💡Okay, so I'm trying to figure out which statement about flowcharts is correct. There are four options: A, B, C, and D. Let me think about what I know about flowcharts and see which one makes sense.First, I remember that a flowchart is a diagram that shows the steps in a process. It usually starts with a "start point" and ends with an "end point." But I'm not entirely sure if it can have more than one start or end point. Maybe in some complex processes, there could be multiple starts or ends?Let me look at the options one by one.Option A says: "A flowchart usually has one 'start point' and one 'end point'." That sounds familiar. I think most flowcharts I've seen have just one start and one end. But I'm not sure if that's always the case.Option B says: "A flowchart usually has one or more 'start points' and one 'end point'." Hmm, this suggests that there could be multiple start points but only one end. I'm not sure if that's common. I think flowcharts typically have a single starting point.Option C says: "A flowchart usually has one 'start point' and one or more 'end points'." This is interesting. It allows for multiple end points but still only one start. I can see how a process might have different outcomes leading to different end points, so this might make sense.Option D says: "A flowchart usually has one or more 'start points' and one or more 'end points'." This seems like it's allowing for multiple starts and multiple ends. I'm not sure if that's standard. I think flowcharts usually have a single start point, but maybe in some cases, they could have multiple ends.Now, I'm a bit confused. I think most flowcharts have one start and one end, but I'm not entirely sure. Maybe in some complex processes, there are multiple ends. Let me try to recall if I've ever seen a flowchart with multiple ends. I think in decision-making processes, depending on the choices made, there could be different outcomes, leading to different end points. So, maybe having one start and multiple ends is possible.On the other hand, having multiple start points seems less common. I think flowcharts usually begin at a single point and then branch out based on decisions or conditions. So, multiple start points might not be standard.Given that, Option C seems plausible because it allows for one start and multiple ends, which could happen in processes with different outcomes. Option A is too restrictive because it only allows for one end, which might not always be the case. Options B and D introduce multiple start points, which I don't think are typical.So, after thinking about it, I believe Option C is the correct answer because flowcharts usually have one start point and can have one or more end points depending on the process.

💡Okay, so I have this problem about a cubic function, f(x) = x³ - 3a x² + 2b x. It says that this function has a local minimum of -1 at x = 1. I need to find the values of a and b, and then figure out where the function is increasing or decreasing. After that, part II asks about the range of a for which the equation f(x) = a has three distinct real roots. Hmm, okay, let's take it step by step.Starting with part I: I know that if a function has a local minimum at a certain point, then the derivative at that point should be zero. Also, the function value at that point is given as -1. So, I can set up two equations: one for f(1) = -1 and another for f'(1) = 0. That should help me find