answer：Alright, so I have this problem here: I need to find the sum of the first six terms of an arithmetic sequence where the sum of any number of terms is equal to four times the square of that number. Hmm, okay, let me try to unpack this step by step.First off, let's recall what an arithmetic sequence is. It's a sequence where each term after the first is obtained by adding a constant difference. So, if the first term is 'a' and the common difference is 'd', then the sequence looks like a, a + d, a + 2d, a + 3d, and so on.Now, the problem says that the sum of any number of terms is equal to four times the square of that number. Let me denote the number of terms as 'n'. So, the sum of the first 'n' terms, which I'll call S_n, is given by S_n = 4n².Wait a minute, I remember that the sum of the first 'n' terms of an arithmetic sequence can also be expressed using the formula S_n = n/2 [2a + (n - 1)d]. So, if I set these two expressions for S_n equal to each other, I can probably solve for 'a' and 'd'.Let me write that down:n/2 [2a + (n - 1)d] = 4n²Okay, so if I simplify this equation, I can maybe find expressions for 'a' and 'd'. Let's see:First, multiply both sides by 2 to eliminate the fraction:n [2a + (n - 1)d] = 8n²Now, divide both sides by 'n' (assuming n ≠ 0, which makes sense because we're talking about the sum of terms):2a + (n - 1)d = 8nHmm, so that gives me an equation involving 'a', 'd', and 'n'. But I need to find specific values for 'a' and 'd', so I need another equation or a way to find these constants.Maybe I can plug in specific values for 'n' to create a system of equations. Let's try n = 1 and n = 2.For n = 1:2a + (1 - 1)d = 8(1)2a + 0 = 82a = 8a = 4Okay, so the first term 'a' is 4. That's straightforward.Now, let's use n = 2:2a + (2 - 1)d = 8(2)2a + d = 16We already know that a = 4, so substitute that in:2(4) + d = 168 + d = 16d = 16 - 8d = 8Great, so the common difference 'd' is 8.Now that I have both 'a' and 'd', I can write out the first six terms of the arithmetic sequence:1st term: a = 42nd term: a + d = 4 + 8 = 123rd term: a + 2d = 4 + 16 = 204th term: a + 3d = 4 + 24 = 285th term: a + 4d = 4 + 32 = 366th term: a + 5d = 4 + 40 = 44Let me double-check these terms to make sure they make sense. The difference between each term is 8, which matches our common difference 'd'. So, 4, 12, 20, 28, 36, 44 – yes, that looks correct.Now, to find the sum of these six terms, I can use the sum formula again:S_6 = 6/2 [2a + (6 - 1)d]S_6 = 3 [2(4) + 5(8)]S_6 = 3 [8 + 40]S_6 = 3 [48]S_6 = 144Alternatively, I could just add up the six terms I listed:4 + 12 = 1616 + 20 = 3636 + 28 = 6464 + 36 = 100100 + 44 = 144Yep, that checks out too. So, the sum of the first six terms is 144.Wait, just to make sure I didn't make any mistakes, let me verify using the original condition given in the problem. The sum of any number of terms should be equal to four times the square of that number. So, for n = 6, the sum should be 4*(6)^2 = 4*36 = 144. Exactly what I got. So, that confirms that my answer is correct.I think I've covered all the steps thoroughly. I found the first term and the common difference by plugging in specific values of 'n', then used those to find the sum of the first six terms. Everything seems consistent and checks out with the given condition.**Final Answer**The sum of the first six terms is boxed{144}.

answer：First, calculate the total amount earned from selling Samoas to the green house:3 boxes × 4 per box = 12.Next, calculate the total amount earned from selling sugar cookies to the brown house:9 boxes × 2 per box = 18.Then, calculate the total amount earned from selling Fudge Delights to the yellow house:1 box × 5 per box = 5.Add up the amounts from Samoas, sugar cookies, and Fudge Delights:12 + 18 + 5 = 35.Subtract this total from the overall earnings to find the amount earned from Thin Mints:42 (total) - 35 (other cookies) = 7.Finally, determine the number of Thin Mints boxes sold by dividing the amount earned from Thin Mints by the price per box:7 ÷ 3.50 per box = 2 boxes.

