💡Okay, so I have this problem about an arithmetic sequence. Let me try to understand what it's asking. First, it says that the arithmetic sequence is denoted as {a_n}, and the sum of its first n terms is S_n. They give me two pieces of information: a_3 = 5 and S_5 = 3S_3 - 2. I need to find the general term formula for {a_n}, which is part 1. Then, for part 2, they define another sequence {b_n} where each term is 2 raised to the power of a_n, and I need to find the sum of the first n terms of this new sequence, denoted as T_n.Alright, let's start with part 1. I remember that in an arithmetic sequence, each term is the previous term plus a common difference, d. So, the general term formula for an arithmetic sequence is a_n = a_1 + (n - 1)d, where a_1 is the first term and d is the common difference.They told me that a_3 = 5. Let me write that down:a_3 = a_1 + 2d = 5.So, that's one equation involving a_1 and d. Now, I need another equation to solve for both a_1 and d. The other piece of information is about the sums: S_5 = 3S_3 - 2.I recall that the sum of the first n terms of an arithmetic sequence is given by S_n = n/2 * (2a_1 + (n - 1)d) or also S_n = n/2 * (a_1 + a_n). Maybe I can use this to write expressions for S_5 and S_3.Let me write expressions for S_3 and S_5 using the formula S_n = n/2 * (2a_1 + (n - 1)d).So, S_3 = 3/2 * (2a_1 + 2d) = 3/2 * (2a_1 + 2d) = 3(a_1 + d).Similarly, S_5 = 5/2 * (2a_1 + 4d) = 5/2 * (2a_1 + 4d) = 5(a_1 + 2d).Now, according to the problem, S_5 = 3S_3 - 2. So, substituting the expressions I found:5(a_1 + 2d) = 3 * [3(a_1 + d)] - 2.Let me compute the right side:3 * [3(a_1 + d)] = 9(a_1 + d), so the equation becomes:5(a_1 + 2d) = 9(a_1 + d) - 2.Now, let's expand both sides:Left side: 5a_1 + 10d.Right side: 9a_1 + 9d - 2.So, putting it all together:5a_1 + 10d = 9a_1 + 9d - 2.Let me bring all terms to the left side:5a_1 + 10d - 9a_1 - 9d + 2 = 0.Simplify:(5a_1 - 9a_1) + (10d - 9d) + 2 = 0.Which is:-4a_1 + d + 2 = 0.So, we have:-4a_1 + d = -2.But from earlier, we have another equation: a_1 + 2d = 5.So now, I have a system of two equations:1) a_1 + 2d = 5.2) -4a_1 + d = -2.I need to solve for a_1 and d. Let me use substitution or elimination. Maybe elimination is easier here.Let me multiply the second equation by 2 to make the coefficients of d the same:-8a_1 + 2d = -4.Now, subtract the first equation from this:(-8a_1 + 2d) - (a_1 + 2d) = -4 - 5.Simplify:-8a_1 + 2d - a_1 - 2d = -9.Which is:-9a_1 = -9.Divide both sides by -9:a_1 = 1.Now, substitute a_1 = 1 into the first equation:1 + 2d = 5.Subtract 1:2d = 4.Divide by 2:d = 2.So, the first term a_1 is 1, and the common difference d is 2. Therefore, the general term formula is:a_n = a_1 + (n - 1)d = 1 + (n - 1)*2.Simplify:a_n = 1 + 2n - 2 = 2n - 1.So, that's part 1 done. The general term is a_n = 2n - 1.Now, moving on to part 2. They define b_n = 2^{a_n}. So, each term of the new sequence is 2 raised to the power of the corresponding term in the arithmetic sequence.Given that a_n = 2n - 1, then b_n = 2^{2n - 1}.I need to find the sum of the first n terms of {b_n}, which is T_n.Let me write out the first few terms of {b_n} to see if I can recognize a pattern.For n=1: b_1 = 2^{2*1 - 1} = 2^{1} = 2.n=2: b_2 = 2^{2*2 - 1} = 2^{3} = 8.n=3: b_3 = 2^{2*3 - 1} = 2^{5} = 32.n=4: b_4 = 2^{2*4 - 1} = 2^{7} = 128.Hmm, so the terms are 2, 8, 32, 128, ... It seems like each term is multiplied by 4 to get the next term. Let me check:8 / 2 = 4.32 / 8 = 4.128 / 32 = 4.Yes, so it's a geometric sequence with the first term b_1 = 2 and common ratio r = 4.The sum of the first n terms of a geometric sequence is given by T_n = b_1 * (r^n - 1)/(r - 1).Plugging in the values:T_n = 2 * (4^n - 1)/(4 - 1) = 2 * (4^n - 1)/3.Simplify:T_n = (2/3)(4^n - 1).So, that's the sum of the