a and b.First, let's find the derivative of f(x). The derivative of x³ is 3x², the derivative of -3a x² is -6a x, and the derivative of 2b x is 2b. So, f'(x) = 3x² - 6a x + 2b.Now, plugging x = 1 into f(x):f(1) = (1)³ - 3a(1)² + 2b(1) = 1 - 3a + 2b. And this is equal to -1. So, equation 1: 1 - 3a + 2b = -1.Next, plugging x = 1 into f'(x):f'(1) = 3(1)² - 6a(1) + 2b = 3 - 6a + 2b. Since it's a local minimum, this derivative is zero. So, equation 2: 3 - 6a + 2b = 0.Now, I have a system of two equations:1) 1 - 3a + 2b = -12) 3 - 6a + 2b = 0Let me write them more clearly:1) 1 - 3a + 2b = -12) 3 - 6a + 2b = 0Let me simplify equation 1 first. Subtract 1 from both sides:-3a + 2b = -2Equation 1: -3a + 2b = -2Equation 2: 3 - 6a + 2b = 0Let me subtract equation 1 from equation 2 to eliminate 2b.Equation 2: 3 - 6a + 2b = 0Minus equation 1: (-3a + 2b = -2)So, subtracting term by term:3 - 6a + 2b - (-3a + 2b) = 0 - (-2)Simplify:3 - 6a + 2b + 3a - 2b = 2Combine like terms:3 - 3a = 2Subtract 3 from both sides:-3a = -1Divide both sides by -3:a = 1/3Okay, so a is 1/3. Now, plug this back into equation 1 to find b.Equation 1: -3a + 2b = -2Plug a = 1/3:-3*(1/3) + 2b = -2Simplify:-1 + 2b = -2Add 1 to both sides:2b = -1Divide by 2:b = -1/2So, a is 1/3 and b is -1/2. Got that.Now, the next part is to identify the intervals of monotonicity for f(x). That means where the function is increasing or decreasing. To do that, I need to analyze the derivative f'(x).We have f'(x) = 3x² - 6a x + 2b. Since we found a and b, let's substitute them in.f'(x) = 3x² - 6*(1/3)x + 2*(-1/2)Simplify:3x² - 2x - 1So, f'(x) = 3x² - 2x - 1To find where the function is increasing or decreasing, we need to find the critical points by setting f'(x) = 0.So, 3x² - 2x - 1 = 0Let me solve this quadratic equation. The quadratic formula is x = [2 ± sqrt(4 + 12)] / 6Wait, discriminant D = b² - 4ac = (-2)² - 4*3*(-1) = 4 + 12 = 16So, sqrt(16) = 4Thus, x = [2 ± 4]/6So, two solutions:x = (2 + 4)/6 = 6/6 = 1x = (2 - 4)/6 = (-2)/6 = -1/3Therefore, the critical points are at x = -1/3 and x = 1.Now, to determine where the function is increasing or decreasing, we can test intervals around these critical points.The critical points divide the real line into three intervals:1) (-∞, -1/3)2) (-1/3, 1)3) (1, ∞)We can pick test points in each interval to see the sign of f'(x).First interval: Let's pick x = -1.f'(-1) = 3*(-1)² - 2*(-1) -1 = 3*1 + 2 -1 = 3 + 2 -1 = 4, which is positive. So, f is increasing on (-∞, -1/3).Second interval: Let's pick x = 0.f'(0) = 3*0 - 2*0 -1 = -1, which is negative. So, f is decreasing on (-1/3, 1).Third interval: Let's pick x = 2.f'(2) = 3*(4) - 2*(2) -1 = 12 -4 -1 = 7, which is positive. So, f is increasing on (1, ∞).Therefore, the function is increasing on (-∞, -1/3] and [1, ∞), and decreasing on [-1/3, 1].Wait, but the question says "intervals of monotonicity". So, I think that's the answer for part I.Now, moving on to part II: If the equation f(x) = a has three distinct real roots, find the range of values for the real number a.Hmm, okay. So, f(x) = a is equivalent to x³ - 3a x² + 2b x - a = 0. But since we already found a and b in terms of the function, maybe it's better to think graphically.Given that f(x) is a cubic function, it tends to -∞ as x approaches -∞ and +∞ as x approaches +∞. So, it must have a local maximum and a local minimum. The number of real roots of f(x) = a depends on the value of a relative