answer:Given two sequences of letters (A) and (B), each containing 100 characters, we need to prove that it is possible to transform the first sequence into the second sequence using no more than 100 operations. Each operation allows us to either insert or delete one or more consecutive identical letters at any position within the sequence. Let's first consider a simpler case: sequences consisting of just two letters. This will help build the foundation for our solution. Step-by-Step Transformation for Simpler Case: 1. Suppose we have two sequences: - Sequence 1: "AA" - Sequence 2: "BB" 2. Here is one strategy: - Remove both 'A's from sequence 1. - Insert two 'B's to form sequence 2. 3. This transformation takes a total of 2 operations: - 1 operation to remove all 'A's. - 1 operation to insert all 'B's. If we have sequences where some letters are common but they are in different positions, the steps would similarly be: 1. Identify common letters between sequences and keep them while removing others. 2. Remove unmatched letters. 3. Insert missing letters to match the target sequence. General Case for Sequences with 100 Letters: For more complex sequences, the idea remains similar. We will work in blocks of consecutive letters to minimize the number of operations required. Let us describe this process step by step: 1. **Breaking Down Sequences into Blocks**: - Break down each sequence into blocks of consecutive identical letters. - For instance, sequence 1: "AAABBB" would be broken into [AAA, BBB]. 2. **Matching Corresponding Blocks**: - Start from the beginning of both sequences. - Compare blocks between sequences and transform them as needed. 3. **Detailed Block Operations**: - Suppose you have sequence 1: (text{"AAABB"}) and sequence 2: (text{"AABB"}). - First block comparison: - Sequence 1 has 'AAA' and Sequence 2 has 'AA'. - An operation to remove 1 'A' makes the blocks identical: (text{"AA"}). - Second block comparison: - Sequence 1 now has 'BB' and Sequence 2 also has 'BB'. - No operation needed, blocks match. Thus, a total of 1 operation is needed. 4. **Calculating the Total Operations**: - Each block transformation takes at most 2 operations: One for removals and one for additions. - If each sequence can be divided into at most 50 blocks (as each block has at least two letters), the worst-case scenario would involve transforming each block individually. - Therefore, transforming 50 blocks would require at most (2 times 50 = 100) operations. Conclusion: In the general scenario, after breaking both sequences into up to 50 blocks, we perform up to 1 operation to remove unmatched letters and 1 operation to add new matching letters for each block. Hence, transforming the first sequence into the second one requires **no more than 100 operations**. [ boxed{text{The first sequence can be transformed into the second sequence in no more than 100 operations.}} ]

answer:To find out when A and B meet, we need to calculate the time it takes for them to cover the distance between them. The combined speed at which A and B are approaching each other is the sum of their individual speeds, which is 6 kmph + 4 kmph = 10 kmph. They are 50 km apart, so the time it takes for them to meet is the distance divided by their combined speed: Time = Distance / Speed Time = 50 km / 10 kmph Time = 5 hours Since they start walking towards each other at 6 pm, we add the 5 hours to this start time to find out when they meet: 6 pm + 5 hours = 11 pm So, A and B will meet at boxed{11} pm.

answer:First, we find the derivative of the function f(x). f'(x) = [x^2 - 2(a-1)x - 2a] cdot e^x Since f(x) is monotonically decreasing on [-1, 1], f'(x) leqslant 0, x in [-1, 1] This implies that x^2 - 2(a-1)x - 2a leqslant 0, x in [-1, 1] Let g(x) = x^2 - 2(a-1)x - 2a, Then we have the following system of inequalities: begin{cases} g(-1) leqslant 0 g(1) leqslant 0 end{cases} Solving this system, we get: begin{cases} 1 + 2(a-1) - 2a leqslant 0 1 - 2(a-1) - 2a leqslant 0 end{cases} Solving these inequalities, we find that: a geqslant frac{3}{4} Thus, the range of values for a is boxed{a geqslant frac{3}{4}}. This problem primarily tests the application of derivatives in determining the monotonicity of a function. By using equivalent transformations, the problem can be converted into a quadratic function problem. Pay attention to the flexible use of the relationship between numbers and shapes, which is a moderately difficult problem.

answer:1. From a_{n+1}= frac {a_n}{2}+ frac {1}{2^{n}}, we have 2^{n}a_{n+1}=2^{n-1}a_n+1. Since b_n=2^{n-1}a_n, it follows that b_{n+1}=b_n+1. Given that b_1=a_1=1, we conclude that the sequence {b_n} is an arithmetic sequence with the first term 1 and common difference 1. Its general term formula is b_n=n. 2. Since a_n= frac {n}{2^{n-1}}, the sum S_n can be written as: S_n= frac {1}{2^{0}}+ frac {2}{2^{1}}+ frac {3}{2^{2}}+…+ frac {n}{2^{n-1}} qquad (1) Multiplying both sides of (1) by frac{1}{2}, we obtain: frac{1}{2}S_n= frac {1}{2^{1}}+ frac {2}{2^{2}}+ frac {3}{2^{3}}+…+ frac {n}{2^{n}} qquad (2) Subtracting (2) from (1), we get: frac{1}{2}S_n = 1+ frac {1}{2^{1}}+ frac {1}{2^{2}}+…+ frac {1}{2^{n-1}}- frac {n}{2^{n}} The sum of the geometric series on the right-hand side is: 1+ frac {1}{2^{1}}+ frac {1}{2^{2}}+…+ frac {1}{2^{n-1}} = frac{1 - (frac{1}{2})^n}{1 - frac{1}{2}} = 2 - (frac{1}{2})^{n-1} Thus, we have: frac{1}{2}S_n = 2 - (frac{1}{2})^{n-1} - frac{n}{2^{n}} Multiplying both sides by 2, we obtain the formula for S_n: S_n = 4 - frac{2+n}{2^{n-1}} 3. Squaring d_n, we have: d_n^2 = 1 + frac{1}{b_n^2} + frac{1}{b_{n+1}^2} = frac{b_n^2b_{n+1}^2 + b_{n+1}^2 + b_n^2}{b_n^2b_{n+1}^2} Since b_n = n and b_{n+1} = n+1, we can simplify d_n as follows: d_n = frac{n(n+1)+1}{n(n+1)} = 1 + frac{1}{n(n+1)} = 1 + frac{1}{n} - frac{1}{n+1} Now, let's calculate D_{100}: D_{100} = (1 + frac{1}{1} - frac{1}{2}) + (1 + frac{1}{2} - frac{1}{3}) + (1 + frac{1}{3} - frac{1}{4}) + ... + (1 + frac{1}{100} - frac{1}{101}) Notice that most fractions cancel out, leaving us with: D_{100} = 101 - frac{1}{101} Therefore, the maximum integer value not exceeding D_{100} is boxed{100}.