answer：Okay, so I need to figure out the third number in the row of Pascal's triangle that has 52 numbers. Hmm, let me think about how Pascal's triangle works. Each row in Pascal's triangle corresponds to the coefficients of the binomial expansion, right? And the number of entries in each row is equal to the row number plus one. So, if a row has 52 numbers, that must mean it's the 51st row because 51 + 1 equals 52. Alright, so we're looking at the 51st row. Now, the question is asking for the third number in that row. In Pascal's triangle, each number is a binomial coefficient. The first number is always 1, which is the combination of 51 choose 0. The second number is 51 choose 1, and the third number would be 51 choose 2. Let me recall the formula for combinations: n choose k is equal to n! divided by (k! times (n - k)!). So, applying that to 51 choose 2, it would be 51 factorial divided by (2 factorial times (51 - 2) factorial). Simplifying that, 51 choose 2 is 51 times 50 divided by 2, because the factorials cancel out a lot. Calculating that, 51 times 50 is 2550. Then, dividing that by 2 gives me 1275. So, the third number in the 51st row, which has 52 numbers, should be 1275. Wait, let me double-check that. If I have 51 choose 2, that's indeed (51 × 50)/2, which is 1275. Yeah, that seems right. I don't think I made any mistakes in the calculations. So, I'm pretty confident that the third number is 1275.

answer：Okay, so I'm trying to figure out for which values of ( n ) a square can be divided into ( n ) smaller squares. This seems like a classic tiling problem, but I'm not entirely sure where to start. Let me think through this step by step.First, I know that dividing a square into smaller squares isn't as straightforward as dividing it into rectangles because squares have equal sides. So, if I try to divide a square into smaller squares, the sizes of these smaller squares might vary, but they all have to fit perfectly within the original square without overlapping or leaving gaps.Let me start with small values of ( n ) and see what happens.For ( n = 1 ), it's trivial because the square itself is just one square. So, ( n = 1 ) works.For ( n = 2 ), I'm not sure. If I try to divide a square into two smaller squares, how would that look? If I place one square in a corner, the remaining space isn't a square—it's a rectangle. So, I don't think ( n = 2 ) is possible.What about ( n = 3 )? Similar to ( n = 2 ), if I try to place three squares inside the original square, I end up with some rectangular spaces that can't be filled with squares without overlapping or leaving gaps. So, ( n = 3 ) might not be possible either.Moving on to ( n = 4 ). This seems straightforward because I can divide the original square into four smaller squares of equal size by cutting it into halves both horizontally and vertically. So, ( n = 4 ) works.For ( n = 5 ), I'm not sure. I recall that sometimes you can have squares of different sizes to fit into the original square. Maybe if I have one larger square and then smaller squares around it. Let me try to visualize this. If I place a smaller square in one corner, the remaining space is an L-shape, which isn't a square. Maybe if I divide the L-shape further into smaller squares, but I'm not sure if that would result in exactly five squares. It feels tricky, so I'm leaning towards ( n = 5 ) not being possible.For ( n = 6 ), I think it's possible. If I divide the original square into a grid of smaller squares, but not all equal. Maybe have some rows with more squares and some with fewer. For example, if I have two rows with three squares each, that gives me six squares in total. Wait, but that would actually be six squares if they're all equal, but I think you can have unequal squares too. So, ( n = 6 ) should work.Continuing this way, I think for ( n = 7 ), ( n = 8 ), and so on, it's possible by further subdividing the squares. But I need to find a general