💡Okay, so I have this equation here: 1/j + 1/k = 1/4, and both j and k are positive integers. I need to find all possible values of k and then add them up. Hmm, let's see how to approach this.First, I remember that when dealing with equations involving reciprocals, it's often helpful to find a common denominator or manipulate the equation to make it easier to solve for the variables. Let me try that.Starting with the equation:1/j + 1/k = 1/4I can combine the fractions on the left side by finding a common denominator, which would be jk. So, rewriting the left side:(k + j) / (jk) = 1/4Now, I can cross-multiply to eliminate the fractions. That would give me:4(k + j) = jkExpanding the left side:4k + 4j = jkHmm, this looks a bit complicated. Maybe I can rearrange the terms to make it more manageable. Let's bring all terms to one side:jk - 4j - 4k = 0This still looks a bit tricky. I remember a technique called factoring by grouping, which might help here. To do that, I can add a constant term to both sides so that the left side can be factored into a product of two binomials.Looking at the equation:jk - 4j - 4k = 0If I add 16 to both sides, it might help:jk - 4j - 4k + 16 = 16Now, let's see if the left side can be factored. It looks like it can be factored into (j - 4)(k - 4). Let me check:(j - 4)(k - 4) = jk - 4j - 4k + 16Yes, that's exactly the left side. So, the equation becomes:(j - 4)(k - 4) = 16Now, since j and k are positive integers, (j - 4) and (k - 4) must also be positive integers. So, I need to find all pairs of positive integers that multiply to 16. These pairs are the factor pairs of 16.Let me list the factor pairs of 16:1 and 16,2 and 8,4 and 4,8 and 2,16 and 1.Each of these pairs corresponds to (j - 4, k - 4). So, I can find the corresponding j and k values by adding 4 to each factor.Let's go through each pair:1. (1, 16): j - 4 = 1 ⇒ j = 5 k - 4 = 16 ⇒ k = 202. (2, 8): j - 4 = 2 ⇒ j = 6 k - 4 = 8 ⇒ k = 123. (4, 4): j - 4 = 4 ⇒ j = 8 k - 4 = 4 ⇒ k = 84. (8, 2): j - 4 = 8 ⇒ j = 12 k - 4 = 2 ⇒ k = 65. (16, 1): j - 4 = 16 ⇒ j = 20 k - 4 = 1 ⇒ k = 5So, the possible values for k are 20, 12, 8, 6, and 5. Now, I need to sum these values.Let me add them up:20 + 12 = 3232 + 8 = 4040 + 6 = 4646 + 5 = 51So, the sum of all possible values for k is 51.Wait a second, let me double-check to make sure I didn't miss any factor pairs or make a calculation error. The factor pairs of 16 are indeed (1,16), (2,8), (4,4), (8,2), and (16,1). Each of these gives a unique k value, and adding them up gives 51. I think that's correct.Just to be thorough, let me plug each (j, k) pair back into the original equation to ensure they satisfy it.1. For (5, 20): 1/5 + 1/20 = 4/20 + 1/20 = 5/20 = 1/4 ✔️2. For (6, 12): 1/6 + 1/12 = 2/12 + 1/12 = 3/12 = 1/4 ✔️3. For (8, 8): 1/8 + 1/8 = 2/8 = 1/4 ✔️4. For (12, 6): 1/12 + 1/6 = 1/12 + 2/12 = 3/12 = 1/4 ✔️5. For (20, 5): 1/20 + 1/5 = 1/20 + 4/20 = 5/20 = 1/4 ✔️All pairs satisfy the original equation, so my solutions are correct. Therefore, the sum of all possible k values is indeed 51.

