💡Alright, so I've got this problem about a sequence defined by some recurrence relation. Let me try to unpack it step by step. First, the sequence is given as {a_n} with a_1 = 1, a_2 = 3, and for any natural number n, a_{n+2} = 3a_{n+1} - 2a_n. There are two parts to this problem. Part (1) asks me to prove that the sequence {a_{n+1} - a_n} is a geometric sequence. Hmm, okay. So, a geometric sequence is one where each term is a constant multiple of the previous term. That means if I can show that the ratio of consecutive terms in {a_{n+1} - a_n} is constant, then I've proven it's geometric.Let me start by computing the first few terms of {a_n} to see if I can spot a pattern. Given a_1 = 1 and a_2 = 3. Then, a_3 = 3a_2 - 2a_1 = 3*3 - 2*1 = 9 - 2 = 7.a_4 = 3a_3 - 2a_2 = 3*7 - 2*3 = 21 - 6 = 15.a_5 = 3a_4 - 2a_3 = 3*15 - 2*7 = 45 - 14 = 31.a_6 = 3a_5 - 2a_4 = 3*31 - 2*15 = 93 - 30 = 63.Okay, so the sequence is 1, 3, 7, 15, 31, 63,... That looks familiar. Each term seems to be one less than a power of 2. Let me check:2^1 - 1 = 1, which is a_1.2^2 - 1 = 3, which is a_2.2^3 - 1 = 7, which is a_3.2^4 - 1 = 15, which is a_4.2^5 - 1 = 31, which is a_5.2^6 - 1 = 63, which is a_6.So, it seems like a_n = 2^n - 1. Let me test this hypothesis with the recurrence relation.Assume a_n = 2^n - 1. Then, a_{n+1} = 2^{n+1} - 1, and a_{n+2} = 2^{n+2} - 1.Let's plug into the recurrence:a_{n+2} = 3a_{n+1} - 2a_nLeft side: 2^{n+2} - 1Right side: 3*(2^{n+1} - 1) - 2*(2^n - 1) = 3*2^{n+1} - 3 - 2*2^n + 2Simplify right side:3*2^{n+1} = 3*2*2^n = 6*2^n2*2^n = 2^{n+1}So, right side becomes 6*2^n - 3 - 2^{n+1} + 2But 6*2^n - 2^{n+1} = 6*2^n - 2*2^n = 4*2^n = 2^{n+2}Then, constants: -3 + 2 = -1So, right side is 2^{n+2} - 1, which equals the left side. Perfect, so a_n = 2^n - 1.Therefore, the difference a_{n+1} - a_n is (2^{n+1} - 1) - (2^n - 1) = 2^{n+1} - 2^n = 2^n(2 - 1) = 2^n.So, the sequence {a_{n+1} - a_n} is 2, 4, 8, 16, 32,... which is clearly a geometric sequence with common ratio 2.Wait, that seems straightforward. So, part (1) is done.But let me think again. The problem didn't tell me to find a closed-form expression for a_n, just to show that the difference is geometric. Maybe I can do it without finding a_n.Given the recurrence a_{n+2} = 3a_{n+1} - 2a_n.Let me consider the difference d_n = a_{n+1} - a_n.Then, d_{n+1} = a_{n+2} - a_{n+1} = (3a_{n+1} - 2a_n) - a_{n+1} = 2a_{n+1} - 2a_n = 2(a_{n+1} - a_n) = 2d_n.So, d_{n+1} = 2d_n. That shows that {d_n} is a geometric sequence with common ratio 2.Given that d_1 = a_2 - a_1 = 3 - 1 = 2, so the first term is 2 and ratio is 2. Hence, d_n = 2^n.So, that's a more straightforward way without finding a_n. Maybe that's what the problem expects.Alright, moving on to part (2). Let me