( frac{1}{8} ), achieved when ( 2a = b ), which, given ( 2a + b =1 ), implies ( 2a = b ), so ( 2a +2a =1 ), so ( 4a =1 ), ( a = frac{1}{4} ), ( b = frac{1}{2} ).So, ( ab leq frac{1}{8} ). Therefore, ( y in (0, frac{1}{8}] ).So, now, ( f(y) = 2sqrt{y} +4y - frac{1}{2} ). I need to find the maximum of this function for ( y in (0, frac{1}{8}] ).Let me compute ( f(y) ):( f(y) = 2sqrt{y} +4y - frac{1}{2} )To find its maximum, I can take the derivative with respect to ( y ):( f'(y) = frac{2}{2sqrt{y}} +4 = frac{1}{sqrt{y}} +4 )Set derivative equal to zero to find critical points:( frac{1}{sqrt{y}} +4 =0 )But ( frac{1}{sqrt{y}} ) is always positive for ( y >0 ), so ( frac{1}{sqrt{y}} +4 ) is always positive. Therefore, the function ( f(y) ) is increasing on ( y in (0, frac{1}{8}] ).Therefore, the maximum of ( f(y) ) occurs at ( y = frac{1}{8} ).Compute ( f(frac{1}{8}) ):( f(frac{1}{8}) = 2sqrt{frac{1}{8}} +4 cdot frac{1}{8} - frac{1}{2} )Simplify:( sqrt{frac{1}{8}} = frac{1}{2sqrt{2}} ), so ( 2 cdot frac{1}{2sqrt{2}} = frac{1}{sqrt{2}} ).( 4 cdot frac{1}{8} = frac{1}{2} ).So,( f(frac{1}{8}) = frac{1}{sqrt{2}} + frac{1}{2} - frac{1}{2} = frac{1}{sqrt{2}} ).Simplify ( frac{1}{sqrt{2}} ) as ( frac{sqrt{2}}{2} ).Therefore, the maximum value of ( f(y) ) is ( frac{sqrt{2}}{2} ).Therefore, ( t ) must be greater than or equal to ( frac{sqrt{2}}{2} ) to satisfy the inequality for all positive ( a ) and ( b ) with ( 2a + b =1 ).Thus, the range of ( t ) is ( [frac{sqrt{2}}{2}, +infty) ), which corresponds to option B.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I started by expressing ( b ) in terms of ( a ): ( b =1 -2a ).2. Then, I substituted into the expression to get ( f(a) ).3. I realized that ( 4a^2 + b^2 =1 -4ab ), which simplified the expression.4. This led me to express the inequality in terms of ( ab ).5. Using AM-GM, I found that ( ab leq frac{1}{8} ).6. Then, I considered ( f(y) = 2sqrt{y} +4y - frac{1}{2} ) and found its maximum by taking the derivative.7. Since the derivative was always positive, the function is increasing, so the maximum occurs at ( y = frac{1}{8} ).8. Calculating ( f(frac{1}{8}) ) gave me ( frac{sqrt{2}}{2} ).Everything seems to check out. Therefore, I'm confident that the correct answer is option B: ( [frac{sqrt{2}}{2}, +infty) ).</think>

🧠:<think>Okay, so Mary used 15 gallons of fuel this week. Last week, she used 20% less. I need to find out how much fuel she used in total for both weeks.First, I should figure out how much fuel she used last week. Since it's 20% less than this week, I'll calculate 20% of 15 gallons. To find 20% of 15, I can multiply 15 by 0.20. Let me do that: 15 * 0.20 equals 3 gallons. So, last week she used 3 gallons less than this week.Now, subtracting that from this week's usage: 15 gallons minus 3 gallons equals 12 gallons. So, last week she used 12 gallons.To find the total fuel used over the two weeks, I'll add this week's and last week's usage together: 15 gallons plus 12 gallons equals 27 gallons.Wait, let me double-check that. If last week was 20% less, and 20% of 15 is 3, then 15 minus 3 is indeed 12. Adding 15 and 12 gives 27. That seems right.I think I've got it. Mary used a total of 27 gallons of fuel over the two weeks.</think>