pattern or rule that tells me for which ( n ) it's possible.I remember hearing something about this problem before, and I think it has to do with whether ( n ) is a perfect square or not, but that doesn't seem right because we've already seen that ( n = 5 ) might not work, and 5 isn't a perfect square. Maybe it's more about the arrangement of squares rather than their sizes.Perhaps there's a formula or a theorem that states the possible values of ( n ). I think it's something like all integers greater than or equal to 6, but I'm not entirely sure. Let me try to verify this.If ( n = 6 ) works, then by adding more squares by subdividing existing squares, I can get higher values of ( n ). For example, if I have a square divided into six smaller squares, I can further divide one of those six into four even smaller squares, increasing the total count by three to make it nine. Wait, that doesn't seem consistent.Maybe a better approach is to consider that once you can divide a square into ( n ) squares, you can always add more squares by subdividing one of the existing squares into four smaller squares, which increases the total count by three. So, if ( n ) is possible, then ( n + 3 ) is also possible.Given that, if ( n = 4 ) is possible, then ( n = 7 ), ( n = 10 ), etc., are also possible. Similarly, if ( n = 6 ) is possible, then ( n = 9 ), ( n = 12 ), etc., are possible. But what about the numbers in between?I think there's a result that says that any ( n ) greater than or equal to 6 is possible, except for some specific cases. But I'm not sure about the exact exceptions. Maybe ( n = 5 ) is the only exception?Wait, earlier I thought ( n = 5 ) wasn't possible, but I'm not entirely certain. Maybe there's a way to divide a square into five smaller squares. Let me try to think of a configuration.If I have one large square taking up most of the space, and then four smaller squares filling in the remaining corners, that might work. But when I try to visualize it, the sizes don't seem to fit perfectly. The remaining spaces after placing the large square are rectangles, not squares, so I can't just fill them with squares without overlapping or leaving gaps.So, I think ( n = 5 ) is indeed not possible. Therefore, the possible values of ( n ) are all integers greater than or equal to 6, plus ( n = 1 ) and ( n = 4 ).But wait, ( n = 2 ) and ( n = 3 ) are not possible, as I initially thought. So, the complete set of possible ( n ) is ( n = 1 ), ( n = 4 ), and all integers ( n geq 6 ).To summarize:- ( n = 1 ): Trivial case, the square itself.- ( n = 2 ): Not possible.- ( n = 3 ): Not possible.- ( n = 4 ): Possible by dividing into four equal smaller squares.- ( n geq 6 ): Possible by various configurations and subdivisions.Therefore, the values of ( n ) for which a square can be divided into ( n ) squares are ( n = 1 ), ( n = 4 ), and all integers ( n geq 6 ).

answer：First, I need to determine the amount of cooking oil each person receives based on the given conditions.Let’s denote the amount given to person C as ( x ) milliliters.According to the problem, person B receives 200 milliliters more than person C. Therefore, the amount given to person B is ( x + 200 ) milliliters.Similarly, person A receives 200 milliliters more than person B. So, the amount given to person A is ( (x + 200) + 200 = x + 400 ) milliliters.The total amount of cooking oil distributed to all three people is 3 liters, which is equivalent to 3000 milliliters.Now, I can set up the equation:[x + (x + 200) + (x + 400) = 3000]Combining like terms:[3x + 600 = 3000]Subtracting 600 from both sides:[3x = 2400]Dividing both sides by 3:[x = 800]So, person C receives 800 milliliters. Therefore, person B receives:[800 + 200 = 1000 text{ milliliters}]

