💡To estimate the year when the population of Zalotis will be approximately 15,000, we start with the initial population of 750 in 1950 and apply the doubling rule every 35 years.1. **Year 1950:** Population = 7502. **Year 1985:** Population doubles to 1,5003. **Year 2020:** Population doubles to 3,0004. **Year 2055:** Population doubles to 6,0005. **Year 2090:** Population doubles to 12,0006. **Year 2125:** Population doubles to 24,000The population first exceeds 15,000 in the year 2125. Therefore, the estimated year is 2125.

💡Okay, so I have this problem here: find the sum of the series (-39) + (-37) + ... + (-1). Hmm, let me think about how to approach this. I remember that when you're dealing with a series of numbers that increase by a constant difference, it's called an arithmetic series. In this case, each term increases by 2, right? So, from -39 to -37 is +2, then -37 to -35 is another +2, and so on, all the way up to -1.Alright, so first things first, I need to figure out how many terms are in this series. I know the first term is -39, and the last term is -1. The common difference, which is the amount each term increases by, is 2. There's a formula to find the number of terms in an arithmetic sequence: n = ((last term - first term) / common difference) + 1Let me plug in the numbers:n = ((-1 - (-39)) / 2) + 1Simplifying inside the parentheses first:-1 - (-39) is the same as -1 + 39, which equals 38.So now we have:n = (38 / 2) + 138 divided by 2 is 19, so:n = 19 + 1 = 20Okay, so there are 20 terms in this series. That makes sense because starting from -39 and adding 2 each time, it should take 20 steps to get to -1.Now, to find the sum of an arithmetic series, there's another formula:Sum = (number of terms / 2) * (first term + last term)Plugging in the values I have:Sum = (20 / 2) * (-39 + (-1))Simplifying:20 divided by 2 is 10.-39 + (-1) is -40.So now:Sum = 10 * (-40) = -400Wait, let me double-check that. If I have 20 terms, each pair of terms adds up to the same thing. For example, the first term is -39 and the last term is -1, which adds up to -40. The second term is -37 and the second-to-last term is -3, which also adds up to -40. This pattern continues all the way through the series. Since there are 20 terms, there are 10 such pairs, each adding up to -40. So, 10 pairs times -40 equals -400. Yeah, that checks out.I think I've got it right. The sum of the series is -400.

💡Okay, so I have this geometry problem here about an equilateral triangle ABC. On side AB, there's a point M chosen. Then, on segment AM, outside of triangle ABC, another equilateral triangle AMN is constructed. K is the intersection of lines CM and AN, and D is the intersection of lines AC and BN. I need to prove that triangle AKD is equilateral. Hmm, that sounds a bit complicated, but let me try to break it down step by step.First, I should probably draw a diagram to visualize what's going on. Let me sketch triangle ABC with AB as the base. Point M is somewhere on AB. Then, from A to M, I construct another equilateral triangle AMN outside of ABC. So, triangle AMN is equilateral, meaning all its sides are equal, and all its angles are 60 degrees. That should help with some angle chasing later.Now, K is the intersection of CM and AN. So, I need to draw lines CM and AN and see where they cross. Similarly, D is where AC and BN intersect. So, I'll draw BN and see