read it again.Let b_n = (2^{n-1}) / (a_n * a_{n+1}), and T_n is the sum of the first n terms of {b_n}. Prove that T_n < 1/2.Hmm, okay. So, I need to find T_n = b_1 + b_2 + ... + b_n and show it's less than 1/2.Given that a_n = 2^n - 1, as I found earlier, so let's plug that into b_n.b_n = 2^{n-1} / [(2^n - 1)(2^{n+1} - 1)]Hmm, that seems a bit complicated. Maybe I can simplify it.Alternatively, since I know that a_{n+1} - a_n = 2^n, perhaps I can express b_n in terms of a_n and a_{n+1}.Wait, let me see:Given that a_{n+1} - a_n = 2^n, so 2^n = a_{n+1} - a_n.So, b_n = (2^{n-1}) / (a_n * a_{n+1}) = (1/2) * (2^n) / (a_n * a_{n+1}) = (1/2) * (a_{n+1} - a_n) / (a_n * a_{n+1})Simplify that:(1/2) * [ (a_{n+1} - a_n) / (a_n * a_{n+1}) ) ] = (1/2) [ (1/a_n - 1/a_{n+1}) ]Ah, that's a telescoping series! So, each b_n can be written as (1/2)(1/a_n - 1/a_{n+1}).Therefore, when we sum b_1 to b_n, we get:T_n = (1/2)[(1/a_1 - 1/a_2) + (1/a_2 - 1/a_3) + ... + (1/a_n - 1/a_{n+1})]All the middle terms cancel out, leaving:T_n = (1/2)(1/a_1 - 1/a_{n+1})Given that a_1 = 1, so 1/a_1 = 1.Thus, T_n = (1/2)(1 - 1/a_{n+1})Since a_{n+1} = 2^{n+1} - 1, which is always greater than 1 for n ≥ 1, so 1/a_{n+1} is positive and less than 1.Therefore, 1 - 1/a_{n+1} is less than 1, so T_n = (1/2)(something less than 1) < 1/2.Hence, T_n < 1/2.Wait, that seems to work. Let me verify with n=1.For n=1, T_1 = b_1 = 2^{0}/(a_1 a_2) = 1/(1*3) = 1/3 ≈ 0.333 < 0.5. Good.For n=2, T_2 = b_1 + b_2 = 1/3 + 2^{1}/(3*7) = 1/3 + 2/21 = 7/21 + 2/21 = 9/21 = 3/7 ≈ 0.428 < 0.5.n=3: T_3 = 3/7 + 4/(7*15) = 3/7 + 4/105 = 45/105 + 4/105 = 49/105 ≈ 0.466 < 0.5.n=4: T_4 = 49/105 + 8/(15*31) = 49/105 + 8/465 ≈ 0.466 + 0.0172 ≈ 0.483 < 0.5.n=5: T_5 ≈ 0.483 + 16/(31*63) ≈ 0.483 + 0.008 ≈ 0.491 < 0.5.It's approaching 0.5 but never reaching it. So, the sum is always less than 1/2.Therefore, the proof seems solid.But let me think again. The problem didn't specify that a_n = 2^n -1, but I used that to compute b_n. Maybe I should do it without assuming a_n.Wait, but in part (1), I showed that a_{n+1} - a_n = 2^n, so I can express b_n in terms of that difference.As I did earlier, b_n = (2^{n-1}) / (a_n a_{n+1}) = (1/2)(a_{n+1} - a_n)/(a_n a_{n+1}) = (1/2)(1/a_n - 1/a_{n+1}).So, regardless of the explicit form of a_n, this telescoping works, and T_n = (1/2)(1/a_1 - 1/a_{n+1}).Given that a_1 = 1, so T_n = (1/2)(1 - 1/a_{n+1}).Since a_{n+1} is increasing and a_{n+1} > a_n ≥ 1, so 1/a_{n+1} < 1, hence T_n < 1/2.Therefore, the conclusion holds without needing the explicit form of a_n beyond knowing it's increasing and a_1=1.So, that's a more general approach, which is probably better since it doesn't rely on knowing a_n = 2^n -1, which wasn't given.Thus, both parts are proven.