💡Let's denote the number of cards Linlin has originally by ( L ), and the number of cards Tongtong has originally by ( T ).1. **First Scenario:** - If Tongtong gives 6 cards to Linlin, then: - Linlin's cards become ( L + 6 ). - Tongtong's cards become ( T - 6 ). - According to the problem, Linlin will have 3 times as many cards as Tongtong: [ L + 6 = 3(T - 6) ] - Simplify the equation: [ L + 6 = 3T - 18 ] [ L = 3T - 24 quad text{(Equation 1)} ]2. **Second Scenario:** - If Linlin gives 2 cards to Tongtong, then: - Linlin's cards become ( L - 2 ). - Tongtong's cards become ( T + 2 ). - According to the problem, Tongtong will have 2 times as many cards as Linlin: [ T + 2 = 2(L - 2) ] - Simplify the equation: [ T + 2 = 2L - 4 ] [ T = 2L - 6 quad text{(Equation 2)} ]3. **Solving the System of Equations:** - Substitute Equation 1 into Equation 2: [ T = 2(3T - 24) - 6 ] [ T = 6T - 48 - 6 ] [ T - 6T = -54 ] [ -5T = -54 ] [ T = frac{54}{5} = 10.8 ] - This result is not an integer, which is not possible since the number of cards must be an integer. Therefore, we need to re-examine the equations.4. **Re-examining the Equations:** - Substitute Equation 2 into Equation 1: [ L = 3(2L - 6) - 24 ] [ L = 6L - 18 - 24 ] [ L = 6L - 42 ] [ -5L = -42 ] [ L = frac{42}{5} = 8.4 ] - Again, this result is not an integer. This suggests there might be an error in the setup or interpretation of the problem.5. **Revisiting the Problem Statement:** - Upon re-reading the problem, it is clear that both scenarios must result in integer values for the number of cards. Therefore, we need to find values of ( L ) and ( T ) that satisfy both equations and result in integer solutions.6. **Alternative Approach:** - Let's assume the number of cards must be integers. Therefore, we need to find integer solutions for the system: [ L = 3T - 24 ] [ T = 2L - 6 ] - Substitute ( L ) from Equation 1 into Equation 2: [ T = 2(3T - 24) - 6 ] [ T = 6T - 48 - 6 ] [ T - 6T = -54 ] [ -5T = -54 ] [ T = frac{54}{5} = 10.8 ] - This still results in a non-integer value. Therefore, we need to check if there is a mistake in the initial setup.7. **Correcting the Setup:** - Upon closer inspection, the second scenario states that if Linlin gives 2 cards to Tongtong, Linlin will have 2 times as many cards as Tongtong. This translates to: [ L - 2 = 2(T + 2) ] - Simplifying this: [ L - 2 = 2T + 4 ] [ L = 2T + 6 quad text{(Equation 3)} ] - Now, we have: [ L = 3T - 24 quad text{(Equation 1)} ] [ L = 2T + 6 quad text{(Equation 3)} ] - Set Equation 1 equal to Equation 3: [ 3T - 24 = 2T + 6 ] [ T = 30 ] - Substitute ( T = 30 ) into Equation 3: [ L = 2(30) + 6 = 66 ]8. **Verification:** - **First Scenario:** - Tongtong gives 6 cards to Linlin: - Tongtong: ( 30 - 6 = 24 ) - Linlin: ( 66 + 6 = 72 ) - Check: ( 72 = 3 times 24 ) ✔️ - **Second Scenario:** - Linlin gives 2 cards to Tongtong: - Linlin: ( 66 - 2 = 64 ) - Tongtong: ( 30 + 2 = 32 ) - Check: ( 64 = 2 times 32 ) ✔️Both scenarios are satisfied with integer values, confirming the solution.