where it crosses AC. Once I have points K and D, I need to show that triangle AKD is equilateral. That means I need to show that all sides AK, KD, and DA are equal, or that all angles are 60 degrees.Let me think about the properties of equilateral triangles. Since ABC is equilateral, all its sides are equal, and all its angles are 60 degrees. Similarly, triangle AMN is equilateral, so angles at A, M, and N are all 60 degrees. Maybe I can use some congruency or similarity of triangles here.Looking at triangle AMC and triangle ANB. Hmm, do they have any common sides or angles? Well, they both share side AM. Also, since AMN is equilateral, angle NAM is 60 degrees. In triangle ABC, angle BAC is also 60 degrees. So, maybe triangles AMC and ANB are congruent? If they are, that would mean some sides and angles are equal, which might help in proving AKD is equilateral.Wait, let me check that. For triangles AMC and ANB to be congruent, they need to satisfy some congruence criteria like SAS, SSS, ASA, etc. They share side AM. Angle at A is 60 degrees for both triangles. But what about the other sides? In triangle AMC, the sides are AM, MC, and AC. In triangle ANB, the sides are AN, NB, and AB. Since AMN is equilateral, AN = AM. But AB is a side of the original triangle ABC, which is also equilateral, so AB = AC. So, if AM is a part of AB, then AM is less than AB, which is equal to AC. Hmm, so AM is not equal to AB or AC unless M is at B, which it's not necessarily. So, maybe triangles AMC and ANB aren't congruent after all.Maybe I need a different approach. Let's consider the angles. Since AMN is equilateral, angle MAN is 60 degrees. In triangle ABC, angle BAC is also 60 degrees. So, if I look at point A, the angles around it are 60 degrees from both triangles ABC and AMN. Maybe that can help me find some similar triangles or something.Looking at point K, which is the intersection of CM and AN. So, K is inside the figure somewhere. Similarly, D is the intersection of AC and BN. So, D is also inside the figure. Maybe I can use some properties of intersecting lines and triangles.Another idea: Maybe using coordinate geometry. If I assign coordinates to the points, I can calculate the coordinates of K and D and then compute the distances AK, KD, and DA to see if they are equal. That might be a bit involved, but it could work.Let me try that. Let's place point A at (0, 0), point B at (1, 0), and since ABC is equilateral, point C will be at (0.5, sqrt(3)/2). Now, point M is somewhere on AB. Let's say M is at (m, 0), where 0 < m < 1. Then, triangle AMN is equilateral, constructed outside ABC. So, point N should be somewhere such that AMN is equilateral.To find the coordinates of N, since AM is from (0,0) to (m,0), the length of AM is m. To construct an equilateral triangle outside ABC, point N will be either above or below the line AM. Since ABC is above AB, and we need to construct AMN outside ABC, so N should be below AB.The coordinates of N can be found using rotation. If we rotate point M around point A by -60 degrees, we'll get point N. The rotation matrix for -60 degrees is:[ cos(-60) -sin(-60) ][ sin(-60) cos(-60) ]Which is:[ 0.5 sqrt(3)/2 ][ -sqrt(3)/2 0.5 ]So, applying this to point M (m, 0):x' = 0.5*m - sqrt(3)/2*0 = 0.5my' = -sqrt(3)/2*m + 0.5*0 = -sqrt(3)/2 mSo, point N is at (0.5m, -sqrt(3)/2 m).Now, let's find the equations of lines CM and AN