💡Alright, so I'm trying to figure out this problem about angles in the Cartesian coordinate system. The question is asking for the set of angles α whose terminal sides lie on the coordinate axes. Hmm, okay, let me break this down step by step.First, I remember that in the Cartesian coordinate system, we have the x-axis and the y-axis. The terminal side of an angle in standard position is the ray where the angle ends after rotation from the initial side, which is along the positive x-axis. So, if the terminal side lies on one of the coordinate axes, it means the angle has rotated to either the x-axis or the y-axis.Let me think about the angles that point directly along the x-axis. When you rotate 0 radians, you're still on the positive x-axis. If you rotate π radians, you end up on the negative x-axis. Rotating another π radians brings you back to the positive x-axis, and so on. So, it seems like all these angles can be represented as multiples of π. In other words, the set of angles whose terminal sides lie on the x-axis is {α | α = kπ, where k is an integer}.Now, what about the y-axis? If I rotate π/2 radians from the positive x-axis, I end up on the positive y-axis. Rotating another π radians would bring me to the negative y-axis, and another π/2 radians would bring me back to the positive x-axis. So, the angles pointing to the y-axis are π/2 more than the angles pointing to the x-axis. Therefore, the set of angles whose terminal sides lie on the y-axis is {α | α = kπ + π/2, where k is an integer}.To find the set of all angles whose terminal sides lie on either the x-axis or the y-axis, I need to combine these two sets. That means I should consider all angles that are either multiples of π or multiples of π plus π/2. If I think about it, these can be represented as multiples of π/2. Because every multiple of π/2 will either land on the x-axis or the y-axis. For example:- 0 radians: x-axis- π/2 radians: y-axis- π radians: x-axis- 3π/2 radians: y-axis- 2π radians: x-axisAnd this pattern continues indefinitely in both the positive and negative directions. So, generalizing this, the set of all such angles can be written as {α | α = nπ/2, where n is an integer}.Let me double-check this. If n is even, say n = 2k, then α = (2k)π/2 = kπ, which are the angles on the x-axis. If n is odd, say n = 2k + 1, then α = (2k + 1)π/2 = kπ + π/2, which are the angles on the y-axis. So, yes, combining both cases gives me all the angles on the coordinate axes.Therefore, the set of angles α whose terminal sides lie on the coordinate axes is {α | α = nπ/2, n ∈ ℤ}.**Final Answer**The set of angles is boxed{{alpha|alpha=frac{npi}{2}, nin mathbb{Z}}}.

💡Okay, so I've got this trigonometric equation here that I need to prove. It says:[ 2 sin^{2}(3pi - 2alpha) cos^{2}(5pi + 2alpha) = frac{1}{4} - frac{1}{4} sinleft(frac{5pi}{2} - 8alpharight)]Hmm, alright. Let me try to break this down step by step. First, I remember that trigonometric functions have periodicity and certain identities that can help simplify expressions. Maybe I can use those to make the left side look like the right side.Starting with the left side:[ 2 sin^{2}(3pi - 2alpha) cos^{2}(5pi + 2alpha) ]I know that sine and cosine have periods of (2pi), so maybe I can subtract multiples of (2pi) from the angles to simplify them. Let's look at each angle separately.For (sin(3pi - 2alpha)), since the sine function has a period of (2pi), I can subtract (2pi) from (3pi) to get (pi). So:[sin(3pi - 2alpha) = sin(pi - 2alpha)]I remember that (sin(pi - x) = sin x), but wait, actually, (sin(pi - x) = sin x). Hmm, no, that's not quite right. Let me think. Actually, (sin(pi - x) = sin x), but with a sign. Since (pi - x) is in the second quadrant where sine is positive, so it's just (sin x). Wait, no, that's not correct. Let me recall the exact identity.Yes, actually, (sin(pi - x) = sin x). So, (sin(pi - 2alpha) = sin(2alpha)). But wait, the angle is (3pi - 2alpha), which is more than (pi). Let me see. Since (3pi - 2alpha = pi + (2pi - 2alpha)), but that might not help. Alternatively, since sine has a period of (2pi), (sin(3pi - 2alpha) = sin(3pi - 2alpha - 2pi) = sin(pi - 2alpha)). Okay, so that's (sin(pi - 2alpha)), which is (sin(2alpha)). Wait, but actually, (sin(pi - x) = sin x), so (sin(pi - 2alpha) = sin(2alpha)). So, (sin(3pi - 2alpha) = sin(2alpha)).But wait, actually, let me double-check. Because (3pi - 2alpha) is in the third quadrant if (alpha) is between 0 and (pi/2), for example. In the third quadrant, sine is negative. So, (sin(3pi - 2alpha) = -sin(2alpha)). Hmm, that might be important. Let me confirm.Yes, because (sin(pi + x) = -sin x). So, (3pi - 2alpha = pi + (2pi - 2alpha)), but that might not be the right way to break it down. Alternatively, (3pi - 2alpha = 2pi + (pi - 2alpha)). Since sine has a period of (2pi), (sin(3pi - 2alpha) = sin(pi - 2alpha)). And as I said, (sin(pi - 2alpha) = sin(2alpha)). But