💡Okay, so I have this problem where a function f(x) satisfies a recurrence relation. The relation is given by f(n+1) = (2f(n) + n)/2 for n in the set of positive integers, and they tell me that f(1) = 2. I need to find the value of f(20). The options are (A) 95, (B) 97, (C) 105, and (D) 192.Hmm, recurrence relations can sometimes be tricky, but I remember that they often involve finding a pattern or solving them using methods like characteristic equations or iteration. Let me think about how to approach this.First, let me write down the recurrence relation again to make sure I have it correctly:f(n+1) = (2f(n) + n)/2I can simplify this equation a bit. Let's multiply both sides by 2 to eliminate the denominator:2f(n+1) = 2f(n) + nThen, subtract 2f(n) from both sides:2f(n+1) - 2f(n) = nFactor out the 2 on the left side:2(f(n+1) - f(n)) = nDivide both sides by 2:f(n+1) - f(n) = n/2Okay, so now I have a simpler recurrence relation: the difference between consecutive terms is n/2. That is, each term is the previous term plus half of the current index. This seems like it might be a linear recurrence, and perhaps I can express f(n) as a sum of these differences.Let me try to write out the first few terms to see if I can spot a pattern.Given f(1) = 2.Then, f(2) = f(1) + 1/2 = 2 + 0.5 = 2.5f(3) = f(2) + 2/2 = 2.5 + 1 = 3.5f(4) = f(3) + 3/2 = 3.5 + 1.5 = 5f(5) = f(4) + 4/2 = 5 + 2 = 7f(6) = f(5) + 5/2 = 7 + 2.5 = 9.5f(7) = f(6) + 6/2 = 9.5 + 3 = 12.5f(8) = f(7) + 7/2 = 12.5 + 3.5 = 16f(9) = f(8) + 8/2 = 16 + 4 = 20f(10) = f(9) + 9/2 = 20 + 4.5 = 24.5Hmm, okay, so the function is increasing, and the increments are getting larger each time, which makes sense because each step adds n/2, where n increases each time.But calculating up to f(20) manually would take a while, and I might make a mistake. Maybe I can find a general formula for f(n).Since the difference f(n+1) - f(n) = n/2, this is a first-order linear recurrence. I think I can solve it using the method of summing the differences.In general, for a recurrence relation like f(n+1) - f(n) = g(n), the solution can be written as:f(n) = f(1) + sum_{k=1}^{n-1} g(k)In this case, g(k) = k/2. So,f(n) = f(1) + sum_{k=1}^{n-1} (k/2)Given that f(1) = 2, let's plug that in:f(n) = 2 + (1/2) * sum_{k=1}^{n-1} kI know that the sum of the first m integers is given by m(m+1)/2. So, sum_{k=1}^{n-1} k = (n-1)n/2.Therefore,f(n) = 2 + (1/2) * [(n-1)n/2] = 2 + [(n-1)n]/4Simplify this:f(n) = 2 + (n^2 - n)/4I can write this as:f(n) = (n^2 - n)/4 + 2To combine the terms, let's express 2 as 8/4:f(n) = (n^2 - n + 8)/4So, that's the general formula for f(n). Let me test this formula with the values I calculated earlier to make sure it's correct.For n=1:f(1) = (1 - 1 + 8)/4 = 8/4 = 2. Correct.For n=2:f(2) = (4 - 2 + 8)/4 = 10/4 = 2.5. Correct.For n=3:f(3) = (9 - 3 + 8)/4 = 14/4 = 3.5. Correct.For n=4:f(4) = (16 - 4 + 8)/4 = 20/4 = 5. Correct.Okay, so the formula seems to be working for the first few terms. That's a good sign.Now, I need to find f(20). Let's plug n=20 into the formula:f(20) = (20^2 - 20 + 8)/4Calculate 20^2: 400Subtract 20: 400 - 20 = 380Add 8: 380 + 8 = 388Divide by 4: 388 / 4 = 97So, f(20) = 97. Looking back at the options, that's option (B).Just to double-check, let me compute f(20) using the recursive formula step by step, although it might take a while.Starting from f(1)=2:f(2)=2 + 1/2=2.5f(3)=2.5 + 2/2=2.5+1=3.5f(4)=3.5 + 3/2=3.5+1.5=5f(5)=5 + 4/2=5+2=7f(6)=7 + 5/2=7+2.5=9.5f(7)=9.5 + 6/2=9.5+3=12.5f(8)=12.5 + 7/2=12.5+3.5=16f(9)=16 + 