to find point K.First, line CM connects point C (0.5, sqrt(3)/2) to point M (m, 0). The slope of CM is (0 - sqrt(3)/2)/(m - 0.5) = (-sqrt(3)/2)/(m - 0.5). Let's denote this slope as s1.So, the equation of CM is y - sqrt(3)/2 = s1(x - 0.5).Similarly, line AN connects point A (0,0) to point N (0.5m, -sqrt(3)/2 m). The slope of AN is (-sqrt(3)/2 m - 0)/(0.5m - 0) = (-sqrt(3)/2 m)/(0.5m) = -sqrt(3). So, the equation of AN is y = -sqrt(3)x.Now, to find point K, we need to solve the equations of CM and AN.Equation of CM: y = s1(x - 0.5) + sqrt(3)/2Equation of AN: y = -sqrt(3)xSet them equal:-sqrt(3)x = s1(x - 0.5) + sqrt(3)/2Let me compute s1:s1 = (-sqrt(3)/2)/(m - 0.5) = (-sqrt(3)/2)/(m - 0.5)So, plugging back in:-sqrt(3)x = [(-sqrt(3)/2)/(m - 0.5)](x - 0.5) + sqrt(3)/2Let me multiply both sides by (m - 0.5) to eliminate the denominator:-sqrt(3)x(m - 0.5) = (-sqrt(3)/2)(x - 0.5) + sqrt(3)/2(m - 0.5)Let me expand both sides:Left side: -sqrt(3)x(m - 0.5)Right side: (-sqrt(3)/2)x + (sqrt(3)/4) + (sqrt(3)/2)m - (sqrt(3)/4)Simplify the right side:(-sqrt(3)/2)x + (sqrt(3)/4 + sqrt(3)/4) + (sqrt(3)/2)m - (sqrt(3)/4)Wait, that seems messy. Maybe I made a miscalculation. Let me re-express the right side:(-sqrt(3)/2)(x - 0.5) + sqrt(3)/2(m - 0.5) =(-sqrt(3)/2)x + (sqrt(3)/4) + (sqrt(3)/2)m - (sqrt(3)/4)Ah, the sqrt(3)/4 and -sqrt(3)/4 cancel out. So, right side simplifies to:(-sqrt(3)/2)x + (sqrt(3)/2)mSo, now we have:-sqrt(3)x(m - 0.5) = (-sqrt(3)/2)x + (sqrt(3)/2)mLet me divide both sides by sqrt(3) to simplify:-x(m - 0.5) = (-1/2)x + (1/2)mExpand the left side:- m x + 0.5x = -0.5x + 0.5mBring all terms to the left side:- m x + 0.5x + 0.5x - 0.5m = 0Combine like terms:(-m x + x) + (-0.5m) = 0Factor x:x(-m + 1) - 0.5m = 0So,x(1 - m) = 0.5mTherefore,x = (0.5m)/(1 - m)Then, y = -sqrt(3)x = -sqrt(3)*(0.5m)/(1 - m) = (-sqrt(3)/2)m/(1 - m)So, point K is at ((0.5m)/(1 - m), (-sqrt(3)/2)m/(1 - m))Now, let's find point D, which is the intersection of AC and BN.First, find the equation of AC. Since AC connects A(0,0) to C(0.5, sqrt(3)/2). The slope is (sqrt(3)/2 - 0)/(0.5 - 0) = (sqrt(3)/2)/0.5 = sqrt(3). So, equation of AC is y = sqrt(3)x.Next, find the equation of BN. Point B is at (1,0), and point N is at (0.5m, -sqrt(3)/2 m). So, the slope of BN is (-sqrt(3)/2 m - 0)/(0.5m - 1) = (-sqrt(3)/2 m)/(0.5m - 1)Let me denote this slope as s2.So, s2 = (-sqrt(3)/2 m)/(0.5m - 1) = [(-sqrt(3)/2 m)]/[0.5(m - 2)]Simplify denominator: 0.5(m - 2) = (m - 2)/2So, s2 = [(-sqrt(3)/2 m)] / [(m - 2)/2] = (-sqrt(3)/2 m) * (2/(m - 2)) = (-sqrt(3) m)/(m - 2)So, the equation of BN is y - 0 = s2(x - 1), which is y = [(-sqrt(3) m)/(m - 2)](x - 1)Now, to find point D, we need to solve the equations of AC and BN.Equation of AC: y = sqrt(3)xEquation of BN: y = [(-sqrt(3) m)/(m - 2)](x - 1)Set them equal:sqrt(3)x = [(-sqrt(3) m)/(m - 2)](x - 1)Divide both sides by sqrt(3):x = [(-m)/(m - 2)](x - 1)Multiply both sides by (m - 2):x(m - 2) = -m(x - 1)Expand both sides:m x - 2x = -m x + mBring all terms to the left:m x - 2x + m x - m = 0Combine like terms:(2m x - 2x) - m = 0Factor x:x(2m - 2) - m = 0Factor 2 from x term:2x(m - 1) - m = 0So,2x(m - 1) = mTherefore,x = m / [2(m - 1)] = m / [2(m - 1)]Then, y = sqrt(3)x = sqrt(3)*m / [2(m - 1)]So, point D is at (m / [2(m - 1)], sqrt(3)*m / [2(m - 1)])Now, I have coordinates for points A(0,0), K((0.5m)/(1 - m), (-sqrt(3)/2)m/(1 - m)), and D(m / [2(m - 1)], sqrt(3)*m / [2(m - 1)])I need to compute the distances AK, KD, and DA to see if they are equal.First, compute AK.Coordinates of A: (0,0)Coordinates of K: (0.5m/(1 - m), (-sqrt(3)/2 m)/(1 - m))Distance AK:sqrt[(0.5m/(1 - m) - 0)^2 + ((-sqrt(3)/2 m)/(1 - m) - 0)^2]= sqrt[(0.25m²)/(1 - m)^2 + (3/4 m²)/(1 - m)^2]= sqrt[(0.25 + 0.75)m²/(1 - m)^2]= sqrt[m²/(1 - m)^2]= |m| / |1 - m|Since m is between 0 and 1, 1 - m is positive, so:AK = m / (1 - m)Next, compute DA.Coordinates of D: (m / [2(m - 1)], sqrt(3)*m / [2(m - 1)])Coordinates of A: (0,0)Distance DA:sqrt[(m / [2(m - 1)] - 0)^2 + (sqrt(3)*m / [2(m - 1)] - 0)^2]= sqrt[(m² / [4(m - 1)^2]) + (3m² / [4(m - 1)^2])]= sqrt[(4m²) / [4(m - 1)^2]]= sqrt[m² / (m - 1)^2]= |m| / |m - 1|Since m is between 0 and 1, m - 1 is negative, so |m - 1| = 1 - m. Therefore,DA = m / (1 - m)So, AK = DA = m / (1 - m)Now, compute KD.Coordinates of K: (0.5m/(1 - m), (-sqrt(3)/2 m)/(1 - m))Coordinates of D: (m / [2(m - 1)], sqrt(3)*m / [2(m - 1)])First, let's note that m / [2(m - 1)] = -m / [2(1 - m)]Similarly, sqrt(3)*m / [2(m - 1)] = -sqrt(3)*m / [2(1 - m)]So, coordinates of D can be written as (-m / [2(1 - m)], -sqrt(3)*m / [2(1 - m)])So, coordinates of K: (0.5m/(1 - m), (-sqrt(3)/2 m)/(1 - m)) = (m/(2(1 - m)), -sqrt(3)m/(2(1 - m)))Coordinates of D: (-m/(2(1 - m)), -sqrt(3)m/(2(1 - m)))So, the distance KD is the distance between (m/(2(1 - m)), -sqrt(3)m/(2(1 - m))) and (-m/(2(1 - m)), -sqrt(3)m/(2(1 - m)))Compute the differences:Δx = (-m/(2(1 - m)) - m/(2(1 - m))) = (-2m)/(2(1 - m)) = -m/(1 - m)Δy = (-sqrt(3)m/(2(1 - m)) - (-sqrt(3)m/(2(1 - m)))) = 0So, distance KD is sqrt[(-m/(1 - m))² + 0²] = |m/(1 - m)|Since m is between 0 and 1, 1 - m is positive, so KD = m/(1 - m)Therefore, all sides AK, KD, and DA are equal to m/(1 - m). Thus, triangle AKD is equilateral.Wait, that seems straightforward. So, by calculating the coordinates and distances, I showed that all sides are equal. Therefore, triangle AKD is equilateral. That should be the proof.**Final Answer**boxed{triangle AKD text{ is equilateral}}

💡Okay, so I have this function f(x) = sin²(2x) + √3 sin(2x)cos(2x). I need to find its smallest positive period and then solve for x when f(x) = 1 within a specific interval. Hmm, let's start with part (1).First, I remember that the period of a function is the smallest positive number T such that f(x + T) = f(x) for all x. For sine and cosine functions, the basic period is 2π, but when there's a coefficient inside the function, like sin(kx), the period becomes 2π/k. So, for sin(2x), the period would be π, right? Similarly, cos(2x) also has a period of π.But in this function, we have sin²(2x) and sin(2x)cos(2x). I think I need to simplify f(x) first to see if it can be written in a more manageable form, maybe using some trigonometric identities.Let me recall that sin²(θ) can be written as (1 - cos(2θ))/2. So, applying that to sin²(2x), we get:sin²(2x) = (1 - cos(4x))/2.Okay, that's one part. Now, the other term is √3 sin(2x)cos(2x). I remember that sin(2θ) = 2 sinθ cosθ, but here it's sin(2x)cos(2x). Wait, maybe I can use the identity for sin(A)cos(A). Let me think. There's an identity that says sin(A)cos(A) = (sin(2A))/2. So, in this case, sin(2x)cos(2x) would be (sin(4x))/2.So, substituting that into the second term:√3 sin(2x)cos(2x) = √3*(sin(4x)/2) = (√3/2) sin(4x).So now, putting it all together, f(x) becomes:f(x) = (1 - cos(4x))/2 + (√3/2) sin(4x).Let me write that as:f(x) = 1/2 - (cos(4x))/2 + (√3/2) sin(4x).Hmm, this looks like a combination of sine and cosine terms with the same argument, 4x. I think I can combine these into a single sine or cosine function using the amplitude-phase form. The general formula is A sin(θ) + B cos(θ) = C sin(θ + φ), where C = √(A² + B²) and tanφ = B/A.Wait, actually, it's either sine or cosine, depending on the phase shift. Let me recall the exact identity.I think it's A sinθ + B cosθ = C sin(θ + φ), where C = √(A² + B²) and φ = arctan(B/A) or something like that. Let me verify.Yes, the identity is:A sinθ + B cosθ = C sin(θ + φ),where C = √(A² + B²),and φ = arctan(B/A) if A ≠ 0.Alternatively, it can also be written as C cos(θ - φ'), depending on the phase shift.But let me stick with sine for now.In our case, the coefficients are:A = √3/2,B = -1/2.Wait, because we have (√3/2) sin(4x) - (1/2) cos(4x).So, A = √3/2,B = -1/2.Therefore, C = √[(√3/2)² + (-1/2)²] = √[(3/4) + (1/4)] = √[1] = 1.Okay, so the amplitude C is 1.Now, φ is given by arctan(B/A). So, φ = arctan[(-1/2)/(√3/2)] = arctan(-1/√3).I know that arctan(-1/√3) is -π/6 because tan(-π/6) = -1/√3.So, φ = -π/6.Therefore, we can write:(√3/2) sin(4x) - (1/2) cos(4x) = sin(4x - π/6).Wait, let me check that.Using the identity:A sinθ + B cosθ = C sin(θ + φ),where φ = arctan(B/A).Wait, actually, I think I might have mixed up the formula. Let me double-check.The identity is:A sinθ + B cosθ = C sin(θ + φ),where C = √(A² + B²),and φ = arctan(B/A) if A ≠ 0.But in our case, it's A sinθ + B cosθ, so:(√3/2) sin(4x) + (-1/2) cos(4x) = sin(4x - π/6).Wait, let me verify this.Using the sine addition formula:sin(4x - π/6) = sin(4x)cos(π/6) - cos(4x)sin(π/6).We know that cos(π/6) = √3/2,and sin(π/6) = 1/2.So,sin(4x - π/6) = sin(4x)*(√3/2) - cos(4x)*(1/2).Which is exactly (√3/2) sin(4x) - (1/2) cos(4x).So, yes, that's correct.Therefore, f(x) can be rewritten as:f(x) = 1/2 + sin(4x - π/6).So, f(x) = sin(4x - π/6) + 1/2.Okay, that simplifies the function a lot.Now, to find the period of f(x). Since f(x) is expressed as sin(4x - π/6) + 1/2, the period of the sine function is 2π divided by the coefficient of x inside the sine, which is 4.So, the period T = 2π / 4 = π/2.Therefore, the smallest positive period of f(x) is π/2.Wait, but let me make sure. The function f(x) is a combination of sin²(2x) and sin(2x)cos(2x). Each of these terms individually has a period of π, right? Because sin²(2x) has period π, and sin(2x)cos(2x) also has period π.But when we combine them, the resulting function f(x) has a period of π/2. That seems correct because when we simplified it, the argument inside the sine became 4x, which has a smaller period.So, yes, the smallest positive period is π/2.Okay, that was part (1). Now, moving on to part (2).We need to find x in the interval [π/8, π/4] such that f(x) = 1.From part (1), we have f(x) = sin(4x - π/6) + 1/2.So, setting f(x) = 1:sin(4x - π/6) + 1/2 = 1.Subtracting 1/2 from both sides:sin(4x - π/6) = 1/2.So, we need to solve sin(θ) = 1/2, where θ = 4x - π/6.The general solution for sin(θ) = 1/2 is θ = π/6 + 2πk or θ = 5π/6 + 2πk, where k is any integer.So, substituting back θ = 4x - π/6:4x - π/6 = π/6 + 2πk,or4x - π/6 = 5π/6 + 2πk.Let's solve for x in both cases.First case:4x - π/6 = π/6 + 2πk,Adding π/6 to both sides:4x = π/6 + π/6 + 2πk = π/3 + 2πk,Dividing both sides by 4:x = π/12 + (π/2)k.Second case:4x - π/6 = 5π/6 + 2πk,Adding π/6 to both sides:4x = 5π/6 + π/6 + 2πk = 6π/6 + 2πk = π + 2πk,Dividing both sides by 4:x = π/4 + (π/2)k.Now, we need to find all x in the interval [π/8, π/4].Let's consider the first case: x = π/12 + (π/2)k.We need to find integer k such that x is in [π/8, π/4].Let's compute x for k = 0: x = π/12 ≈ 0.2618 radians.