wait, actually, (sin(pi - x) = sin x), but in this case, since (3pi - 2alpha) is more than (pi), it's actually in the third quadrant where sine is negative. So, (sin(3pi - 2alpha) = -sin(2alpha)). That makes sense because (sin(pi + x) = -sin x), so if we write (3pi - 2alpha = pi + (2pi - 2alpha)), then (sin(3pi - 2alpha) = sin(pi + (2pi - 2alpha)) = -sin(2pi - 2alpha)). But (sin(2pi - 2alpha) = -sin(2alpha)), so overall, (sin(3pi - 2alpha) = -(-sin(2alpha)) = sin(2alpha)). Wait, that contradicts my earlier thought. Hmm, maybe I'm confusing something.Let me try a different approach. Let's compute (sin(3pi - 2alpha)). Since (3pi) is an odd multiple of (pi), and sine is an odd function, we can write:[sin(3pi - 2alpha) = sin(3pi)cos(2alpha) - cos(3pi)sin(2alpha)]We know that (sin(3pi) = 0) and (cos(3pi) = -1), so:[sin(3pi - 2alpha) = 0 cdot cos(2alpha) - (-1) cdot sin(2alpha) = sin(2alpha)]Wait, so that's positive (sin(2alpha)). Hmm, so maybe my initial thought about it being negative was wrong. So, (sin(3pi - 2alpha) = sin(2alpha)). Okay, that's good to know.Now, moving on to (cos(5pi + 2alpha)). Cosine has a period of (2pi), so let's subtract multiples of (2pi) to simplify the angle.[5pi + 2alpha = 2pi times 2 + (pi + 2alpha) = pi + 2alpha]So, (cos(5pi + 2alpha) = cos(pi + 2alpha)). I remember that (cos(pi + x) = -cos x), so:[cos(pi + 2alpha) = -cos(2alpha)]Therefore, (cos(5pi + 2alpha) = -cos(2alpha)).So, putting it all together, the left side becomes:[2 sin^{2}(3pi - 2alpha) cos^{2}(5pi + 2alpha) = 2 [sin(2alpha)]^{2} [-cos(2alpha)]^{2}]Since squaring a negative makes it positive, this simplifies to:[2 sin^{2}(2alpha) cos^{2}(2alpha)]Okay, so now I have:[2 sin^{2}(2alpha) cos^{2}(2alpha)]I need to simplify this expression. I recall that there's a double-angle identity for sine: (sin(2x) = 2sin x cos x). Maybe I can use that here. Let me see.If I let (x = 2alpha), then (sin(4alpha) = 2sin(2alpha)cos(2alpha)). So, (sin(4alpha) = 2sin(2alpha)cos(2alpha)). Therefore, (sin(2alpha)cos(2alpha) = frac{1}{2}sin(4alpha)).So, (sin^{2}(2alpha)cos^{2}(2alpha) = [sin(2alpha)cos(2alpha)]^{2} = left(frac{1}{2}sin(4alpha)right)^{2} = frac{1}{4}sin^{2}(4alpha)).Therefore, the expression becomes:[2 times frac{1}{4}sin^{2}(4alpha) = frac{1}{2}sin^{2}(4alpha)]So, now we have:[frac{1}{2}sin^{2}(4alpha)]I need to express this in terms of a sine function with a different angle, as in the right side of the equation. Let me recall that (sin^{2}x) can be expressed using the double-angle identity for cosine:[sin^{2}x = frac{1 - cos(2x)}{2}]Applying this to (sin^{2}(4alpha)):[sin^{2}(4alpha) = frac{1 - cos(8alpha)}{2}]So, substituting back into our expression:[frac{1}{2} times frac{1 - cos(8alpha)}{2} = frac{1}{4}(1 - cos(8alpha))]So, now we have:[frac{1}{4}(1 - cos(8alpha))]Hmm, the right side of the original equation is:[frac{1}{4} - frac{1}{4} sinleft(frac{5pi}{2} - 8alpharight)]So, I need to show that:[frac{1}{4}(1 - cos(8alpha)) = frac{1}{4} - frac{1}{4} sinleft(frac{5pi}{2} - 8alpharight)]Let me simplify the right side:[frac{1}{4} - frac{1}{4} sinleft(frac{5pi}{2} - 8alpharight) = frac{1}{4}left(1 - sinleft(frac{5pi}{2} - 8alpharight)right)]So, I need to show that:[1 - cos(8alpha) = 1 - sinleft(frac{5pi}{2} - 8alpharight)]Which would mean:[cos(8alpha) = sinleft(frac{5pi}{2} - 8alpharight)]Is this true? Let me check. I know that (sinleft(frac{pi}{2} - xright) = cos x), but here we have (frac{5pi}{2}). Let me see:[sinleft(frac{5pi}{2} - 8alpharight) = sinleft(2pi + frac{pi}{2} - 8alpharight)]Since sine has a period of (2pi), this is equal to:[sinleft(frac{pi}{2} - 8alpharight)]And as I mentioned earlier, (sinleft(frac{pi}{2} - xright) = cos x). So:[sinleft(frac{pi}{2} - 8alpharight) = cos(8alpha)]Therefore:[sinleft(frac{5pi}{2} - 8alpharight) = cos(8alpha)]So, substituting back:[1 - cos(8alpha) = 1 - sinleft(frac{5pi}{2} - 8alpharight)]Which means:[frac{1}{4}(1 - cos(8alpha)) = frac{1}{4} - frac{1}{4} sinleft(frac{5pi}{2} - 8alpharight)]Therefore, the left side equals the right side, and the equation is proven.Wait, let me double-check all the steps to make sure I didn't make any mistakes.1. Simplified (sin(3pi - 2alpha)) to (sin(2alpha)). Checked using angle addition and periodicity, seems correct.2. Simplified (cos(5pi + 2alpha)) to (-cos(2alpha)). That also seems correct using periodicity and identity for (cos(pi + x)).3. Squared both sine and cosine terms, correctly noting that the negative becomes positive.4. Used the double-angle identity to express (sin(4alpha)) in terms of (sin(2alpha)cos(2alpha)), then squared it to get (sin^{2}(4alpha)).5. Applied the identity for (sin^{2}x) in terms of cosine, leading to (frac{1}{4}(1 - cos(8alpha))).6. Compared with the right side, realized that (cos(8alpha)) equals (sinleft(frac{5pi}{2} - 8alpharight)) by using periodicity and co-function identity.Everything seems to check out. I think the proof is solid.