8/2=16+4=20f(10)=20 + 9/2=20+4.5=24.5f(11)=24.5 + 10/2=24.5+5=29.5f(12)=29.5 + 11/2=29.5+5.5=35f(13)=35 + 12/2=35+6=41f(14)=41 + 13/2=41+6.5=47.5f(15)=47.5 + 14/2=47.5+7=54.5f(16)=54.5 + 15/2=54.5+7.5=62f(17)=62 + 16/2=62+8=70f(18)=70 + 17/2=70+8.5=78.5f(19)=78.5 + 18/2=78.5+9=87.5f(20)=87.5 + 19/2=87.5+9.5=97Yes, that matches the result from the formula. So, f(20)=97, which is option (B).I think that's solid. I used two methods: deriving the general formula and then verifying by computing step by step. Both gave me the same answer, so I'm confident that 97 is correct.**Final Answer**boxed{B}

💡Alright, so I'm trying to figure out this math problem here. It says that the concept of negative numbers first appeared in an ancient Chinese math book called "Nine Chapters on the Mathematical Art." Okay, that's interesting, but I'm not sure how relevant that part is to solving the problem. Maybe it's just giving some historical context?Anyway, the main part of the problem is about income and expenses. It says that if an income of 5 yuan is denoted as +5 yuan, then what would expenses of 5 yuan be denoted as? The options are A: -5 yuan, B: 0 yuan, C: +5 yuan, and D: +10 yuan.So, let's break this down. Income is money coming in, and expenses are money going out. In accounting or finance, income is usually considered a positive number because it's adding to your total, while expenses are negative because they're subtracting from your total. That makes sense because if you earn money, your net worth increases, and if you spend money, it decreases.Given that, if +5 yuan represents income, then expenses should be the opposite. So, instead of a positive number, it should be a negative number. That would mean expenses of 5 yuan would be -5 yuan. Looking at the options, that's option A.But wait, let me make sure I'm not missing something. Could there be a reason why expenses wouldn't be negative? Maybe in some contexts, people represent expenses differently? I don't think so. In standard accounting practices, expenses are definitely negative numbers when income is positive. It's a way to show the flow of money in and out.Also, the other options don't make much sense. Option B is 0 yuan, which would mean no change, but that's not the case here because there is an expense. Option C is +5 yuan, which is the same as income, so that's not right. Option D is +10 yuan, which is even more income, so that's definitely not expenses.So, yeah, I'm pretty confident that the correct answer is A: -5 yuan. It aligns with how positive and negative numbers are used to represent income and expenses in financial contexts. Plus, it matches the basic understanding that income increases your total and expenses decrease it.I guess another way to think about it is using a simple equation. If I have an income of +5 yuan and an expense of -5 yuan, my net result would be zero. That makes sense because earning and spending the same amount would cancel each other out. If I used any other number for expenses, like +5 or +10, that wouldn't make sense because it wouldn't balance the equation properly.Also, considering the historical context mentioned in the problem, the ancient Chinese were using negative numbers, so they must have had a system where positive and negative quantities represented opposite concepts like income and expenses. It's cool to think that this basic principle has been around for so long and is still applicable today.In summary, income is positive, expenses are negative, and the amount is the same, so -5 yuan is the right answer.