π/8 ≈ 0.3927 radians.So, π/12 is less than π/8, so it's not in the interval.For k = 1: x = π/12 + π/2 = π/12 + 6π/12 = 7π/12 ≈ 1.8326 radians, which is greater than π/4 ≈ 0.7854 radians. So, not in the interval.Similarly, negative k would give x less than π/12, which is also less than π/8. So, no solutions in the first case.Now, the second case: x = π/4 + (π/2)k.Again, let's find k such that x is in [π/8, π/4].For k = 0: x = π/4 ≈ 0.7854 radians, which is the upper bound of the interval. So, x = π/4 is included.For k = -1: x = π/4 - π/2 = -π/4, which is negative, so not in the interval.For k = 1: x = π/4 + π/2 = 3π/4 ≈ 2.3562 radians, which is greater than π/4. So, not in the interval.Therefore, the only solution in the interval [π/8, π/4] is x = π/4.Wait, but let me check if x = π/4 is indeed a solution.Plugging x = π/4 into f(x):f(π/4) = sin²(2*(π/4)) + √3 sin(2*(π/4))cos(2*(π/4)).Simplify:2*(π/4) = π/2.So,sin²(π/2) + √3 sin(π/2)cos(π/2).sin(π/2) = 1,cos(π/2) = 0.Therefore,f(π/4) = 1² + √3*1*0 = 1 + 0 = 1.Yes, that checks out.But wait, I also need to check if there's another solution in the interval. The general solution for sin(θ) = 1/2 is θ = π/6 + 2πk or θ = 5π/6 + 2πk.So, when solving for x, we have two cases:Case 1: 4x - π/6 = π/6 + 2πk,Case 2: 4x - π/6 = 5π/6 + 2πk.We found that in Case 1, x = π/12 + (π/2)k, which for k=0 gives x=π/12, which is less than π/8, so not in the interval.In Case 2, x = π/4 + (π/2)k, which for k=0 gives x=π/4, which is in the interval.But wait, let's check if there's another solution in the interval for k=0 in Case 2.Wait, x=π/4 is the upper limit of the interval. Is there another solution when k=0 in Case 2? No, because for k=0, x=π/4 is the only solution.But let me think again. The interval is [π/8, π/4], which is approximately [0.3927, 0.7854].We found x=π/4 is a solution. Is there another solution within this interval?Let me consider the equation sin(4x - π/6) = 1/2.We can also think about the range of θ = 4x - π/6 when x is in [π/8, π/4].Compute θ:When x = π/8,θ = 4*(π/8) - π/6 = π/2 - π/6 = (3π/6 - π/6) = 2π/6 = π/3 ≈ 1.0472 radians.When x = π/4,θ = 4*(π/4) - π/6 = π - π/6 = 5π/6 ≈ 2.61799 radians.So, θ ranges from π/3 to 5π/6.We need to find θ in [π/3, 5π/6] such that sinθ = 1/2.We know that sinθ = 1/2 at θ = π/6 and θ = 5π/6 in the interval [0, 2π].But in our case, θ is in [π/3, 5π/6].So, θ = π/6 is less than π/3, so it's not in the interval.θ = 5π/6 is exactly the upper bound of θ, which corresponds to x=π/4.Therefore, the only solution in θ is 5π/6, which corresponds to x=π/4.Hence, the only solution in the interval [π/8, π/4] is x=π/4.Wait, but let me double-check. Is there another angle in [π/3, 5π/6] where sinθ=1/2?No, because sinθ=1/2 only at θ=π/6 and θ=5π/6 in [0, 2π]. Since π/6 is less than π/3, it's not in our interval. So, only θ=5π/6 is in [π/3, 5π/6].Therefore, x=π/4 is the only solution.So, summarizing:(1) The smallest positive period of f(x) is π/2.(2) The value of x in [π/8, π/4] where f(x)=1 is π/4.

