💡First, I recognize that a regular hexagon has six equal sides. Knowing that the length of one side is 5 centimeters, I can calculate the total circumference by multiplying the side length by the number of sides.So, the circumference is 5 cm multiplied by 6, which equals 30 centimeters.

💡Okay, so I have this problem where I need to find the minimum value of the sum ( S = sum_{i=1}^{6} a_{i} b_{i} c_{i} ), where each of ( a_i ), ( b_i ), and ( c_i ) are permutations of the numbers 1 through 6. Hmm, that sounds a bit tricky, but let me try to break it down.First, I remember that permutations mean each number from 1 to 6 appears exactly once in each sequence ( a ), ( b ), and ( c ). So, each of these sequences is just a rearrangement of 1, 2, 3, 4, 5, 6. My goal is to arrange these numbers in such a way that when I multiply the corresponding elements from each sequence and add them all up, the total is as small as possible.I think about the AM-GM inequality because it relates sums and products, which seems relevant here since I'm dealing with a sum of products. The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is always greater than or equal to the geometric mean. Maybe I can apply this to each term ( a_i b_i c_i ).Let me write down the AM-GM inequality for the six terms ( a_1 b_1 c_1, a_2 b_2 c_2, ldots, a_6 b_6 c_6 ):[frac{a_1 b_1 c_1 + a_2 b_2 c_2 + ldots + a_6 b_6 c_6}{6} geq sqrt[6]{(a_1 b_1 c_1)(a_2 b_2 c_2) ldots (a_6 b_6 c_6)}]Multiplying both sides by 6 gives:[S geq 6 sqrt[6]{prod_{i=1}^{6} a_i b_i c_i}]Now, since each of ( a_i ), ( b_i ), and ( c_i ) are permutations of 1 through 6, the product of each sequence is ( 6! ) (which is 720). Therefore, the product of all ( a_i b_i c_i ) is ( (6!)^3 ). Plugging this into the inequality:[S geq 6 sqrt[6]{(6!)^3} = 6 sqrt[6]{720^3}]Calculating ( sqrt[6]{720^3} ) is the same as ( (720^3)^{1/6} = 720^{1/2} = sqrt{720} ). So,[S geq 6 sqrt{720}]Approximating ( sqrt{720} ), since ( 26^2 = 676 ) and ( 27^2 = 729 ), it's somewhere between 26 and 27. Let's say approximately 26.83. Then,[6 times 26.83 approx 160.98]Hmm, so the sum ( S ) is at least approximately 161. But this is just a lower bound. I need to check if this bound can actually be achieved or if I need to find a better arrangement.I recall that the AM-GM inequality gives equality when all the terms are equal. So, for ( S ) to be minimized, each ( a_i b_i c_i ) should be as equal as possible. But since ( a_i ), ( b_i ), and ( c_i ) are integers from 1 to 6, it's impossible for all products ( a_i b_i c_i ) to be exactly the same. So, I need to arrange the numbers such that the products are as balanced as possible.Maybe I can try to pair the largest numbers with the smallest numbers in different sequences to balance the products. For example, if I have a large number in one sequence, I can pair it with smaller numbers in the other sequences to keep the product from getting too big.Let me try constructing such permutations. Let's consider arranging ( a ), ( b ), and ( c ) such that when one sequence has a high number, the others have low numbers in the same position.Suppose I set ( a = [1, 2, 3, 4, 5, 6] ). Then, for ( b ) and ( c ), I can try to arrange them so that when ( a_i ) is large, ( b_i ) and ( c_i ) are small, and vice versa.For instance, let me try:- ( a = [1, 2, 3, 4, 5, 6] )- ( b = [6, 5, 4, 3, 2, 1] ) (reverse order)- ( c = [6, 5, 4, 3, 2, 1] ) (reverse order)Calculating the products:1. ( 1 times 6 times 6 = 36 )2. ( 2 times 5 times 5 = 50 )3. ( 3 times 4 times 4 = 48 )4. ( 4 times 3 times 3 = 36 )5. ( 5 times 2 times 2 = 20 )6. ( 6 times 1 times 1 = 6 )Adding these up: ( 36 + 50 + 48 + 36 + 20 + 6 = 196 ). Hmm, that's higher than the lower bound I found earlier. Maybe reversing both ( b ) and ( c ) isn't the best approach.Let me try a different arrangement. Perhaps instead of reversing both, I can stagger them so that the high numbers in one sequence are paired with medium numbers in the others.What if I set:- ( a = [1, 2, 3, 4, 5, 6] )- ( b = [5, 4, 3, 6, 1, 2] )- ( c = [5, 4, 3, 1, 6, 2] )Let me calculate each product:1. ( 1 times 5 times 5 = 25 )2. ( 2 times 4 times 4 = 32 )3. ( 3 times 3 times 3 = 27 )4. ( 4 times 6 times 1 = 24 )5. ( 5 times 1 times 6 = 30 )6. ( 6 times 2 times 2 = 24 )Adding these up: ( 25 + 32 + 27 + 24 + 30 + 24 = 162 ). That's better! It's exactly the lower bound I calculated earlier, which was approximately 161. So, this arrangement gives me a sum of 162, which seems to be the minimum.Wait, but how did I come up with that specific arrangement? Maybe I should check if there's a systematic way to arrange the permutations to achieve this balance.I think the key is to pair the largest numbers in one sequence with the smallest numbers in the other sequences. For example, if ( a_i ) is large, then ( b_i ) and ( c_i ) should be small, and if ( a_i ) is small, then ( b_i ) and ( c_i ) can be larger. This way, the products ( a_i b_i c_i ) are kept as balanced as possible, preventing any single term from being too large and thus minimizing the overall sum.Let me verify this arrangement again:- ( a = [1, 2, 3, 4, 5, 6] )- ( b = [5, 4, 3, 6, 1, 2] )- ( c = [5, 4, 3, 1, 6, 2] )Calculating each product:1. ( 1 times 5 times 5 = 25 )2. ( 2 times 4 times 4 = 32 )3. ( 3 times 3 times 3 = 27 )4. ( 4 times 6 times 1 = 24 )5. ( 5 times 1 times 6 = 30 )6. ( 6 times 2 times 2 = 24 )Adding them up: 25 + 32 + 27 + 24 + 30 + 24 = 162. Yep, that's consistent.Is there a way to get a sum lower than 162? Let me try another arrangement to