💡Okay, so I have this problem about a sequence defined by some initial terms and a recurrence relation. Let me try to understand it step by step.The sequence is given by ( x_0 = 1 ), ( x_1 = k ), and for each ( n ), ( x_{n+2} = x_n - x_{n+1} ). I need to show that there's only one value of ( k ) such that all the terms in the sequence are positive. Hmm, interesting.First, I should probably write out the first few terms to see what's going on. Let's see:- ( x_0 = 1 )- ( x_1 = k )- ( x_2 = x_0 - x_1 = 1 - k )- ( x_3 = x_1 - x_2 = k - (1 - k) = 2k - 1 )- ( x_4 = x_2 - x_3 = (1 - k) - (2k - 1) = 1 - k - 2k + 1 = 2 - 3k )- ( x_5 = x_3 - x_4 = (2k - 1) - (2 - 3k) = 2k - 1 - 2 + 3k = 5k - 3 )- ( x_6 = x_4 - x_5 = (2 - 3k) - (5k - 3) = 2 - 3k - 5k + 3 = 5 - 8k )- ( x_7 = x_5 - x_6 = (5k - 3) - (5 - 8k) = 5k - 3 - 5 + 8k = 13k - 8 )Wait a minute, these coefficients look familiar. 1, 1, 2, 3, 5, 8, 13... These are Fibonacci numbers! Interesting. So, the coefficients of ( k ) and the constants seem to follow the Fibonacci sequence.Let me note that down. The Fibonacci sequence is defined by ( F_0 = 0 ), ( F_1 = 1 ), and ( F_{n+2} = F_{n+1} + F_n ). So, ( F_2 = 1 ), ( F_3 = 2 ), ( F_4 = 3 ), ( F_5 = 5 ), ( F_6 = 8 ), ( F_7 = 13 ), etc.Looking back at the terms I calculated:- ( x_2 = 1 - k = F_2 - F_1 k )- ( x_3 = 2k - 1 = F_3 k - F_2 )- ( x_4 = 2 - 3k = F_4 - F_3 k )- ( x_5 = 5k - 3 = F_5 k - F_4 )- ( x_6 = 5 - 8k = F_6 - F_5 k )- ( x_7 = 13k - 8 = F_7 k - F_6 )So, it seems like there's a pattern here where each term alternates between being expressed as ( F_{n} - F_{n-1}k ) and ( F_{n}k - F_{n-1} ). That is, for even ( n ), it's ( F_n - F_{n-1}k ), and for odd ( n ), it's ( F_n k - F_{n-1} ).Let me generalize this:For ( n geq 2 ),- If ( n ) is even, ( x_n = F_n - F_{n-1}k )- If ( n ) is odd, ( x_n = F_n k - F_{n-1} )Okay, so to ensure all terms are positive, both expressions must be positive for all ( n ). That gives us inequalities:For even ( n geq 2 ):( F_n - F_{n-1}k > 0 ) Which implies ( k < frac{F_n}{F_{n-1}} )For odd ( n geq 3 ):( F_n k - F_{n-1} > 0 ) Which implies ( k > frac{F_{n-1}}{F_n} )So, for each ( n ), we have bounds on ( k ). As ( n ) increases, these bounds get tighter because the ratio ( frac{F_{n}}{F_{n-1}} ) approaches the golden ratio ( phi = frac{1+sqrt{5}}{2} approx 1.618 ).Wait, so as ( n ) increases, the upper bound for ( k ) approaches ( phi ), and the lower bound approaches ( frac{1}{phi} = frac{sqrt{5}-1}{2} approx 0.618 ). Therefore, the only value of ( k ) that satisfies all these inequalities for all ( n ) is the limit of these bounds as ( n ) approaches infinity, which is ( frac{sqrt{5}-1}{2} ).But let me verify this. If ( k = frac{sqrt{5}-1}{2} ), then let's compute some terms:- ( x_0 = 1 )- ( x_1 = frac{sqrt{5}-1}{2} approx 0.618 )- ( x_2 = 1 - 0.618 = 0.382 )- ( x_3 = 0.618 - 0.382 = 0.236 )- ( x_4 = 0.382 - 0.236 = 0.146 )- ( x_5 = 0.236 - 0.146 = 0.090 )- ( x_6 = 0.146 - 0.090 = 0.056 )- ( x_7 = 0.090 - 0.056 = 0.034 )- And so on...All these terms are positive, and they seem to be decreasing. So, it looks like with ( k = frac{sqrt{5}-1}{2} ), all terms are positive.But what if ( k ) is slightly larger than ( frac{sqrt{5}-1}{2} )? Let's say ( k = 0.7 ).Compute the terms:- ( x_0 = 1 )- ( x_1 = 0.7 )- ( x_2 = 1 - 0.7 = 0.3 )- ( x_3 = 0.7 - 0.3 = 0.4 )- ( x_4 = 0.3 - 0.4 = -0.1 )Oops, ( x_4 ) is negative. So, ( k = 0.7 ) doesn't work because ( x_4 ) becomes negative.Similarly, if ( k ) is slightly less than ( frac{sqrt{5}-1}{2} ), say ( k = 0.5 ):- ( x_0 = 1 )- ( x_1 = 0.5 )- ( x_2 = 1 - 0.5 = 0.5 )- ( x_3 = 0.5 - 0.5 = 0 )- ( x_4 = 0.5 - 0 = 0.5 )- ( x_5 = 0 - 0.5 = -0.5 )Again, ( x_5 ) is negative. So, ( k = 0.5 ) also doesn't work.Therefore, it seems that ( k ) must be exactly ( frac{sqrt{5}-1}{2} ) to ensure all terms are positive. If ( k ) is any larger or smaller, eventually some term becomes negative.To make this more rigorous, I should probably consider the general solution of the recurrence relation. The recurrence is linear and homogeneous, so we can solve it using characteristic equations.The recurrence is ( x_{n+2} = x_n - x_{n+1} ). Let's write the characteristic equation:( r^{n+2} = r^n - r^{n+1} )Divide both sides by ( r^n ) (assuming ( r neq 0 )):( r^2 = 1 - r )Bring all terms to one side:( r^2 + r - 1 = 0 )Solve this quadratic equation:( r = frac{-1 pm sqrt{1 + 4}}{2} = frac{-1 pm sqrt{5}}{2} )So, the roots are ( r_1 = frac{-1 + sqrt{5}}{2} ) and ( r_2 = frac{-1 - sqrt{5}}{2} ).Therefore, the general solution is:( x_n = A left( frac{-1 + sqrt{5}}{2} right)^n + B left( frac{-1 - sqrt{5}}{2} right)^n )Now, apply the initial conditions to find ( A ) and ( B ).For ( n = 0 ):( x_0 = 1 = A + B )For ( n = 1 ):( x_1 = k = A left( frac{-1 + sqrt{5}}{2} right) + B left( frac{-1 - sqrt{5}}{2} right) )So, we have a system of equations:1. ( A + B = 1 )2. ( A left( frac{-1 + sqrt{5}}{2} right) + B left( frac{-1 - sqrt{5}}{2} right) = k )Let me solve this system.From equation 1: ( B = 1 - A )Substitute into equation 2:( A left( frac{-1 + sqrt{5}}{2} right) + (1 - A) left( frac{-1 - sqrt{5}}{2} right) = k )Expand:( A left( frac{-1 + sqrt{5}}{2} right) + frac{-1 - sqrt{5}}{2} - A left( frac{-1 - sqrt{5}}{2} right) = k )Combine like terms:( A left( frac{-1 + sqrt{5}}{2} - frac{-1 - sqrt{5}}{2} right) + frac{-1 - sqrt{5}}{2} = k )Simplify the coefficient of ( A ):( frac{-1 + sqrt{5} + 1 + sqrt{5}}{2} = frac{2sqrt{5}}{2} = sqrt{5} )So, we have:( A sqrt{5} + frac{-1 - sqrt{5}}{2} = k )Solve for ( A ):( A sqrt{5} = k + frac{1 + sqrt{5}}{2} )( A = frac{k + frac{1 + sqrt{5}}{2}}{sqrt{5}} )Similarly, since ( B = 1 - A ):( B = 1 - frac{k + frac{1 + sqrt{5}}{2}}{sqrt{5}} )Now, for the sequence to have all positive terms, both ( A ) and ( B ) must be such that the terms don't become negative. However, the term with ( r_2 = frac{-1 - sqrt{5}}{2} ) is negative and its magnitude is greater than 1, so as ( n ) increases, ( r_2^n ) will alternate in sign and grow in magnitude if ( B neq 0 ). Therefore, to prevent the sequence from becoming negative, we must have ( B = 0 ).So, set ( B = 0 ):( 1 - A = 0 Rightarrow A = 1 )Substitute ( A = 1 ) into the expression for ( A ):( 1 = frac{k + frac{1 + sqrt{5}}{2}}{sqrt{5}} )Multiply both sides by ( sqrt{5} ):( sqrt{5} = k + frac{1 + sqrt{5}}{2} )Solve for ( k ):( k = sqrt{5} - frac{1 + sqrt{5}}{2} = frac{2sqrt{5} - 1 - sqrt{5}}{2} = frac{sqrt{5} - 1}{2} )Therefore, the only value of ( k ) that ensures all terms are positive is ( frac{sqrt{5} - 1}{2} ).To confirm, let's plug ( k = frac{sqrt{5} - 1}{2} ) back into the general solution:( x_n = A r_1^n + B r_2^n )Since ( B = 0 ), we have:( x_n = r_1^n )Given that ( r_1 = frac{sqrt{5} - 1}{2} approx 0.618 ), which is less than 1, so ( x_n ) will decrease but remain positive for all ( n ).Thus, this value of ( k ) ensures all terms are positive, and any other value would result in some term becoming negative due to the influence of the other root ( r_2 ).**Final Answer**The unique value of ( k ) is boxed{dfrac{sqrt{5} - 1}{2}}.