💡Okay, so I have this problem about M-class sequences. Let me try to understand what an M-class sequence is first. It says that for a sequence {c_n}, if there exist real constants p and q such that c_{n+1} = p*c_n + q for any n in natural numbers, then it's called an M-class sequence. So, basically, it's a linear recurrence relation of order 1 with constant coefficients. Got it.Now, moving on to the first part of the problem. It asks whether the sequences {a_n} where a_n = 2n and {b_n} where b_n = 3*2^n are M-class sequences. Hmm, okay. Let me think about each one separately.Starting with {a_n} = 2n. So, the sequence is 2, 4, 6, 8, 10, and so on. Let's see if this can be written in the form c_{n+1} = p*c_n + q. Let's compute a_{n+1} and see if it can be expressed in terms of a_n.a_{n+1} = 2*(n+1) = 2n + 2. But a_n = 2n, so a_{n+1} = a_n + 2. So, comparing this to the M-class definition, we have p = 1 and q = 2. Since p is not zero, this satisfies the condition. So, {a_n} is indeed an M-class sequence.Now, let's check {b_n} = 3*2^n. The sequence is 3, 6, 12, 24, 48, etc. Let's compute b_{n+1} and see if it can be expressed as p*b_n + q.b_{n+1} = 3*2^{n+1} = 3*2*2^n = 2*(3*2^n) = 2*b_n. So, b_{n+1} = 2*b_n + 0. Therefore, p = 2 and q = 0. Since p ≠ 0, this also satisfies the M-class condition. So, {b_n} is also an M-class sequence.Alright, so both sequences are M-class sequences. That was straightforward. I think I got part 1.Moving on to part 2. It says, if {a_n} is an M-class sequence, are the sequences {a_n + a_{n+1}} and {a_n * a_{n+1}} necessarily M-class sequences? Hmm, interesting. Let me break this down.First, since {a_n} is an M-class sequence, we know that a_{n+1} = p*a_n + q for some constants p and q. So, let's see what happens when we consider the sum a_n + a_{n+1}.Let me denote s_n = a_n + a_{n+1}. Then, s_{n+1} = a_{n+1} + a_{n+2}. But since a_{n+2} = p*a_{n+1} + q, we can substitute that in:s_{n+1} = a_{n+1} + (p*a_{n+1} + q) = (1 + p)*a_{n+1} + q.But s_n = a_n + a_{n+1}, so a_{n+1} = s_n - a_n. Wait, maybe another approach. Let's express s_{n+1} in terms of s_n.From s_n = a_n + a_{n+1}, and a_{n+1} = p*a_n + q, so s_n = a_n + p*a_n + q = (1 + p)*a_n + q.Similarly, s_{n+1} = a_{n+1} + a_{n+2} = a_{n+1} + p*a_{n+1} + q = (1 + p)*a_{n+1} + q.But a_{n+1} = p*a_n + q, so substituting back:s_{n+1} = (1 + p)*(p*a_n + q) + q = (1 + p)*p*a_n + (1 + p)*q + q.Hmm, this seems a bit messy. Maybe instead, let's express s_{n+1} in terms of s_n.We have s_n = a_n + a_{n+1} = a_n + (p*a_n + q) = (1 + p)*a_n + q.Similarly, s_{n+1} = a_{n+1} + a_{n+2} = a_{n+1} + (p*a_{n+1} + q) = (1 + p)*a_{n+1} + q.But a_{n+1} = p*a_n + q, so:s_{n+1} = (1 + p)*(p*a_n + q) + q = (1 + p)*p*a_n + (1 + p)*q + q.But from s_n, we have (1 + p)*a_n = s_n - q. So, substituting:s_{n+1} = (1 + p)*p*a_n + (1 + p)*q + q = p*(s_n - q) + (1 + p)*q + q.Expanding this:s_{n+1} = p*s_n - p*q + (1 + p)*q + q = p*s_n - p*q + q + p*q + q = p*s_n + 2*q.Wait, so s_{n+1} = p*s_n + 2*q. So, that's a linear recurrence relation with constants p and 2q. Since p ≠ 0, this means {s_n} is also an M-class sequence. So, the sum {a_n + a_{n+1}} is an M-class sequence. That seems to work.Now, what about the product {a_n * a_{n+1}}? Let me denote p_n = a_n * a_{n+1}. Let's see if p_{n+1} can be expressed as some linear function of p_n.We have p_{n+1} = a_{n+1} * a_{n+2}. But a_{n+2} = p*a_{n+1} + q. So,p_{n+1} = a_{n+1}*(p*a_{n+1} + q) = p*(a_{n+1})^2 + q*a_{n+1}.But p_n = a_n * a_{n+1}, so a_{n+1} = p_n / a_n. Hmm, but this might complicate things because it introduces a division by a_n, which could be zero or vary with n.Alternatively, let's express a_{n+1} in terms of a_n: a_{n+1} = p*a_n + q. So, a_{n+2} = p*a_{n+1} + q = p*(p*a_n + q) + q = p^2*a_n + p*q + q.Therefore, p_{n+1} = a_{n+1}*a_{n+2} = (p*a_n + q)*(p^2*a_n + p*q + q).Expanding