💡First, I need to identify the common difference of the arithmetic series. The series starts at 1 and each subsequent term increases by 2, so the common difference ( d ) is 2.Next, I'll determine the number of terms in the series. The first term ( a_1 ) is 1, and the last term ( a_n ) is 21. Using the formula for the ( n )-th term of an arithmetic sequence:[a_n = a_1 + (n - 1)d]I can solve for ( n ):[21 = 1 + (n - 1) times 2][20 = (n - 1) times 2][10 = n - 1][n = 11]There are 11 terms in the series.Finally, I'll calculate the sum of the series using the formula for the sum of an arithmetic series:[S_n = frac{n}{2} times (a_1 + a_n)]Plugging in the values:[S_{11} = frac{11}{2} times (1 + 21)][S_{11} = frac{11}{2} times 22][S_{11} = 11 times 11][S_{11} = 121]

💡Alright, so I'm trying to figure out this probability problem about a cube with stripes on each face. Each face has four possible orientations for the stripe: either from the center of one edge to the center of the opposite edge or along a diagonal. The question is asking for the probability that there's a continuous stripe encircling the cube. The options are A) 1/1024, B) 3/1024, C) 1/256, and D) 1/64.First, I need to understand the problem better. Each face of the cube has four possible stripe orientations. Since the cube has six faces, the total number of possible stripe combinations is 4^6, which is 4096. So, the total number of possible outcomes is 4096.Now, I need to find how many of these combinations result in a continuous stripe encircling the cube. To do that, I should think about what it means for a stripe to encircle the cube. An encircling stripe would mean that there's a path that goes around the cube without breaking, connecting through the edges or corners.I remember that on a cube, you can have stripes that go along the edges or along the diagonals. If a stripe is along an edge, it connects the centers of two opposite edges. If it's along a diagonal, it connects two opposite corners. For a continuous stripe to encircle the cube, it needs to form a loop that goes around the cube, which probably means that the stripes on adjacent faces need to align in a specific way.Let me try to visualize this. If I pick one face, say the top face, and choose a stripe that goes from the center of the front edge to the center of the back edge, then for the stripe to continue on the front face, the stripe on the front face should also go from the center of the top edge to the center of the bottom edge. Similarly, this pattern should continue around the cube.But wait, the problem says that each face has four possible orientations, which includes both edge-to-edge and diagonal stripes. So, not all stripes are edge-to-edge; some are diagonals. That complicates things because a diagonal stripe might not connect in the same way as an edge-to-edge stripe.I think that for a continuous stripe to encircle the cube, the stripes on adjacent faces need to align edge-to-edge. If a face has a diagonal stripe, it might not contribute to a continuous encircling stripe. So, maybe only the edge-to-edge stripes can form such a loop.Given that, I should consider only the edge-to-edge stripes for forming the continuous loop. Each face has two edge-to-edge orientations and two diagonal orientations. So, for each face, the probability of having an edge-to-edge stripe is 2/4, which is 1/2.But I need more than just the probability for one face; I need the probability that all the necessary faces have edge-to-edge stripes aligned in a way that forms a loop. Hmm, how many such loops are possible?I recall that a cube has three pairs of parallel faces: front-back, left-right, and top-bottom. For each pair, there are two possible orientations for the stripes to form a loop around the cube. So, for each pair, there are two favorable orientations.Wait, let me think again. If I fix a pair of parallel faces, say front and back, then to form a loop around the cube, the stripes on the front and back faces need to be oriented in a specific way, and the stripes on the top, bottom, left, and right faces also need to align accordingly.Actually, for each pair of parallel faces, there are two possible ways to orient the stripes so that they form a loop. Since there are three pairs of parallel faces, there are 3 * 2 = 6 possible favorable configurations.But hold on, each face has two edge-to-edge orientations, so for each pair, there are 2 * 2 = 4 ways to orient the stripes on the front and back faces, and similarly for the other pairs. But not all of these will result in a continuous loop.I think I need to count the number of favorable outcomes where the stripes on all six faces are aligned to form a loop. For each pair of parallel faces, there are two possible