to the local maximum and minimum.From part I, we found that f(x) has a local minimum at x = 1 with value f(1) = -1, and a local maximum at x = -1/3. Let's compute f(-1/3) to find the local maximum value.Compute f(-1/3):f(-1/3) = (-1/3)³ - 3a*(-1/3)² + 2b*(-1/3)We know a = 1/3 and b = -1/2.So, plug in:(-1/3)^3 = -1/27-3a*(-1/3)^2 = -3*(1/3)*(1/9) = - (1)*(1/9) = -1/92b*(-1/3) = 2*(-1/2)*(-1/3) = ( -1 )*(-1/3) = 1/3So, adding them up:-1/27 - 1/9 + 1/3Convert to 27 denominator:-1/27 - 3/27 + 9/27 = (-1 -3 +9)/27 = 5/27So, the local maximum is 5/27 at x = -1/3.Therefore, the function f(x) has a local maximum of 5/27 and a local minimum of -1.For the equation f(x) = a to have three distinct real roots, the horizontal line y = a must intersect the graph of f(x) three times. This happens when a is between the local minimum and the local maximum.So, the range of a is (-1, 5/27). That is, a must be greater than -1 and less than 5/27.Wait, let me think again. Since the function goes from -∞ to +∞, and it has a local maximum at 5/27 and a local minimum at -1. So, if a is between -1 and 5/27, the line y = a will cross the graph three times: once on the left of the local maximum, once between the local maximum and local minimum, and once on the right of the local minimum.Yes, that makes sense. So, the range of a is -1 < a < 5/27.Let me just double-check my calculations for f(-1/3):f(-1/3) = (-1/3)^3 - 3*(1/3)*(-1/3)^2 + 2*(-1/2)*(-1/3)Compute each term:First term: (-1/3)^3 = -1/27Second term: -3*(1/3)*(-1/3)^2 = -1*(1/9) = -1/9Third term: 2*(-1/2)*(-1/3) = ( -1 )*(-1/3) = 1/3So, adding them: -1/27 -1/9 +1/3Convert to 27 denominator:-1/27 -3/27 +9/27 = ( -1 -3 +9 ) /27 = 5/27Yes, that's correct.So, the local maximum is indeed 5/27, and the local minimum is -1. Therefore, for f(x) = a to have three distinct real roots, a must lie between these two values.Therefore, the range of a is (-1, 5/27).**Final Answer**(I) The values are ( a = boxed{dfrac{1}{3}} ) and ( b = boxed{-dfrac{1}{2}} ). The function is increasing on ( (-infty, -dfrac{1}{3}] ) and ( [1, infty) ), and decreasing on ( [-dfrac{1}{3}, 1] ).(II) The range of values for ( a ) is ( boxed{(-1, dfrac{5}{27})} ).

💡Alright, so I have this problem about triangle ABC. The sides opposite angles A, B, and C are a, b, and c respectively. It's given that ( b^2 + c^2 = a^2 + bc ). I need to find the measure of angle A and then find the maximum value of the function ( f(x) = sin(x - A) + sqrt{3}cos x ).Starting with part (1). Hmm, the equation given is ( b^2 + c^2 = a^2 + bc ). I remember the Law of Cosines, which relates the sides of a triangle to the cosine of one of its angles. The formula is ( a^2 = b^2 + c^2 - 2bccos A ). Maybe I can use this to find angle A.Let me write down the Law of Cosines formula again:( a^2 = b^2 + c^2 - 2bccos A ).But in the problem, it's given that ( b^2 + c^2 = a^2 + bc ). Let me rearrange this equation to see if I can express it in terms similar to the Law of Cosines.Starting with ( b^2 + c^2 = a^2 + bc ), I can subtract ( a^2 ) from both sides:( b^2 + c^2 - a^2 = bc ).Now, from the Law of Cosines, I know that ( b^2 + c^2 - a^2 = 2bccos A ). So substituting that into the equation above:( 2bccos A = bc ).Hmm, okay, so if I divide both sides by bc (assuming bc ≠ 0, which they can't be in a triangle), I get:( 2cos A = 1 ).Therefore, ( cos A = frac{1}{2} ).Now, I need to find the angle A where the cosine is 1/2. I