this:p_{n+1} = p*(p^2*a_n^2 + p*q*a_n + q*a_n) + q*(p^2*a_n^2 + p*q*a_n + q*a_n)= p^3*a_n^2 + p^2*q*a_n + p*q*a_n + p^2*q*a_n^2 + p*q^2*a_n + q^2*a_n.Hmm, this is getting complicated. It seems like p_{n+1} is a quadratic function of a_n, not linear. So, unless p or q is zero, it's not going to be linear in p_n.Wait, let's see. If q = 0, then a_{n+1} = p*a_n. So, a_n = a_1*p^{n-1}. Then, p_n = a_n * a_{n+1} = a_1*p^{n-1} * a_1*p^n = a_1^2*p^{2n - 1}. Then, p_{n+1} = a_1^2*p^{2n + 1}. So, p_{n+1} = p^2 * p_n. So, in this case, p_{n+1} = p^2 * p_n, which is an M-class sequence with q = 0.But if q ≠ 0, then as we saw earlier, p_{n+1} is a quadratic function of a_n, which doesn't fit the linear form required for an M-class sequence. Therefore, {a_n * a_{n+1}} is an M-class sequence only if q = 0, otherwise, it's not necessarily.So, to summarize part 2: The sum {a_n + a_{n+1}} is always an M-class sequence, but the product {a_n * a_{n+1}} is only an M-class sequence if q = 0. Otherwise, it isn't. So, it's not necessarily an M-class sequence.Alright, moving on to part 3. This seems more involved. We have a sequence {a_n} with a_1 = 1, and for each n, a_n + a_{n+1} = 3*2^n. We need to find the sum S_n of the first n terms and determine if {a_n} is an M-class sequence.First, let's try to find a general formula for a_n. Given the recurrence relation a_n + a_{n+1} = 3*2^n, we can write a_{n+1} = 3*2^n - a_n.This is a linear recurrence relation. Let me try to solve it. The homogeneous solution comes from the equation a_{n+1} + a_n = 0, which has characteristic equation r + 1 = 0, so r = -1. So, the homogeneous solution is A*(-1)^n.For the particular solution, since the nonhomogeneous term is 3*2^n, we can try a particular solution of the form B*2^n. Let's substitute into the recurrence:a_{n+1} + a_n = 3*2^nB*2^{n+1} + B*2^n = 3*2^nB*2*2^n + B*2^n = 3*2^n(2B + B)*2^n = 3*2^n3B*2^n = 3*2^nSo, 3B = 3 => B = 1.Therefore, the general solution is a_n = A*(-1)^n + 1*2^n.Now, apply the initial condition a_1 = 1.a_1 = A*(-1)^1 + 2^1 = -A + 2 = 1So, -A + 2 = 1 => -A = -1 => A = 1.Thus, the general term is a_n = (-1)^n + 2^n.Wait, let me check this. For n=1: a_1 = (-1)^1 + 2^1 = -1 + 2 = 1, which matches. For n=2: a_2 = (-1)^2 + 2^2 = 1 + 4 = 5. Let's check the recurrence: a_1 + a_2 = 1 + 5 = 6, which should be 3*2^1 = 6. Correct. For n=3: a_3 = (-1)^3 + 2^3 = -1 + 8 = 7. Then, a_2 + a_3 = 5 + 7 = 12, which is 3*2^2 = 12. Correct. Seems good.So, a_n = (-1)^n + 2^n.Now, let's find S_n, the sum of the first n terms. S_n = sum_{k=1}^n a_k = sum_{k=1}^n [(-1)^k + 2^k] = sum_{k=1}^n (-1)^k + sum_{k=1}^n 2^k.Compute each sum separately.First, sum_{k=1}^n (-1)^k. This is a finite geometric series with ratio -1. The sum is [(-1)(1 - (-1)^n)] / (1 - (-1)) = [(-1)(1 - (-1)^n)] / 2.Simplify: [ -1 + (-1)^{n+1} ] / 2.Second, sum_{k=1}^n 2^k. This is a geometric series with ratio 2. The sum is 2*(2^n - 1)/(2 - 1) = 2^{n+1} - 2.Therefore, S_n = [ -1 + (-1)^{n+1} ] / 2 + 2^{n+1} - 2.Simplify:S_n = 2^{n+1} - 2 + [ (-1)^{n+1} - 1 ] / 2.Let me combine the constants:= 2^{n+1} - 2 - 1/2 + [ (-1)^{n+1} ] / 2= 2^{n+1} - 5/2 + [ (-1)^{n+1} ] / 2.Alternatively, factor 1/2:= 2^{n+1} - 5/2 + ( (-1)^{n+1} ) / 2= 2^{n+1} + [ (-1)^{n+1} - 5 ] / 2.Hmm, maybe another approach. Let me write it as:S_n = 2^{n+1} - 2 + [ (-1)^{n+1} - 1 ] / 2= 2^{n+1} - 2 - 1/2 + [ (-1)^{n+1} ] / 2= 2^{n+1} - 5/2 + [ (-1)^{n+1} ] / 2.Alternatively, factor 1/2:= (2^{n+2} - 5 + (-1)^{n+1}) / 2.But maybe it's better to write it in terms of cases for even and odd n.Notice that (-1)^{n+1} is 1 when n is even and -1 when n is odd.So, let's split into two cases:Case 1: n is even, say n = 2k.Then, (-1)^{n+1} = (-1)^{2k + 1} = -1.So, S_n = 2^{n+1} - 2 + [ -1 - 1 ] / 2 = 2^{n+1} - 2 - 1 = 2^{n+1} - 3.Wait, hold on. Wait, let's recast the earlier expression:S_n = 2^{n+1} - 2 + [ (-1)^{n+1} - 1 ] / 