first n terms of {b_n}.Wait, let me double-check my calculations to make sure I didn't make a mistake.First, for the arithmetic sequence:a_3 = 5: a_1 + 2d = 5.S_5 = 3S_3 - 2: 5(a_1 + 2d) = 3*3(a_1 + d) - 2.Which led to 5a_1 + 10d = 9a_1 + 9d - 2.Simplifying: -4a_1 + d = -2.Solving with a_1 + 2d = 5, we found a_1 = 1 and d = 2.So, a_n = 2n - 1. That seems correct.For the geometric sequence:b_n = 2^{2n - 1}.So, b_1 = 2^{1} = 2, b_2 = 2^{3} = 8, etc.Common ratio r = b_{n+1}/b_n = 2^{2(n+1)-1}/2^{2n -1} = 2^{2n + 2 -1}/2^{2n -1} = 2^{2n +1}/2^{2n -1} = 2^{2} = 4.Yes, so r = 4.Sum formula: T_n = b_1*(r^n -1)/(r -1) = 2*(4^n -1)/3.Yes, that looks correct.I think I did everything right. So, the general term is 2n -1, and the sum T_n is (2/3)(4^n -1).**Final Answer**1. The general term formula for {a_n} is boxed{2n - 1}.2. The sum of the first n terms of {b_n} is boxed{dfrac{2}{3}(4^n - 1)}.

💡Okay, so I have this function f(x) given by a fraction, and I need to find f(-11π/12) and then analyze another function g(x) which is a combination of f(x) and sin2x. Let me try to break this down step by step.First, let me write down the function again to make sure I have it correctly:f(x) = [4cos⁴x - 2cos2x - 1] / [sin(π/4 + x) * sin(π/4 - x)]Hmm, that looks a bit complicated. Maybe I can simplify the numerator and the denominator separately.Starting with the numerator: 4cos⁴x - 2cos2x - 1.I remember that cos²x can be expressed using the double-angle identity: cos²x = (1 + cos2x)/2. Maybe I can use that to rewrite cos⁴x.So, cos⁴x = (cos²x)² = [(1 + cos2x)/2]^2. Let me compute that:(1 + cos2x)² / 4 = (1 + 2cos2x + cos²2x)/4.So, 4cos⁴x would be 4 * [(1 + 2cos2x + cos²2x)/4] = 1 + 2cos2x + cos²2x.Therefore, the numerator becomes:4cos⁴x - 2cos2x - 1 = (1 + 2cos2x + cos²2x) - 2cos2x - 1.Simplify that:1 - 1 cancels out, 2cos2x - 2cos2x cancels out, so we're left with cos²2x.So, numerator simplifies to cos²2x.Now, the denominator: sin(π/4 + x) * sin(π/4 - x).I recall that sinA * sinB can be expressed using the identity:sinA sinB = [cos(A - B) - cos(A + B)] / 2.Let me apply that here.Let A = π/4 + x and B = π/4 - x.So, A - B = (π/4 + x) - (π/4 - x) = 2x.A + B = (π/4 + x) + (π/4 - x) = π/2.Therefore, sin(π/4 + x) sin(π/4 - x) = [cos(2x) - cos(π/2)] / 2.But cos(π/2) is 0, so this simplifies to [cos2x - 0]/2 = cos2x / 2.So, denominator is cos2x / 2.Putting numerator and denominator together:f(x) = [cos²2x] / [cos2x / 2] = [cos²2x] * [2 / cos2x] = 2cos2x.Oh, that's a big simplification! So f(x) simplifies to 2cos2x. That makes things much easier.So, part (I) asks for f(-11π/12). Since f(x) = 2cos2x, let's compute that.First, compute 2x when x = -11π/12:2x = 2*(-11π/12) = -11π/6.But cosine is an even function, so cos(-θ) = cosθ. Therefore, cos(-11π/6) = cos(11π/6).11π/6 is in the fourth quadrant, and cos(11π/6) = cos(2π - π/6) = cos(π/6) = √3/2.Therefore, f(-11π/12) = 2*(√3/2) = √3.Alright, that seems straightforward.Now, moving on to part (II). We need to find the maximum and minimum values of g(x) = (1/2)f(x) + sin2x when x is in [0, π/4).Since f(x) = 2cos2x, let's substitute that into g(x):g(x) = (1/2)*(2cos2x) + sin2x = cos2x + sin2x.So, g(x) = cos2x + sin2x.Hmm, this is a combination of sine and cosine functions with the same argument. Maybe I can express this as a single sine or cosine function using the amplitude-phase form.I remember that A cosθ + B sinθ can be written as C cos(θ - φ) or C sin(θ + φ), where C = √(A² + B²) and tanφ = B/A.In this case, A = 1 (coefficient of cos2x) and B = 1 (coefficient of sin2x). So,C = √(1² + 1²) = √2.And tanφ = B/A = 1/1 = 1, so φ = π/4.Therefore, g(x) can be written as √2 sin(2x + π/4). Wait, let me verify that.Alternatively, since A and B are both positive, it's equivalent to √2 sin(2x + π/4). Let me check:√2 sin(2x + π/4) = √2 [sin2x cosπ/4 + cos2x sinπ/4] = √2 [sin2x*(√2/2) + cos2x*(√2/2)] = √2*(√2/2)(sin2x + cos2x) = (2/2)(sin2x + cos2x) = sin2x + cos2x.Yes, that works. So, g(x) = √2 sin(2x + π/4).Now, to find the maximum and minimum values of g(x) over x ∈ [0, π/4).First, let's consider the range of 2x + π/4 when x is in [0, π/4):When x = 0: 2x + π/4 = 0 + π/4 = π/4.When x approaches π/4: 2x approaches π/2, so 2x + π/4 approaches π/2 + π/4 = 3π/4.Therefore, the argument of the sine function, 2x + π/4, ranges from π/4 to 3π/4.We know that sinθ reaches its maximum at θ = π/2, which is 1, and its minimum at θ = π/4 and θ = 3π/4, which are both √2/2.But wait, let's think carefully. The function sinθ increases from π/4 to π/2, reaching maximum at π/2, then decreases from π/2 to 3π/4.So, in the interval [π/4, 3π/4], the maximum value of sinθ is 1, and the minimum is √2/2.Therefore, the maximum value of g(x) is √2 * 1 = √2, and the minimum value is √2 * (√2/2) = 1.Now, let's find the x values where these extrema occur.Maximum occurs when 2x + π/4 = π/2, so 2x = π/2 - π/4 = π/4, so x = π/8.Minimum occurs at the endpoints of the interval. Let's check x=0 and x approaching π/4.At x=0: g(0) = cos0 + sin0 = 1 + 0 = 1.As x approaches π/4: 2x approaches π/2, so 2x + π/4 approaches 3π/4. sin(3π/4) = √2/2, so g(x) approaches √2*(√2/2) = 1.Therefore, the minimum value is 1, achieved at x=0, and the maximum value is √2, achieved at x=π/8.So, summarizing:(I) f(-11π/12) = √3.(II) The maximum value of g(x) is √2, and the minimum value is 1.I think that makes sense. Let me just double-check my steps.For part (I), I simplified f(x) correctly using double-angle identities, and then evaluated at x = -11π/12, which gave me √3. That seems right.For part (II), I expressed g(x) as √2 sin(2x + π/4), found the range of the argument, and determined the maximum and minimum values accordingly. The calculations seem consistent.Yes, I feel confident about these answers.**Final Answer**(I) boxed{sqrt{3}} (II) Maximum value: boxed{sqrt{2}}, Minimum value: boxed{1}

💡Okay, so I have this problem about a digital clock that lights up every 9 minutes and chimes every hour. At exactly 12:00 PM, it both chimes and lights up. I need to figure out when it will next both chime and light up. Hmm, let's break this down.First, I need to understand what's happening here. The clock has two functions: one that lights up every 9 minutes and another that chimes every hour. At 12:00 PM, both functions happen at the same time. I guess this means that 12:00 PM is a common multiple of both 9 minutes and 60 minutes (since an hour is 60 minutes). So, I need to find the next time after 12:00 PM when both these events coincide again.I remember that when you have two events happening at regular intervals, the next time they coincide is given by the least common multiple (LCM) of their intervals. So, I should probably find the LCM of 9 minutes and 60 minutes. That should give me the number of minutes after 12:00 PM when both the light and the chime will happen together again.Okay, let's recall how to find the LCM of two numbers. One way to do it is by listing the multiples of each number until I find the smallest common one. But that might take a while, especially since 60 is a larger number. Maybe there's a better way.I think another method involves prime factorization. If I can express both numbers as products of their prime factors, then the LCM is the product of the highest powers of all primes that appear in either number. Let's try that.First, let's factorize 9 and 60.Starting with 9:- 9 divided by 3 is 3.- 3 divided by 3 is 1.So, 9 = 3 × 3 = 3².Now, 60:- 60 divided by 2 is 30.- 30 divided by 2 is 15.- 15 divided by 3 is 5.