answer：Alright, so I have this problem here about two non-coincident lines on a plane. The original proposition is: "If two lines are parallel, then their slopes are equal." I need to figure out how many of this proposition, its converse, inverse, and contrapositive are true. Hmm, okay, let's break this down step by step.First off, I remember that in geometry, parallel lines are lines that never intersect, right? And on a plane, if two lines are parallel, they must have the same slope. That makes sense because the slope determines how steep a line is, and if they're parallel, they should rise and run at the same rate. So, the original proposition seems true to me. If two lines are parallel, their slopes must be equal.Now, let's think about the converse of this proposition. The converse would switch the hypothesis and the conclusion. So, the converse would be: "If the slopes of two lines are equal, then the lines are parallel." Hmm, does this hold true? Well, if two lines have the same slope, they should never intersect because they rise at the same rate. So, yes, they should be parallel. That seems true too.Next up is the inverse. The inverse of a proposition negates both the hypothesis and the conclusion. So, the inverse would be: "If two lines are not parallel, then their slopes are not equal." Let's see, if two lines are not parallel, that means they intersect at some point. If they intersect, their slopes must be different because if they had the same slope, they'd be parallel and never intersect. So, this inverse statement also seems true.Finally, the contrapositive. The contrapositive negates both the hypothesis and conclusion and switches them. So, the contrapositive of the original proposition would be: "If the slopes of two lines are not equal, then the lines are not parallel." This makes sense because if the slopes are different, the lines must intersect at some point, meaning they aren't parallel. So, the contrapositive is also true.Wait a minute, so all four statements—the original proposition, its converse, inverse, and contrapositive—are true? That seems a bit surprising because I thought usually only the original and contrapositive are logically equivalent, and the converse and inverse might not hold. But in this specific case, since we're dealing with lines on a plane and their slopes, it seems like all four hold true.Let me double-check to make sure I'm not missing something. For the converse: equal slopes imply parallel lines. Yeah, that's a standard result in coordinate geometry. For the inverse: not parallel implies different slopes. That also aligns with what I know. And the contrapositive is logically equivalent to the original, so if the original is true, the contrapositive must be true as well.So, yeah, it looks like all four propositions are true in this scenario. I guess when dealing with lines and their slopes, the relationships are pretty straightforward and symmetric in this case.**Final Answer**The number of true propositions is boxed{4}.

answer：Alright, let's tackle this problem step by step. So, we have a sequence {a_n} where the first term a₁ is 2. The difference sequence, which is defined as {a_{n+1} - a_n}, has a general formula of 2^n. We need to find the sum of the first n terms of the sequence {a_n}, denoted as S_n.First, let me make sure I understand the problem correctly. We have a sequence where each term is defined by the previous term plus 2 raised to the power of n. So, starting from a₁ = 2, each subsequent term is the previous term plus 2^n. That means:a₂ = a₁ + 2¹ a₃ = a₂ + 2² a₄ = a₃ + 2³ ... and so on.So, each term is built by adding a power of 2 to the previous term. That makes sense. Now, our goal is to find S_n, the sum of the first n terms. To do that, I think we first need to find a general formula for a_n, and then we can sum those terms.Let me write down the given information:a₁ = 2 a_{n+1} - a_n = 2^n for n ≥ 1So, we can express a_{n} in terms of a_{n-1}:a_n = a_{n-1} + 2^{n-1}This recursive relation suggests that each term is the sum of all previous 2^k terms. So, maybe we can write a_n as a sum of these differences.Let me try to write out the first few terms to see the pattern:a₁ = 2 a₂ = a₁ + 2¹ = 2 + 2 = 4 a₃ = a₂ + 2² = 4 + 4 = 8 a₄ = a₃ + 2³ = 8 + 8 = 16 a₅ = a₄ + 2⁴ = 16 + 16 = 32 ...Wait a minute, this looks familiar. The terms are doubling each time: 2, 4, 8, 16, 32... That's a geometric sequence where each term is 2 times the previous term. So, is a_n = 2^n?Let me check:If a₁ = 2 = 2¹ a₂ = 4 = 2² a₃ = 8 = 2³ a₄ = 16 = 2⁴ ... Yes, it seems that a_n = 2^n.But wait, let me verify this with the recursive formula. If a_n = 2^n, then:a_{n+1} - a_n = 2^{n+1} - 2^n = 2^n(2 - 1) = 2^nWhich matches the given difference sequence. So, that checks out. Therefore, the general term a_n is indeed 2^n.Now, we need to find the sum of the first n terms, S_n. Since each term is 2^k, where k goes from 1 to n, this is a geometric series.The formula for the sum of a geometric series is:S_n = a₁ * (r^n - 1) / (r - 1)Where a₁ is the first term, r is the common ratio, and n is the number of terms.In our case, a₁ = 2, r = 2, and the number of terms is n. Plugging these into the formula:S_n = 2 * (2^n - 1) / (2 - 1) S_n = 2 * (2^n - 1) / 1 S_n = 2*(2^n - 1) S_n = 2^{n+1} - 2So, the sum of the first n terms is 2^{n+1} - 2.Let me double-check this with the first few terms to make sure.For n=1: S₁ = a₁ = 2 Using the formula: 2^{1+1} - 2 = 4 - 2 = 2. Correct.For n=2: S₂ = a₁ + a₂ = 2 + 4 = 6 Formula: 2^{2+1} - 2 = 8 - 2 = 6. Correct.For n=3: S₃ = 2 + 4 + 8 = 14 Formula: 2^{3+1} - 2 = 16 - 2 = 14. Correct.For n=4: S₄ = 2 + 4 + 8 + 16 = 30 Formula: 2^{4+1} - 2 = 32 - 2 = 30. Correct.Looks like the formula works for these cases. So, I feel confident that S_n = 2^{n+1} - 2 is the correct answer.**Final Answer**boxed{D}

