💡Okay, so I have this math problem here: I need to calculate the value of the expression 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + ... + 2005 - 2006 - 2007 + 2008. And the options are A: 0, B: 1, C: -1, D: 2008. Hmm, interesting. Let me try to figure this out step by step.First, I notice that the sequence seems to have a pattern. Let me write down the first few terms to see if I can spot it:1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + 9 - 10 - 11 + 12 + ... and so on, up to 2008.Looking at these terms, it seems like every four numbers form a group. Let me check:- The first group: 1 - 2 - 3 + 4- The second group: 5 - 6 - 7 + 8- The third group: 9 - 10 - 11 + 12- And so on...So each group has four numbers, and the operations alternate between subtraction and addition. Let me calculate the sum of each group to see if there's a pattern.First group: 1 - 2 - 3 + 4Calculating step by step:1 - 2 = -1-1 - 3 = -4-4 + 4 = 0Okay, so the first group sums to 0.Second group: 5 - 6 - 7 + 8Calculating step by step:5 - 6 = -1-1 - 7 = -8-8 + 8 = 0Again, the second group also sums to 0.Third group: 9 - 10 - 11 + 12Calculating step by step:9 - 10 = -1-1 - 11 = -12-12 + 12 = 0Same result, the third group is 0.Hmm, so it seems like every group of four numbers adds up to 0. That's a useful pattern. So, if every four numbers sum to 0, then the entire expression can be broken down into multiple such groups, each contributing 0 to the total sum.Now, I need to figure out how many such groups there are in the entire sequence up to 2008.Since each group has four numbers, the total number of groups would be the total number of terms divided by 4.Let's find out how many terms there are in the sequence. The sequence starts at 1 and goes up to 2008, so there are 2008 terms in total.Therefore, the number of groups is 2008 divided by 4, which is 502 groups.Since each group sums to 0, the total sum of all groups would be 502 times 0, which is 0.Wait, but let me double-check to make sure I didn't miss anything. Sometimes, when dealing with sequences, especially if the number of terms isn't a multiple of the group size, there might be a leftover group that doesn't complete the pattern. But in this case, 2008 is perfectly divisible by 4, so there are exactly 502 complete groups with no leftover terms.Therefore, the entire sum should indeed be 0.Just to be thorough, let me verify with a few more groups.Fourth group: 13 - 14 - 15 + 1613 - 14 = -1-1 - 15 = -16-16 + 16 = 0Yep, still 0.Fifth group: 17 - 18 - 19 + 2017 - 18 = -1-1 - 19 = -20-20 + 20 = 0Consistently, each group adds up to 0.So, unless there's a mistake in my grouping or calculation, the total sum should be 0.But wait, let me check the signs again to make sure I didn't misinterpret the pattern.Looking at the original sequence: 1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + ...So, the pattern is: +, -, -, +, +, -, -, +, etc.So, every four terms, the operations are: +, -, -, +.Therefore, each group of four terms follows the same operation pattern.So, if I group them as (1 - 2 - 3 + 4), (5 - 6 - 7 + 8), etc., each group indeed sums to 0.Therefore, the entire sequence can be divided into 502 such groups, each summing to 0, so the total sum is 0.Hence, the answer should be A: 0.But just to be absolutely sure, let me think about another way to approach this problem.Alternatively, I could try to represent the sequence as a series and see if I can find a general formula or pattern.Let me denote the nth term of the sequence. But since the operations are alternating in a specific pattern, it might be a bit more complex.Alternatively, I can think of the sequence as a combination of positive and negative terms, and see if I can pair them in a way that simplifies the calculation.Looking back at the sequence:1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + ... + 2005 - 2006 - 2007 + 2008If I group them as (1 - 2 - 3 + 4), (5 - 6 - 7 + 8), etc., each group of four terms sums to 0, as I found earlier.Alternatively, if I think about the sequence in terms of positive and negative terms:Positive terms: 1, 4, 5, 8, 9, 12, 13, 16, ..., 2005, 2008Negative terms: -2, -3, -6, -7, -10, -11, -14, -15, ..., -2006, -2007So, let's calculate the sum of positive terms and the sum of negative terms separately.First, let's find how many positive