- 5 divided by 5 is 1.So, 60 = 2 × 2 × 3 × 5 = 2² × 3 × 5.Now, to find the LCM, I take the highest power of each prime number that appears in the factorizations:- For 2: the highest power is 2².- For 3: the highest power is 3².- For 5: the highest power is 5.So, LCM = 2² × 3² × 5.Let's calculate that:- 2² = 4- 3² = 9- 4 × 9 = 36- 36 × 5 = 180So, the LCM of 9 and 60 is 180 minutes.But wait, 180 minutes is equal to 3 hours because 60 minutes make an hour. So, 180 minutes is 3 hours.Since the clock coincides at 12:00 PM, adding 3 hours to that time will give me the next time both events happen together.12:00 PM plus 3 hours is 3:00 PM.Let me double-check to make sure I didn't make a mistake. If I list the multiples of 9 minutes:9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99, 108, 117, 126, 135, 144, 153, 162, 171, 180,...And the multiples of 60 minutes:60, 120, 180, 240,...The smallest common multiple after 0 is 180 minutes, which is 3 hours. So, yes, 3:00 PM seems correct.I think that makes sense. The clock will next both chime and light up at 3:00 PM.

💡Okay, so I have this problem here about a circle with a radius of 15 cm, and there's a sector with an arc length of π/3 cm. I need to find the area of this sector, and the options are given as A: 5 cm², B: 5π cm², C: 5/2 cm², and D: 5π/2 cm². Hmm, let me think about how to approach this.First, I remember that the area of a sector can be calculated if we know the radius and the central angle, or if we know the arc length. Since the problem gives me the arc length, I should use that. I think the formula for the area of a sector when you know the arc length is something like (1/2) times the arc length times the radius. Let me write that down to make sure: Area = (1/2) * arc length * radius.So, plugging in the values I have, the arc length is π/3 cm, and the radius is 15 cm. Let me substitute those into the formula. That would be Area = (1/2) * (π/3) * 15. Let me compute that step by step.First, multiply (1/2) and (π/3). That would be (1/2) * (π/3) = π/6. Then, multiply that by 15. So, π/6 * 15. Let me do the multiplication: 15 divided by 6 is 2.5, which is the same as 5/2. So, 5/2 times π is (5π)/2. Wait, let me double-check that. 15 divided by 6 is indeed 2.5, which is 5/2. So, multiplying that by π gives (5π)/2. That seems right.Let me think if there's another way to approach this problem. Maybe by finding the central angle first and then using the area formula in terms of the angle. The arc length formula is arc length = radius * central angle in radians. So, if I have the arc length (π/3) and the radius (15), I can find the central angle θ.So, θ = arc length / radius = (π/3) / 15 = π/(3*15) = π/45 radians. Hmm, that seems quite small. Is that correct? Let me check: 15 cm radius, and the arc length is π/3 cm. Since π is approximately 3.14, π/3 is about 1.05 cm. So, an arc length of about 1.05 cm on a circle with a 15 cm radius would correspond to a small angle, which makes sense.Now, if I have the central angle θ in radians, the area of the sector can also be calculated using the formula Area = (1/2) * r² * θ. Let me try that with θ = π/45.So, Area = (1/2) * (15)² * (π/45). Let's compute that. 15 squared is 225. So, Area = (1/2) * 225 * (π/45). Simplify that: (1/2) * (225/45) * π. 225 divided by 45 is 5. So, (1/2) * 5 * π = (5/2) * π, which is the same as (5π)/2.Okay, so both methods give me the same result: (5π)/2 cm². That must be the correct answer. Let me look back at the options: A is 5 cm², B is 5π cm², C is 5/2 cm², and D is (5π)/2 cm². So, D is the one that matches my calculation.Just to make sure I didn't make any mistakes, let me go through the steps again. The arc length is π/3, radius is 15. Using the area formula with arc length: (1/2)*arc length*radius. So, (1/2)*(π/3)*15. Simplify: (1/2)*(15/3)*π = (1/2)*5*π = (5π)/2. Yep, that's