answer:Let's denote the length of the first platform as ( L ) meters. When the train crosses a platform, it has to cover a distance equal to its own length plus the length of the platform. The speed of the train remains constant, so we can set up a ratio using the time it takes to cross each platform. For the first platform: Time taken to cross = 15 sec Distance covered = Length of train + Length of first platform = ( 100 + L ) meters For the second platform: Time taken to cross = 20 sec Distance covered = Length of train + Length of second platform = ( 100 + 300 ) meters = 400 meters Since the speed of the train is constant, we can say that the distances covered are proportional to the times taken to cover them. Therefore, we can set up the following proportion: [ frac{100 + L}{15} = frac{400}{20} ] Now we can solve for ( L ): [ frac{100 + L}{15} = frac{400}{20} implies 100 + L = frac{400 times 15}{20} ] [ 100 + L = frac{400 times 15}{20} = 400 times frac{3}{4} = 400 times 0.75 = 300 ] Now, subtract 100 from both sides to find ( L ): [ L = 300 - 100 = 200 ] So, the length of the first platform is boxed{200} meters.

answer:**Answer**: For a number to be divisible by 11, the difference between the sum of the digits in odd positions and the sum of the digits in even positions must also be a multiple of 11. The sum of these 9 digits is 45, meaning the number of digits in odd and even positions must be either 4 or 5. Assuming the sum of the digits in odd positions is a and the sum of the digits in even positions is b, we have a + b = 45. It is known that a and b must be one odd and one even, and the minimum possible value for a or b is 1 + 2 + 3 + 4 = 10. Therefore, the only possible solution for a and b is 28 and 17. To make the nine-digit number as large as possible, the digits 9, 8, 7, 6, 5 should be placed as early as possible, and the digits 4, 3, 2, 1 should be placed later. In other words, the first three odd positions should ideally be filled with 9, 7, 5, and the first two even positions should ideally be filled with 8, 6. Next, we check if it is possible to combine them to get 28 and 17 respectively: 1. For the odd positions: 28 - (9 + 7 + 5) = 7, which can only be split into 3 + 4; 2. For the even positions: 17 - (8 + 6) = 3, which can only be split into 1 + 2; Therefore, the first five digits of the nine-digit number should be 98765, and to make the last four digits as large as possible, they can only be 2413. In conclusion, the largest nine-digit number divisible by 11 is boxed{987652413} .

answer:Given the curve y=x+3ln x, We find the derivative: y'=frac{3}{x}+1, Thus, the slope of the tangent line to the curve y=3ln x+x at the point (1,1) is: 4. The equation of the tangent line to the curve y=3ln x+x at the point (1,1) is: y-1=4(x-1), or boxed{4x-y-3=0}. To find this, we use the geometric meaning of a function's derivative. We compute the derivative and then substitute the x-coordinate of the given point to find the slope of the tangent line. Finally, we use the point-slope form of a linear equation to find the equation of the tangent line. This problem tests understanding of basic operations with function derivatives, the geometric meaning of derivatives, and methods for finding tangent line equations, as well as computational skills.