terms there are.Looking at the sequence, the positive terms are in positions 1, 4, 5, 8, 9, 12, 13, ..., 2005, 2008.Similarly, the negative terms are in positions 2, 3, 6, 7, 10, 11, ..., 2006, 2007.So, let's see if we can find the number of positive and negative terms.Since the sequence goes up to 2008, and every four terms consist of two positive terms and two negative terms.Wait, actually, in each group of four terms, there are two positive terms and two negative terms.Wait, no, let's see:In the first group: 1 (positive), -2 (negative), -3 (negative), +4 (positive). So, two positives and two negatives.Similarly, the next group: +5, -6, -7, +8: again, two positives and two negatives.So, in each group of four terms, there are two positive terms and two negative terms.Therefore, in total, since there are 502 groups, the number of positive terms is 502 * 2 = 1004, and the number of negative terms is also 1004.Wait, but the total number of terms is 2008, which is 502 * 4, so that makes sense.Therefore, the number of positive terms is 1004, and the number of negative terms is 1004.Now, let's calculate the sum of positive terms and the sum of negative terms.First, the positive terms:They are 1, 4, 5, 8, 9, 12, 13, ..., 2005, 2008.Looking at this sequence, it seems like every four numbers, the positive terms are the first and the fourth numbers.So, in the first group: 1 and 4Second group: 5 and 8Third group: 9 and 12And so on.So, the positive terms can be represented as:For each group k (starting from 0), the positive terms are 4k + 1 and 4k + 4.Similarly, the negative terms are 4k + 2 and 4k + 3.Therefore, the positive terms are 1, 4, 5, 8, 9, 12, ..., 2005, 2008.So, the positive terms can be split into two separate arithmetic sequences:First sequence: 1, 5, 9, ..., 2005Second sequence: 4, 8, 12, ..., 2008Similarly, the negative terms can be split into two arithmetic sequences:First negative sequence: 2, 6, 10, ..., 2006Second negative sequence: 3, 7, 11, ..., 2007So, let's calculate the sum of positive terms.First, the first positive sequence: 1, 5, 9, ..., 2005This is an arithmetic sequence where the first term a1 = 1, common difference d = 4.We can find the number of terms n in this sequence.The nth term of an arithmetic sequence is given by:a_n = a1 + (n - 1) * dSo, 2005 = 1 + (n - 1) * 42005 - 1 = (n - 1) * 42004 = (n - 1) * 4n - 1 = 2004 / 4 = 501n = 502So, there are 502 terms in this sequence.The sum of an arithmetic sequence is given by:S = n/2 * (a1 + a_n)So, sum of first positive sequence:S1 = 502/2 * (1 + 2005) = 251 * 2006Similarly, the second positive sequence: 4, 8, 12, ..., 2008This is also an arithmetic sequence with a1 = 4, d = 4.Number of terms n:2008 = 4 + (n - 1) * 42008 - 4 = (n - 1) * 42004 = (n - 1) * 4n - 1 = 501n = 502So, sum of second positive sequence:S2 = 502/2 * (4 + 2008) = 251 * 2012Therefore, total sum of positive terms:S_positive = S1 + S2 = 251 * 2006 + 251 * 2012Factor out 251:S_positive = 251 * (2006 + 2012) = 251 * 4018Similarly, let's calculate the sum of negative terms.First negative sequence: 2, 6, 10, ..., 2006Arithmetic sequence with a1 = 2, d = 4.Number of terms n:2006 = 2 + (n - 1) * 42006 - 2 = (n - 1) * 42004 = (n - 1) * 4n - 1 = 501n = 502Sum of first negative sequence:S3 = 502/2 * (2 + 2006) = 251 * 2008Second negative sequence: 3, 7, 11, ..., 2007Arithmetic sequence with a1 = 3, d = 4.Number of terms n:2007 = 3 + (n - 1) * 42007 - 3 = (n - 1) * 42004 = (n - 1) * 4n - 1 = 501n = 502Sum of second negative sequence:S4 = 502/2 * (3 + 2007) = 251 * 2010Therefore, total sum of negative terms:S_negative = S3 + S4 = 251 * 2008 + 251 * 2010Factor out 251:S_negative = 251 * (2008 + 2010) = 251 * 4018Wait a minute, that's interesting. Both S_positive and S_negative are equal to 251 * 4018.Therefore, the total sum of the sequence is S_positive + S_negative = 251 * 4018 + (-251 * 4018) = 0.So, that confirms my initial thought that the total sum is 0.Alternatively, another way to think about it is that for every positive term, there's a corresponding negative term that cancels it out, but in this case, it's more about the grouping leading to each group summing to 0.But regardless of the approach, whether grouping or splitting into positive and negative sequences, the result is the same: the total sum is 0.Therefore, the answer is A: 0.