2.If n is even:[ (-1)^{n+1} - 1 ] / 2 = [ (-1)^{odd} - 1 ] / 2 = [ -1 - 1 ] / 2 = (-2)/2 = -1.Thus, S_n = 2^{n+1} - 2 - 1 = 2^{n+1} - 3.If n is odd:[ (-1)^{n+1} - 1 ] / 2 = [ (-1)^{even} - 1 ] / 2 = [ 1 - 1 ] / 2 = 0/2 = 0.Thus, S_n = 2^{n+1} - 2 + 0 = 2^{n+1} - 2.Wait, that contradicts my initial thought. Wait, let me check with n=1:n=1: S_1 = a_1 = 1. According to the formula, if n is odd, S_n = 2^{2} - 2 = 4 - 2 = 2. But S_1 should be 1. Hmm, something's wrong.Wait, maybe I made a mistake in splitting the cases. Let me re-examine.Wait, when n is even, n = 2k, then:S_n = 2^{n+1} - 2 + [ (-1)^{n+1} - 1 ] / 2.But n is even, so n+1 is odd, so (-1)^{n+1} = -1.Thus, [ (-1)^{n+1} - 1 ] / 2 = (-1 - 1)/2 = -1.So, S_n = 2^{n+1} - 2 - 1 = 2^{n+1} - 3.When n is odd, n = 2k - 1, then n+1 is even, so (-1)^{n+1} = 1.Thus, [1 - 1]/2 = 0.So, S_n = 2^{n+1} - 2 + 0 = 2^{n+1} - 2.Wait, but for n=1, which is odd, S_1 = 2^{2} - 2 = 4 - 2 = 2, but a_1 = 1. So, discrepancy here.Wait, maybe my initial splitting is wrong. Let me compute S_n for small n to see.Compute S_1: a_1 = 1. So, S_1 = 1.Compute S_2: a_1 + a_2 = 1 + 5 = 6.Compute S_3: 1 + 5 + 7 = 13.Compute S_4: 1 + 5 + 7 + 13 = 26.Wait, let's see:Using the formula:For n=1 (odd): S_1 = 2^{2} - 2 = 4 - 2 = 2. But actual S_1 is 1. So, wrong.For n=2 (even): S_2 = 2^{3} - 3 = 8 - 3 = 5. But actual S_2 is 6. Hmm, discrepancy.Wait, perhaps I made a mistake in the sum.Wait, let's recompute S_n.Given a_n = (-1)^n + 2^n.So, S_n = sum_{k=1}^n [ (-1)^k + 2^k ] = sum_{k=1}^n (-1)^k + sum_{k=1}^n 2^k.Compute sum_{k=1}^n (-1)^k:This is a geometric series with first term -1, ratio -1, n terms.Sum = (-1)*(1 - (-1)^n)/(1 - (-1)) = (-1)*(1 - (-1)^n)/2 = [ (-1) + (-1)^{n+1} ] / 2.Similarly, sum_{k=1}^n 2^k = 2*(2^n - 1)/(2 - 1) = 2^{n+1} - 2.Thus, S_n = [ (-1) + (-1)^{n+1} ] / 2 + 2^{n+1} - 2.Simplify:= 2^{n+1} - 2 + [ (-1)^{n+1} - 1 ] / 2.Let me write this as:= 2^{n+1} - 2 - 1/2 + [ (-1)^{n+1} ] / 2= 2^{n+1} - 5/2 + [ (-1)^{n+1} ] / 2.Alternatively, factor 1/2:= (2^{n+2} - 5 + (-1)^{n+1}) / 2.But let's test this formula with n=1:(2^{3} - 5 + (-1)^2)/2 = (8 - 5 + 1)/2 = 4/2 = 2. But S_1 should be 1. Hmm, still discrepancy.Wait, maybe my initial sum is incorrect. Let me compute S_n manually for small n.n=1: a1=1, S1=1.n=2: a1=1, a2=5, S2=6.n=3: a3=7, S3=13.n=4: a4=13, S4=26.n=5: a5=25, S5=51.Wait, let's compute S_n using the formula:For n=1:S1 = 2^{2} - 2 + [ (-1)^2 - 1 ] / 2 = 4 - 2 + (1 - 1)/2 = 2 + 0 = 2. Not matching.Wait, perhaps the formula is wrong. Let me re-examine the sum.Wait, sum_{k=1}^n (-1)^k is equal to [ (-1)(1 - (-1)^n) ] / (1 - (-1)) = [ (-1 + (-1)^{n+1}) ] / 2.So, S_n = [ (-1 + (-1)^{n+1}) ] / 2 + 2^{n+1} - 2.Thus, S_n = 2^{n+1} - 2 + [ (-1)^{n+1} - 1 ] / 2.Let me compute for n=1:2^{2} - 2 + [ (-1)^2 - 1 ] / 2 = 4 - 2 + (1 - 1)/2 = 2 + 0 = 2. Still wrong.Wait, maybe the initial term in the sum is wrong. The sum from k=1 to n of (-1)^k is:When n=1: -1.When n=2: -1 + 1 = 0.When n=3: -1 + 1 -1 = -1.When n=4: -1 +1 -1 +1=0.So, it alternates between -1 and 0 depending on whether n is odd or even.Thus, for n odd: sum = -1.For n even: sum = 0.So, sum_{k=1}^n (-1)^k = (-1)^{n} * (n mod 2). Wait, not exactly. It's -1 if n is odd, 0 if n is even.So, we can write:sum_{k=1}^n (-1)^k = (-1)^n * (1 - (n mod 2)).But perhaps a better way is to express it as:sum_{k=1}^n (-1)^k = [ (-1)^n - 1 ] / 2.Wait, let's test:For n=1: [ (-1) - 1 ] / 2 = (-2)/2 = -1. Correct.For n=2: [1 - 1]/2 = 0. Correct.For n=3: [ (-1)^3 -1 ] /2 = (-2)/2 = -1. Correct.Yes, so sum_{k=1}^n (-1)^k = [ (-1)^n - 1 ] / 2.Therefore, S_n = [ (-1)^n - 1 ] / 2 + 2^{n+1} - 2.Simplify:= 2^{n+1} - 2 + [ (-1)^n - 1 ] / 2= 2^{n+1} - 2 - 1/2 + [ (-1)^n ] / 2= 2^{n+1} - 5/2 + [ (-1)^n ] / 2.Alternatively, factor 1/2:= (2^{n+2} - 5 + (-1)^n ) / 2.Wait, let's test this formula:For n=1:(2^{3} -5 + (-1)^1)/2 = (8 -5 -1)/2 = 2/2 =1. Correct.For n=2:(2^{4} -5 + (-1)^2)/2 = (16 -5 +1)/2 =12/2=6. Correct.For n=3:(2^{5} -5 + (-1)^3)/2 = (32 -5 -1)/2=26/2=13. Correct.For n=4:(2^{6} -5 + (-1)^4)/2=(64 -5 +1)/2=60/2=30. Wait, but earlier I thought S4=26. Wait, let me compute S4 manually:a1=1, a2=5, a3=7, a4=13. So, S4=1+5+7+13=26. But according to the formula, it's 30. Hmm, discrepancy.Wait, what's a4? a4 = (-1)^4 + 2^4 =1 +16=17. Wait, earlier I thought a4=13, but according to the general formula, a4=17. Wait, that's a mistake.Wait, let's recompute a_n:Given a_n = (-1)^n + 2^n.So,a1= (-1)^1 +2^1= -1+2=1.a2= (-1)^2 +2^2=1+4=5.a3= (-1)^3 +2^3=-1+8=7.a4= (-1)^4 +2^4=1+16=17.a5= (-1)^5 +2^5=-1+32=31.So, S1=1.S2=1+5=6.S3=1+5+7=13.S4=1+5+7+17=20+7=27? Wait, 1+5=6, 6+7=13, 13+17=30. Wait, no, 1+5+7+17=30.Wait, but earlier I thought a4=13, which was wrong. It's actually 17. So, S4=30.Wait, but according to the formula:For n=4, which is even:S4=2^{5} -3=32 -3=29. Wait, but according to the formula I derived earlier, S_n=2^{n+1} -3 when n is even.But 2^{5} -3=32-3=29, but actual S4=30. Hmm, discrepancy.Wait, maybe my splitting into cases was wrong. Let me re-examine.Earlier, I thought:sum_{k=1}^n (-1)^k = [ (-1)^n -1 ] / 2.Thus, S_n = [ (-1)^n -1 ] / 2 + 2^{n+1} - 2.So, for n=4:[ (-1)^4 -1 ] /2 + 2^5 -2 = [1 -1]/2 +32 -2=0 +30=30. Correct.For n=3:[ (-1)^3 -1 ] /2 +2^4 -2= [ -1 -1 ]/2 +16 -2= (-2)/2 +14= -1 +14=13. Correct.For n=2:[1 -1]/2 +8 -2=0 +6=6. Correct.For n=1:[ -1 -1 ]/2 +4 -2= (-2)/2 +2= -1 +2=1. Correct.So, the formula S_n = 2^{n+1} - 2 + [ (-1)^n -1 ] / 2 is correct.But when I tried to split into cases, I messed up. Let me try again.Express S_n as:If n is even, then (-1)^n =1, so [1 -1]/2=0. Thus, S_n=2^{n+1} -2.If n is odd, then (-1)^n=-1, so [ -1 -1 ] /2= -1. Thus, S_n=2^{n+1} -2 -1=2^{n+1} -3.Wait, but for n=4 (even), S4=30=2^{5} -2=32-2=30. Correct.For n=3 (odd), S3=13=2^{4} -3=16-3=13. Correct.For n=2 (even), S2=6=2^{3} -2=8-2=6. Correct.For n=1 (odd), S1=1=2^{2} -3=4-3=1. Correct.So, the correct expression is:S_n = 2^{n+1} -2 when n is even,S_n = 2^{n+1} -3 when n is odd.Therefore, S_n can be written as:S_n = begin{cases}2^{n+1} - 2, & text{if } n text{ is even}, 2^{n+1} - 3, & text{if } n text{ is odd}.end{cases}Alright, that seems correct.Now, the second part of question 3 is to determine whether {a_n} is an M-class sequence. Recall that {a_n} is given by a_n = (-1)^n + 2^n.An M-class sequence satisfies a_{n+1} = p*a_n + q for some constants p and q.Let me check if such p and q exist.Compute a_{n+1}:a_{n+1} = (-1)^{n+1} + 2^{n+1}.Express this in terms of a_n:a_n = (-1)^n + 2^n.So, let's see:a_{n+1} = (-1)*(-1)^n + 2*2^n = (-1)*(-1)^n + 2*2^n.Compare this to p*a_n + q:p*a_n + q = p*[ (-1)^n + 2^n ] + q = p*(-1)^n + p*2^n + q.So, we have:a_{n+1} = (-1)*(-1)^n + 2*2^n = (-1)^{n+1} + 2^{n+1}.And p*a_n + q = p*(-1)^n + p*2^n + q.For these to be equal for all n, the coefficients of (-1)^n and 2^n must match, and the constants must match.So, equate coefficients:Coefficient of (-1)^n: -1 = p.Coefficient of 2^n: 2 = p.Constant term: 0 = q.But from the coefficients, we have p = -1 and p = 2 simultaneously, which is impossible. Therefore, there are no constants p and q such that a_{n+1} = p*a_n + q for all n. Hence, {a_n} is not an M-class sequence.Alternatively, let's try to see if such p and q exist. Suppose p and q exist.Then, for all n:(-1)^{n+1} + 2^{n+1} = p*(-1)^n + p*2^n + q.Let me rearrange:(-1)^{n+1} - p*(-1)^n + 2^{n+1} - p*2^n - q =0.Factor terms:(-1)^n*(-1 - p) + 2^n*(2 - p) - q =0.For this to hold for all n, each coefficient must be zero:-1 - p =0 => p = -1.2 - p =0 => p=2.But p cannot be both -1 and 2. Contradiction. Therefore, no such p and q exist. Hence, {a_n} is not an M-class sequence.So, summarizing part 3: S_n is equal to 2^{n+1} -2 when n is even, and 2^{n+1} -3 when n is odd. And {a_n} is not an M-class sequence.