see if I can get a lower sum.Suppose I set:- ( a = [1, 2, 3, 4, 5, 6] )- ( b = [6, 1, 2, 3, 4, 5] )- ( c = [6, 1, 2, 3, 4, 5] )Calculating the products:1. ( 1 times 6 times 6 = 36 )2. ( 2 times 1 times 1 = 2 )3. ( 3 times 2 times 2 = 12 )4. ( 4 times 3 times 3 = 36 )5. ( 5 times 4 times 4 = 80 )6. ( 6 times 5 times 5 = 150 )Adding these up: 36 + 2 + 12 + 36 + 80 + 150 = 316. That's way higher, so this arrangement is worse.What if I try to distribute the larger numbers more evenly across the sequences? Maybe not putting all the large numbers in one sequence.Let me try:- ( a = [1, 2, 3, 4, 5, 6] )- ( b = [3, 6, 2, 5, 1, 4] )- ( c = [4, 5, 6, 1, 2, 3] )Calculating the products:1. ( 1 times 3 times 4 = 12 )2. ( 2 times 6 times 5 = 60 )3. ( 3 times 2 times 6 = 36 )4. ( 4 times 5 times 1 = 20 )5. ( 5 times 1 times 2 = 10 )6. ( 6 times 4 times 3 = 72 )Adding them up: 12 + 60 + 36 + 20 + 10 + 72 = 210. Still higher than 162.Hmm, maybe the initial arrangement I found is actually the optimal one. Let me see if I can find another arrangement that gives a sum lower than 162.What if I adjust the permutations slightly:- ( a = [1, 2, 3, 4, 5, 6] )- ( b = [5, 4, 3, 6, 2, 1] )- ( c = [5, 4, 3, 1, 6, 2] )Calculating the products:1. ( 1 times 5 times 5 = 25 )2. ( 2 times 4 times 4 = 32 )3. ( 3 times 3 times 3 = 27 )4. ( 4 times 6 times 1 = 24 )5. ( 5 times 2 times 6 = 60 )6. ( 6 times 1 times 2 = 12 )Adding them up: 25 + 32 + 27 + 24 + 60 + 12 = 180. That's higher than 162.Wait, maybe the key is to have the largest numbers in each sequence multiplied by the smallest numbers in the other sequences as much as possible. So, for the largest ( a_i = 6 ), pair it with the smallest ( b_i = 1 ) and ( c_i = 2 ), giving ( 6 times 1 times 2 = 12 ). Similarly, for the next largest ( a_i = 5 ), pair it with ( b_i = 2 ) and ( c_i = 6 ), giving ( 5 times 2 times 6 = 60 ). Hmm, but that might not be the most balanced.Wait, in the initial arrangement that gave 162, the products were:25, 32, 27, 24, 30, 24.These are all relatively close to each other, which is good for minimizing the sum. If I can keep the products as close as possible, the sum will be minimized.Let me try another arrangement where the products are even more balanced.Suppose:- ( a = [1, 2, 3, 4, 5, 6] )- ( b = [4, 5, 6, 1, 2, 3] )- ( c = [5, 4, 3, 6, 1, 2] )Calculating the products:1. ( 1 times 4 times 5 = 20 )2. ( 2 times 5 times 4 = 40 )3. ( 3 times 6 times 3 = 54 )4. ( 4 times 1 times 6 = 24 )5. ( 5 times 2 times 1 = 10 )6. ( 6 times 3 times 2 = 36 )Adding them up: 20 + 40 + 54 + 24 + 10 + 36 = 184. Still higher than 162.I think the initial arrangement is indeed the best I can do. Let me try to see if there's a mathematical way to confirm that 162 is the minimum.I remember that for three sequences, the minimum sum of products is achieved when the sequences are similarly ordered, but in this case, since we're dealing with permutations, it's a bit different. Wait, actually, for two sequences, the minimum sum is achieved when one is sorted in ascending order and the other in descending order (this is called the rearrangement inequality). But with three sequences, it's more complicated.However, in this case, since we have three sequences, perhaps arranging them such that the largest numbers are paired with the smallest numbers in the other sequences as much as possible will give the minimum sum.In the arrangement that gave 162, the products are:25, 32, 27, 24, 30, 24.These are all relatively close, with the largest being 32 and the smallest being 24. This balance is likely the key to achieving the minimum sum.Let me try one more arrangement to see if I can get a lower sum.Suppose:- ( a = [1, 2, 3, 4, 5, 6] )- ( b = [3, 6, 2, 5, 1, 4] )- ( c = [4, 5, 6, 1, 2, 3] )Wait, I think I tried this earlier and got a sum of 210. Yeah, that's higher.Another idea: Maybe instead of keeping ( a ) in order, I can permute ( a ) as well to better balance the products.Let me try:- ( a = [3, 2, 6, 1, 4, 5] )- ( b = [5, 4, 3, 6, 1, 2] )- ( c = [5, 4, 3, 1, 6, 2] )Calculating the products:1. ( 3 times 5 times 5 = 75 )2. ( 2 times 4 times 4 = 32 )3. ( 6 times 3 times 3 = 54 )4. ( 1 times 6 times 1 = 6 )5. ( 4 times 1 times 6 = 24 )6. ( 5 times 2 times 2 = 20 )Adding them up: 75 + 32 + 54 + 6 + 24 + 20 = 211. That's even higher.Hmm, maybe permuting ( a ) isn't helping. Let me stick with ( a ) in order and focus on arranging ( b ) and ( c ) to balance the products.Another attempt:- ( a = [1, 2, 3, 4, 5, 6] )- ( b = [4, 3, 6, 5, 2, 1] )- ( c = [5, 4, 3, 1, 6, 2] )Calculating the products:1. ( 1 times 4 times 5 = 20 )2. ( 2 times 3 times 4 = 24 )3. ( 3 times 6 times 3 = 54 )4. ( 4 times 5 times 1 = 20 )5. ( 5 times 2 times 6 = 60 )6. ( 6 times 1 times 2 = 12 )Adding them up: 20 + 24 + 54 + 20 + 60 + 12 = 190. Still higher than 162.I think I'm convinced that the arrangement where ( a ) is in order, ( b ) is [5, 4, 3, 6, 1, 2], and ( c ) is [5, 4, 3, 1, 6, 2] gives the minimum sum of 162. Let me just double-check the calculations for that arrangement:1. ( 1 times 5 times 5 = 25 )2. ( 2 times 4 times 4 = 32 )3. ( 3 times 3 times 3 = 27 )4. ( 4 times 6 times 1 = 24 )5. ( 5 times 1 times 6 = 30 )6. ( 6 times 2 times 2 = 24 )Adding them: 25 + 32 = 57; 57 + 27 = 84; 84 + 24 = 108; 108 + 30 = 138; 138 + 24 = 162. Yep, that's correct.I don't think I can find a better arrangement because any other permutation either results in some products being too large or too small, which increases the overall sum. Therefore, I conclude that the minimum value of ( S ) is 162.