💡Okay, so I have this problem: I need to find the coefficient of the (x^3) term in the expansion of ((x^2 - 1)(x - 2)^7). Hmm, that seems a bit tricky at first glance, but maybe I can break it down step by step.First, I remember that when you have two polynomials multiplied together, like ((x^2 - 1)) and ((x - 2)^7), the resulting expansion will have terms that are combinations of the terms from each polynomial. So, to find the coefficient of (x^3), I need to figure out which combinations of terms from ((x^2 - 1)) and ((x - 2)^7) will result in an (x^3) term when multiplied together.Let me write this out more formally. If I have two polynomials (A(x)) and (B(x)), then the product (A(x) cdot B(x)) will have terms where each term from (A(x)) is multiplied by each term from (B(x)). So, in this case, (A(x) = x^2 - 1) and (B(x) = (x - 2)^7). Therefore, the product will be:[(x^2 - 1)(x - 2)^7]I need to find the coefficient of (x^3) in this product. To do that, I can consider each term in (A(x)) and see how it interacts with the terms in (B(x)) to produce an (x^3) term.So, let's break down (A(x)):1. The first term is (x^2).2. The second term is (-1).Now, I need to find the terms in (B(x) = (x - 2)^7) that, when multiplied by (x^2) and (-1), will result in an (x^3) term.Let me denote the expansion of (B(x)) as:[B(x) = sum_{k=0}^{7} binom{7}{k} x^{7 - k} (-2)^k]This is the binomial expansion of ((x - 2)^7). Each term in this expansion is of the form (binom{7}{k} x^{7 - k} (-2)^k), where (k) ranges from 0 to 7.Now, let's consider how each term in (A(x)) interacts with (B(x)):1. The term (x^2) in (A(x)) will multiply each term in (B(x)). So, when (x^2) is multiplied by a term (x^{7 - k}) in (B(x)), the resulting term will be (x^{2 + (7 - k)} = x^{9 - k}). We want this to be (x^3), so we set up the equation:[9 - k = 3 implies k = 6]Therefore, the term in (B(x)) that, when multiplied by (x^2), gives an (x^3) term is the term where (k = 6). Let's find the coefficient of this term:[binom{7}{6} (-2)^6 = 7 times 64 = 448]So, the contribution to the (x^3) term from the (x^2) term in (A(x)) is (448x^3).2. The second term in (A(x)) is (-1). When (-1) is multiplied by each term in (B(x)), the resulting term will be (-x^{7 - k}). We want this to be (x^3), so we set up the equation:[7 - k = 3 implies k = 4]Therefore, the term in (B(x)) that, when multiplied by (-1), gives an (x^3) term is the term where (k = 4). Let's find the coefficient of this term:[binom{7}{4} (-2)^4 = 35 times 16 = 560]But since we're multiplying by (-1), the contribution to the (x^3) term is (-560x^3).Now, to find the total coefficient of (x^3) in the expansion of ((x^2 - 1)(x - 2)^7), we add the contributions from both terms in (A(x)):[448x^3 - 560x^3 = (448 - 560)x^3 = -112x^3]Therefore, the coefficient of the (x^3) term is (-112).Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, for the term where (k = 6):[binom{7}{6} = 7 quad text{and} quad (-2)^6 = 64][7 times 64 = 448]That seems correct.Next, for the term where (k = 4):[binom{7}{4} = 35 quad text{and} quad (-2)^4 = 16][35 times 16 = 560]Yes, that's correct as well.Then, subtracting the two contributions:[448 - 560 = -112]Yes, that checks out.So, I think my final answer is correct. The coefficient of the (x^3) term is (-112).**Final Answer**The coefficient of the (x^3) term is boxed{-112}.