💡First, I recognize that the sequence is defined by ( a_{n+1} = a_n - 2 ), which means each term decreases by 2 from the previous term. This indicates that the sequence is arithmetic with a common difference ( d = -2 ).Given that the first term ( a_1 = 3 ), I can use the formula for the ( n )-th term of an arithmetic sequence:[a_n = a_1 + (n - 1)d]To find ( a_{100} ), I substitute ( n = 100 ), ( a_1 = 3 ), and ( d = -2 ) into the formula:[a_{100} = 3 + (100 - 1) times (-2) = 3 + 99 times (-2) = 3 - 198 = -195]Therefore, the value of ( a_{100} ) is (-195).

💡First, I need to determine the effective width Carl mows with each pass, considering the overlap. The swath width is 30 inches, and he overlaps by 6 inches, so the effective width per pass is 24 inches, which is 2 feet.Next, I calculate the number of passes Carl needs to make to cover the entire 120-foot width of the lawn. Dividing the total width by the effective width per pass gives me 60 passes.Each pass is 100 feet long, so the total distance Carl walks is 60 passes multiplied by 100 feet per pass, resulting in 6000 feet.Finally, I determine the time it takes Carl to mow the lawn by dividing the total distance by his walking speed of 4000 feet per hour. This calculation shows that it takes Carl 1.5 hours to mow the lawn.

💡Alright, so I've got this problem here: there's a competition with 16 teams, and they've played a total of 55 games. I need to prove that among these 16 teams, there exist 3 teams that haven't played any matches against each other. Hmm, okay. Let me try to wrap my head around this.First off, let's think about what this means. We have 16 teams, and each game is between two teams. So, the total number of possible games is the number of ways to choose 2 teams out of 16, right? That would be the combination formula: C(16, 2). Let me calculate that real quick. C(16, 2) is (16 × 15)/2 = 120. So, there are 120 possible games in total. But they've only played 55 games. That means there are 120 - 55 = 65 games that haven't been played yet.Okay, so there are 65 games that haven't happened. Now, I need to show that among these 16 teams, there are 3 teams where none of the three possible games between them have been played. In other words, there's a trio of teams where each pair hasn't played each other. That sounds like we're looking for an "independent set" of size 3 in the graph where teams are vertices and games are edges.Wait, maybe I should model this as a graph. Let's consider each team as a vertex, and each game as an edge connecting two vertices. So, we have a graph with 16 vertices and 55 edges. The problem is asking us to prove that there exists an independent set of size 3 in this graph. An independent set is a set of vertices with no edges connecting them, which in this context means three teams that haven't played each other.Now, I remember something about Ramsey numbers, which deal with conditions under which order must appear. Specifically, the Ramsey number R(s, t) is the smallest number of vertices such that all undirected graphs of that size contain a clique of size s or an independent set of size t. In our case, we're dealing with an independent set of size 3. I think the Ramsey number R(3, n) might be relevant here, but I'm not entirely sure.Alternatively, maybe Turán's theorem could be useful. Turán's theorem gives the maximum number of edges a graph can have without containing a complete subgraph of a certain size. But in our case, we're looking for an independent set, which is the complement of a complete subgraph. So, perhaps we can use the complement graph.Let me think about the complement graph. If our original graph has 55 edges, then the complement graph will have 120 - 55 = 65 edges. So, the complement graph has 65 edges. Now, if the complement graph has an independent set of size 3, that would correspond to a clique of size 3 in the original graph. Wait, no, actually, an independent set in the complement graph corresponds to a clique in the original graph. So, if the complement graph has a clique of size 3, that means the original graph has an independent set of size 3.But I'm not sure if that helps directly. Maybe I should approach this differently. Let's consider the total number of possible triples of teams. The number of ways to choose 3 teams out of 16 is C(16, 3). Let me calculate that: (16 × 15 × 14)/(3 × 2 × 1) = 560. So, there are 560 possible triples.Now, each game played affects the number of triples that could potentially be independent sets. Specifically, each game played reduces the number of possible independent sets by the number of triples that include that game. But I'm not sure how to quantify that exactly.Wait, maybe I can use the probabilistic method or some combinatorial argument. Let's think about the expected number of independent sets of size 3. If we randomly select a triple, what's the probability that none of the three possible games have been played? Well, the total number of possible games is 120, and 55 have been played, so the probability that a specific game hasn't been played is (120 - 55)/120 = 65/120 ≈ 0.5417.But the probability that none of the three games in a triple have been played would be (65/120)^3 ≈ 0.158. So, the expected number of independent sets of size 3 would be 560 × 0.158 ≈ 88.48. That's a rough estimate, but it suggests that there are likely many independent sets of size 3. However, this is just an expectation, and it doesn't guarantee that at least one exists.Hmm, maybe I need a more deterministic approach. Let's consider the number of non-edges in the graph. We have 65 non-edges. Now, each non-edge can be part of multiple triples. Specifically, each non-edge can be part of (16 - 2) = 14 triples, since for each non-edge (u, v), we can pair it with any of the remaining 14 teams to form a triple.But wait, if we count the number of triples that include at least one non-edge, we might be overcounting. Because a triple could have multiple non-edges. For example, a triple with two non-edges would be counted twice, and a triple with all three non-edges would be counted three times.This seems complicated. Maybe instead of counting the number of triples with at least one non-edge, I should count the number of triples with all three edges missing. That's exactly what we're looking for: an independent set of size 3.Let me denote the number of such triples as X. We need to show that X ≥ 1.To find X, we can use the principle of inclusion-exclusion. The total number of triples is C(16, 3) = 560. The number