💡Okay, so I have this problem where I need to rewrite the general equations of a line in canonical form. The given system of equations is:[left{begin{array}{l}2x - 3y - 3z - 9 = 0 x - 2y + z + 3 = 0end{array}right.]Alright, first, I need to recall what canonical form means for a line in three-dimensional space. From what I remember, the canonical form of a line is usually expressed as:[frac{x - x_0}{a} = frac{y - y_0}{b} = frac{z - z_0}{c}]where ((x_0, y_0, z_0)) is a point on the line, and (a), (b), and (c) are the direction ratios of the line. So, essentially, I need to find a point that lies on the line and the direction vector of the line.To find a point on the line, I can solve the system of equations by assigning a specific value to one of the variables and then solving for the other two. Let me try setting (z = 0) first, as that might simplify things.So, substituting (z = 0) into both equations:1. (2x - 3y - 9 = 0)2. (x - 2y + 3 = 0)From the second equation, I can solve for (x):[x - 2y + 3 = 0 implies x = 2y - 3]Now, substitute this expression for (x) into the first equation:[2(2y - 3) - 3y - 9 = 0 4y - 6 - 3y - 9 = 0 y - 15 = 0 implies y = 15]Now that I have (y = 15), I can plug this back into the expression for (x):[x = 2(15) - 3 = 30 - 3 = 27]So, one point on the line is ((27, 15, 0)). Let's call this point (M_1(27, 15, 0)).Next, I need another point on the line to determine the direction vector. Let me choose a different value for (z). Maybe (z = -3). Let's see what happens.Substituting (z = -3) into both equations:1. (2x - 3y - 9 = 0)2. (x - 2y + (-3) + 3 = 0 implies x - 2y = 0)From the second equation, (x = 2y). Substitute this into the first equation:[2(2y) - 3y - 9 = 0 4y - 3y - 9 = 0 y - 9 = 0 implies y = 9]Wait, that doesn't seem right. If (y = 9), then (x = 2(9) = 18). But let me check the first equation:[2(18) - 3(9) - 9 = 36 - 27 - 9 = 0]Okay, that works. So, another point on the line is ((18, 9, -3)). Let's call this point (M_2(18, 9, -3)).Now, with two points on the line, (M_1(27, 15, 0)) and (M_2(18, 9, -3)), I can find the direction vector by subtracting the coordinates of (M_2) from (M_1):[text{Direction vector} = (27 - 18, 15 - 9, 0 - (-3)) = (9, 6, 3)]So, the direction vector is ((9, 6, 3)). I can simplify this by dividing each component by 3:[left(frac{9}{3}, frac{6}{3}, frac{3}{3}right) = (3, 2, 1)]Therefore, the direction ratios are (3), (2), and (1).Now, using point (M_1(27, 15, 0)) and the direction ratios, I can write the canonical form of the line:[frac{x - 27}{3} = frac{y - 15}{2} = frac{z - 0}{1}]Simplifying, this becomes:[frac{x - 27}{3} = frac{y - 15}{2} = frac{z}{1}]Alternatively, I can write it using point (M_2(18, 9, -3)):[frac{x - 18}{3} = frac{y - 9}{2} = frac{z + 3}{1}]Both forms are correct, but typically, the canonical form is written using the simplest point. Let me check if there's a point with smaller coordinates. Maybe I made a mistake earlier when choosing (z = -3). Let me try (z = -3) again.Wait, when I set (z = -3), I got (y = 9) and (x = 18). But let me see if there's a simpler point. Maybe setting (y = 0). Let's try that.Setting (y = 0) in both equations:1. (2x - 0 - 3z - 9 = 0 implies 2x - 3z = 9)2. (x - 0 + z + 3 = 0 implies x + z = -3)From the second equation, (x = -3 - z). Substitute into the first equation:[2(-3 - z) - 3z = 9 -6 - 2z - 3z = 9 -6 - 5z = 9 -5z = 15 z = -3]Then, (x = -3 - (-3) = 0). So, another point is ((0, 0, -3)). Let's call this point (M_3(0, 0, -3)).Now, using (M_3(0, 0, -3)) and the direction vector ((3, 2, 1)), the canonical form is:[frac{x - 0}{3} = frac{y - 0}{2} = frac{z + 3}{1}]Simplifying, this is:[frac{x}{3} = frac{y}{2} = frac{z + 3}{1}]This seems simpler. So, I think this is the canonical form. Let me double-check.If I use point (M_3(0, 0, -3)) and direction vector ((3, 2, 1)), then any point on the line can be expressed as:[x = 0 + 3t y = 0 + 2t z = -3 + t]Substituting these into the original equations:First equation:[2(3t) - 3(2t) - 3(-3 + t) - 9 = 0 6t - 6t + 9 - 3t - 9 = 0 -3t = 0 implies t = 0]Wait, that doesn't seem right. It should hold for all (t). Maybe I made a mistake in substitution.Let me try again. Substitute (x = 3t), (y = 2t), (z = -3 + t) into the first equation:[2(3t) - 3(2t) - 3(-3 + t) - 9 = 0 6t - 6t + 9 - 3t - 9 = 0 -3t = 0 implies t = 0]Hmm, that's not good. It should satisfy the equation for all (t). Maybe my direction vector is incorrect.Wait, let's find the direction vector again. The direction vector of the line of intersection of two planes can be found by taking the cross product of the normal vectors of the planes.The given planes are:1. (2x - 3y - 3z - 9 = 0) with normal vector (mathbf{n_1} = (2, -3, -3))2. (x - 2y + z + 3 = 0) with normal vector (mathbf{n_2} = (1, -2, 1))The direction vector (mathbf{d}) is (mathbf{n_1} times mathbf{n_2}):[mathbf{d} = begin{vmatrix}mathbf{i} & mathbf{j} & mathbf{k} 2 & -3 & -3 1 & -2 & 1end{vmatrix}]Calculating the determinant:[mathbf{d} = mathbf{i}((-3)(1) - (-3)(-2)) - mathbf{j}((2)(1) - (-3)(1)) + mathbf{k}((2)(-2) - (-3)(1)) = mathbf{i}(-3 - 6) - mathbf{j}(2 + 3) + mathbf{k}(-4 + 3) = -9mathbf{i} - 5mathbf{j} - 1mathbf{k}]So, the direction vector is ((-9, -5, -1)). I can also write it as ((9, 5, 1)) by multiplying by -1.Wait, earlier I had direction vector ((3, 2, 1)), which is different. So, I must have made a mistake in finding the direction vector by subtracting points. Let me see.Earlier, I found points (M_1(27, 15, 0)) and (M_2(18, 9, -3)), and the direction vector was ((9, 6, 3)), which simplifies to ((3, 2, 1)). But according to the cross product, it's ((9, 5, 1)). There's a discrepancy here.I think the error is in the calculation of the second point. When I set (z = -3), I got (y = 9) and (x = 18). But let me verify this.From the second equation with (z = -3):[x - 2y + (-3) + 3 = 0 implies x - 2y = 0 implies x = 2y]Substituting into the first equation:[2(2y) - 3y - 3(-3) - 9 = 0 4y - 3y + 9 - 9 = 0 y = 0]Wait, that's different from what I did earlier. So, if (y = 0), then (x = 2(0) = 0). So, the point should be ((0, 0, -3)), not ((18, 9, -3)). I must have made a mistake earlier when I thought (y = 9).So, the correct second point is (M_3(0, 0, -3)). Therefore, the direction vector from (M_1(27, 15, 0)) to (M_3(0, 0, -3)) is:[(0 - 27, 0 - 15, -3 - 0) = (-27, -15, -3)]Simplifying by dividing by -3:[(9, 5, 1)]Ah, that matches the cross product result. So, the correct direction vector is ((9, 5, 1)).Therefore, using point (M_3(0, 0, -3)) and direction vector ((9, 5, 1)), the canonical form is:[frac{x - 0}{9} = frac{y - 0}{5} = frac{z + 3}{1}]Simplifying, this is:[frac{x}{9} = frac{y}{5} = frac{z + 3}{1}]Let me verify this. If I take a parameter (t), then:[x = 9t y = 5t z = -3 + t]Substituting into the first equation:[2(9t) - 3(5t) - 3(-3 + t) - 9 = 0 18t - 15t + 9 - 3t - 9 = 0 0 = 0]And into the second equation:[9t - 2(5t) + (-3 + t) + 3 = 0 9t - 10t - 3 + t + 3 = 0 0 = 0]Perfect, it satisfies both equations for all (t). So, the canonical form is correct.Therefore, the canonical form of the line is:[frac{x}{9} = frac{y}{5} = frac{z + 3}{1}]

💡Okay, so I have this problem where I need to find the ratio of the sum of the odd divisors of M to the sum of the even divisors of M, where M is given as 39 multiplied by 48 multiplied by 77 multiplied by 150. The answer choices are given as A: 1:64, B:1:62, C:1:60, D:1:58.First, I think I need to factorize each of these numbers into their prime factors because that will help me figure out the divisors. Let me start by breaking down each number:- 39: I know that 39 is 3 times 13 because 3 times 13 is 39.- 48: Hmm, 48 is a multiple of 16, which is 2^4, and 16 times 3 is 48, so 48 is 2^4 times 3.- 77: 77 is 7 times 11 because 7 times 11 is 77.