remember that ( cos frac{pi}{3} = frac{1}{2} ), and since A is an angle in a triangle, it must be between 0 and π radians (or 0 and 180 degrees). So, the angle A must be ( frac{pi}{3} ) radians or 60 degrees.Okay, that seems straightforward. So, part (1) is solved, angle A is ( frac{pi}{3} ).Moving on to part (2). I need to find the maximum value of the function ( f(x) = sin(x - A) + sqrt{3}cos x ). Since we found that A is ( frac{pi}{3} ), I can substitute that into the function:( f(x) = sinleft(x - frac{pi}{3}right) + sqrt{3}cos x ).I need to simplify this expression to find its maximum value. Let me recall some trigonometric identities that might help. The sine of a difference can be expanded using the identity:( sin(a - b) = sin a cos b - cos a sin b ).Applying this to ( sinleft(x - frac{pi}{3}right) ):( sin x cos frac{pi}{3} - cos x sin frac{pi}{3} ).I know that ( cos frac{pi}{3} = frac{1}{2} ) and ( sin frac{pi}{3} = frac{sqrt{3}}{2} ). Substituting these values in:( sin x cdot frac{1}{2} - cos x cdot frac{sqrt{3}}{2} ).So, ( sinleft(x - frac{pi}{3}right) = frac{1}{2}sin x - frac{sqrt{3}}{2}cos x ).Now, substituting this back into the function f(x):( f(x) = left( frac{1}{2}sin x - frac{sqrt{3}}{2}cos x right) + sqrt{3}cos x ).Let me combine like terms. The terms with ( sin x ) and the terms with ( cos x ):- The ( sin x ) term: ( frac{1}{2}sin x ).- The ( cos x ) terms: ( -frac{sqrt{3}}{2}cos x + sqrt{3}cos x ).Simplify the ( cos x ) terms:( -frac{sqrt{3}}{2}cos x + sqrt{3}cos x = left( -frac{sqrt{3}}{2} + sqrt{3} right)cos x ).To combine these, I can write ( sqrt{3} ) as ( frac{2sqrt{3}}{2} ):( -frac{sqrt{3}}{2} + frac{2sqrt{3}}{2} = frac{sqrt{3}}{2} ).So, the ( cos x ) term simplifies to ( frac{sqrt{3}}{2}cos x ).Putting it all together, the function f(x) becomes:( f(x) = frac{1}{2}sin x + frac{sqrt{3}}{2}cos x ).Now, this looks like a linear combination of sine and cosine functions. I remember that any expression of the form ( asin x + bcos x ) can be written as a single sine (or cosine) function with a phase shift. The amplitude of such a function is ( sqrt{a^2 + b^2} ), which gives the maximum value of the function.So, let me compute the amplitude of ( f(x) ):( sqrt{left( frac{1}{2} right)^2 + left( frac{sqrt{3}}{2} right)^2} ).Calculating each term:( left( frac{1}{2} right)^2 = frac{1}{4} ).( left( frac{sqrt{3}}{2} right)^2 = frac{3}{4} ).Adding them together:( frac{1}{4} + frac{3}{4} = 1 ).So, the amplitude is ( sqrt{1} = 1 ).Therefore, the maximum value of ( f(x) ) is 1.Alternatively, I could have written ( f(x) ) as a single sine function. Let me try that to confirm.The expression ( frac{1}{2}sin x + frac{sqrt{3}}{2}cos x ) can be written as ( sin(x + phi) ), where ( phi ) is the phase shift.Using the identity ( sin(x + phi) = sin x cos phi + cos x sin phi ), we can equate coefficients:( cos phi = frac{1}{2} ) and ( sin phi = frac{sqrt{3}}{2} ).These correspond to ( phi = frac{pi}{3} ), since ( cos frac{pi}{3} = frac{1}{2} ) and ( sin frac{pi}{3} = frac{sqrt{3}}{2} ).Therefore, ( f(x) = sinleft(x + frac{pi}{3}right) ).The maximum value of the sine function is 1, so the maximum value of ( f(x) ) is indeed 1.So, both methods confirm that the maximum value is 1.In summary:1. Angle A is ( frac{pi}{3} ) radians.2. The maximum value of ( f(x) ) is 1.**Final Answer**(1) The measure of angle ( A ) is (boxed{dfrac{pi}{3}}).(2) The maximum value of ( f(x) ) is (boxed{1}).