💡Alright, so I have this problem here: Given the polynomial ((1 + x)^{2008} = a_0 + a_1 x + a_2 x^2 + cdots + a_{2008} x^{2008}), I need to find the number of even coefficients among (a_0, a_1, ldots, a_{2008}). The options are (A) 127, (B) 1003, (C) 1005, and (D) 1881.Hmm, okay. So, first, I remember that the coefficients (a_k) in the expansion of ((1 + x)^n) are given by the binomial coefficients, right? So, (a_k = binom{2008}{k}). That makes sense.Now, the question is about determining how many of these coefficients are even. I recall that there's a theorem related to binomial coefficients modulo 2, which might help here. I think it's called Lucas' Theorem. Let me try to recall what it says.From what I remember, Lucas' Theorem states that the binomial coefficient (binom{n}{k}) modulo a prime number (p) can be determined by looking at the base-(p) expansions of (n) and (k). Specifically, if (n) and (k) are written in base (p) as (n = n_m n_{m-1} ldots n_0) and (k = k_m k_{m-1} ldots k_0), then:[binom{n}{k} equiv prod_{i=0}^m binom{n_i}{k_i} mod p]Since we're dealing with modulo 2 here, (p = 2). So, for each bit in the binary representations of (n) and (k), the binomial coefficient modulo 2 is the product of the binomial coefficients of each corresponding pair of bits. Importantly, (binom{n_i}{k_i}) is 0 if (k_i > n_i), and 1 otherwise.So, applying this theorem, (binom{2008}{k}) is odd (i.e., congruent to 1 modulo 2) if and only if for every bit position, the bit of (k) is less than or equal to the corresponding bit of 2008. If any bit of (k) is greater than the corresponding bit of 2008, then the binomial coefficient is even.Alright, so to find the number of even coefficients, I need to find the number of (k) such that (binom{2008}{k}) is even. Alternatively, since the total number of coefficients is 2009 (from (k = 0) to (k = 2008)), if I can find the number of odd coefficients, I can subtract that from 2009 to get the number of even coefficients.So, first, let's find the number of odd coefficients. To do this, I need to look at the binary representation of 2008.Let me convert 2008 to binary. I can do this by dividing by 2 repeatedly:2008 ÷ 2 = 1004, remainder 01004 ÷ 2 = 502, remainder 0502 ÷ 2 = 251, remainder 0251 ÷ 2 = 125, remainder 1125 ÷ 2 = 62, remainder 162 ÷ 2 = 31, remainder 031 ÷ 2 = 15, remainder 115 ÷ 2 = 7, remainder 17 ÷ 2 = 3, remainder 13 ÷ 2 = 1, remainder 11 ÷ 2 = 0, remainder 1So, writing the remainders from last to first, 2008 in binary is 11111011000.Let me count the number of 1s in this binary representation. Starting from the left: 1, 1, 1, 1, 1, 0, 1, 1, 0, 0, 0. So, that's 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 bits. The number of 1s is 7.According to Lucas' Theorem, the number of odd coefficients is (2^{text{number of 1s in binary representation of } n}). So, in this case, it's (2^7 = 128).Wait, but hold on. The problem is asking for the number of even coefficients. So, if there are 128 odd coefficients, then the number of even coefficients is the total number of coefficients minus the number of odd coefficients.The total number of coefficients is 2009 (since it's from (k = 0) to (k = 2008)). So, the number of even coefficients is (2009 - 128 = 1881).But wait, hold on a second. Let me double-check this because 1881 is one of the options, specifically option (D). But I recall that sometimes when dealing with Lucas' Theorem, we have to be careful about how we count the coefficients.Wait, actually, when we have the binary representation of 2008 as 11111011000, which has 7 ones, the number of subsets of these 1s is (2^7 = 128). Each subset corresponds to a number (k) such that (binom{2008}{k}) is odd. So, that's correct.But hold on, I think I might have made a mistake here. Because the number of even coefficients is 2009 minus the number of odd coefficients, which is 2009 - 128 = 1881. But 1881 is option (D). However, I also remember that sometimes the first coefficient (a_0) is 1, which is odd, so maybe that's included in the 128.Wait, no, (a_0) is 1, which is odd, so it's included in the 128. So, the number of even coefficients is indeed 2009 - 128 = 1881.But wait, looking back at the options, (D) is 1881, which would be the answer. But in the initial thought process, I thought the answer was 127, which is option (A). Hmm, now I'm confused.Wait, perhaps I made a mistake in the initial thought process. Let me go back.Wait, in the initial thought process, I concluded that the number of odd coefficients is 127, but according to Lucas' Theorem, it's 128. So, which one is correct?Wait, let's think carefully. The number of subsets of the 1s in the