- 150: 150 is 15 times 10, which is 3 times 5 times 2 times 5, so that's 2 times 3 times 5 squared.So putting it all together, M is equal to 3 times 13 times 2^4 times 3 times 7 times 11 times 2 times 3 times 5 squared. Let me write that out:M = 3 * 13 * 2^4 * 3 * 7 * 11 * 2 * 3 * 5^2.Now, let me combine like terms. The powers of 2: I have 2^4 and another 2, so that's 2^(4+1) = 2^5. For the powers of 3: I have three 3s, so that's 3^3. The 5s: I have 5 squared. And then the other primes: 7, 11, and 13.So, M can be written as:M = 2^5 * 3^3 * 5^2 * 7 * 11 * 13.Okay, now I need to find the sum of the odd divisors and the sum of the even divisors.I remember that the sum of the divisors of a number can be found using the formula where for each prime factor, you take the sum of the powers from 0 up to the exponent in the prime factorization.For example, for a number N = p^a * q^b, the sum of the divisors is (1 + p + p^2 + ... + p^a)(1 + q + q^2 + ... + q^b).So, for the sum of the odd divisors, I need to exclude the factor of 2 because even divisors include at least one 2. Therefore, the sum of the odd divisors would be the product of the sums of the other prime factors.Let me calculate that:Sum of odd divisors = (1 + 3 + 3^2 + 3^3) * (1 + 5 + 5^2) * (1 + 7) * (1 + 11) * (1 + 13).Let me compute each part step by step:First, (1 + 3 + 9 + 27) for the 3s: 1 + 3 is 4, plus 9 is 13, plus 27 is 40.Next, (1 + 5 + 25) for the 5s: 1 + 5 is 6, plus 25 is 31.Then, (1 + 7) is 8.(1 + 11) is 12.(1 + 13) is 14.So, multiplying all these together: 40 * 31 * 8 * 12 * 14.Wait, that seems like a lot, but maybe I can compute it step by step.First, 40 * 31: 40 times 30 is 1200, plus 40 is 1240.Then, 1240 * 8: 1240 * 8 is 9920.Next, 9920 * 12: Let's see, 9920 * 10 is 99200, plus 9920 * 2 is 19840, so total is 119040.Then, 119040 * 14: 119040 * 10 is 1,190,400, plus 119040 * 4 is 476,160, so total is 1,666,560.So, the sum of the odd divisors is 1,666,560.Now, for the sum of the even divisors, I think I can use the total sum of all divisors and subtract the sum of the odd divisors.The total sum of all divisors would include all the factors, including the powers of 2.So, the total sum of divisors is:(1 + 2 + 4 + 8 + 16 + 32) * (1 + 3 + 9 + 27) * (1 + 5 + 25) * (1 + 7) * (1 + 11) * (1 + 13).We already calculated the part without the 2s as 1,666,560. Now, let's compute the sum for the 2s:(1 + 2 + 4 + 8 + 16 + 32) is a geometric series. The sum is (2^6 - 1)/(2 - 1) = 63.So, the total sum of divisors is 63 * 1,666,560.Wait, that would be 63 times 1,666,560. Let me compute that:First, 60 * 1,666,560 is 99,993,600.Then, 3 * 1,666,560 is 4,999,680.Adding them together: 99,993,600 + 4,999,680 = 104,993,280.So, the total sum of all divisors is 104,993,280.Therefore, the sum of the even divisors is total sum minus the sum of the odd divisors:104,993,280 - 1,666,560 = 103,326,720.Wait, let me check that subtraction:104,993,280 minus 1,666,560.Subtracting 1,666,560 from 104,993,280:104,993,280 - 1,666,560 = 103,326,720.Yes, that seems correct.Now, the ratio of the sum of the odd divisors to the sum of the even divisors is 1,666,560 : 103,326,720.To simplify this ratio, I can divide both numbers by 1,666,560.So, 1,666,560 divided by 1,666,560 is 1.103,326,720 divided by 1,666,560: Let's compute that.First, let's see how many times 1,666,560 goes into 103,326,720.Divide 103,326,720 by 1,666,560.Let me compute 1,666,560 * 60: 1,666,560 * 60 is 99,993,600.Subtract that from 103,326,720: 103,326,720 - 99,993,600 = 3,333,120.Now, how many times does 1,666,560 go into 3,333,120? It goes exactly 2 times because 1,666,560 * 2 = 3,333,120.So, total is 60 + 2 = 62.Therefore, the ratio is 1:62.So, the answer should be option B: 1:62.Wait, let me double-check my calculations to make sure I didn't make any errors.First, the prime factorization of M: 2^5 * 3^3 * 5^2 * 7 * 11 * 13. That seems correct.Sum of odd divisors: (1 + 3 + 9 + 27) * (1 + 5 + 25) * (1 + 7) * (1 + 11) * (1 + 13) = 40 * 31 * 8 * 12 * 14 = 1,666,560. That seems correct.Total sum of divisors: (1 + 2 + 4 + 8 + 16 + 32) * sum of odd divisors = 63 * 1,666,560 = 104,993,280. Correct.Sum of even divisors: 104,993,280 - 1,666,560 = 103,326,720. Correct.Ratio: 1,666,560 : 103,326,720 simplifies to 1:62. Correct.Yes, I think that's right. So, the answer is B: 1:62.

