answer:① If a parallel b and b parallel c, then a parallel c is a true proposition, because two lines parallel to the same plane are parallel to each other; ② If a perp b and b perp c, then a perp c is a false proposition, because two lines perpendicular to the same line can be parallel, perpendicular, or skew to each other; ③ If a parallel gamma and b parallel gamma, then a parallel b is a false proposition, because two lines parallel to the same plane can be parallel, intersect, or be skew to each other; ④ If a perp gamma and b perp gamma, then a parallel b is correct, because two lines perpendicular to the same plane are parallel to each other. Therefore, the answer is: boxed{①④}.

answer:Solution: (1) According to the problem, for the geometric sequence {a_n}, we have S_n=2a_n−2. When n=1, we have S_1=2a_1−2=a_1, solving this gives a_1=2. When ngeq2, a_n=S_n−S_{n−1}=(2a_n−2)−(2a_{n−1}−2), transforming this gives a_n=2a_{n−1}, thus, for the geometric sequence {a_n}, we have a_1=2, and the common ratio q=2, hence, the general formula for the sequence {a_n} is a_n=2×2^{n−1}=2^n, For the arithmetic sequence {b_n}, since b_3=a_2=4, and b_2+b_6=2b_4=10, this implies b_4=5, thus, its common difference d=b_4−b_3=1, hence, its general formula is b_n=b_3+(n−3)×d=n+1. (2) From the conclusion of (1): a_n=2^n, b_n=n+1, a_n(2b_n−3)=(2n−1)⋅2^n, thus, T_n=1×2^1+3×2^2+5×2^3+⋯+(2n-1)×2^n, (①) and 2T_n=1×2^2+3×2^3+5×2^4+⋯+(2n-1)×2^{n+1}, (②) Subtracting (②) from (①) gives: −T_n=2+2(2^2+2^3+…+2^n)−(2n−1)×2^{n+1}, =2+2× dfrac{2^2(1-2^{n-1})}{1-2}-(2n-1)×2^{n+1} =-6+(3-2n)·2^{n+1}, Therefore, T_n=(2n−3)⋅2^{n+1}+6. Thus, the final answers are: (1) The general formula for {a_n} is a_n=2^n, and for {b_n} is b_n=n+1. (2) The sum of the first n terms T_n of the sequence {a_n(2b_n-3)} is T_n=(2n−3)⋅2^{n+1}+6. Therefore, the final answers in the required format are: (1) boxed{a_n=2^n, b_n=n+1} (2) boxed{T_n=(2n−3)⋅2^{n+1}+6}

answer:# Problem: 8.9 (1) Two people, A and B, play a game: A writes two rows of 10 numbers each, making them satisfy the following rule: if b is below a and d is below c, then a + d = b + c. After knowing this rule, B wants to determine all the numbers written by asking questions like: "What is the number in the third position of the first row?" or "What is the number in the ninth position of the second row?" etc. How many such questions does B need to ask to be sure to know all the numbers written? (2) In an m times n grid filled with numbers, make any two rows and any two columns form a rectangle whose sum of numbers on opposing corners are equal. After erasing some of the numbers, show that at least m + n - 1 numbers remain to recover the erased numbers. 1. **Part (1):** Let's analyze how many questions B needs to ask in order to determine all the numbers based on the given rule, ( a + d = b + c ). - Suppose B knows all 10 numbers in the first row and 1 number in the second row. - Using the rule, we can determine the remaining 9 numbers in the second row. - If B asks about fewer than 10 numbers in total, there could be multiple possible sets of numbers for the unknown spots which can still satisfy the rule. Thus, B would need to ask at least 11 questions to determine all the numbers with certainty, because: - If only 10 numbers are known, there will be ambiguity about the numbers in the columns. - Knowing 11 numbers guarantees that all other numbers can be determined without any ambiguity following the rule, ( a + d = b + c ). **Conclusion:** B needs to ask at least ( boxed{11} ) questions. 2. **Part (2):** To show that at least ( m + n - 1 ) numbers must remain to recover the erased numbers in an (m times n) grid which satisfies the stated sum property of opposite corners, we can use mathematical induction and the properties of linear independence. **Start with the Base Case:** - For a 2x2 grid, the rule implies that ( a + d = b + c ). - This means any one number can be determined if the other three are known. **Inductive Step:** - Assume for an ( (m-1) times n ) or ( m times (n-1) ) grid, the minimal number of remaining numbers to recover the grid is ( (m-1) + n - 1 = m + n - 2 ) or ( m + (n-1) - 1 = m + n - 2 ), respectively. **To prove for ( m times n ) Grid:** - Suppose ( m + n - 1 ) remaining numbers are known. - Removing an entire column or row reduces the grid to ( (m-1) times n ) or ( m times (n-1) ). - Induction assumption ensures that either ( m times (n-1) ) or ( (m-1) times n ) can determine itself with ( m+n-2 ) numbers. So: - **Number of unknown numbers** (total minus known) must be recoverable with ( m+n-1 ). - This means ( m+n-1 ) ensure that there is always non-ambiguous information to recover all entries in the grid. **Conclusion:** At least ( m + n - 1 ) numbers must remain to restore the grid: [ boxed{m + n - 1} ]

answer:Since a_3a_4a_5=3^{pi}=a_4^3, we have a_4=3^{frac{pi}{3}}. log _{3}a_{1}+log _{3}a_{2}+cdots+log _{3}a_{7}=log _{3}(a_{1}a_{2}cdots a_{7})=log _{3}a_4^7=7log _{3}3^{frac{pi}{3}}=frac{7pi}{3}, Therefore, sin (log _{3}a_{1}+log _{3}a_{2}+cdots+log _{3}a_{7})=frac{sqrt{3}}{2}. The correct answer is boxed{text{B: }frac{sqrt{3}}{2}}.