ways to orient the stripes to form a loop. Since there are three pairs, the total number of favorable outcomes is 3 * 2 = 6.But wait, each face has two edge-to-edge orientations, so for each pair, there are 2 * 2 = 4 ways to orient the stripes on the front and back faces, and similarly for the other pairs. But not all of these will result in a continuous loop.Actually, for each pair of parallel faces, there are two possible ways to orient the stripes so that they form a loop. Since there are three pairs, the total number of favorable outcomes is 3 * 2 = 6.But each face has four possible orientations, so the total number of possible stripe combinations is 4^6 = 4096.Therefore, the probability is 6 / 4096, which simplifies to 3 / 2048. But that's not one of the options. Hmm, maybe I'm missing something.Wait, perhaps I need to consider that for each pair of parallel faces, there are more than two favorable orientations. Let me think again.Each pair of parallel faces can have their stripes oriented in two ways to form a loop, but the remaining four faces also need to have their stripes oriented in a way that continues the loop. So, for each pair, there are 2 * 2 = 4 ways to orient the stripes on the pair, and the remaining four faces have only one way to orient their stripes to continue the loop.Therefore, for each pair, there are 4 favorable outcomes, and since there are three pairs, the total number of favorable outcomes is 3 * 4 = 12.So, the probability is 12 / 4096, which simplifies to 3 / 1024. That matches option B.Wait, but I thought earlier that for each pair, there are two favorable orientations, but now I'm considering four. Maybe I was confused before.Let me clarify. For each pair of parallel faces, say front and back, the stripes can be oriented in two ways: either both horizontal or both vertical. But actually, each face has two edge-to-edge orientations, so for front and back, there are 2 * 2 = 4 ways to orient the stripes. However, only two of these will result in a continuous loop because the stripes need to align with the adjacent faces.Wait, no, actually, for each pair, there are two possible loops: one going around the cube in one direction and the other in the opposite direction. So, for each pair, there are two favorable orientations.But then, considering the remaining four faces, their stripes need to align accordingly. So, for each pair, there are 2 * 1 = 2 favorable outcomes, and since there are three pairs, the total is 3 * 2 = 6.But earlier, I thought it was 12. I'm getting confused here.Maybe I should think differently. Let's fix a pair of parallel faces, say front and back. For the stripes to form a loop around the cube, the front and back faces must have their stripes oriented in a way that connects to the left, right, top, and bottom faces.Each of the front and back faces has two edge-to-edge orientations. If we choose one orientation for the front face, the back face must have the corresponding orientation to continue the loop. So, for each pair, there are 2 favorable orientations.Now, for the remaining four faces (left, right, top, bottom), their stripes must also be oriented in a way that continues the loop. Each of these faces has two edge-to-edge orientations, but only one of them will align with the loop.Therefore, for each of these four faces, there's only one favorable orientation. So, the number of favorable outcomes for the remaining four faces is 1^4 = 1.Therefore, for each pair of parallel faces, the number of favorable outcomes is 2 (for the pair) * 1 (for the remaining four faces) = 2.Since there are three pairs of parallel faces, the total number of favorable outcomes is 3 * 2 = 6.Therefore, the probability is 6 / 4096 = 3 / 2048, which is not one of the options. Hmm, something's wrong here.Wait, maybe I'm not considering that the remaining four faces can have their stripes oriented in two ways each, but only one way continues the loop. So, for each of the remaining four faces, there's only one favorable orientation, making the total number of favorable outcomes for the remaining four faces 1^4 = 1.But then, for each pair, it's 2 * 1 = 2, and with three pairs, it's 6. So, 6 / 4096 = 3 / 2048.But the options are 1/1024, 3/1024, 1/256, and 1/64. 3/2048 is approximately 0.00146, which is less than 1/1024 (which is approximately 0.000976). Wait, no, 1/1024 is approximately 0.000976, and 3/2048 is approximately 0.00146, which is actually larger than 1/1024.Wait, maybe I made a mistake in simplifying. 