💡Okay, so I have this problem where I'm given that α is between 0 and π, and that cos α plus sin α equals -1/5. I need to find tan 2α. Hmm, let me think about how to approach this.First, I remember that there are some trigonometric identities that relate sums of sine and cosine to other expressions. Maybe I can square both sides of the equation to use the Pythagorean identity. Let me try that.So, starting with cos α + sin α = -1/5. If I square both sides, I get:(cos α + sin α)^2 = (-1/5)^2Expanding the left side, I get cos²α + 2 sin α cos α + sin²α. And the right side is 1/25.Now, I know that cos²α + sin²α equals 1, so substituting that in, the equation becomes:1 + 2 sin α cos α = 1/25Subtracting 1 from both sides gives:2 sin α cos α = 1/25 - 1Calculating the right side, 1/25 - 1 is the same as 1/25 - 25/25, which equals -24/25.So, 2 sin α cos α = -24/25.Wait a second, 2 sin α cos α is equal to sin 2α, right? So, that means sin 2α = -24/25.But I need tan 2α, not sin 2α. Hmm, how do I get tan 2α from sin 2α?I recall that tan 2α is sin 2α divided by cos 2α. So, I have sin 2α, but I need cos 2α.How can I find cos 2α? Maybe I can use another identity. Let me think.I know that cos 2α can be expressed in terms of sin²α or cos²α. Since I have sin 2α already, maybe I can find cos 2α using the Pythagorean identity.The identity is sin² 2α + cos² 2α = 1. So, if I know sin 2α is -24/25, then sin² 2α is (24/25)^2, which is 576/625.Therefore, cos² 2α = 1 - 576/625 = (625 - 576)/625 = 49/625.Taking the square root, cos 2α is either 7/25 or -7/25. Hmm, which one is it?I need to determine the sign of cos 2α. Since α is between 0 and π, 2α is between 0 and 2π. But let's think about the specific range.Given that cos α + sin α is negative, which is -1/5. Since α is between 0 and π, cos α is positive in the first quadrant and negative in the second. Similarly, sin α is positive in both the first and second quadrants.So, if cos α + sin α is negative, that suggests that cos α is negative because sin α is positive. Therefore, α must be in the second quadrant, between π/2 and π.So, α is in the second quadrant, which means 2α is between π and 2π. Therefore, 2α is in the third or fourth quadrant.But since sin 2α is negative (-24/25), that means 2α is in the third or fourth quadrant where sine is negative. But since 2α is between π and 2π, it's either in the third or fourth.But in the third quadrant, both sine and cosine are negative, and in the fourth quadrant, sine is negative and cosine is positive.So, we have sin 2α negative. If 2α is in the third quadrant, cos 2α would be negative, and if it's in the fourth quadrant, cos 2α would be positive.But how do we determine which one it is?Wait, let's think about the original equation: cos α + sin α = -1/5. Since α is in the second quadrant, cos α is negative and sin α is positive. So, cos α is negative, sin α is positive, and their sum is negative, which makes sense because the magnitude of cos α is larger than sin α.So, let's try to find cos α and sin α individually. Maybe that can help us figure out where 2α is.Let me denote cos α = c and sin α = s. So, c + s = -1/5.We also know that c² + s² = 1.So, we have two equations:1. c + s = -1/52. c² + s² = 1Let me try to solve for c and s.From equation 1, I can write c = -1/5 - s.Substitute this into equation 2:(-1/5 - s)^2 + s² = 1Expanding the first term:(1/25 + 2*(1/5)*s + s²) + s² = 1Simplify:1/25 + (2/5)s + s² + s² = 1Combine like