answer:We are given the sequence ({a_n}), where the sum of the first (n) terms is (S_n), and (a_n = 2 sqrt{S_n} - 1). Part (1): Finding the general term 1. Given the relationship (a_n = 2 sqrt{S_n} - 1), where (S_n) is the sum of the first (n) terms of the sequence ({a_n}). 2. We know the sum (S_n = a_1 + a_2 + cdots + a_n). 3. Substituting (a_n) in the sum: [ S_n = a_1 + a_2 + cdots + a_n = left(2 sqrt{S_n} - 1right) + left(2 sqrt{S_{n-1}} - 1right) + cdots + left(2 sqrt{S_1} - 1right) ] 4. To ensure this identity holds true, let's assume (S_n = n^2). 5. Then, [ a_n = 2sqrt{S_n} - 1 = 2sqrt{n^2} - 1 = 2n - 1 ] 6. We have the general term of the sequence: [ a_n = 2n - 1 ] Conclusion for part (1): [ boxed{a_n = 2n - 1} ] Part (2): Proving (R_n < T_n) Given sequences: [ T_n = sum_{k=1}^n frac{2}{sqrt{a_k} + sqrt{a_{k+1}}} ] [ R_n = frac{a_1 a_2 cdots a_n}{(a_1+1)(a_2+1)cdots(a_n+1)} ] 1. Consider the expression for (R_n): [ R_n = prod_{k=1}^n frac{a_k}{a_k + 1} ] 2. Given (a_n = 2n - 1), we have: [ R_n = prod_{k=1}^n frac{2k - 1}{2k - 1 + 1} = prod_{k=1}^n frac{2k - 1}{2k} ] 3. Simplifying: [ R_n = prod_{k=1}^n frac{2k - 1}{2k} = prod_{k=1}^n frac{2k-1}{2k} ] 4. Now consider (T_n): [ T_n = sum_{k=1}^n frac{2}{sqrt{a_k} + sqrt{a_{k+1}}} ] 5. We know (a_n = 2n - 1), hence: [ T_n = sum_{k=1}^n frac{2}{sqrt{2k-1} + sqrt{2k+1}} ] 6. For each (k), note that the product: [ frac{a_k}{a_k + 1} < frac{2}{sqrt{a_k} + sqrt{a_{k+1}}} ] 7. Since this holds for each term, summing over (k = 1, 2, ldots, n) gives: [ sum_{k=1}^n frac{a_k}{a_k + 1} < sum_{k=1}^n frac{2}{sqrt{a_k} + sqrt{a_{k+1}}} ] 8. Hence: [ R_n < T_n ] Conclusion for part (2): [ blacksquare ]

answer:To solve for the intersection A cap B given the sets A={xleft|right.x lt -3 or x gt 1} and B={xleft|right.0 lt xleqslant 4}, we need to find the set of elements that are common to both A and B. - Set A includes all x such that x < -3 or x > 1. - Set B includes all x such that 0 < x leqslant 4. For x to be in both A and B, it must satisfy the conditions of both sets simultaneously. - The condition x < -3 from set A does not overlap with any condition in set B because all elements in B are greater than 0. - The condition x > 1 from set A overlaps with the condition 0 < x leqslant 4 from set B. Specifically, the overlap occurs for 1 < x leqslant 4. Therefore, the intersection A cap B is the set of all x such that 1 < x leqslant 4, which can be written as (1,4]. Hence, the correct answer is boxed{A}.

answer:1. **Initial Setup:** - Let's consider there are 15 boys and 15 girls. - Initially, they are paired randomly, and the maximum height difference in any pair is observed to be 10 cm. 2. **Goal:** - We need to prove that if the boys and girls are lined up in descending order of height and then paired, the maximum height difference in any pair will not exceed 10 cm. 3. **Assumption for Contradiction:** - Assume that in one of the pairs, say the (k)-th pair, the boy's height (B_k) differs from the girl's height (G_k) by more than 10 cm. - Without loss of generality, suppose (B_k > G_k) and the height difference (B_k - G_k > 10) cm. 4. **Height Order:** - Since both the boys and girls are sorted in descending order of height: [ B_1 geq B_2 geq cdots geq B_{15} ] [ G_1 geq G_2 geq cdots geq G_{15} ] 5. **Implication of the Assumption:** - If (B_k - G_k > 10), then for all pairs (j < k): [ B_j > B_k ] and [ G_j < G_k. ] 6. **Contradiction Derived:** - Since (G_j < G_k), and pairs are formed such that girls' heights are sorted in descending order, (G_1, G_2, ldots, G_{k-1}) are all larger than (G_k). - However, by our assumption, we know the maximum initial difference was only 10 cm. Therefore, if at least (G_{k-1} < G_k + 10) or otherwise all these previous (k-1) pairs would exceed the 10 cm difference which contradicts our assumption of maximum difference being 10 cm. 7. **Proof Conclusion:** - Hence, the assumption that (B_k - G_k > 10) cm leads to a contradiction. - Therefore, the maximum height difference in the rearranged ordered pairs cannot be more than 10 cm. [ boxed{10 text{ cm}} ]

answer:To find all two-digit positive integers x such that for all three-digit (base 10) positive integers underline{a}underline{b}underline{c}, if underline{a}underline{b}underline{c} is a multiple of x, then the three-digit (base 10) number underline{b}underline{c}underline{a} is also a multiple of x, we need to follow several steps. 1. **Represent the Numbers**: - A three-digit number underline{a}underline{b}underline{c} can be represented as: [ 100a + 10b + c ] - The number underline{b}underline{c}underline{a} can be represented as: [ 100b + 10c + a ] 2. **Condition for Multiples**: - We need to find x such that if x divides 100a + 10b + c, then x must also divide 100b + 10c + a. 3. **Establish the Relationship**: - Let: [ N_1 = 100a + 10b + c ] [ N_2 = 100b + 10c + a ] - We need: [ x mid N_1 implies x mid N_2 ] - Subtracting N_1 from N_2 we get: [ N_2 - N_1 = (100b + 10c + a) - (100a + 10b + c) = 99b - 99a + 9c - 9c ] Simplifying this we get: [ N_2 - N_1 = 99(b - a) ] 4. **Divisibility Condition**: - Therefore, for x to satisfy the condition, x must divide 99(b - a). This must hold for any a, b in {0, 1, 2, ldots, 9}. - Hence, x must divide 99. 5. **Two-Digit Factors of 99**: - The factors of 99 are 1, 3, 9, 11, 33, 99. - Among these, the two-digit factors of 99 are 11, 33, and 99. 6. **Sum of the Two-Digit Factors**: - However, 99 is not a two-digit number; it is three digits. The two-digit factors of 99 are 11 and 33. - Therefore, adding these together gives us [ 11 + 33 = 44 ] Conclusion: The sum of all two-digit positive integers x that satisfy the given conditions is: [ boxed{44} ]