6 / 4096 is equal to 3 / 2048, which is indeed approximately 0.00146. But 3 / 1024 is approximately 0.00293, which is double that.So, perhaps I missed a factor of 2 somewhere. Maybe for each pair, there are two possible loops, one in each direction, so 2 * 2 = 4 per pair, leading to 12 total.Wait, let's think about it again. For each pair of parallel faces, there are two possible directions for the loop: clockwise and counterclockwise. So, for each pair, there are two favorable orientations.But also, for each pair, the front and back faces can have their stripes oriented in two ways each, but only two of those combinations will align to form a loop. So, 2 favorable orientations for the pair.Then, for the remaining four faces, each has two edge-to-edge orientations, but only one will align with the loop. So, 1^4 = 1.Therefore, for each pair, it's 2 * 1 = 2, and with three pairs, it's 6.But maybe I'm not considering that the loop can go around the cube in different ways, not just along the three pairs. Maybe there are more possible loops.Wait, actually, a cube has more than three possible loops. For example, you can have a loop that goes around the cube along the equator, but also loops that go around the cube in different orientations.But I think the standard loops are the three pairs of parallel faces. So, maybe only three possible loops.But then, why is the answer 3/1024 instead of 3/2048?Wait, maybe I'm not considering that for each loop, there are two possible directions, so each loop has two favorable orientations.So, for each pair, there are two loops (clockwise and counterclockwise), so 2 * 3 = 6.But then, for each loop, the number of favorable outcomes is 2 (for the pair) * 1 (for the remaining four faces) = 2.So, total favorable outcomes are 6.Wait, but 6 / 4096 is 3 / 2048, which is still not matching the options.Alternatively, maybe for each loop, there are 4 favorable outcomes because each face in the pair has two orientations, and the remaining four faces each have one orientation.So, for each loop, it's 2 * 2 * 1 * 1 * 1 * 1 = 4.With three loops, it's 12.So, 12 / 4096 = 3 / 1024.Ah, that makes sense. So, for each loop, there are four favorable outcomes: two for the front face, two for the back face, and one for each of the remaining four faces.Therefore, total favorable outcomes are 3 * 4 = 12.So, the probability is 12 / 4096 = 3 / 1024.Yes, that seems right. So, the answer should be B) 3/1024.

💡To determine the distance from the origin to the point (-12, 16), I can use the distance formula. The distance formula between two points (x1, y1) and (x2, y2) is given by:d = √[(x2 - x1)² + (y2 - y1)²]Since one of the points is the origin (0, 0), the formula simplifies to:d = √[(-12 - 0)² + (16 - 0)²] = √[144 + 256] = √400 = 20Alternatively, recognizing that the legs of the right triangle formed by the coordinates are 12 and 16, which are multiples of the 3-4-5 Pythagorean triplet, the hypotenuse (distance) is 5 multiplied by 4, resulting in 20.Therefore, the distance from the origin to the point (-12, 16) is 20 units.

💡First, identify the total number of marbles Tom has. There are 4 non-yellow marbles (red, green, blue, orange) and 4 identical yellow marbles, making a total of 8 marbles.Next, calculate the total number of ways to choose any 3 marbles from the 8 marbles without any restrictions. This is given by the combination formula C(8, 3), which equals 56.Then, determine the number of ways to choose 3 marbles with no yellow marbles. This involves selecting all 3 marbles from the 4 non-yellow marbles, which is C(4, 3) = 4.Finally, subtract the number of groups without any yellow marbles from the total number of groups to find the number of groups with at least one yellow marble: 56 - 4 = 52.

💡First, I need to understand the structure of the larger cube and how it is divided into smaller cubes. The larger cube is painted on all its faces and then cut into 27 smaller cubes, which means it's divided into a 3x3x3 grid.Next, I'll categorize the smaller cubes based on their positions in the larger cube and the number of painted faces they have:- **Corner Cubes**: There are 8 corner cubes in a cube, and each corner cube has 3 painted faces.- **Edge Cubes**: Each edge of the cube has 3 smaller cubes, but excluding the corners, there is 1 cube per edge with 2 painted faces. Since a cube has 12 edges, there are 12 edge cubes.- **Face Center Cubes**: Each face of the cube has 1 smaller cube at its center with only 1 painted face. With 6 faces, there are 6 face center cubes.- **Inner Cube**: There is 1 cube at the very center of the larger cube that has no painted faces.Given that there are 12 smaller cubes with the same number of painted faces, these must be the edge cubes, each having 2 painted faces.Finally, to find the total number of painted faces on these 12 smaller cubes, I'll multiply the number of cubes by the number of painted faces per cube:12 cubes * 2 painted faces per cube = 24 painted faces.