🧠:<think>Okay, so I have this problem about an arithmetic sequence. Let me try to figure it out step by step. First, part (I) says: Given an arithmetic sequence {a_n}, where a_3 = 6 and a_5 + a_8 = 26. I need to find the general formula for the sequence. Hmm, arithmetic sequences have a common difference, right? So, the nth term of an arithmetic sequence can be written as a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference.Given that a_3 = 6, so plugging into the formula: a_3 = a_1 + 2d = 6. That's one equation. Then, a_5 + a_8 = 26. Let me write expressions for a_5 and a_8. a_5 = a_1 + 4d and a_8 = a_1 + 7d. So, adding them together: (a_1 + 4d) + (a_1 + 7d) = 26. That simplifies to 2a_1 + 11d = 26. So now I have two equations:1. a_1 + 2d = 62. 2a_1 + 11d = 26I need to solve this system of equations to find a_1 and d. Let me use substitution or elimination. Maybe elimination is easier here. If I multiply the first equation by 2, I get 2a_1 + 4d = 12. Then subtract this from the second equation: (2a_1 + 11d) - (2a_1 + 4d) = 26 - 12. That gives 7d = 14, so d = 2.Now plug d = 2 back into the first equation: a_1 + 2*2 = 6 => a_1 + 4 = 6 => a_1 = 2. So the first term is 2 and the common difference is 2. Therefore, the general formula is a_n = a_1 + (n-1)d = 2 + (n-1)*2. Let me simplify that: 2 + 2n - 2 = 2n. So, a_n = 2n. That seems straightforward.Now, part (II) says: Let b_n = 2^{a_n} + n. Find the sum of the first n terms of the sequence {b_n}, denoted as S_n.Okay, so first, from part (I), we know a_n = 2n. Therefore, b_n = 2^{2n} + n. Simplify 2^{2n}: that's (2^2)^n = 4^n. So, b_n = 4^n + n.Now, to find S_n, which is the sum of the first n terms of {b_n}. So, S_n = sum_{k=1}^n b_k = sum_{k=1}^n (4^k + k). I can split this into two separate sums: sum_{k=1}^n 4^k + sum_{k=1}^n k.The first sum is a geometric series. The formula for the sum of a geometric series sum_{k=1}^n r^k is r(1 - r^n)/(1 - r), where r is the common ratio. Here, r = 4, so sum_{k=1}^n 4^k = 4(1 - 4^n)/(1 - 4) = 4(1 - 4^n)/(-3) = (4^{n+1} - 4)/3.The second sum is the sum of the first n natural numbers. The formula for that is n(n + 1)/2.So, putting it all together, S_n = (4^{n+1} - 4)/3 + n(n + 1)/2.Let me check if that makes sense. For the geometric series part, when n=1, sum is 4. Plugging into the formula: (4^{2} - 4)/3 = (16 - 4)/3 = 12/3 = 4. Correct. For n=2, sum is 4 + 16 = 20. Formula: (4^{3} - 4)/3 = (64 - 4)/3 = 60/3 = 20. Correct. So that part seems right.For the arithmetic series part, when n=1, sum is 1. Formula: 1(1 + 1)/2 = 1. Correct. For n=2, sum is 1 + 2 = 3. Formula: 2(2 + 1)/2 = 6/2 = 3. Correct. So that part is also right.Therefore, combining both, S_n = (4^{n+1} - 4)/3 + n(n + 1)/2. I think that's the final answer.Wait, let me see if I can write it in a more simplified form or factor something out. Maybe not necessary, but let me check:(4^{n+1} - 4)/3 + (n^2 + n)/2. It might be acceptable as is, but sometimes people prefer to have a common denominator. Let me try that.The denominators are 3 and 2, so common denominator is 6. So:[(4^{n+1} - 4)/3]*(2/2) + [(n^2 + n)/2]*(3/3) = [2(4^{n+1} - 4) + 3(n^2 + n)] / 6.Simplify numerator: 2*4^{n+1} - 8 + 3n^2 + 3n. So, S_n = (2*4^{n+1} + 3n^2 + 3n - 8)/6. Hmm, not sure if that's any better, but it's another way to write it.Alternatively, maybe factor out something else? Not sure. The original expression is probably fine.So, in conclusion, part (I) gives a_n = 2n, and part (II) gives S_n = (4^{n+1} - 4)/3 + (n^2 + n)/2.Final Answer(I) The general formula is boxed{a_n = 2n}.(II) The sum of the first ( n ) terms is boxed{S_n = frac{4^{n+1} - 4}{3} + frac{n^2 + n}{2}}.</think>