of triples with at least one edge is equal to the total number of triples minus the number of triples with no edges, which is X. So, the number of triples with at least one edge is 560 - X.But how can we relate this to the number of edges? Each edge is part of (16 - 2) = 14 triples. So, the total number of edge-triple incidences is 55 × 14 = 770. On the other hand, the number of triples with at least one edge can also be expressed as the sum over all triples of the number of edges in each triple. But this counts each triple multiple times depending on how many edges it has.Wait, maybe I can use the following identity: the sum over all triples of the number of edges in each triple is equal to the number of edge-triple incidences, which is 770. Let me denote Y as the number of triples with exactly one edge, Z as the number with exactly two edges, and W as the number with all three edges. Then, we have Y + 2Z + 3W = 770.But we also know that the total number of triples is Y + Z + W + X = 560. So, we have two equations:1. Y + Z + W + X = 5602. Y + 2Z + 3W = 770Subtracting the first equation from the second, we get Z + 2W = 210.Hmm, interesting. So, Z + 2W = 210. But I'm not sure how this helps us find X. Maybe I need another equation or a different approach.Alternatively, perhaps I can use the fact that the number of non-edges is 65. Each non-edge can be part of 14 triples, as I thought earlier. So, the total number of non-edge-triple incidences is 65 × 14 = 910. But each triple with exactly one non-edge contributes 1 to this count, each triple with exactly two non-edges contributes 2, and each triple with all three non-edges contributes 3. So, we have:Y' + 2Z' + 3X = 910Where Y' is the number of triples with exactly one non-edge, Z' is the number with exactly two non-edges, and X is the number with all three non-edges.But we also know that the total number of triples is Y' + Z' + X + W' = 560, where W' is the number of triples with all three edges. Wait, but this seems similar to the previous notation, just from the perspective of non-edges.I'm getting a bit tangled here. Maybe I should try a different approach altogether.Let me think about the average number of games played per team. There are 16 teams and 55 games. Each game involves two teams, so the total number of games per team is 55 × 2 = 110. Therefore, the average number of games per team is 110 / 16 ≈ 6.875.So, on average, each team has played about 6.875 games. That means, on average, each team has not played about 16 - 1 - 6.875 = 8.125 games. Wait, no, that's not quite right. Each team can play up to 15 games (against the other 15 teams). So, if they've played an average of 6.875 games, they've not played 15 - 6.875 = 8.125 games on average.So, each team has not played against approximately 8.125 other teams. That suggests that there are quite a few non-edges in the graph. Now, if each team has about 8 non-edges, maybe we can find a trio of teams where each pair hasn't played each other.But how can we formalize this? Maybe we can use the pigeonhole principle. If each team has a certain number of non-edges, then there must be some overlap where three teams all have non-edges between each other.Let me try to think about it this way: suppose we pick a team, say Team A. Team A has not played against approximately 8 other teams. Let's call this set S, which has 8 teams. Now, within set S, each team has also not played against some number of teams. If we can find two teams in S that haven't played each other, then together with Team A, we have our trio.But how many games have been played within set S? Set S has 8 teams, so the total number of possible games within S is C(8, 2) = 28. Now, how many games have actually been played within S? Well, the total number of games played is 55, but we don't know how many of those are within S.Wait, maybe we can bound the number of games played within S. Since each team in S has played an average of about 6.875 games, but some of those games could be against teams outside of S. Let me denote the number of games played within S as G. Then, the total number of games played by teams in S is 8 × 6.875 = 55. But this counts both games within S and games against teams outside S.However, the total number of games played by teams in S against teams outside S is equal to the number of non-edges from S to the rest of the graph. Wait, this is getting too convoluted.Maybe I should use a more systematic approach. Let's consider the complement graph, which has 65 edges. We need to show that the complement graph contains a triangle (a set of three mutually connected vertices), which would correspond to an independent set of size 3 in the original graph.There's a theorem called Turán's theorem which gives the maximum number of edges a graph can have without containing a complete subgraph of a certain size. Specifically, Turán's theorem states that the maximum number of edges in an n-vertex graph without a (r+1)-clique is given by a certain formula. In our case, we're looking for a 3-clique in the complement graph, so we can set r = 2.Turán's theorem for r = 2 gives the maximum number of edges without a triangle as floor(n²/4). For n = 16, this would be floor(16²/4) = floor(256/4) = 64. So, the maximum number of edges a graph on 16 vertices can have without containing a triangle is 64.But our complement graph has 65 edges, which is one more than the Turán number. Therefore, by Turán's theorem, the complement graph must contain at least one triangle. Hence, the original graph must contain an independent set of size 3.Wait, that seems like a solid argument. Let me recap:1. The original graph has 16 vertices and 55 edges.2. The complement graph has 16 vertices and 65 edges.3. Turán's theorem states that any graph with more than floor(n²/4) edges must contain a triangle.4. For n = 16, floor(16²/4) = 64.5. Since the complement graph has 65 edges, it must contain at least one triangle.6. Therefore, the original graph must contain an independent set of size 3.Yes, that makes sense. So, by using Turán's theorem on the complement graph, we've shown that there must be a triangle in the complement, which corresponds to an independent set of size 3 in the original graph. Therefore, there exist 3 teams that haven't played any matches against each other.I think that wraps it up. I used Turán's theorem to analyze the complement graph and showed that the number of edges forces the existence of a triangle, which translates back to the original problem as the desired independent set.