💡Alright, so I have this math problem here that Galya thought of a number, did some operations on it, and ended up with a number that's 2021 less than the original. I need to find out what N is. Okay, let's break this down step by step.First, let me try to understand the problem. Galya starts with a number, let's call it k. Then she multiplies it by N, adds N to that result, divides the whole thing by N, and then subtracts N. After all these operations, she ends up with a number that's 2021 less than the original number k. So, I need to find the value of N that makes this happen.Let me write down the operations step by step to make it clearer:1. Start with the number k.2. Multiply by N: k * N.3. Add N to the result: (k * N) + N.4. Divide by N: [(k * N) + N] / N.5. Subtract N: [(k * N) + N] / N - N.6. The final result is k - 2021.Okay, so now I can translate these steps into an equation. Let's do that.Starting with k, after multiplying by N, we have kN. Adding N gives us kN + N. Dividing by N, we get (kN + N)/N. Then subtracting N, the expression becomes (kN + N)/N - N. According to the problem, this equals k - 2021.So, the equation is:(kN + N)/N - N = k - 2021Now, let's simplify this equation step by step.First, let's simplify (kN + N)/N. I can factor out N from the numerator:(kN + N) = N(k + 1)So, (kN + N)/N = N(k + 1)/N = k + 1That simplifies nicely. So now, the equation becomes:(k + 1) - N = k - 2021Now, let's subtract k from both sides to simplify further:(k + 1) - N - k = -2021This simplifies to:1 - N = -2021Now, I can solve for N. Let's add N to both sides:1 = N - 2021Then, add 2021 to both sides:1 + 2021 = NWhich gives:2022 = NSo, N is 2022.Wait, let me double-check my steps to make sure I didn't make any mistakes.Starting with the operations:1. Multiply by N: k * N2. Add N: kN + N3. Divide by N: (kN + N)/N = k + 14. Subtract N: (k + 1) - NSet equal to k - 2021:(k + 1) - N = k - 2021Subtract k:1 - N = -2021Add N:1 = N - 2021Add 2021:2022 = NYes, that seems correct. So, N is indeed 2022.I think that's the solution. Let me just recap to ensure everything makes sense.If N is 2022, then starting with any number k, multiplying by 2022, adding 2022, dividing by 2022, and subtracting 2022 should give k - 2021.Let's test it with a simple number, say k = 2022.1. Multiply by N: 2022 * 2022 = 2022²2. Add N: 2022² + 20223. Divide by N: (2022² + 2022)/2022 = 2022 + 1 = 20234. Subtract N: 2023 - 2022 = 1Now, k - 2021 = 2022 - 2021 = 1Yes, it works for k = 2022.Let's try another number, say k = 100.1. Multiply by N: 100 * 2022 = 2022002. Add N: 202200 + 2022 = 2042223. Divide by N: 204222 / 2022 = 101 (since 2022 * 101 = 204222)4. Subtract N: 101 - 2022 = -1921Now, k - 2021 = 100 - 2021 = -1921It matches again. So, N = 2022 is indeed the correct answer.I think I've got it right. The key was to translate each operation into an algebraic expression and then solve for N step by step. It's important to carefully simplify each part and verify the solution with different numbers to ensure it works universally, not just for specific cases.

💡Okay, so I have this triangle ABC where the sides opposite angles A, B, and C are a, b, c respectively. The problem says that angles A, B, C form an arithmetic sequence, and that c = 2a. I need to find the measure of angle A. Then, there's a second part about a sequence involving cosine terms and finding n when the sum is 20. Let me focus on the first part first.Alright, arithmetic sequence. That means the angles increase by a common difference. So, if I let B be the middle term, then A = B - d and C = B + d, where d is the common difference. Since it's a triangle, the sum of angles A + B + C = 180 degrees or π radians. So, substituting, (B - d) + B + (B + d) = 3B = π. Therefore, B = π/3 or 60 degrees. That simplifies things because now I know angle B is 60 degrees.Now, c = 2a. Using the Law of Sines, which says that a/sin A = b/sin B = c/sin C. So, c = 2a implies that sin C = 2 sin A. Hmm, interesting. Also, since angles A, B, C are in arithmetic sequence, and B is 60 degrees, then A = 60 - d and C = 60 + d. So, angle C is 60 + d.So, sin C = sin(60 + d). And from the Law of Sines, sin C = 2 sin A. So, sin(60 + d) = 2 sin(60 - d). Let me write that down:sin(60 + d) = 2 sin(60 - d)I can use the sine addition formula here. Remember that sin(A + B) = sin A cos B + cos A sin B. So, expanding both sides:sin 60 cos d + cos 60 sin d = 2 [sin 60 cos d - cos 60 sin d]Let me compute the values. sin 60 is √3/2, cos 60 is 1/2. Plugging these in:(√3/2) cos d + (1/2) sin d = 2 [ (√3/2) cos d - (1/2) sin d ]Simplify the right side:2*(√3/2 cos d) = √3 cos d2*(-1/2 sin d) = -sin dSo, right side is √3 cos d - sin dLeft side is (√3/2 cos d + 1/2 sin d)So, equation becomes:(√3/2 cos d + 1/2 sin d) = √3 cos d - sin dLet me bring all terms to one side:√3/2 cos d + 1/2 sin d - √3 cos d + sin d = 0Combine like terms:(√3/2 - √3) cos d + (1/2 + 1) sin d = 0Simplify coefficients:√3/2 - √3 = -√3/21/2 + 1 = 3/2So, equation is:-√3/2 cos d + 3/2 sin d = 0Multiply both sides by 2 to eliminate denominators:-√3 cos d + 3 sin d = 0Let me rearrange:3 