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question:Given the sequence {a_n} that satisfies 2S_n = 4a_n 1, find the sum of the first 100 terms of the sequence {frac{1}{log_2 a_{n+3} log_2 a_{n+2}}}. A: frac{97}{100} B: frac{98}{99} C: frac{99}{100} D: frac{100}{101}

answer:Since 2S_n = 4a_n - 1, we can also write 2S_{n-1} = 4a_{n-1} - 1 for n geq 2. By subtracting these two equations, we obtain 2a_n = 4a_n - 4a_{n-1}, which simplifies to a_n = 2a_{n-1} for n geq 2. Also, from 2S_1 = 4a_1 - 1, we have a_1 = frac{1}{2}. Therefore, {a_n} is a geometric sequence with the first term frac{1}{2} and the common ratio 2. Thus, a_n = frac{1}{2} cdot 2^{n-1} = 2^{n-2}. Consequently, frac{1}{log_2 a_{n+3} log_2 a_{n+2}} = frac{1}{n(n+1)} = frac{1}{n} - frac{1}{n+1}. Hence, the required sum is (1 - frac{1}{2} + frac{1}{2} - frac{1}{3} + ldots + frac{1}{99} - frac{1}{100} + frac{1}{100} - frac{1}{101}) = boxed{frac{100}{101}}.

question:Given a nonempty set ( M ) satisfying [ M subseteq {0,1, cdots, n} quad text{(where ( n geq 2 ), ( n ) is a positive integer)}. ] If there exists a nonnegative integer ( k ) (( k leq n )) such that whenever ( a in M ), ( 2k a in M ) holds, then the set ( M ) is said to have the property ( P ). Let ( f(n) ) represent the number of sets ( M ) that have the property ( P ). Find the value of ( f(9) f(8) ).

answer:Given a non-empty set M which satisfies the condition: [ M subseteq {0,1, cdots, n } quad (n geqslant 2, n in mathbf{Z}_{+}) ] If there exists a nonnegative integer (k (k leqslant n)) such that for all (a in M), (2k - a in M) holds, the set M is said to possess property P. Let (f(n)) be the number of sets M with property P for a given n. We need to determine the value of (f(9) - f(8)). 1. Base Case Calculation: For (n = 2), the sets satisfying the property (P) are: [ M = {0}, {1}, {2}, {0, 2}, {0, 1, 2} ] For these sets, the corresponding values of (k) are (0, 1, 2, 1, 1) respectively. Thus, (f(2) = 5). 2. Recursive Formula: For general (n = t), if (f(t)) is the number of sets M with property P, when (n = t + 1), the total count of sets becomes: [ f(t+1) = f(t) + g(t+1) ] where (g(t+1)) represents the number of sets M including (t+1) and having property (P). We need to express (g(t+1)) in terms of (t). 3. Conditions for (g(t+1)): When (t) is even, (t = 2m): [ k geqslant frac{t + 2}{2} = m + 1 ] Thus, the sets must include all combinations of (k) where (k = m + 1, m + 2, ldots, t+1). The number of such sets for each (k) is given by: [ 2^{ t+1 - k } ] Summing over (k): [ g(t+1) = sum_{k = m+1}^{2m+1} 2^{ t+1 - k } = 2 + 2^1 + 2^0 = 2^{m+1} ] When (t) is odd, (t = 2m + 1): [ k geqslant frac{t+1}{2} = m + 1 ] Approaching similarly: [ g(t+1) = sum_{k = m+1}^{2m+1} 2^{t+1 - k} = 2^{m+1} ] Thus, the recurrence relation is: [ f(t+1) = f(t) + 2^{leftlfloor frac{t+2}{2} rightrfloor } ] 4. Simplifying for Specific Cases: Using induction or direct computation, we find: For (n) even: [ f(n) = 6 cdot 2^{frac{n}{2}} - n - 5 ] For (n) odd: [ f(n) = 4 cdot 2^{frac{n+1}{2}} - n - 5 ] 5. Final Calculation: To find (f(9) - f(8)), [ f(9) = 4 cdot 2^5 - 9 - 5 = 4 cdot 32 - 14 = 128 - 14 = 114 ] [ f(8) = 6 cdot 2^4 - 8 - 5 = 6 cdot 16 - 13 = 96 - 13 = 83 ] Thus, [ f(9) - f(8) = 114 - 83 = 31 ] # Conclusion: [ boxed{31} ]

question:A rectangle has a ratio of length to width of 5:2. If the diagonal of the rectangle is 13 units, determine k such that the area of the rectangle may be expressed as kd^2, where d is the diagonal. **(A)** frac{10}{25} **(B)** frac{10}{29} **(C)** frac{10}{23} **(D)** frac{8}{29} **(E)** frac{12}{29}

answer:1. **Assign Variables:** Let the length of the rectangle be (5x) and the width (2x). Since the ratio of the length to the width is 5:2. 2. **Use The Pythagorean Theorem:** With (d = 13) (the diagonal), the Pythagorean theorem gives: [ (5x)^2 + (2x)^2 = 13^2 ] [ 25x^2 + 4x^2 = 169 ] [ 29x^2 = 169 ] [ x^2 = frac{169}{29} ] [ x = sqrt{frac{169}{29}} ] 3. **Calculate the Rectangle's Area:** [ A = (5x) times (2x) = 10x^2 ] [ A = 10 times frac{169}{29} = frac{1690}{29} ] [ A = frac{1690}{29} ] 4. **Express Area in Relation to (d^2):** Since (d^2 = 169), we rewrite area (A) as: [ A = frac{1690}{29} = frac{10 times 169}{29} = frac{10 times d^2}{29} ] Set (k = frac{10}{29}). Conclusion with boxed answer: [ k = frac{10{29}} ] The correct answer is **(B)** boxed{frac{10}{29}}.

question:a bag containing one red ball and one blue ball of the same size, if drawing a ball and recording its color is considered one experiment, and the experiment is conducted three times with replacement, then the probability of drawing a red ball at least once is ( ) A: dfrac {1}{8} B: dfrac {7}{8} C: dfrac {3}{8} D: dfrac {5}{8}

answer:Solution: Since the experiment is conducted three times, and the probability of drawing a blue ball each time is dfrac {1}{2}, the probability of drawing a blue ball all three times is ( dfrac {1}{2})^{3}= dfrac {1}{8}. Therefore, the probability of drawing a red ball at least once is 1- dfrac {1}{8}= dfrac {7}{8}. Hence, the correct answer is: boxed{B}. First, calculate the probability of drawing a blue ball all three times, then subtract this probability from 1 to find the desired probability. This question examines the probability multiplication formula for mutually independent events and the probability formula for exactly k occurrences in n independent repeated experiments. It involves the relationship between the probability of an event and its complementary event, making it a basic question.

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