terms:1/25 + (2/5)s + 2s² = 1Subtract 1 from both sides:2s² + (2/5)s + 1/25 - 1 = 0Simplify 1/25 - 1:1/25 - 25/25 = -24/25So, equation becomes:2s² + (2/5)s - 24/25 = 0Multiply all terms by 25 to eliminate denominators:50s² + 10s - 24 = 0Simplify by dividing all terms by 2:25s² + 5s - 12 = 0Now, we have a quadratic equation in terms of s: 25s² + 5s - 12 = 0Let me solve for s using the quadratic formula.s = [-b ± sqrt(b² - 4ac)] / (2a)Here, a = 25, b = 5, c = -12.So, discriminant D = b² - 4ac = 25 - 4*25*(-12) = 25 + 1200 = 1225sqrt(D) = sqrt(1225) = 35So, s = [-5 ± 35]/(2*25) = (-5 ± 35)/50So, two solutions:1. s = (-5 + 35)/50 = 30/50 = 3/52. s = (-5 - 35)/50 = (-40)/50 = -4/5But since α is in the second quadrant, sin α is positive, so s = 3/5.So, sin α = 3/5.Then, from equation 1, c = -1/5 - s = -1/5 - 3/5 = -4/5.So, cos α = -4/5.Therefore, tan α = sin α / cos α = (3/5)/(-4/5) = -3/4.Now, to find tan 2α, I can use the double angle formula:tan 2α = 2 tan α / (1 - tan² α)Plugging in tan α = -3/4:tan 2α = 2*(-3/4) / [1 - (-3/4)^2] = (-6/4) / [1 - 9/16] = (-3/2) / (7/16)Dividing by 7/16 is the same as multiplying by 16/7:tan 2α = (-3/2)*(16/7) = (-48)/14 = -24/7So, tan 2α is -24/7.Wait, let me double-check my steps to make sure I didn't make a mistake.Starting from cos α + sin α = -1/5, squared to get 1 + 2 sin α cos α = 1/25, leading to sin 2α = -24/25. Then, found cos 2α by using sin² 2α + cos² 2α = 1, so cos 2α = ±7/25.But then, instead of directly using that, I solved for sin α and cos α individually, found sin α = 3/5, cos α = -4/5, then tan α = -3/4, then used the double angle formula to get tan 2α = -24/7.Alternatively, could I have used the identity for tan 2α in terms of sin 2α and cos 2α?Yes, since tan 2α = sin 2α / cos 2α. We have sin 2α = -24/25, and cos 2α is either 7/25 or -7/25.But earlier, we determined that 2α is in the third or fourth quadrant. Since sin 2α is negative, 2α is in the third or fourth. But from α being in the second quadrant, 2α is between π and 2π.But is 2α in the third or fourth? Let's see.If α is between π/2 and π, then 2α is between π and 2π.But to know whether it's in the third or fourth, we need more information.Wait, from sin 2α = -24/25, which is negative, so 2α is in the third or fourth quadrant.But from the original equation, cos α + sin α = -1/5, which is negative. Since α is in the second quadrant, cos α is negative and sin α is positive. So, their sum is negative, which tells us that the magnitude of cos α is greater than sin α.So, cos α is -4/5, sin α is 3/5, so indeed, |cos α| > sin α, which makes sense.So, 2α is in the third or fourth quadrant. But how do we know which one?Wait, if α is in the second quadrant, specifically, let's find the approximate value of α.Given that sin α = 3/5, so α is arcsin(3/5). Since α is in the second quadrant, α = π - arcsin(3/5).Calculating arcsin(3/5) is approximately 0.6435 radians, so α ≈ π - 0.6435 ≈ 2.498 radians.So, 2α ≈ 4.996 radians, which is just a bit less than 2π (which is about 6.283). So, 2α is in the fourth quadrant because it's between 3π/2 (≈4.712) and 2π.Therefore, cos 2α is positive because in the fourth quadrant, cosine is positive.So, cos 2α = 7/25.Therefore, tan 2α = sin 2α / cos 2α = (-24/25)/(7/25) = -24/7.So, that confirms the earlier result.Alternatively, if I had used the double angle formula for tangent directly, I would have gotten the same answer.So, I think my answer is correct: tan 2α is -24/7.