💡Okay, so I have this problem about selecting workers for two different jobs: pliers workers and car workers. There are 11 workers in total, and they have different capabilities. Let me try to break this down step by step.First, let me categorize the workers based on their skills:- 5 workers can only work as pliers workers. Let's call them P-only.- 4 workers can only work as car workers. Let's call them C-only.- 2 workers can work both as pliers and car workers. Let's call them Both.So, in total, we have 5 P-only, 4 C-only, and 2 Both, making 11 workers.The task is to select 4 workers for pliers jobs and 4 workers for car jobs. So, we need to choose 4 pliers workers and 4 car workers from these 11. But since some workers can do both, we have to be careful about how we count the selections to avoid overcounting or undercounting.Let me think about how to approach this. Since the Both workers can be assigned to either job, their assignment affects how many P-only and C-only workers we need to select.I think the best way is to consider different cases based on how many Both workers are assigned to car jobs. Because the C-only workers are limited to 4, and we need to select 4 car workers, the number of Both workers assigned to car jobs can vary, which will affect how many P-only workers we need to select for pliers jobs.Let me outline the possible cases:1. **Case 1:** All 4 car workers are C-only. That means we don't use any Both workers for car jobs. Then, we need to select all 4 pliers workers from the P-only and Both workers.2. **Case 2:** 3 car workers are C-only, and 1 is a Both worker. Then, we need to select 3 C-only and 1 Both for car jobs, and then select 4 pliers workers from the remaining P-only and Both workers.3. **Case 3:** 2 car workers are C-only, and 2 are Both workers. Then, we need to select 2 C-only and 2 Both for car jobs, and then select 4 pliers workers from the remaining P-only workers.Wait, can we have more than 2 Both workers assigned to car jobs? Since there are only 2 Both workers, the maximum number we can assign to car jobs is 2. So, these three cases cover all possibilities.Let me formalize each case.**Case 1: All 4 car workers are C-only.**- Number of ways to choose 4 C-only workers from 4 available: Since we need all 4, there's only 1 way.- Then, we need to choose 4 pliers workers from the remaining workers. The remaining workers are 5 P-only and 2 Both, so 7 workers in total.- The number of ways to choose 4 pliers workers from 7 is C(7,4).So, total ways for Case 1: 1 * C(7,4).**Case 2: 3 car workers are C-only, and 1 is a Both worker.**- Number of ways to choose 3 C-only workers from 4: C(4,3).- Number of ways to choose 1 Both worker from 2: C(2,1).- Then, we need to choose 4 pliers workers from the remaining workers. After selecting 3 C-only and 1 Both for car jobs, we have 5 P-only and 1 Both left (since 1 Both was used for car).- So, total pliers workers available: 5 + 1 = 6.- Number of ways to choose 4 pliers workers from 6: C(6,4).So, total ways for Case 2: C(4,3) * C(2,1) * C(6,4).**Case 3: 2 car workers are C-only, and 2 are Both workers.**- Number of ways to choose 2 C-only workers from 4: C(4,2).- Number of ways to choose 2 Both workers from 2: C(2,2).- Then, we need to choose 4 pliers workers from the remaining workers. After selecting 2 C-only and 2 Both for car jobs, we have 5 P-only left (since all Both workers are used for car).- So, total pliers workers available: 5.- Number of ways to choose 4 pliers workers from 5: C(5,4).So, total ways for Case 3: C(4,2) * C(2,2) * C(5,4).Now, let me calculate each case separately.**Calculations:**First, let me recall that C(n,k) is the combination formula, which is n! / (k! * (n - k)!).**Case 1:**- C(7,4) = 35.So, total ways: 1 * 35 = 35.**Case 2:**- C(4,3) = 4.- C(2,1) = 2.- C(6,4) = 15.So, total ways: 4 * 2 * 15 = 120.**Case 3:**- C(4,2) = 6.- C(2,2) = 1.- C(5,4) = 5.So, total ways: 6 * 1 * 5 = 30.Now, add up all the cases:35 (Case 1) + 120 (Case 2) + 30 (Case 3) = 185.So, the total number of different selection methods is 185.Wait, let me double-check my calculations to make sure I didn't make any mistakes.**Double-Checking:**- Case 1: C(7,4) is indeed 35. Correct.- Case 2: C(4,3)=4, C(2,1)=2, C(6,4)=15. 4*2=8, 8*15=120. Correct.- Case 3: C(4,2)=6, C(2,2)=1, C(5,4)=5. 6*1=6, 6*5=30. Correct.Adding them up: 35 + 120 = 155; 155 + 30 = 185. Correct.Hmm, so 185 is the total number of ways. Looking back at the options, D is 185. So, the answer should be D.But just to make sure, let me think if there's another way to approach this problem, maybe using the principle of inclusion-exclusion or something else.Alternatively, we can think of it as assigning the Both workers first and then selecting the rest. But that might complicate things more. The case method seems straightforward and covers all possibilities without overlap.Another way: Total ways without considering overlaps would be C(11,4) for pliers and C(7,4) for car, but that would be incorrect because it doesn't account for the overlapping workers and the specific constraints on P-only and C-only workers.So, the case method is appropriate here.I think my initial approach is correct, and the answer is 185, which is option D.**Final Answer**boxed{D}