answer：First, I need to subtract (28^circ 36') from (72^circ 24').I notice that (24') is less than (36'), so I can't subtract (36') directly from (24'). To solve this, I'll borrow (1^circ) from the degrees, which is equivalent to (60'), and add it to the minutes.Borrowing (1^circ) from (72^circ) gives me (71^circ), and adding (60') to (24') results in (84').Now, I can subtract the degrees and minutes separately:- Subtracting the degrees: (71^circ - 28^circ = 43^circ)- Subtracting the minutes: (84' - 36' = 48')Combining these results, the final answer is (43^circ 48').

answer：Okay, so I'm trying to figure out this game where Petya and the computer take turns removing stones from a heap. There are 16 stones to start with, and each player can take 1, 2, 3, or 4 stones on their turn. The person who takes the last stone wins. Petya goes first, but she's not very experienced and just picks randomly each time. The computer, on the other hand, is using some sort of optimal strategy to always leave itself in the best position. I need to find the probability that Petya will win.Alright, let's break this down. First, it's a turn-based game with a fixed number of stones. The key here is that the computer is using an optimal strategy, which probably means it's trying to force Petya into a losing position. Since Petya is just choosing randomly, her chances depend on how often she can make the right moves to counter the computer's strategy.I remember something about these types of games being related to modular arithmetic, specifically modulo 5. The idea is that if you can always leave the number of stones as a multiple of 5 after your turn, you can force the opponent into a losing position. Let me think about that.If the heap has 16 stones, and Petya goes first, she can take 1 to 4 stones. If she takes 1 stone, leaving 15, which is a multiple of 5, then the computer is in a tough spot because whatever number of stones it takes (1-4), Petya can take the complement to 5 and maintain the multiple of 5. For example, if the computer takes 2 stones, Petya can take 3, keeping the heap at 10, and so on.But wait, Petya isn't following this strategy; she's just picking randomly. So, the probability that she takes exactly the right number of stones to leave a multiple of 5 is 1 out of 4 each time she has to make a move. Since she needs to do this multiple times to keep the heap at multiples of 5, the probability would be (1/4) raised to the number of times she needs to make such a move.Let me see, starting from 16 stones, if Petya takes 1, leaving 15, then the computer will take some number, say 'k', and Petya needs to take '5 - k' to get back to 10. Then the computer takes 'm', and Petya takes '5 - m' to get to 5, and finally, the computer is forced to take the last stone, but wait, no, if Petya follows this perfectly, she can take the last stone.But since Petya isn't following this strategy, she might mess up at any step. So, each time she has to make a move that keeps the heap at a multiple of 5, she has a 1/4 chance of doing it correctly. How many such moves does she need to make?Starting from 16, she needs to get to 15, then 10, then 5, and then 0. So that's four moves where she needs to take the correct number of stones. Each time, she has a 1/4 chance. So, the probability that she wins is (1/4)^4, which is 1/256.But wait, is that accurate? Let me double-check. If she fails at any point, the computer can take over and force her into a losing position. So, yes, she needs to make four correct moves in a row to win. Each move has a 1/4 chance, so multiplying them together gives 1/256.Hmm, that seems really low. Is there another way to think about this? Maybe considering all possible sequences of moves and calculating the probability accordingly. But that might be more complicated. I think the initial reasoning holds because the optimal strategy requires precise moves, and since Petya is random, her chances are minimal.Another angle: the game is essentially a series of forced moves if both players play optimally. Since the computer is playing optimally, Petya has to counter each move perfectly to win. Any mistake, and the computer can capitalize on it. So, the only way Petya can win is if she makes all the necessary correct moves, which is a 1/4 chance each time, four times in a row.Yeah, I think 1/256 is the right answer here.