sin d = √3 cos dDivide both sides by cos d (assuming cos d ≠ 0):3 tan d = √3So, tan d = √3 / 3Which is tan d = 1/√3So, d = arctan(1/√3) = 30 degrees or π/6 radians.So, angle d is 30 degrees. Therefore, angle A is 60 - d = 60 - 30 = 30 degrees. So, angle A is 30 degrees or π/6 radians.Wait, let me just verify that. If d is 30 degrees, then angle C is 60 + 30 = 90 degrees. So, triangle ABC has angles 30, 60, 90. That makes it a right-angled triangle at C. So, side c is opposite angle C, which is 90 degrees, so c is the hypotenuse. Given that c = 2a, and in a 30-60-90 triangle, the sides are in the ratio 1 : √3 : 2. So, if a is the side opposite 30 degrees, which is 1, then c is 2, which is 2a. That checks out. So, yes, angle A is 30 degrees.Okay, so that's the first part. Now, moving on to the second part.We have a sequence {a_n} where a_n = 2^n |cos(nC)|. S_n is the sum of the first n terms, and S_n = 20. We need to find n.First, let's recall that in the first part, angle C was 90 degrees or π/2 radians. So, nC = n*(π/2). So, cos(nC) = cos(nπ/2). The absolute value of that is |cos(nπ/2)|.So, let's think about the values of |cos(nπ/2)| for integer n.When n is 1: cos(π/2) = 0, so |cos(π/2)| = 0n=2: cos(π) = -1, absolute value is 1n=3: cos(3π/2) = 0, absolute value is 0n=4: cos(2π) = 1, absolute value is 1n=5: cos(5π/2) = 0, absolute value is 0n=6: cos(3π) = -1, absolute value is 1So, the pattern is that for even n, |cos(nπ/2)| alternates between 1 and 0? Wait, no. Wait, let's see:Wait, n=1: 0n=2: 1n=3: 0n=4: 1n=5: 0n=6: 1So, it's 0,1,0,1,0,1,... So, for odd n, it's 0, and for even n, it's 1.Therefore, a_n = 2^n * |cos(nπ/2)| = 2^n * 0 when n is odd, and 2^n *1 when n is even.So, the sequence a_n is 0, 2^2, 0, 2^4, 0, 2^6, etc.So, the non-zero terms are at even n: 2^2, 2^4, 2^6, etc.Therefore, the sum S_n depends on whether n is even or odd.If n is even, say n=2k, then S_n = sum_{i=1}^{2k} a_i = sum of the first k even terms: 2^2 + 2^4 + ... + 2^{2k}If n is odd, say n=2k+1, then S_n = sum_{i=1}^{2k+1} a_i = sum_{i=1}^{2k} a_i + a_{2k+1} = same as S_{2k} + 0 = S_{2k}So, whether n is even or odd, S_n is the same for n=2k and n=2k+1. So, S_n is equal to S_{2k} when n is 2k or 2k+1.So, let's compute S_{2k} = 2^2 + 2^4 + ... + 2^{2k}This is a geometric series with first term 4 (2^2) and common ratio 4 (since each term is 4 times the previous). The number of terms is k.The sum of a geometric series is a*(r^n - 1)/(r - 1). So, here, a=4, r=4, number of terms=k.Therefore, S_{2k} = 4*(4^k - 1)/(4 - 1) = (4^{k+1} - 4)/3We are given that S_n = 20. So, (4^{k+1} - 4)/3 = 20Multiply both sides by 3: 4^{k+1} - 4 = 60So, 4^{k+1} = 6464 is 4^3, since 4^3 = 64. Therefore, k+1 = 3 => k=2So, k=2. Therefore, n=2k or n=2k+1, which is n=4 or n=5.So, the value of n is either 4 or 5.Let me verify that.Compute S_4: a1 + a2 + a3 + a4 = 0 + 4 + 0 + 16 = 20Compute S_5: a1 + a2 + a3 + a4 + a5 = 0 + 4 + 0 + 16 + 0 = 20So, both n=4 and n=5 give S_n=20.Therefore, the answer is n=4 or 5.**Final Answer**1. The measure of angle A is boxed{dfrac{pi}{6}}.2. The value of ( n ) is boxed{4} or boxed{5}.

💡Alright, so I have this problem about sequences and whether they belong to a set W. The set W has two conditions: first, that the average of a_n and a_{n+2} is less than a_{n+1}, and second, that the sequence is bounded above by some real number M. I need to check four given sequences to see which ones satisfy both conditions.Let me start by understanding the first condition better. The inequality is (a_n + a_{n+2}) / 2 < a_{n+1}. If I rearrange this, it becomes a_n + a_{n+2} < 2a_{n+1}, which simplifies to a_{n+2} - 2a_{n+1} + a_n < 0. Hmm, that looks familiar—it's similar to the second difference in sequences. In calculus, the second derivative tells us about concavity, so maybe this is about concave sequences? If the second difference is negative, the sequence is concave down, which would mean it's curving downward.Okay, so the first condition is about the concavity of the sequence. The second condition is just that the sequence doesn't go to infinity; it's bounded above. So I need to check both concavity and boundedness for each sequence.Let me go through each sequence one by one.1. The first sequence is {n² + 1}. Let's see. Is this bounded? Well, as n increases, n² grows without bound, so this sequence isn't bounded above. Therefore, it doesn't satisfy the second condition, so it can't be in set W. That was straightforward.2. The second sequence is {(2n + 9)/(2n + 11)}. Let me simplify this. If I divide numerator and denominator by n, it becomes (2 + 9/n)/(2 + 11/n). As n approaches infinity, this approaches 2/2 = 1. So the sequence converges to 1, which means it's bounded above by, say, 1 or something slightly larger. So the second condition is satisfied.Now, let's check the first condition. I need to compute a_n + a_{n+2} - 2a_{n+1} and see if it's less than 0. Let's compute each term:a_n = (2n + 9)/(2n + 11)a_{n+1} = (2(n+1) + 9)/(2(n+1) + 11) = (2n + 11)/(2n + 13)a_{n+2} = (2(n+2) + 9)/(2(n+2) + 11) = (2n + 13)/(2n + 15)So, a_n + a_{n+2} - 2a_{n+1} = [(2n + 9)/(2n + 