answer:To convert the given complex number -(2+2i) from rectangular form to polar form, we first find its magnitude r and then its argument theta. 1. Find the magnitude r: The magnitude r is given by the distance of the point from the origin in the complex plane. Using the distance formula, we have: r = sqrt{(-2)^2 + (-2)^2} = sqrt{4+4} = sqrt{8} = 2sqrt{2} 2. Find the argument theta: The argument theta is the angle made by the line joining the point to the origin and the positive real axis. Since the given point lies in the second quadrant, we find the reference angle first and then add pi radians to it. The reference angle is given by: text{reference angle} = arctanleft(frac{text{imaginary part}}{text{real part}}right) = arctanleft(frac{-2}{-2}right) = arctan(1) = frac{pi}{4} Adding pi radians to the reference angle, we get the argument: theta = pi + frac{pi}{4} = frac{5pi}{4} However, since polar coordinates are unique up to integer multiples of 2pi, we can simplify this by subtracting 2pi: theta = frac{5pi}{4} - 2pi = -frac{3pi}{4} Thus, the polar coordinates of the point P are (2sqrt{2}, -frac{3pi}{4}). But since the problem asks for the polar axis to be the positive half of the real axis, we add 2pi to the angle to get it within the range [0, 2pi): theta = -frac{3pi}{4} + 2pi = frac{5pi}{4} Therefore, the correct answer in the given options is not present. However, if we consider the polar coordinates as (r, theta + 2kpi) for any integer k, the closest option would be: boxed{left( 2sqrt{2},frac{5pi }{4}+2kpi right)left( kin mathbb{Z} right)}

answer:1. **Understand the Problem Statement**: - There are 30 palm trees, each with a sign. - We have the following signs: - 15 trees: "There is a treasure under exactly 15 signs." - 8 trees: "There is a treasure under exactly 8 signs." - 4 trees: "There is a treasure under exactly 4 signs." - 3 trees: "There is a treasure under exactly 3 signs." 2. **Observe the Conditions**: - Only the signs under which there is no treasure are truthful. 3. **Assume that there is a treasure under less than or equal to 15 signs**: - If a treasure were hidden under 16 or more signs, at least one of the true signs would indicate a number that does not match with the actual count of treasures, leading to contradictions. 4. **Check the possible contradictions**: - If a sign is true, then it means the statement it makes should be considered. This implies that for all the false signs (under which treasures are hidden) the count of treasures under signs must not match the statements on true signs. 5. **Identify the Possession Without a Contradiction**: - Choose a scenario where the no. of treasures hidden and the statements can coexist without contradictions. In-depth Scenario Analysis: - If there were treasures under more than 15 signs, at least one sign would state something that would be impossible because multiple statements cannot be true simultaneously as per problem constraints. - If we consider it accurately for fewer ones and get an accurate minimal number and eliminate false cases, we conclude: - If treasures are exactly under 15 signs, it's the maximum non-contradictory situation satisfying no more than 15 signs statement. Conclusion: The scenario with exactly 15 treasures buried follows every constraint provided. Hence, under every truthful assurance of this would not lead to contradiction else. # Conclusion: Thus, the minimum number of signs under which the treasure is buried that fits all conditions is: [ boxed{15} ]

answer:Observe that the power 4 changes the property of the series. We have: - For any integer a, (-a)^4 = a^4. This is because the negative sign is eliminated by the even power. Given this, the series (-1)^4 + (-2)^4 + (-3)^4 + dots + (-49)^4 + (-50)^4 simplifies to 1^4 + 2^4 + 3^4 + dots + 49^4 + 50^4. Adding the two series: (1^4 + 2^4 + 3^4 + dots + 49^4 + 50^4) + (1^4 + 2^4 + 3^4 + dots + 49^4 + 50^4) = 2(1^4 + 2^4 + 3^4 + dots + 49^4 + 50^4). Now, using the formula for the sum of the fourth powers of the first n natural numbers: sum_{k=1}^{n} k^4 = frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} for n=50, the sum becomes: frac{50 times 51 times 101 times (3 times 50^2 + 3 times 50 - 1)}{30} = frac{50 times 51 times 101 times 7600}{30} = frac{19525500}{30} = 650850. Thus, the final sum is: 2 times 650850 = 1301700. Therefore, the final answer is boxed{1301700}.

answer:Solution: (I) When n=1, 4S_1=4a_1=a_1^2+2a_1-3, we get a_1^2-4a_1-3=0, a_1=3 or a_1=-1. Given that a_n > 0, we have a_1=3. (II) For n geqslant 2, 4S_n=a_n^2+2a_n-3, 4S_{n-1}=a_{n-1}^2+2a_{n-1}-3; Thus, 4S_n-4S_{n-1}=a_n^2+2a_n-3-a_{n-1}^2-2a_{n-1}+3, which gives 4a_n=a_n^2+2a_n-a_{n-1}^2-2a_{n-1}, a_n^2-2a_n-a_{n-1}^2-2a_{n-1}=0, (a_n+a_{n-1})(a_n-a_{n-1}-2)=0, Given a_n+a_{n-1} > 0, we have a_n-a_{n-1}=2, Therefore, the sequence {a_n} is an arithmetic sequence with the first term 3 and common difference 2, thus a_n=2n+1. (III) From (I), b_n= sqrt{2^{a_n-1}}= sqrt{2^{2n+1-1}}=2^n, frac{a_n}{b_n}= frac{2n+1}{2^n}, Therefore, T_n= frac{3}{2}+ frac{5}{4}+ldots+ frac{2n-1}{2^{n-1}}+ frac{2n+1}{2^n}, Multiplying both sides of the equation by frac{1}{2}, we get frac{1}{2}T_n= frac{3}{4}+ frac{5}{8}+ldots+ frac{2n-1}{2^n}+ frac{2n+1}{2^{n+1}}, Subtracting the second equation from the first, we obtain frac{1}{2}T_n= frac{3}{2}+ frac{2}{4}+ frac{2}{8}+ldots+ frac{2}{2^n}- frac{2n+1}{2^{n+1}}= frac{5}{2}- frac{2n+5}{2^{n+1}}, thus T_n=5- frac{2n+5}{2^n}. Since n in mathbb{N}^*, frac{2n+5}{2^n} > 0, therefore, T_n < 5. So, the final answers are: (I) a_1=boxed{3} (II) a_n=boxed{2n+1} (III) T_n < boxed{5}