answer:1. **Formulation of the problem**: Given a geometric sequence a_1, a_2, cdots, a_9 where: [ a_1 + a_2 = frac{3}{4} ] and [ a_3 + a_4 + a_5 + a_6 = 15. ] 2. **Express terms of the geometric progression**: Let the common ratio be q. The terms can be written as a_1, a_1q, a_1q^2, a_1q^3, cdots. 3. **Set equations based on given conditions**: [ a_1 + a_1q = frac{3}{4} implies a_1 (1 + q) = frac{3}{4} ] 4. **For the second condition**: [ a_3 + a_4 + a_5 + a_6 = 15 implies a_1 q^2 + a_1 q^3 + a_1 q^4 + a_1 q^5 = 15 ] Factor out (a_1 q^2): [ a_1 q^2 (1 + q + q^2 + q^3) = 15 ] 5. **Evaluate ( frac{a_1 q^2 (1 + q + q^2 + q^3)}{a_1 (1 + q)} )**: [ frac{a_1 q^2 (1 + q + q^2 + q^3)}{a_1 (1 + q)} = frac{15}{frac{3}{4}} = 20 ] 6. **Simplify the ratio**: [ frac{q^2 (1 + q + q^2 + q^3)}{1 + q} = 20 ] This simplifies further: [ q^2 cdot (1 + q^2) = 20 ] 7. **Solve for (q)**: [ q^4 + q^2 = 20 implies q^4 + q^2 - 20 = 0 ] Let (x = q^2), then: [ x^2 + x - 20 = 0 ] Solving this quadratic equation using the quadratic formula: (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}) where (a = 1, b = 1, c = -20): [ x = frac{-1 pm sqrt{1 + 80}}{2} = frac{-1 pm 9}{2} ] Hence: [ x = 4 quad text{or} quad x = -5 ] Since (x = q^2) and it must be positive: [ q^2 = 4 implies q = 2 quad (text{since } q > 0) ] 8. **Determine (a_1)**: Using (a_1 (1 + q) = frac{3}{4}): [ a_1 (1 + 2) = frac{3}{4} implies a_1 cdot 3 = frac{3}{4} implies a_1 = frac{1}{4} ] 9. **Calculate (a_7 + a_8 + a_9)**: [ a_7 + a_8 + a_9 = a_1 q^6 + a_1 q^7 + a_1 q^8 ] Factor out (a_1 q^6): [ a_7 + a_8 + a_9 = a_1 q^6 (1 + q + q^2) ] Substituting (a_1 = frac{1}{4}) and (q = 2): [ a_7 + a_8 + a_9 = frac{1}{4} cdot 2^6 (1 + 2 + 4) ] Simplify: [ = frac{1}{4} cdot 64 cdot 7 = frac{448}{4} = 112 ] Conclusion: [ boxed{112} ]

answer:1. **Understanding Magnanimous Numbers**: A magnanimous number is either a one-digit number or its digits form a strictly increasing or strictly decreasing sequence possibly ending with zero. 2. **Counting One-Digit Magnanimous Numbers**: There are 9 one-digit magnanimous numbers (0 through 8). 3. **Counting Multi-Digit Increasing Magnanimous Numbers**: Choosing n distinct digits from 0 to 8 and arranging them in increasing order gives binom{9}{n} ways. 4. **Counting Multi-Digit Decreasing Magnanimous Numbers**: Choosing n distinct digits from 1 to 8 and arranging them in decreasing order yields binom{8}{n}. 5. **Counting Mult-Digit Decreasing Magnanimous Numbers Ending with Zero**: Choose n-1 digits from 1 to 8 for the non-zero parts and append a zero gives additional binom{8}{n-1} numbers. 6. **Total Counting**: Summing up all the contributions, we have: [ text{Total} = sum_{n=1}^{9} binom{9}{n} + sum_{n=1}^{8} binom{8}{n} + sum_{n=2}^{8} binom{8}{n-1} + 9 ] 7. **Simplifying the Sum**: [ = 2^9 - 1 + 2^8 - 1 + (2^8 - 1) + 9 ] [ = 511 + 255 + 255 + 9 ] [ = 1030 ] 8. **Final Calculation**: [ 1030 ] Conclusion: Since all possible sequences and counting strategies are properly considered and correctly summed, the final result is valid. The final answer is boxed{C) 1030}

answer:(I) The parametric equation of C1 is given by C_{1}:begin{cases} x=2t+1 y=2t-3 end{cases}. Eliminating the parameter t, we get x-y=4. Using the formulas begin{cases} x=rhocostheta y=rhosintheta end{cases}, substitute them into x-y=4, we get ρcosθ-ρsinθ=4, that is, ρ(sinθ-cosθ)+4=0. Thus, sqrt{2}rhosin(theta-frac{pi}{4})+4=0. So the polar coordinate equation of C1 is rhosin(theta-frac{pi}{4})+2sqrt{2}=0. Since the curve C2 is given by ρ=2acosθ (a>0), we have ρ²=2aρcosθ, i.e., x²+y²=2ax, that is, (x-a)²+y²=a² (a>0). So the center of the circle is at (a, 0) and the radius is a. Since C2 is symmetric about C1, the center is on curve C1, so a=4. Hence, C2 is given by (x-4)²+y²=16². (II) Translate C2 four units to the left, the equation of the new curve is x²+y²=a², and then transform it according to begin{cases} x'=x y'=frac{sqrt{3}}{2}y end{cases} to obtain C3, which is given by x^2+(frac{2}{sqrt{3}}y)^2=16. Simplifying, we get frac{x^2}{16}+frac{y^2}{12}=1, that is, C3: frac{x^2}{16}+frac{y^2}{12}=1. C3 intersects with both coordinate axes at points A and B. Let A(4, 0) and B(0, 2sqrt{3}). Then |AB|=2sqrt{7}. Let P be any point on C3, denoted by P(4cosθ, 2sqrt{3}sintheta). The distance d from P to line AB is given by: d = |frac{4sqrt{3}sintheta+4sqrt{3}costheta-4sqrt{3}}{sqrt{7}}| = frac{4sqrt{3}|sqrt{2}sin(theta+frac{pi}{4})-1|}{sqrt{7}} leq frac{4sqrt{3}(sqrt{2}+1)}{sqrt{7}}. So, when θ=frac{5π}{4}, d takes the maximum value frac{4sqrt{3}(sqrt{2}+1)}{sqrt{7}}. Therefore, the maximum area of ΔABP is boxed{frac{1}{2} times |AB|d = 4sqrt{3} + 4sqrt{6}}.

answer:1. **Sequence Definition**: Hulk's jump distance can be defined by a geometric sequence where the first term (a_1 = 3) and each subsequent term triples the previous term. This is expressed as: [ a_n = 3 times 3^{n-1} = 3^n ] where (n) is the jump number. 2. **Set the Condition**: We need to find the smallest (n) such that (a_n > 500) meters. 3. **Inequality Setup**: [ 3^n > 500 ] 4. **Solve the Inequality**: - Approximating powers of 3, we have (3^5 = 243) and (3^6 = 729). - (3^6) is the smallest power of 3 greater than 500. 5. **Determine (n)**: [ n = 6 ] 6. **Conclusion**: Hulk will first be able to jump more than 500 meters on his (6^{text{th}}) jump. The answer is therefore (6^{text{th} text{ jump}}). The final answer is boxed{textbf{(B)} 6^text{th}}