answer:We are tasked with finding the maximum possible number of forced terms in a regular sequence with 1000 terms. We'll analyze the structure of forced terms and use properties of Farey sequences to arrive at the solution. 1. **Definition and Initial Setup**: - A sequence a_1, a_2, ldots, a_n is called regular if there exists an x in mathbb{R} such that a_k = lfloor kx rfloor for 1 leq k leq n. - A term a_k is forced if the sequence a_1, a_2, ldots, a_{k-1}, b is regular only if b = a_k. 2. **Shifting and Initial Condition**: - Without loss of generality (WLOG), we can set a_1 = 0 by shifting the sequence. - The initial interval for x is 0 leq x < 1. 3. **Construction Using Farey Sequences**: - We use the Farey fractions technique to analyze forced terms. Start by considering the interval for x: [ frac{0}{1} leq x < frac{1}{1}. ] 4. **Iterative Procedure**: - For each k = 2, 3, ldots, 1000, check whether there is a fraction frac{m}{k} in the current interval A leq x < B. - If there is no fraction of the form frac{m}{k} in the interval, a_k is forced. - If there is, update the interval by choosing the value of a_k accordingly. 5. **Using Farey Sequences**: - The Farey sequence theory tells us that the next fraction in the interval frac{a}{b} leq x < frac{c}{d} is frac{a+c}{b+d}, and it is the only fraction in that interval at the next stage. - Continue updating one endpoint of the interval with the mediant frac{a+c}{b+d} until one of the denominators exceeds 1000. 6. **Number of Non-Forced Terms**: - To find the maximum number of forced terms, minimize the number of denominators appearing in the sequence. - The optimal configuration occurs by always replacing the smaller denominator. - This generates a sequence where the non-forced terms correspond to Fibonacci numbers since each step introduces a new term in this minimal configuration. 7. **Counting Forced Terms**: - There are 15 Fibonacci numbers less than 1000: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987. - Hence, there are 1000 - 15 = 985 forced terms. # Conclusion: The maximum possible number of forced terms in a regular sequence with 1000 terms is: [ boxed{985} ]

answer:① By the theorem of the property of line perpendicular to plane, it is correct; ② alpha and beta can also intersect, which is incorrect; ③ By the theorem for determining when two planes are perpendicular, it is correct; ④ By the Three Perpendiculars Theorem, it is correct. Therefore, the correct choice is boxed{text{C}}.

answer:First, let's calculate how many pages Arthur has already read from each book. For the 500-page book, Arthur has read 80% of it. To find out how many pages that is, we calculate 80% of 500: 0.80 * 500 = 400 pages For the 1000-page book, Arthur has read 1/5 of it. To find out how many pages that is, we calculate 1/5 of 1000: (1/5) * 1000 = 200 pages Now, let's add the pages he has read from both books: 400 pages (from the 500-page book) + 200 pages (from the 1000-page book) = 600 pages Arthur still needs to read 200 more pages to meet his goal. So, let's add the pages he has already read to the pages he still needs to read: 600 pages (already read) + 200 pages (still needs to read) = 800 pages Therefore, Arthur needs to finish reading a total of boxed{800} pages over the summer.

answer:# Step-by-Step Solution Part 1: Finding the values of a and b Given the function f(x) = |2x + 1| + |x + a|, and the solution set of f(x) leq 3 is [b, 1]. 1. **Finding a:** Since f(1) leq 3 must hold, we substitute x = 1 into the function, which simplifies to 3 + |1 + a| = 3. Solving for a gives: [ |1 + a| = 0 implies a = -1 ] Therefore, a = -1. 2. **Simplifying f(x) with a = -1:** The function becomes f(x) = |2x + 1| + |x - 1|. 3. **Case Analysis:** - **Case 1: x leq -frac{1}{2}** [ -2x - 1 - x + 1 leq 3 implies -3x leq 3 implies x geq -1 ] This gives the interval -1 leq x leq -frac{1}{2}. - **Case 2: -frac{1}{2} < x < 1** [ 2x + 1 - x + 1 leq 3 implies x + 2 leq 3 implies x leq 1 ] This gives the interval -frac{1}{2} < x < 1. - **Case 3: x geq 1** [ 2x + 1 + x - 1 leq 3 implies 3x leq 3 implies x leq 1 ] This gives the point x = 1. Combining these intervals, the solution set is [-1, 1]. Therefore, b = -1. The values of a and b are boxed{a = -1} and boxed{b = -1}. Part 2: Proving the inequality 4m^{2} + n^{2} geq 4 Given frac{1}{{2m}} + frac{2}{n} + 2a = 0 and from part (1), we know a = -1, so frac{1}{{2m}} + frac{2}{n} = 2. 1. **Applying AM-GM Inequality:** [ 2 = frac{1}{2m} + frac{2}{n} geq 2sqrt{frac{1}{2m} cdot frac{2}{n}} = 2sqrt{frac{1}{mn}} ] Simplifying gives: [ 1 geq sqrt{frac{1}{mn}} implies mn geq 1 ] 2. **Using the result mn geq 1:** [ 4m^{2} + n^{2} geq 2sqrt{4m^{2} cdot n^{2}} = 4sqrt{m^{2}n^{2}} geq 4sqrt{1} = 4 ] Therefore, we have shown that 4m^{2} + n^{2} geq 4. The proof is complete, and the inequality is boxed{4m^{2} + n^{2} geq 4}.