answer:Let's proceed step by step to solve the given problem. Part (1): Find the general term formula for the series left{a_nright} and left{b_nright} We are given two sequences satisfying the recurrence relations: [ begin{cases} a_1 = 3, & b_1 = 1 a_{n+1} = a_n + b_n + sqrt{a_n^2 - a_n b_n + b_n^2}, b_{n+1} = a_n + b_n - sqrt{a_n^2 - a_n b_n + b_n^2}. end{cases} ] To find the general term for a_n and b_n, observe the following pattern: 1. Add the expressions for a_{n+1} and b_{n+1}: [ a_{n+1} + b_{n+1} = left(a_n + b_n + sqrt{a_n^2 - a_n b_n + b_n^2}right) + left(a_n + b_n - sqrt{a_n^2 - a_n b_n + b_n^2}right). ] Simplifying this, we get: [ a_{n+1} + b_{n+1} = 2(a_n + b_n). ] So, the sequence a_n + b_n follows a recurrence relation: [ a_{n+1} + b_{n+1} = 2(a_n + b_n). ] Let c_n = a_n + b_n. Hence, we have: [ c_{n+1} = 2c_n. ] Since c_1 = a_1 + b_1 = 3 + 1 = 4, we get: [ c_n = 4 cdot 2^{n-1}. ] 2. Consider the difference a_{n+1} - b_{n+1}: [ a_{n+1} - b_{n+1} = left(a_n + b_n + sqrt{a_n^2 - a_n b_n + b_n^2}right) - left(a_n + b_n - sqrt{a_n^2 - a_n b_n + b_n^2}right). ] Simplifying this, we get: [ a_{n+1} - b_{n+1} = 2sqrt{a_n^2 - a_n b_n + b_n^2}. ] Recall that: [ a_1 - b_1 = 3 - 1 = 2. ] Assuming this pattern holds, we then have: [ a_{n+1} - b_{n+1} = 2 cdot 2^{n-1} = 2^n. ] So, we can represent a_n and b_n in terms of c_n and their initial differences: [ begin{cases} a_n = frac{c_n + d_n}{2}, b_n = frac{c_n - d_n}{2}. end{cases} ] With the relations: [ begin{cases} c_n = 4 cdot 2^{n-1}, d_n = 2^n. end{cases} ] Thus: [ begin{cases} a_n = frac{4 cdot 2^{n-1} + 2^n}{2} = frac{2^n (2 + 1)}{2} = 3 cdot 2^{n-1}, b_n = frac{4 cdot 2^{n-1} - 2^n}{2} = frac{2^n (2 - 1)}{2} = 2^{n-1}. end{cases} ] To conclude part (1): The general terms for the sequences are: [ begin{cases} a_n = 3 cdot 2^{n-1}, b_n = 2^{n-1}. end{cases} ] Part (2): Find the smallest n such that: [ sum_{k=1}^{n} (S_k + T_k) > 2017, ] where S_n = sum_{i=1}^n left[a_iright] and T_n = sum_{i=1}^n left[b_iright]. Given: [ a_n = 3 cdot 2^{n-1} quad text{and} quad b_n = 2^{n-1}, ] Since a_n and b_n are always integers: [ left[a_iright] = a_i = 3 cdot 2^{i-1}, quad left[b_iright] = b_i = 2^{i-1}. ] Thus: [ S_n = sum_{i=1}^n 3 cdot 2^{i-1}, quad T_n = sum_{i=1}^n 2^{i-1}. ] Recall the sum of a geometric series sum_{i=0}^{n-1} 2^i = 2^n - 1: [ S_n = 3 left(sum_{i=0}^{n-1} 2^i right) = 3 left( 2^n - 1 right) = 3 cdot 2^n - 3. ] [ T_n = sum_{i=0}^{n-1} 2^i = 2^n - 1. ] We need to find n such that: [ sum_{k=1}^n (S_k + T_k) > 2017. ] Analyzing: [ S_k + T_k = (3 cdot 2^k - 3) + (2^k - 1) = 4 cdot 2^k - 4. ] [ sum_{k=1}^{n} (4 cdot 2^k - 4) = 4 sum_{k=1}^{n} 2^k - 4n. ] Using the series sum of sum_{k=1}^{n} 2^k = 2(2^n - 1): [ 4 sum_{k=1}^{n} 2^k - 4n = 4 left(2^{n+1} - 2right) - 4n = 2^{n+3} - 8 - 4n. ] We need: [ 2^{n+3} - 4n - 8 > 2017, ] [ 2^{n+3} - 4n > 2025. ] By checking viable values of n: For n = 8, [ 2^{8+3} - 4 cdot 8 = 2^{11} - 32 = 2048 - 32 = 2016 quad < quad 2025, ] So, the next integer n = 9: [ 2^{9+3} - 4 cdot 9 = 2^{12} - 36 = 4096 - 36 = 4060 quad > quad 2025. ] Therefore, the smallest n is: [ boxed{9}. ]