answer：Okay, so I have this problem: In the expansion of (1 + x)(x² + 1/x)^6, the coefficient of x³ is _______. Hmm, I need to find the coefficient of x³. Alright, let's break this down step by step.First, I remember that when you have an expression like (a + b)^n, you can expand it using the binomial theorem. The binomial theorem says that each term in the expansion is of the form C(n, k) * a^(n - k) * b^k, where C(n, k) is the combination of n things taken k at a time. So, maybe I can apply that here.Looking at the expression, it's (1 + x) multiplied by (x² + 1/x)^6. So, I think I should first expand (x² + 1/x)^6 using the binomial theorem, and then multiply the result by (1 + x). After that, I can find the coefficient of x³.Let me write down the expansion for (x² + 1/x)^6. Using the binomial theorem, each term will be C(6, r) * (x²)^(6 - r) * (1/x)^r. Simplifying that, it becomes C(6, r) * x^(12 - 2r) * x^(-r) = C(6, r) * x^(12 - 3r). So, each term is C(6, r) * x^(12 - 3r), where r ranges from 0 to 6.Now, I need to find the terms in this expansion that, when multiplied by (1 + x), will give me an x³ term. Let's think about this. When I multiply (1 + x) by the expanded form of (x² + 1/x)^6, each term in the expansion will be multiplied by 1 and by x. So, the x³ term in the final product can come from two places:1. The term in (x² + 1/x)^6 that is x³ multiplied by 1.2. The term in (x² + 1/x)^6 that is x² multiplied by x.So, I need to find the coefficients of x³ and x² in the expansion of (x² + 1/x)^6 and then add them together to get the coefficient of x³ in the final product.Let's find the coefficient of x³ first. From the general term, we have x^(12 - 3r). We set 12 - 3r = 3 and solve for r:12 - 3r = 3 12 - 3 = 3r 9 = 3r r = 3So, the coefficient of x³ is C(6, 3). Calculating that, C(6, 3) = 6! / (3! * (6 - 3)!) = (720) / (6 * 6) = 720 / 36 = 20. So, the coefficient is 20.Now, let's find the coefficient of x². Again, using the general term x^(12 - 3r), we set 12 - 3r = 2:12 - 3r = 2 12 - 2 = 3r 10 = 3r r = 10/3 ≈ 3.333Wait, r has to be an integer because it's the term number in the binomial expansion. Since 10/3 is not an integer, there is no term with x² in the expansion of (x² + 1/x)^6. That means the coefficient of x² is 0.So, going back to the original expression, the coefficient of x³ is the sum of the coefficient of x³ multiplied by 1 and the coefficient of x² multiplied by x. Since the coefficient of x² is 0, the only contribution comes from the x³ term multiplied by 1, which is 20.Therefore, the coefficient of x³ in the expansion is 20.Wait, let me double-check. Maybe I missed something? Let me verify the steps again.1. Expanded (x² + 1/x)^6 using binomial theorem: each term is C(6, r) * x^(12 - 3r). That seems correct.2. Identified that to get x³ in the final product, we need x³ * 1 and x² * x. That makes sense because (1 + x) is being multiplied.3. Calculated the coefficient for x³: set 12 - 3r = 3, solved for r = 3, which is valid. C(6, 3) = 20. Correct.4. Tried to find the coefficient for x²: set 12 - 3r = 2, got r = 10/3, which is not an integer. So, no such term. Hence, coefficient is 0. That seems right.So, adding them together: 20 + 0 = 20. Therefore, the coefficient is indeed 20.I think that's solid. I don't see any mistakes in the reasoning. So, the answer should be 20.**Final Answer**The coefficient of (x^3) is boxed{20}.

answer：First, I need to determine the number of boys present in the class. The problem states that there are twice as many girls as boys present. Given that there are 140 girls present, I can set up the equation:Number of girls = 2 × Number of boys140 = 2 × Number of boysSolving for the number of boys:Number of boys = 140 / 2 = 70Next, I'll calculate the total number of students present by adding the number of girls and boys:Total students present = Number of girls + Number of boys = 140 + 70 = 210The class has a total of 250 students. To find out how many students are absent:Number of students absent = Total students - Total students present = 250 - 210 = 40Since all the absent students are boys, the number of boys absent is 40.