💡Okay, so I have this problem where α is between 0 and π/3, and it satisfies the equation √3 sin α + cos α = √6 / 2. I need to find two things: first, the value of cos(α + π/6), and second, the value of cos(2α + 7π/12). Hmm, let me try to figure this out step by step.Starting with the first part: finding cos(α + π/6). The given equation is √3 sin α + cos α = √6 / 2. I remember that expressions like a sin α + b cos α can be rewritten using a sine addition formula. Specifically, a sin α + b cos α can be expressed as R sin(α + φ), where R is the amplitude and φ is the phase shift. Let me recall the exact formula.Yes, the formula is: a sin α + b cos α = R sin(α + φ), where R = √(a² + b²) and tan φ = b/a. Alternatively, it can also be written as R cos(α - θ), depending on how you set it up. Wait, maybe I should double-check which form is more appropriate here.Looking at the given equation: √3 sin α + cos α. So, a is √3 and b is 1. Let me compute R first. R = √( (√3)² + 1² ) = √(3 + 1) = √4 = 2. Okay, so R is 2. Now, to find φ, we can use tan φ = b/a = 1/√3. So, tan φ = 1/√3, which means φ is π/6 because tan(π/6) = 1/√3. So, that means √3 sin α + cos α can be written as 2 sin(α + π/6). Let me verify that.Using the sine addition formula: sin(α + π/6) = sin α cos(π/6) + cos α sin(π/6). Cos(π/6) is √3/2 and sin(π/6) is 1/2. So, 2 sin(α + π/6) = 2 [ sin α (√3/2) + cos α (1/2) ] = √3 sin α + cos α. Perfect, that matches the given equation. So, √3 sin α + cos α = 2 sin(α + π/6) = √6 / 2.Therefore, 2 sin(α + π/6) = √6 / 2. Dividing both sides by 2, we get sin(α + π/6) = √6 / 4. Okay, so sin(α + π/6) is √6 / 4. Now, since α is between 0 and π/3, let's find the range of α + π/6. If α is 0, then α + π/6 is π/6. If α is π/3, then α + π/6 is π/3 + π/6 = π/2. So, α + π/6 is between π/6 and π/2. That means α + π/6 is in the first quadrant, so both sine and cosine are positive there.Since we need to find cos(α + π/6), and we know sin(α + π/6) = √6 / 4, we can use the Pythagorean identity: sin²θ + cos²θ = 1. So, cos²θ = 1 - sin²θ. Plugging in sinθ = √6 / 4, we get cos²θ = 1 - (6/16) = 1 - 3/8 = 5/8. Therefore, cosθ = √(5/8) = (√10)/4. Since we're in the first quadrant, cosine is positive, so cos(α + π/6) = √10 / 4. That should be the answer for part (1).Moving on to part (2): finding cos(2α + 7π/12). Hmm, 2α + 7π/12. Let me see if I can express this angle in terms of something I already know. I know that 2α + 7π/12 can be written as 2α + π/3 + π/4, because π/3 is 4π/12 and π/4 is 3π/12, so 4π/12 + 3π/12 = 7π/12. Alternatively, maybe I can express 2α + 7π/12 as (2α + π/3) + π/4. That might be useful because I already have information about α + π/6, so 2α + π/3 is just 2*(α + π/6). Let me check that.Yes, 2*(α + π/6) is 2α + π/3. So, 2α + π/3 is 2*(α + π/6). Therefore, 2α + 7π/12 is equal to 2*(α + π/6) + π/4. So, that's (2α + π/3) + π/4. So, maybe I can use the angle addition formula for cosine: cos(A + B) = cos A cos B - sin A sin B. So, if I let A = 2α + π/3 and B = π/4, then cos(2α + 7π/12) = cos(A + B) = cos A cos B - sin A sin B.So, I need to find cos A and sin A, where A = 2α + π/3. To find cos A and sin A, I can use double-angle formulas because A is 2*(α + π/6). Let me denote θ = α + π/6, so A = 2θ. Then, cos A = cos(2θ) and sin A = sin(2θ). From part (1), we already know cos θ = √10 / 4 and sin θ = √6 / 4. So, let's compute cos(2θ) and sin(2θ).Using the double-angle formulas:cos(2θ) = 2 cos²θ - 1 = 2*(10/16) - 1 = 20/16 - 1 = 20/16 - 16/16 = 4/16 = 1/4.sin(2θ) = 2 sinθ cosθ = 2*(√6 / 4)*(√10 / 4) = 2*(√60)/16 = (2*2√15)/16 = (4√15)/16 = √15 / 4.So, cos A = 1/4 and sin A = √15 / 4. Now, we need to compute cos(A + B) where B = π/4. So, cos(π/4) = √2 / 2 and sin(π/4) = √2 / 2.Putting it all together:cos(2α + 7π/12) = cos(A + B) = cos A cos B - sin A sin B= (1/4)*(√2 / 2) - (√15 / 4)*(√2 / 2)= (√2 / 8) - (√30 / 8)= (√2 - √30) / 8.So, the value of cos(2α + 7π/12) is (√2 - √30)/8.Let me just recap to make sure I didn't make any mistakes. For part (1), I converted the given equation into a sine addition formula, found sin(α + π/6), then used the Pythagorean identity to find cos(α + π/6). That seems solid.For part (2), I recognized that 2α + 7π/12 can be expressed as 2*(α + π/6) + π/4, which allowed me to use the double-angle formulas on θ = α + π/6. Then, I used the angle addition formula for cosine to combine the results. The calculations for cos(2θ) and sin(2θ) seem correct, and then plugging into the addition formula also looks right. I think I did everything properly.Just to double-check, let me verify the double-angle calculations:cos(2θ) = 2*(√10 / 4)^2 - 1 = 2*(10/16) - 1 = 20/16 - 16/16 = 4/16 = 1/4. Correct.sin(2θ) = 2*(√6 / 4)*(√10 / 4) = 2*(√60)/16 = 2*(2√15)/16 = √15 / 4. Correct.Then, cos(A + B) = (1/4)*(√2 / 2) - (√15 / 4)*(√2 / 2) = √2/8 - √30/8. Yes, that's correct. So, I think both parts are correct.**Final Answer**(1) boxed{dfrac{sqrt{10}}{4}} (2) boxed{dfrac{sqrt{2} - sqrt{30}}{8}}

💡Okay, so I have this arithmetic sequence problem here. Let me try to figure it out step by step. The problem says:Given the arithmetic sequence {a_n}, where a₁ = -3 and 11a₅ = 5a₈. Let S_n denote the sum of the first n terms.1. Find a_n;2. For what value of n is S_n minimum? And find the minimum value of S_n.Alright, let's start with part 1: finding a_n.First, I remember that in an arithmetic sequence, each term is obtained by adding a common difference, d, to the previous term. So, the general formula for the nth term of an arithmetic sequence is:a_n = a₁ + (n - 1)dWe know that a₁ is -3. So, we can write:a_n = -3 + (n - 1)dBut we don't know the common difference, d. So, we need to find d using the given condition: 11a₅ = 5a₈.Let me write expressions for a₅ and a₈ using the general formula.a₅ = a₁ + (5 - 1)d = a₁ + 4da₈ = a₁ + (8 - 1)d = a₁ + 7dGiven that 11a₅ = 5a₈, let's substitute the expressions:11(a₁ + 4d) = 5(a₁ + 7d)Now, substitute a₁ = -3:11(-3 + 4d) = 5(-3 + 7d)Let me compute both sides:Left side: 11*(-3) + 11*4d = -33 + 44dRight side: 5*(-3) + 5*7d = -15 + 35dSo, the equation becomes:-33 + 44d = -15 + 35dLet me solve for d. I'll subtract 35d from both sides:-33 + 44d - 35d = -15-33 + 9d = -15Now, add 33 to both sides:9d = -15 + 339d = 18Divide both sides by 9:d = 18 / 9d = 2Okay, so the common difference d is 2. Now, let's write the general term a_n.a_n = -3 + (n - 1)*2Simplify that:a_n = -3 + 2n - 2a_n = 2n - 5So, part 1 is done. a_n is 2n - 5.Now, moving on to part 2: finding the value of n for which S_n is minimum and the minimum value of S_n.First, I recall that the sum of the first n terms of an arithmetic sequence is given by:S_n = n/2 * (a₁ + a_n)We already have a₁ = -3 and a_n = 2n - 5. So, let's plug these into the formula:S_n = n/2 * (-3 + (2n - 5))Simplify inside the parentheses:-3 + 2n - 5 = 2n - 8So, S_n = n/2 * (2n - 8)Multiply through:S_n = (n * 2n)/2 - (n * 8)/2S_n = (2n²)/2 - (8n)/2S_n = n² - 4nSo, S_n is a quadratic function in terms of n: S_n = n² - 4n.Now, since this is a quadratic function, its graph is a parabola. The coefficient of n² is positive (1), so the parabola opens upwards. That means the vertex is the minimum point.To find the value of n where S_n is minimum, we can use the vertex formula for a parabola. The vertex occurs at n = -b/(2a), where the quadratic is in the form an² + bn + c.In our case, a = 1, b = -4.So, n = -(-4)/(2*1) = 4/2 = 2.Wait, n = 2. So, the sum S_n is minimum when n = 2.But let me think about this. n is the number of terms, so it has to be a positive integer. So, n = 2 is valid.Now, let's compute S_2 to find the minimum value.S_2 = 2² - 4*2 = 4 - 8 = -4.So, the minimum value of S_n is -4, occurring at n = 2.Wait, but let me double-check. Maybe