💡Alright, so I have this problem here about sequences and determining which general term formula corresponds to an increasing sequence. Let me try to figure this out step by step.First, let me recall what an increasing sequence is. An increasing sequence is one where each term is greater than the previous term. In other words, for all n, a_{n+1} > a_n. So, I need to check each of the given options to see if they satisfy this condition.The options are:A: a_n = -2n + 3B: a_n = n² - 3n + 1C: a_n = 1/(2^n)D: a_n = 1 + log₂nLet me go through each option one by one.Starting with Option A: a_n = -2n + 3.This is a linear function in terms of n. The coefficient of n is -2, which is negative. That means as n increases, the value of a_n decreases. So, this is a decreasing sequence, not an increasing one. Therefore, Option A is not the answer.Moving on to Option B: a_n = n² - 3n + 1.This is a quadratic function in n. Quadratic functions have a parabolic shape. The coefficient of n² is positive (1), so the parabola opens upwards. The vertex of this parabola is at n = -b/(2a) = 3/(2*1) = 1.5. Since n is a positive integer, the sequence will first decrease until it reaches the vertex and then start increasing. Let me compute the first few terms to check:For n=1: a₁ = 1 - 3 + 1 = -1For n=2: a₂ = 4 - 6 + 1 = -1For n=3: a₃ = 9 - 9 + 1 = 1For n=4: a₄ = 16 - 12 + 1 = 5So, the sequence goes -1, -1, 1, 5, ... Wait, the first two terms are the same. So, it's not strictly increasing because a₂ is not greater than a₁. It actually stays the same. Then, it starts increasing after that. But since the first two terms are equal, the sequence isn't strictly increasing from the start. Therefore, Option B is not the answer either.Next, Option C: a_n = 1/(2^n).This is an exponential function with base 1/2. Since the base is between 0 and 1, the function decreases as n increases. Let's verify:For n=1: a₁ = 1/2For n=2: a₂ = 1/4For n=3: a₃ = 1/8Clearly, each term is smaller than the previous one. So, this is a decreasing sequence, not increasing. Option C is out.Finally, Option D: a_n = 1 + log₂n.This is a logarithmic function with base 2. Logarithmic functions increase as n increases, but they do so at a decreasing rate. Let me compute the first few terms:For n=1: a₁ = 1 + log₂1 = 1 + 0 = 1For n=2: a₂ = 1 + log₂2 = 1 + 1 = 2For n=3: a₃ = 1 + log₂3 ≈ 1 + 1.58496 ≈ 2.58496For n=4: a₄ = 1 + log₂4 = 1 + 2 = 3So, the terms are 1, 2, approximately 2.585, 3, and so on. Each term is indeed larger than the previous one. Therefore, this is an increasing sequence.To make sure, let me check the difference between consecutive terms:a_{n+1} - a_n = [1 + log₂(n+1)] - [1 + log₂n] = log₂(n+1) - log₂n = log₂((n+1)/n) = log₂(1 + 1/n)Since n is a positive integer, 1/n is positive, so (n+1)/n > 1, which means log₂((n+1)/n) > 0. Therefore, a_{n+1} - a_n > 0 for all n, confirming that the sequence is strictly increasing.So, after checking all the options, only Option D satisfies the condition of being an increasing sequence.**Final Answer**boxed{D}

💡Alright, so I came across this interesting problem that asks to prove that infinitely many positive integers can be represented in the form ((a-1)/b + (b-1)/c + (c-1)/a), where (a), (b), and (c) are pairwise distinct positive integers greater than 1. At first glance, it seems a bit intimidating, but I think if I break it down step by step, I can figure it out.First, I need to understand what the problem is asking. It wants to show that there are infinitely many positive integers that can be expressed as the sum of three fractions: ((a-1)/b), ((b-1)/c), and ((c-1)/a). The variables (a), (b), and (c) have to be distinct and each greater than 1. So, my goal is to find a way to choose (a), (b), and (c) such that this sum is an integer, and that there are infinitely many such integers.I think a good starting point is to try to find a pattern or a specific choice of (a), (b), and (c) that can generate such integers. Maybe if I can find a relationship between (a), (b), and (c) that simplifies the expression, it could lead me somewhere.Let me consider setting (a), (b), and (c) in terms of a single variable, say (k), which can take