answer：Okay, so I have this problem about a geometric progression where any term is equal to the product of the next two terms. I need to find the common ratio. Let me try to understand what this means.First, I remember that in a geometric progression, each term is obtained by multiplying the previous term by a constant called the common ratio, usually denoted as ( r ). So, if the first term is ( a ), the sequence goes ( a, ar, ar^2, ar^3, ) and so on.The problem says that any term is equal to the product of the next two terms. Let me write that down mathematically. If I take the ( n )-th term, which is ( ar^{n-1} ), it should be equal to the product of the ( (n+1) )-th term and the ( (n+2) )-th term. So, that would be:[ar^{n-1} = (ar^n)(ar^{n+1})]Hmm, let me simplify the right side. Multiplying ( ar^n ) and ( ar^{n+1} ) gives me ( a^2 r^{2n+1} ). So, the equation becomes:[ar^{n-1} = a^2 r^{2n+1}]Now, I can divide both sides by ( ar^{n-1} ) to simplify further. Assuming ( a ) is not zero and ( r ) is not zero, which makes sense because we're dealing with a geometric progression with positive terms.Dividing both sides by ( ar^{n-1} ):[1 = a r^{n+2}]Wait, that seems a bit confusing. Let me check my steps again. I had:[ar^{n-1} = a^2 r^{2n+1}]Dividing both sides by ( ar^{n-1} ):[1 = a r^{n+2}]Hmm, this suggests that ( a r^{n+2} = 1 ) for all ( n ). But ( a ) and ( r ) are constants, and ( n ) is a variable. The only way this equation can hold true for all ( n ) is if ( r = 1 ), because if ( r ) were anything other than 1, ( r^{n+2} ) would change with ( n ), making the equation invalid for different values of ( n ).Let me test this. If ( r = 1 ), then every term in the geometric progression is just ( a ). So, the sequence would be ( a, a, a, a, ldots ). Let's check if this satisfies the condition that any term is equal to the product of the next two terms.Taking the first term, ( a ), it should be equal to the product of the second and third terms, which are both ( a ). So, ( a = a times a ), which simplifies to ( a = a^2 ). This implies that ( a^2 - a = 0 ), so ( a(a - 1) = 0 ). Therefore, ( a = 0 ) or ( a = 1 ).But the problem states that the terms are positive, so ( a ) cannot be zero. Hence, ( a = 1 ). So, the sequence is ( 1, 1, 1, 1, ldots ), and indeed, each term is equal to the product of the next two terms because ( 1 = 1 times 1 ).Wait a minute, what if ( r ) is not 1? Let me consider another possibility. Suppose ( r ) is something else, say ( r = frac{1}{2} ). Then the sequence would be ( a, frac{a}{2}, frac{a}{4}, frac{a}{8}, ldots ). Let's check the condition for the first term: ( a = frac{a}{2} times frac{a}{4} ). That simplifies to ( a = frac{a^2}{8} ), which leads to ( 8a = a^2 ), so ( a^2 - 8a = 0 ), meaning ( a(a - 8) = 0 ). Again, ( a ) can't be zero, so ( a = 8 ).But then, checking the second term: ( frac{a}{2} = frac{a}{4} times frac{a}{8} ). Substituting ( a = 8 ), we get ( frac{8}{2} = frac{8}{4} times frac{8}{8} ), which simplifies to ( 4 = 2 times 1 ), so ( 4 = 2 ). That's not true. Therefore, ( r = frac{1}{2} ) doesn't satisfy the condition for all terms.What about ( r = -1 )? If ( r = -1 ), the sequence would alternate signs: ( a, -a, a, -a, ldots ). But the problem states that all terms are positive, so ( r = -1 ) is not acceptable because it would make the terms alternate between positive and negative.What about ( r = 0 )? If ( r = 0 ), the sequence would be ( a, 0, 0, 0, ldots ). But then, the second term is zero, and multiplying it by the third term (which is also zero) gives zero, but the first term is ( a ), which is positive. So, ( a = 0 times 0 = 0 ), which contradicts the fact that the terms are positive. Hence, ( r = 0 ) is not a valid solution.So, the only valid solution is ( r = 1 ), which makes all terms equal to ( a ), and as we saw earlier, ( a = 1 ) to satisfy the condition.Wait, but in the problem, the options are A) 0, B) 1, C) -1, D) (frac{1}{2}). So, based on my reasoning, the answer should be B) 1.But let me double-check. If ( r = 1 ), then every term is the same, say ( a ). Then, the condition is ( a = a times a ), which implies ( a = a^2 ). Since ( a ) is positive, ( a = 1 ). So, the sequence is all ones, and indeed, each term is equal to the product of the next two terms because ( 1 = 1 times 1 ).Yes, that seems correct. So, the common ratio is 1.