I made a mistake somewhere.First, let's verify the common difference d.We had 11a₅ = 5a₈.a₅ = a₁ + 4d = -3 + 4da₈ = a₁ + 7d = -3 + 7d11a₅ = 5a₈:11*(-3 + 4d) = 5*(-3 + 7d)-33 + 44d = -15 + 35d44d - 35d = -15 + 339d = 18d = 2That's correct.So, a_n = -3 + (n - 1)*2 = 2n - 5. Correct.Sum S_n = n/2*(a₁ + a_n) = n/2*(-3 + 2n - 5) = n/2*(2n - 8) = n² - 4n. Correct.Quadratic function n² - 4n. Since the coefficient of n² is positive, it opens upwards, so the vertex is the minimum.Vertex at n = -b/(2a) = 4/(2*1) = 2. So, n = 2.Compute S_2: 2² - 4*2 = 4 - 8 = -4.So, yes, that seems correct.But just to be thorough, let's compute S₁, S₂, S₃, etc., to see if S₂ is indeed the minimum.Compute S₁: sum of first 1 term is just a₁ = -3.Compute S₂: a₁ + a₂ = -3 + (2*2 - 5) = -3 + (-1) = -4.Compute S₃: a₁ + a₂ + a₃ = -3 + (-1) + (1) = -3.Compute S₄: sum up to 4 terms: -3 + (-1) + 1 + 3 = 0.Compute S₅: sum up to 5 terms: -3 + (-1) + 1 + 3 + 5 = 5.So, the sums are:n | S_n1 | -32 | -43 | -34 | 05 | 5So, indeed, the minimum occurs at n = 2 with S_n = -4.Therefore, the answers are:1. a_n = 2n - 52. The minimum S_n occurs at n = 2, and the minimum value is -4.I think that's solid. I don't see any mistakes in my calculations.**Final Answer**1. The nth term is boxed{2n - 5}.2. The sum ( S_n ) is minimum when ( n = boxed{2} ) and the minimum value is boxed{-4}.

💡Okay, so I have this trigonometric expression to evaluate: sin 15° cos 75° + cos 15° sin 105°. Hmm, at first glance, it looks a bit complicated, but maybe I can simplify it using some trigonometric identities. Let me think about what identities I know that could help here.I remember that there are sum and difference formulas for sine and cosine. Specifically, the sine of a sum formula is sin(A + B) = sin A cos B + cos A sin B. Wait a second, that looks a lot like the expression I have here. Let me check:sin 15° cos 75° + cos 15° sin 105°If I compare this to the sine of a sum formula, it seems like A is 15° and B is 75°, but then the second term has sin 105° instead of sin 75°. That's a bit confusing. Maybe I can adjust the angles to make them fit the formula better.I know that 105° is equal to 90° + 15°, so maybe I can express sin 105° in terms of sine and cosine of 15°. Let me recall that sin(90° + θ) = cos θ. So, sin 105° = sin(90° + 15°) = cos 15°. That's helpful!So, substituting that back into the original expression, I get:sin 15° cos 75° + cos 15° * cos 15°Now, let's look at the first term: sin 15° cos 75°. I also know that 75° is equal to 90° - 15°, so cos 75° = cos(90° - 15°). There's an identity for cosine of a complementary angle: cos(90° - θ) = sin θ. Therefore, cos 75° = sin 15°.Substituting that into the first term, I have:sin 15° * sin 15° + cos 15° * cos 15°Which simplifies to:sin²15° + cos²15°Now, I remember the Pythagorean identity, which states that sin²θ + cos²θ = 1 for any angle θ. Applying this identity here, since θ is 15°, I get:sin²15° + cos²15° = 1So, putting it all together, the original expression simplifies to 1.Wait, let me double-check my steps to make sure I didn't make any mistakes. I used the identity sin(90° + θ) = cos θ correctly for sin 105°, and I used cos(90° - θ) = sin θ correctly for cos 75°. Then I substituted those into the expression, which gave me sin²15° + cos²15°, and that indeed equals 1. Yeah, that seems right.I guess another way to think about it is by recognizing that the original expression resembles the sine of a sum formula, but with different angles. If I adjust the angles appropriately using co-function identities, I can simplify the expression to something familiar. It's interesting how these identities can transform the problem into something more manageable.So, after carefully applying the trigonometric identities step by step, I'm confident that the value of sin 15° cos 75° + cos 15° sin 105° is 1.