💡Okay, so I have this problem where I need to prove that the product from i equals 1 to 2023 of (1 + 1/b_i) is less than 7/4. The sequences a_n and b_n are defined as a_n being the sum from i=1 to n of 1/i!, and b_n is (n+1)! times a_n. Hmm, okay, let me try to unpack this step by step.First, I know that a_n is the partial sum of the series for e, right? Because the exponential function e^x is the sum from i=0 to infinity of x^i / i!, so when x=1, e is the sum from i=0 to infinity of 1/i!. Therefore, a_n is just the sum starting from i=1, so it's e minus 1 minus the tail from n+1 to infinity. But maybe that's not directly useful here.Next, b_n is defined as (n+1)! times a_n. So, let's write that out: b_n = (n+1)! * a_n. Since a_n is the sum from i=1 to n of 1/i!, then b_n is (n+1)! times that sum. Maybe I can express b_n in terms of b_{n-1} or something recursive? Let me see.Let me try to write b_n in terms of b_{n-1}. So, b_n = (n+1)! * a_n. But a_n = a_{n-1} + 1/n!. So, substituting that in, b_n = (n+1)! * (a_{n-1} + 1/n!) = (n+1)! * a_{n-1} + (n+1)! / n!.Simplifying that, (n+1)! / n! is just (n+1). So, b_n = (n+1)! * a_{n-1} + (n+1). But wait, (n+1)! * a_{n-1} is actually (n+1) * n! * a_{n-1}, which is (n+1) * b_{n-1}, since b_{n-1} = n! * a_{n-1}. So, putting it all together, b_n = (n+1) * b_{n-1} + (n+1).Hmm, that's interesting. So, b_n = (n+1)(b_{n-1} + 1). Maybe that can help me find a recursive relation or something.Now, the product I need to evaluate is the product from i=1 to 2023 of (1 + 1/b_i). Let me write that out:Product = (1 + 1/b_1)(1 + 1/b_2)...(1 + 1/b_{2023}).I wonder if this product telescopes or if there's some cancellation that can happen. Maybe I can express each term in a way that relates to b_i and b_{i+1} or something like that.Looking back at the recursive relation I found: b_n = (n+1)(b_{n-1} + 1). Let me rearrange that:b_n / (n+1) = b_{n-1} + 1.So, b_{n-1} + 1 = b_n / (n+1).Therefore, 1 + 1/b_{n-1} = (b_{n-1} + 1)/b_{n-1} = (b_n / (n+1)) / b_{n-1} = b_n / [(n+1) b_{n-1}].Wait, that might not be directly helpful. Let me think again.Alternatively, let's consider the term (1 + 1/b_i). Since b_i = (i+1)! a_i, then 1/b_i = 1 / [(i+1)! a_i]. So, 1 + 1/b_i = 1 + 1 / [(i+1)! a_i].But I'm not sure if that helps. Maybe I can express a_i in terms of b_i? Since b_i = (i+1)! a_i, then a_i = b_i / (i+1)!.So, 1 + 1/b_i = 1 + 1 / [(i+1)! a_i] = 1 + 1 / [(i+1)! * (b_i / (i+1)!)] = 1 + (i+1)! / (b_i (i+1)! ) = 1 + 1 / b_i. Wait, that just circles back. Hmm.Maybe I need to find a telescoping product. Let's think about the product:Product = ∏_{i=1}^{2023} (1 + 1/b_i) = ∏_{i=1}^{2023} (b_i + 1)/b_i.So, that's equal to [ (b_1 + 1)/b_1 ] * [ (b_2 + 1)/b_2 ] * ... * [ (b_{2023} + 1)/b_{2023} ].If I can express each (b_i + 1) in terms of b_{i+1} or something, maybe the product will telescope.From the recursive relation, we have b_i = (i+1)(b_{i-1} + 1). So, rearranged, b_{i-1} + 1 = b_i / (i+1).Therefore, (b_i + 1) = (i+2)(b_i / (i+1)) + 1? Wait, no, let's see.Wait, actually, let's consider b_i + 1. From the recursive formula, b_i = (i+1)(b_{i-1} + 1). So, b_i + 1 = (i+1)(b_{i-1} + 1) + 1. Hmm, not