binary representation is (2^7 = 128). Each subset corresponds to a number (k) such that (binom{2008}{k}) is odd. So, that includes (k = 0), which is 1, which is odd. So, the number of odd coefficients is 128, not 127.Therefore, the number of even coefficients is 2009 - 128 = 1881, which is option (D). But in the initial thought process, I concluded 127, which is incorrect.Wait, so why did I initially think 127? Maybe I confused the number of non-zero coefficients or something else. Let me check.Wait, perhaps I miscounted the number of 1s in the binary representation. Let me recount.2008 in binary is 11111011000.Let's write it out:From right to left (LSB to MSB):Position: 10 9 8 7 6 5 4 3 2 1 0Bits: 1 1 1 1 1 0 1 1 0 0 0Wait, actually, positions are usually counted from 0 on the right, so the rightmost bit is position 0.So, writing it out:Position 10: 1Position 9: 1Position 8: 1Position 7: 1Position 6: 1Position 5: 0Position 4: 1Position 3: 1Position 2: 0Position 1: 0Position 0: 0So, the 1s are at positions 10, 9, 8, 7, 6, 4, 3. That's 7 positions. So, 7 ones.Therefore, the number of subsets is (2^7 = 128). So, 128 odd coefficients.Therefore, the number of even coefficients is 2009 - 128 = 1881.So, the correct answer should be (D) 1881.But wait, in the initial thought process, I thought the answer was 127. That must have been a mistake. So, the correct answer is (D) 1881.Wait, but let me double-check this because sometimes when dealing with Lucas' Theorem, especially for exponents, there might be a different approach.Alternatively, I remember that the number of odd coefficients in ((1 + x)^n) is (2^{text{number of 1s in binary representation of } n}). So, for 2008, which has 7 ones, it's (2^7 = 128). So, 128 odd coefficients.Therefore, the number of even coefficients is 2009 - 128 = 1881.Yes, that seems correct.Wait, but let me think again. Is there another way to approach this problem? Maybe using generating functions or something else.Alternatively, I know that the number of even coefficients can be found using the formula:Number of even coefficients = Total coefficients - Number of odd coefficients.Which is exactly what I did.Total coefficients: 2009.Number of odd coefficients: 128.Therefore, even coefficients: 2009 - 128 = 1881.So, the answer should be (D) 1881.But wait, in the initial thought process, I thought it was 127, which is incorrect. So, I must have made a mistake there.Wait, perhaps I confused the number of even coefficients with the number of odd coefficients. Let me see.If the number of odd coefficients is 128, then the number of even coefficients is 2009 - 128 = 1881.Yes, that makes sense.Alternatively, maybe I can think about the binary representation and how the coefficients behave.Each time you multiply by (1 + x), you're essentially shifting and adding. But that might complicate things.Alternatively, I can think about the generating function modulo 2.If I consider the polynomial modulo 2, then ((1 + x)^{2008} mod 2) is equal to the product of ((1 + x)^{2^k}) for each bit set in 2008.Wait, that's another way to think about it. Since 2008 in binary is 11111011000, which is 2^10 + 2^9 + 2^8 + 2^7 + 2^6 + 2^4 + 2^3.So, ((1 + x)^{2008} mod 2 = (1 + x)^{2^{10}} cdot (1 + x)^{2^9} cdot (1 + x)^{2^8} cdot (1 + x)^{2^7} cdot (1 + x)^{2^6} cdot (1 + x)^{2^4} cdot (1 + x)^{2^3} mod 2).But since ((1 + x)^{2^k} mod 2 = 1 + x^{2^k}), because all the middle terms have even coefficients and thus vanish modulo 2.Therefore, the product becomes:[(1 + x^{2^{10}})(1 + x^{2^9})(1 + x^{2^8})(1 + x^{2^7})(1 + x^{2^6})(1 + x^{2^4})(1 + x^{2^3}) mod 2]When you multiply these out, each term corresponds to a subset of the exponents. So, the number of terms is (2^7 = 128), which are all the possible products of these terms. Each term corresponds to a unique exponent, which is the sum of distinct powers of 2, hence each exponent is unique and corresponds to a unique binary number with 1s in the positions corresponding to the exponents.Therefore, the number of non-zero terms (i.e., the number of odd coefficients) is 128, as before.Thus, the number of even coefficients is 2009 - 128 = 1881.Therefore, the correct answer is (D) 1881.Wait, but in the initial thought process, I thought the answer was 127, which is incorrect. So, I must have made a mistake there. Probably, I confused the number of even coefficients with the number of odd coefficients or miscounted the number of 1s in the binary representation.In conclusion, after carefully considering the problem using Lucas' Theorem and verifying through generating functions, the number of even coefficients is indeed 1881, which corresponds to option (D).