11) + (2n + 13)/(2n + 15)] - 2*(2n + 11)/(2n + 13)Let me compute this step by step.First, compute a_n + a_{n+2}:(2n + 9)/(2n + 11) + (2n + 13)/(2n + 15)To add these, I need a common denominator, which is (2n + 11)(2n + 15).So, [(2n + 9)(2n + 15) + (2n + 13)(2n + 11)] / [(2n + 11)(2n + 15)]Let me expand the numerators:First term: (2n + 9)(2n + 15) = 4n² + 30n + 18n + 135 = 4n² + 48n + 135Second term: (2n + 13)(2n + 11) = 4n² + 22n + 26n + 143 = 4n² + 48n + 143Adding them together: 4n² + 48n + 135 + 4n² + 48n + 143 = 8n² + 96n + 278So, a_n + a_{n+2} = (8n² + 96n + 278) / [(2n + 11)(2n + 15)]Now, compute 2a_{n+1}:2*(2n + 11)/(2n + 13) = (4n + 22)/(2n + 13)So, a_n + a_{n+2} - 2a_{n+1} = [ (8n² + 96n + 278) / ( (2n + 11)(2n + 15) ) ] - (4n + 22)/(2n + 13)To subtract these, I need a common denominator, which is (2n + 11)(2n + 13)(2n + 15).So, let's rewrite both terms:First term: (8n² + 96n + 278)*(2n + 13) / [ (2n + 11)(2n + 13)(2n + 15) ]Second term: (4n + 22)*(2n + 11)(2n + 15) / [ (2n + 11)(2n + 13)(2n + 15) ]So, subtracting them:[ (8n² + 96n + 278)(2n + 13) - (4n + 22)(2n + 11)(2n + 15) ] / [ (2n + 11)(2n + 13)(2n + 15) ]This looks complicated, but let me compute the numerator step by step.First, expand (8n² + 96n + 278)(2n + 13):Multiply term by term:8n²*2n = 16n³8n²*13 = 104n²96n*2n = 192n²96n*13 = 1248n278*2n = 556n278*13 = 3614So, adding all together:16n³ + 104n² + 192n² + 1248n + 556n + 3614Combine like terms:16n³ + (104 + 192)n² + (1248 + 556)n + 3614Which is:16n³ + 296n² + 1804n + 3614Now, expand (4n + 22)(2n + 11)(2n + 15). Let's first compute (2n + 11)(2n + 15):(2n + 11)(2n + 15) = 4n² + 30n + 22n + 165 = 4n² + 52n + 165Now, multiply by (4n + 22):(4n + 22)(4n² + 52n + 165)Multiply term by term:4n*4n² = 16n³4n*52n = 208n²4n*165 = 660n22*4n² = 88n²22*52n = 1144n22*165 = 3630So, adding all together:16n³ + 208n² + 660n + 88n² + 1144n + 3630Combine like terms:16n³ + (208 + 88)n² + (660 + 1144)n + 3630Which is:16n³ + 296n² + 1804n + 3630So, now, subtracting the two expanded terms:Numerator = [16n³ + 296n² + 1804n + 3614] - [16n³ + 296n² + 1804n + 3630]Let's subtract term by term:16n³ - 16n³ = 0296n² - 296n² = 01804n - 1804n = 03614 - 3630 = -16So, the numerator is -16, and the denominator is (2n + 11)(2n + 13)(2n + 15), which is always positive for positive n. Therefore, the entire expression is -16 / [positive] = negative. So, a_n + a_{n+2} - 2a_{n+1} < 0, which satisfies the first condition.Since both conditions are satisfied, sequence (2) is in set W.3. The third sequence is {2 + 4/n}. Let's check boundedness first. As n increases, 4/n approaches 0, so the sequence approaches 2. So it's bounded above by, say, 2 + 4/1 = 6. So it's bounded.Now, check the first condition: a_n + a_{n+2} - 2a_{n+1} < 0.Compute a_n = 2 + 4/na_{n+1} = 2 + 4/(n+1)a_{n+2} = 2 + 4/(n+2)So, a_n + a_{n+2} - 2a_{n+1} = [2 + 4/n + 2 + 4/(n+2)] - 2*[2 + 4/(n+1)]Simplify:(4 + 4/n + 4/(n+2)) - (4 + 8/(n+1)) = 4 + 4/n + 4/(n+2) - 4 - 8/(n+1) = 4/n + 4/(n+2) - 8/(n+1)Factor out 4:4[1/n + 1/(n+2) - 2/(n+1)]Let me compute the expression inside the brackets:1/n + 1/(n+2) - 2/(n+1)To combine these, find a common denominator, which is n(n+1)(n+2).So:[ (n+1)(n+2) + n(n+1) - 2n(n+2) ] / [n(n+1)(n+2)]Compute numerator:(n+1)(n+2) = n² + 3n + 2n(n+1) = n² + n2n(n+2) = 2n² + 4nSo, numerator:(n² + 3n + 2) + (n² + n) - (2n² + 4n) = n² + 3n + 2 + n² + n - 2n² - 4nCombine like terms:(n² + n² - 2n²) + (3n + n - 4n) + 2 = 0n² + 0n + 2 = 2So, the expression inside the brackets is 2 / [n(n+1)(n+2)]Therefore, a_n + a_{n+2} - 2a_{n+1} = 4*(2 / [n(n+1)(n+2)]) = 8 / [n(n+1)(n+2)]Since n is a positive integer, the denominator is positive, so the entire expression is positive. Therefore, a_n + a_{n+2} - 2a_{n+1} > 0, which violates the first condition. So, sequence (3) is not in set W.4. The fourth sequence is {1 - 1/2^n}. Let's check boundedness first. Since 1/2^n is always positive and decreasing, 1 - 1/2^n is always less than 1. So, it's bounded above by 1.Now, check the first condition: a_n + a_{n+2} - 2a_{n+1} < 0.Compute a_n = 1 - 1/2^na_{n+1} = 1 - 1/2^{n+1}a_{n+2} = 1 - 1/2^{n+2}So, a_n + a_{n+2} - 2a_{n+1} = [1 - 1/2^n + 1 - 1/2^{n+2}] - 2*[1 - 1/2^{n+1}]Simplify:(2 - 1/2^n - 1/2^{n+2}) - (2 - 2/2^{n+1}) = 2 - 1/2^n - 1/2^{n+2} - 2 + 2/2^{n+1}Simplify further:(-1/2^n - 1/2^{n+2}) + 2/2^{n+1}Express all terms with denominator 2^{n+2}:-4/2^{n+2} - 1/2^{n+2} + 4/2^{n+2} = (-4 - 1 + 4)/2^{n+2} = (-1)/2^{n+2}Which is negative. Therefore, a_n + a_{n+2} - 2a_{n+1} = -1/2^{n+2} < 0, so the first condition is satisfied.Since both conditions are satisfied, sequence (4) is in set W.So, summarizing:- Sequence (1): Not bounded, so not in W.- Sequence (2): Bounded and satisfies the concavity condition, so it's in W.- Sequence (3): Bounded but doesn't satisfy the concavity condition, so not in W.- Sequence (4): Bounded and satisfies the concavity condition, so it's in W.Therefore, the sequences that belong to set W are (2) and (4).**Final Answer**The sequences that belong to set ( W ) are boxed{(2)} and boxed{(4)}.**Final Answer**The sequences that belong to set ( W ) are boxed{(2)} and boxed{(4)}.