correct.Alternatively, finding the angle: θ = arc length / radius = (π/3)/15 = π/45. Then, area = (1/2)*r²*θ = (1/2)*225*(π/45) = (1/2)*5*π = (5π)/2. Same result.I think I'm confident that D is the right answer. It's always good to check using different methods to ensure consistency. Both approaches lead to the same conclusion, so I don't think I made any calculation errors here.Another thing I can consider is the units. The radius is in centimeters, so the area should be in square centimeters, which all the options have. The numerical value is (5π)/2, which is approximately (5*3.14)/2 ≈ 7.85 cm². Looking at the options, D is the only one that matches this approximate value. The other options are either too small or don't include π, which is necessary since we're dealing with a circle.So, after verifying through multiple methods and checking the units, I'm pretty sure that D: (5π)/2 cm² is the correct answer.**Final Answer**boxed{D}

💡Okay, so I have this problem here where I need to find the unit digit of the sum of powers of 3 from 3^1 up to 3^2023. Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, I remember that the unit digit of powers of numbers often follows a repeating pattern. So maybe if I can figure out the pattern of the unit digits for powers of 3, I can use that to simplify the problem.Let me start by calculating the unit digits of the first few powers of 3:- 3^1 = 3, so the unit digit is 3.- 3^2 = 9, so the unit digit is 9.- 3^3 = 27, so the unit digit is 7.- 3^4 = 81, so the unit digit is 1.- 3^5 = 243, so the unit digit is 3 again.Wait a minute, I see a pattern here. The unit digits are repeating every four powers: 3, 9, 7, 1, and then back to 3. So the cycle is 3, 9, 7, 1, and then repeats.That means every four exponents, the unit digits cycle through these four numbers. So if I can figure out how many complete cycles there are in 2023 exponents, and then handle the remaining exponents, I can find the total sum's unit digit.Let me calculate how many complete cycles there are in 2023. Since each cycle is 4 exponents, I'll divide 2023 by 4.2023 divided by 4 is 505 with a remainder of 3. So there are 505 complete cycles and then 3 additional exponents.Now, for each complete cycle, the unit digits are 3, 9, 7, and 1. Let me add those up: 3 + 9 + 7 + 1 = 20. The unit digit of 20 is 0. So each complete cycle contributes a 0 to the unit digit of the total sum.Since there are 505 complete cycles, each contributing 0, the total contribution from the complete cycles is 0.Now, I need to handle the remaining 3 exponents. These correspond to the first three exponents in the cycle: 3^1, 3^2, and 3^3. Their unit digits are 3, 9, and 7 respectively.Adding these up: 3 + 9 + 7 = 19. The unit digit of 19 is 9.So, combining the contributions from the complete cycles (which is 0) and the remaining exponents (which is 9), the total unit digit of the sum is 9.Wait, let me double-check that. The complete cycles contribute nothing to the unit digit, and the remaining three terms add up to 19, whose unit digit is 9. Yeah, that seems right.I think I got it. The unit digit of the sum from 3^1 to 3^2023 is 9.

💡First, I recognize that each of the three integers leaves a specific remainder when divided by 53. Let's denote these integers as ( a ), ( b ), and ( c ), with remainders 31, 15, and 7 respectively.To find the remainder when the sum of these integers is divided by 53, I can use modular arithmetic. Specifically, I can add the remainders together and then find the remainder of this sum when divided by 53.Adding the remainders:[31 + 15 + 7 = 53]Next, I determine the remainder of 53 when divided by 53:[53 div 53 = 1 text{ with a remainder of } 0]Therefore, the remainder when the sum of the three integers is divided by 53 is 0.