on infinitely many values. If I can express (a), (b), and (c) as functions of (k), then perhaps the expression will simplify to an integer that depends on (k), thereby giving me infinitely many integers.Let me try to define (a), (b), and (c) in a way that might make the fractions simplify nicely. Suppose I set (a = (2k + 1)k), (b = (2k + 1)k - 1), and (c = 2k + 1). Here, (k) is a positive integer greater than 1. I chose these expressions because they seem to create a relationship where (a) is slightly larger than (b), and (c) is a smaller number that could help in simplifying the fractions.Now, let's substitute these into the expression:1. Calculate ((a - 1)/b): [ a - 1 = (2k + 1)k - 1 ] [ b = (2k + 1)k - 1 ] So, ((a - 1)/b = 1), because the numerator and denominator are the same.2. Calculate ((b - 1)/c): [ b - 1 = (2k + 1)k - 2 ] [ c = 2k + 1 ] So, ((b - 1)/c = frac{(2k + 1)k - 2}{2k + 1}). Let's simplify this: [ frac{(2k + 1)k - 2}{2k + 1} = k - frac{2}{2k + 1} ] Because ((2k + 1)k = 2k^2 + k), so subtracting 2 gives (2k^2 + k - 2), and dividing by (2k + 1) gives (k - frac{2}{2k + 1}).3. Calculate ((c - 1)/a): [ c - 1 = 2k ] [ a = (2k + 1)k ] So, ((c - 1)/a = frac{2k}{(2k + 1)k} = frac{2}{2k + 1}).Now, adding all three parts together:[1 + left(k - frac{2}{2k + 1}right) + frac{2}{2k + 1}]Simplify this expression:[1 + k - frac{2}{2k + 1} + frac{2}{2k + 1} = 1 + k]So, the entire expression simplifies to (k + 1), which is clearly an integer for any positive integer (k) greater than 1.This means that for each positive integer (k > 1), we can choose (a = (2k + 1)k), (b = (2k + 1)k - 1), and (c = 2k + 1), and the expression ((a - 1)/b + (b - 1)/c + (c - 1)/a) will equal (k + 1). Since (k) can be any positive integer greater than 1, this gives us infinitely many positive integers that can be represented in the desired form.To make sure this works, let's test it with a specific value of (k). Let's take (k = 2):- (a = (2*2 + 1)*2 = 5*2 = 10)- (b = 10 - 1 = 9)- (c = 2*2 + 1 = 5)Now, compute each term:- ((a - 1)/b = (10 - 1)/9 = 9/9 = 1)- ((b - 1)/c = (9 - 1)/5 = 8/5 = 1.6)- ((c - 1)/a = (5 - 1)/10 = 4/10 = 0.4)Adding them up: (1 + 1.6 + 0.4 = 3), which is indeed (k + 1 = 2 + 1 = 3).Let's try another one, say (k = 3):- (a = (2*3 + 1)*3 = 7*3 = 21)- (b = 21 - 1 = 20)- (c = 2*3 + 1 = 7)Compute each term:- ((a - 1)/b = 20/20 = 1)- ((b - 1)/c = 19/7 ≈ 2.714)- ((c - 1)/a = 6/21 ≈ 0.286)Adding them up: (1 + 2.714 + 0.286 = 4), which is (k + 1 = 3 + 1 = 4).This seems consistent. Therefore, by choosing (a), (b), and (c) in this specific way, we can generate infinitely many positive integers of the form (k + 1), which are clearly positive integers.I should also check if (a), (b), and (c) are pairwise distinct and greater than 1. For (k = 2), (a = 10), (b = 9), (c = 5) – all distinct and greater than 1. For (k = 3), (a = 21), (b = 20), (c = 7) – again, all distinct and greater than 1. This pattern should hold for any (k > 1), ensuring that (a), (b), and (c) are always distinct and greater than 1.One thing I need to be careful about is ensuring that the denominators in the fractions are never zero, but since (a), (b), and (c) are all greater than 1, their denominators (b), (c), and (a) are all at least 2, so we don't have any division by zero issues.Another point to consider is whether this construction covers all positive integers or just an infinite subset. In this case, we're generating integers of the form (k + 1), which are all integers greater than 2. So, while we're not covering every positive integer, we are covering an infinite subset of them, which is sufficient for the problem's requirement.I also wonder if there are other ways to choose (a), (b), and (c) to get different forms of integers, but for the purpose of proving that infinitely many such integers exist, this construction seems sufficient.In summary, by defining (a), (b), and (c) in terms of a parameter (k), we can ensure that the expression ((a - 1)/b + (b - 1)/c + (c - 1)/a) simplifies to an integer (k + 1), and since (k) can be any positive integer greater than 1, this provides an infinite number of positive integers that can be represented in the desired form.