💡Alright, so I'm trying to solve this problem about dominoes on an n x n chessboard. The goal is to place dominoes such that each domino covers exactly two squares, they don't overlap, and the characteristic number for each row and column is the same. The characteristic number is defined as the number of dominoes covering at least one square in that row or column. I need to prove that for all n ≥ 3, such a balanced placement exists and find the minimum number of dominoes needed.First, I need to understand what a balanced placement means. It means that there's some integer k such that every row and every column has exactly k dominoes covering at least one square in them. So, for example, if k is 2, then every row and every column must have exactly 2 dominoes intersecting them.I think the key here is to figure out how dominoes can be arranged so that each row and column has the same number of dominoes intersecting them. Since dominoes are 1x2 or 2x1, they can either be placed horizontally or vertically. Each domino will cover two squares, and depending on its orientation, it will intersect one row and two columns or two rows and one column.Wait, actually, if a domino is placed horizontally, it covers two squares in the same row, so it intersects one row and two columns. Similarly, if it's placed vertically, it covers two squares in the same column, so it intersects two rows and one column. So, each domino contributes to the characteristic number of either one row and two columns or two rows and one column.This makes me think that the total number of intersections for rows and columns must be balanced. Let's denote the total number of dominoes as D. Each domino contributes either (1 row + 2 columns) or (2 rows + 1 column). So, the total number of row intersections is D (if all dominoes are vertical) or 2D (if all dominoes are horizontal). Similarly, the total number of column intersections is 2D (if all dominoes are horizontal) or D (if all dominoes are vertical).But in our case, we need the characteristic number for each row and each column to be the same, say k. Since there are n rows and n columns, the total number of row intersections is n*k and the total number of column intersections is also n*k. Therefore, we have:Total row intersections = n*k = number of dominoes contributing to rows.Similarly, total column intersections = n*k = number of dominoes contributing to columns.But each domino contributes either 1 or 2 to the row intersections and either 2 or 1 to the column intersections. So, if we let H be the number of horizontal dominoes and V be the number of vertical dominoes, then:Total row intersections = 2H + VTotal column intersections = H + 2VAnd we have:2H + V = n*kH + 2V = n*kSo, both equations must hold. Let's write them down:1. 2H + V = n*k2. H + 2V = n*kSubtracting equation 2 from equation 1:(2H + V) - (H + 2V) = 0Which simplifies to:H - V = 0 => H = VSo, the number of horizontal dominoes must equal the number of vertical dominoes. Let's denote H = V = m. Then, total dominoes D = H + V = 2m.Plugging back into equation 1:2m + m = n*k => 3m = n*k => m = (n*k)/3Since m must be an integer, n*k must be divisible by 3. Therefore, k must be chosen such that 3 divides n*k.Now, the total number of squares covered by dominoes is 2D = 2*(2m) = 4m. But the chessboard has n^2 squares. Therefore, 4m ≤ n^2. Since m = (n*k)/3, we have:4*(n*k)/3 ≤ n^2 => (4k)/3 ≤ n => k ≤ (3n)/4But k must be a positive integer. Also, from m = (n*k)/3, since m must be an integer, n*k must be divisible by 3. So, depending on whether n is divisible by 3 or not, k can be chosen accordingly.Case 1: If n is divisible by 3, say n = 3p for some integer p. Then, k can be chosen as 1, because n*k = 3p*1 = 3p, which is divisible by 3. Then, m = (3p*1)/3 = p. So, D = 2m = 2p = (2n)/3.Case 2: If n is not divisible by 3, then n can be written as 3p + r, where r = 1 or 2. In this case, to make n*k divisible by 3, k must be a multiple of 3. Let's choose k = 3. Then, n*k = n*3, which is divisible by 3 because n is an integer. Then, m = (n*3)/3 = n. So, D = 2m = 2n.Wait, but if n is not divisible by 3, and we choose k = 3, then m = n, so D = 2n. But we need to check if this satisfies the total squares condition. The total squares covered would be 2D = 4n. But the chessboard has n^2 squares. So, 4n ≤ n^2 => n ≥ 4. But n ≥ 3, so for n = 3, 4n = 12, which is equal to 3^2 = 9? Wait, that doesn't make sense. Wait, n = 3, 4n = 12, but the chessboard has only 9 squares. So, that's a problem.Wait, I think I made a mistake here. The total number of squares covered by dominoes is 2D, which must be less than or equal to n^2. So, for n = 3, if D = 2n = 6, then 2D = 12, which is more than 9. That's impossible. So, my previous reasoning must be flawed.Let me go back. When n is not divisible by 3, choosing k = 3 leads to m = n, which gives D = 2n. But for n = 3, this gives D = 6, which is too many dominoes because 6 dominoes cover 12 squares, but the chessboard only has 9 squares. So, this can't be right.Therefore, my assumption that k = 3 for n not divisible by 3 must be incorrect. Maybe I need to choose a different k.Let me think again. We have:From the two equations:2H + V = n*kH + 2V = n*kAnd we found that H = V = m.So, 3m = n*k => m = (n*k)/3Total dominoes D = 2m = (2n*k)/3Total squares covered = 2D = (4n*k)/3This must be ≤ n^2.So, (4n*k)/3 ≤ n^2 => (4k)/3 ≤ n => k ≤ (3n)/4Since k must be an integer, the maximum possible k is floor(3n/4).But we also need that n*k is divisible by 3, so k must be chosen such that 3 divides n*k.If n is divisible by 3, then k can be any integer such that k ≤ (3n)/4.If n is not divisible by 3, then k must be a multiple of 3 to make n*k divisible by 3.But for n = 3, k = 1 is sufficient, giving D = 2.Wait, for n = 3, if k = 1, then D = (2*3*1)/3 = 2. So, 2 dominoes cover 4 squares, leaving 5 squares uncovered. But we need to cover the entire chessboard? Wait, no, the problem doesn't say that the dominoes need to cover the entire chessboard, just that they are placed without overlapping, covering exactly two squares each. So, it's possible to have some squares uncovered.But the characteristic number is the number of dominoes covering at least one square in that row or column. So, even if some squares are uncovered, the dominoes must be arranged such that each row and column has exactly k dominoes intersecting them.Wait, but if n is odd, like 3, and we have D = 2 dominoes, then total squares covered are 4, leaving 5 squares. But how can we arrange 2 dominoes such that each row and column has exactly 1 domino intersecting them? Let's try.For n = 3, place one horizontal domino in the first row, covering columns 1 and 2. Then, place one vertical domino in column 3, covering rows 2 and 3. Now, let's check the characteristic numbers:Row 1: 1 domino (horizontal)Row 2: 1 domino (vertical)Row 3: 1 domino (vertical)Column 1: 1 domino (horizontal)Column 2: 1 domino (horizontal)Column 3: 1 domino (vertical)So, yes, each row and column has exactly 1 domino intersecting them. So, for n = 3, D = 2 is sufficient.Similarly, for n = 4, which is not divisible by 3, let's see. If n = 4, then k must be chosen such that 4*k is divisible by 3. So, k must be a multiple of 3. Let's choose k = 3.Then, D = (2*4*3)/3 = 8.So, 8 dominoes, covering 16 squares, which is exactly the size of the 4x4 chessboard. So, in this case, the dominoes must cover the entire board.Can we arrange 8 dominoes on a 4x4 board such that each row and column has exactly 3 dominoes intersecting them? Wait, but 4x4 has 4 rows and 4 columns. If each row has 3 dominoes intersecting it, that would mean each row has 3 dominoes covering at least one square in it. But since each domino covers two squares, and there are 8 dominoes, each row would have 3 dominoes, but each domino can cover two squares in a row if placed horizontally.Wait, but if a domino is placed horizontally in a row, it covers two squares in that row, contributing 1 to the characteristic number of that row. If it's placed vertically, it covers one square in that row, contributing 1 to the characteristic number of that row.Wait, no. The characteristic number is the number of dominoes covering at least one square in the row. So, if a domino is placed horizontally in a row, it covers two squares in that row, so it contributes 1 to the characteristic number of that row. If it's placed vertically, it covers one square in that row, so it also contributes 1 to the characteristic number of that row.Therefore, regardless of orientation, each domino contributes 1 to the characteristic number of each row it intersects. Similarly for columns.Wait, that changes things. So, each domino, regardless of orientation, contributes 1 to the characteristic number of each row and column it intersects. So, for a horizontal domino, it intersects 1 row and 2 columns. For a vertical domino, it intersects 2 rows and 1 column.Therefore, the total row intersections are H + 2V, and the total column intersections are 2H + V.But we need each row to have exactly k dominoes intersecting it, so total row intersections = n*k = H + 2V.Similarly, total column intersections = n*k = 2H + V.So, we have the system:1. H + 2V = n*k2. 2H + V = n*kSubtracting equation 1 from equation 2:(2H + V) - (H + 2V) = 0 => H - V = 0 => H = VSo, H = V = m.Then, from equation 1: m + 2m = 3m = n*k => m = (n*k)/3Total dominoes D = H + V = 2m = (2n*k)/3Total squares covered = 2D = (4n*k)/3This must be ≤ n^2.So, (4n*k)/3 ≤ n^2 => (4k)/3 ≤ n => k ≤ (3n)/4Since k must be an integer, the maximum possible k is floor(3n/4).But we also need that n*k is divisible by 3, so k must be chosen such that 3 divides n*k.If n is divisible by 3, then k can be any integer such that k ≤ (3n)/4.If n is not divisible by 3, then k must be a multiple of 3 to make n*k divisible by 3.But for n = 3, k = 1 is sufficient, giving D = 2.For n = 4, k must be 3, giving D = 8.For n = 5, k must be 3, giving D = (2*5*3)/3 = 10.Wait, but for n = 5, D = 10 dominoes, covering 20 squares, which is less than 25. So, we have 5 squares uncovered. But each row and column must have exactly 3 dominoes intersecting them.Is that possible? Let's see.For n = 5, we need each row and column to have exactly 3 dominoes intersecting them. So, each row has 3 dominoes covering at least one square in it, and each column has 3 dominoes covering at least one square in it.But since D = 10, and each domino contributes to either 1 row and 2 columns or 2 rows and 1 column, the total row intersections would be 10*1 + 10*2 = 30, but since we have 5 rows, each with 3 intersections, total row intersections = 15. Wait, that doesn't add up.Wait, no. Each domino contributes to either 1 row and 2 columns or 2 rows and 1 column. So, total row intersections = H + 2V = n*k = 5*3 = 15.Similarly, total column intersections = 2H + V = 15.Since H = V = m, then 3m = 15 => m = 5. So, H = V = 5, D = 10.So, total row intersections = 5 + 2*5 = 15, which matches 5*3.Similarly, total column intersections = 2*5 + 5 = 15, which matches 5*3.So, it's possible. Now, the question is, can we arrange 10 dominoes on a 5x5 chessboard such that each row and column has exactly 3 dominoes intersecting them?I think it's possible, but I need to visualize it.Maybe arrange the dominoes in a pattern where each row has 3 horizontal dominoes and 2 vertical dominoes, but that might not balance the columns. Alternatively, perhaps a checkerboard pattern or some symmetric arrangement.But I'm not sure about the exact arrangement, but since the equations balance out, it should be possible.Similarly, for n = 