sure.Alternatively, maybe express (b_i + 1) in terms of b_{i+1}. Let's see:From the recursive formula, b_{i+1} = (i+2)(b_i + 1). So, solving for (b_i + 1), we get (b_i + 1) = b_{i+1} / (i+2).Ah, that seems useful! So, (b_i + 1) = b_{i+1} / (i+2). Therefore, (b_i + 1)/b_i = (b_{i+1} / (i+2)) / b_i = b_{i+1} / [b_i (i+2)].So, each term in the product becomes b_{i+1} / [b_i (i+2)]. Therefore, the entire product is:Product = ∏_{i=1}^{2023} [b_{i+1} / (b_i (i+2))].Let's write that out:Product = [b_2 / (b_1 * 3)] * [b_3 / (b_2 * 4)] * [b_4 / (b_3 * 5)] * ... * [b_{2024} / (b_{2023} * 2025)].Now, notice that in this product, the b_2 in the numerator cancels with the b_2 in the denominator of the next term, similarly for b_3, b_4, etc., all the way up to b_{2023}. So, all the intermediate b terms cancel out, leaving us with:Product = b_{2024} / (b_1 * 3 * 4 * 5 * ... * 2025).Simplify the denominator: 3 * 4 * 5 * ... * 2025 is equal to 2025! / (2!) because it's the product from 3 to 2025, which is 2025! divided by 2!.So, 3 * 4 * ... * 2025 = 2025! / 2.Therefore, Product = b_{2024} / (b_1 * (2025! / 2)).Now, let's compute b_1. Since b_n = (n+1)! a_n, then b_1 = 2! a_1 = 2 * (1/1!) = 2.So, b_1 = 2.Thus, Product = b_{2024} / (2 * (2025! / 2)) ) = b_{2024} / (2025! / 1) ) = b_{2024} / 2025!.But wait, b_{2024} = (2025)! a_{2024}.So, substituting that in, Product = (2025! a_{2024}) / 2025! = a_{2024}.Wait, that's interesting. So, the entire product simplifies down to a_{2024}.But a_{2024} is the sum from i=1 to 2024 of 1/i!.I know that the sum from i=0 to infinity of 1/i! is e, so the sum from i=1 to infinity is e - 1. Therefore, a_{2024} is less than e - 1.But e is approximately 2.718, so e - 1 is approximately 1.718, which is less than 7/4 (which is 1.75). So, a_{2024} < e - 1 < 7/4.Therefore, the product is equal to a_{2024}, which is less than 7/4. Hence, the product is less than 7/4.Wait, but let me double-check that step where I said the product simplifies to a_{2024}. Let's go back.We had Product = b_{2024} / 2025!.But b_{2024} = (2025)! a_{2024}, so substituting that in, we get:Product = (2025! a_{2024}) / 2025! = a_{2024}.Yes, that seems correct. So, the product is exactly a_{2024}.Now, since a_{2024} is the partial sum of the series for e - 1, and since the series converges to e - 1, which is approximately 1.718, which is indeed less than 7/4 (1.75). Therefore, a_{2024} < 7/4, so the product is less than 7/4.But wait, is a_{2024} strictly less than 7/4? Let me compute a few terms to see.Compute a_1 = 1.a_2 = 1 + 1/2 = 1.5.a_3 = 1 + 1/2 + 1/6 ≈ 1.6667.a_4 = 1 + 1/2 + 1/6 + 1/24 ≈ 1.7083.a_5 = a_4 + 1/120 ≈ 1.7167.a_6 ≈ 1.7180.And so on, approaching e - 1 ≈ 1.71828.So, a_{2024} is very close to e - 1, but still less than e - 1, which is less than 7/4 (1.75). Therefore, a_{2024} < 7/4, so the product is less than 7/4.Therefore, the inequality holds.I think that makes sense. So, the key was to express the product in terms of a_{2024}, which is a partial sum of the series for e - 1, and since that partial sum is less than e - 1, which is less than 7/4, the inequality follows.**Final Answer**boxed{dfrac{7}{4}}