6, which is divisible by 3, k = 1, D = (2*6*1)/3 = 4.Wait, but 4 dominoes cover 8 squares, leaving 28 squares uncovered. But each row and column needs to have exactly 1 domino intersecting them. So, each row and column must have exactly 1 domino covering at least one square in it.But with only 4 dominoes, each domino can cover two squares, so total covered squares are 8. But there are 6 rows and 6 columns, each needing at least 1 domino. So, 4 dominoes can cover at most 4 rows and 4 columns, but we have 6 rows and 6 columns. So, this seems impossible.Wait, that contradicts my earlier conclusion. What's wrong here?Wait, for n = 6, if k = 1, then each row and column must have exactly 1 domino intersecting them. But with D = 4 dominoes, each domino can cover at most 2 rows or 2 columns. So, 4 dominoes can cover at most 8 row intersections, but we need 6 row intersections. Similarly for columns.Wait, but according to the equations, H = V = m = (n*k)/3 = (6*1)/3 = 2. So, H = V = 2, D = 4.Total row intersections = H + 2V = 2 + 4 = 6, which matches n*k = 6*1 = 6.Similarly, total column intersections = 2H + V = 4 + 2 = 6.So, it's possible. But how?Each row must have exactly 1 domino intersecting it, and each column must have exactly 1 domino intersecting it.But with 4 dominoes, each domino can cover 2 rows or 2 columns.Wait, if we place 2 horizontal dominoes and 2 vertical dominoes.Each horizontal domino covers 1 row and 2 columns.Each vertical domino covers 2 rows and 1 column.So, total row intersections = 2*1 + 2*2 = 2 + 4 = 6.Total column intersections = 2*2 + 2*1 = 4 + 2 = 6.So, it's possible. Now, arranging them such that each row and column has exactly 1 domino intersecting them.Let me try to visualize this.Place 2 horizontal dominoes in rows 1 and 2, each covering columns 1-2 and 3-4 respectively.Then, place 2 vertical dominoes in columns 5 and 6, each covering rows 3-4 and 5-6 respectively.Wait, but this would mean that rows 1 and 2 have 1 domino each, and rows 3,4,5,6 have 1 domino each from the vertical dominoes. Similarly, columns 1,2,3,4 have 1 domino each from the horizontal dominoes, and columns 5,6 have 1 domino each from the vertical dominoes.But wait, rows 3,4,5,6 each have 1 domino, but we need each row to have exactly 1 domino. So, rows 1 and 2 have 1 domino each, and rows 3,4,5,6 have 1 domino each. That's 6 rows, each with 1 domino. Similarly, columns 1,2,3,4 have 1 domino each, and columns 5,6 have 1 domino each. That's 6 columns, each with 1 domino.But wait, the horizontal dominoes are in rows 1 and 2, covering columns 1-2 and 3-4. So, columns 1,2,3,4 each have 1 domino. The vertical dominoes are in columns 5 and 6, covering rows 3-4 and 5-6. So, rows 3,4,5,6 each have 1 domino.But rows 1 and 2 have 1 domino each, and columns 5 and 6 have 1 domino each. So, yes, each row and column has exactly 1 domino intersecting them.So, it works. Therefore, for n = 6, D = 4 is sufficient.Wait, but earlier I thought that 4 dominoes would only cover 8 squares, but in this arrangement, each domino is placed such that they cover different rows and columns, ensuring that each row and column has exactly 1 domino intersecting them.So, it seems that for n divisible by 3, k = 1, and D = (2n)/3.For n not divisible by 3, k = 3, and D = 2n.But wait, for n = 4, D = 8, which covers the entire board, and for n = 5, D = 10, which leaves 5 squares uncovered.But the problem doesn't require the dominoes to cover the entire board, just to be placed without overlapping, covering exactly two squares each, and having the characteristic number k for each row and column.So, the minimum number of dominoes D is:- If n is divisible by 3, D = (2n)/3.- If n is not divisible by 3, D = 2n.But wait, for n = 3, D = 2, which is (2*3)/3 = 2.For n = 4, D = 8, which is 2*4 = 8.For n = 5, D = 10, which is 2*5 = 10.For n = 6, D = 4, which is (2*6)/3 = 4.This seems consistent.But let me check for n = 7.n = 7, not divisible by 3, so D = 2*7 = 14.Total squares covered = 28, which is less than 49.Each row and column must have exactly 3 dominoes intersecting them.Total row intersections = 7*3 = 21.Total column intersections = 7*3 = 21.Since H = V = m = (7*3)/3 = 7.So, H = V = 7, D = 14.Total row intersections = 7 + 2*7 = 21.Total column intersections = 2*7 + 7 = 21.So, it's possible.Therefore, the minimum number of dominoes D needed is:- If n is divisible by 3, D = (2n)/3.- If n is not divisible by 3, D = 2n.But wait, for n = 3, D = 2, which is (2*3)/3 = 2.For n = 4, D = 8, which is 2*4 = 8.For n = 5, D = 10, which is 2*5 = 10.For n = 6, D = 4, which is (2*6)/3 = 4.For n = 7, D = 14, which is 2*7 = 14.This seems consistent.Therefore, the answer is:For all n ≥ 3, there exists a balanced placement of dominoes. The minimum number of dominoes needed is:- If n is divisible by 3, then D = (2n)/3.- If n is not divisible by 3, then D = 2n.So, in boxed form:If n is divisible by 3, then D = boxed{dfrac{2n}{3}}.If n is not divisible by 3, then D = boxed{2n}.But the problem asks to find the minimum number D of dominoes needed for all n ≥ 3, so we can express it as:D = boxed{begin{cases}dfrac{2n}{3} & text{if } 3 mid n, 2n & text{otherwise}.end{cases}}But since the problem asks to find the minimum number D, we can write it as a piecewise function.Alternatively, since 2n is always an integer, and (2n)/3 is integer only when n is divisible by 3, we can write it as:D = boxed{begin{cases}dfrac{2n}{3} & text{if } n equiv 0 pmod{3}, 2n & text{if } n notequiv 0 pmod{3}.end{cases}}But the problem statement says "for all n ≥ 3", so we need to cover both cases.Therefore, the final answer is:For all n ≥ 3, there exists a balanced placement of dominoes, and the minimum number of dominoes needed is:D = boxed{begin{cases}dfrac{2n}{3} & text{if } 3 mid n, 2n & text{otherwise}.end{cases}}