💡First, I need to determine the time it takes to travel 1 km at 80 km/h. To do this, I'll convert 80 km/h to km/s by dividing by 3600, since there are 3600 seconds in an hour.Next, I'll calculate the time it takes to travel 1 km at the unknown speed ( v ) in seconds. This involves setting up an equation where the time at speed ( v ) is equal to the time at 80 km/h plus 4 seconds.After setting up the equation, I'll solve for ( v ) by isolating it on one side of the equation. Finally, I'll perform the division to find the numerical value of ( v ) and present the answer in a boxed format.

💡Okay, so I have this problem where I need to find the sum of WSUMs over all subsets of the set S = {1, 2, ..., 10}. WSUM is defined as 3 times the sum of the elements in the odd positions plus 2 times the sum of the elements in the even positions when the subset is arranged in increasing order. The null set has a WSUM of 0.First, I need to understand what exactly WSUM is. For a given subset A of S, when I arrange its elements in increasing order, say A = {a₁, a₂, ..., a_k}, then WSUM is calculated as 3*(a₁ + a₃ + ...) + 2*(a₂ + a₄ + ...). So, the odd-indexed elements are multiplied by 3, and the even-indexed elements are multiplied by 2.My goal is to compute the sum of WSUMs for all possible subsets of S. Since S has 10 elements, there are 2¹⁰ = 1024 subsets, including the empty set. Calculating WSUM for each subset individually and then summing them up would be impractical because of the sheer number of subsets. So, I need a smarter way to compute this sum without enumerating all subsets.Let me think about how each element of S contributes to the total sum. For each element m in S, I need to determine how many times it appears in an odd position and how many times it appears in an even position across all subsets. Once I know that, I can compute its total contribution to the WSUM as 3*(number of times it's in an odd position) + 2*(number of times it's in an even position), and then sum this over all elements.So, let's fix an element m in S. I need to find the number of subsets where m is in an odd position and the number of subsets where m is in an even position.To do this, I can consider the position of m in a subset. For m to be in position i (where i is 1, 2, ..., k), there must be exactly i-1 elements less than m in the subset and any number of elements greater than m. The number of such subsets is equal to the number of ways to choose i-1 elements from the elements less than m multiplied by the number of ways to choose any number of elements from the elements greater than m.Let me formalize this. Let’s say m is the j-th smallest element in S, so there are j-1 elements less than m and 10 - j elements greater than m. For m to be in position i in a subset, we need to choose i-1 elements from the j-1 elements less than m and any number of elements from the 10 - j elements greater than m.Therefore, the number of subsets where m is in position i is C(j-1, i-1) * 2^(10 - j), where C(n, k) is the combination function.Now, the position i can be either odd or even. So, to find the total number of subsets where m is in an odd position, I need to sum C(j-1, i-1) * 2^(10 - j) over all odd i. Similarly, for even positions, I sum over all even i.But wait, j is fixed for each m. For example, if m = 1, then j = 1, so there are 0 elements less than m. If m = 2, j = 2, so there's 1 element less than m, and so on.Let me consider m = 1 first. Since there are no elements less than 1, the only way 1 can be in a subset is as the first element. So, the number of subsets where 1 is in position 1 is 2^(10 - 1) = 512. Therefore, 1 is always in an odd position whenever it is included in a subset. So, the number of subsets where 1 is in an odd position is 512, and it never appears in an even position.Similarly, for m = 2, j = 2. So, the number of subsets where 2 is in position 1 is C(1, 0) * 2^(10 - 2) = 1 * 256 = 256. The number of subsets where 2 is in position 2 is C(1, 1) * 2^(10 - 2) = 1 * 256 = 256. So, 2 can be in position 1 or position 2. Therefore, it can be in an odd or even position.Wait, but position 1 is odd, position 2 is even. So, the number of subsets where 2 is in an odd position is 256, and the number where it's in an even position is also 256.Similarly, for m = 3, j = 3. The number of subsets where 3 is in position 1 is C(2, 0) * 2^(10 - 3) = 1 * 128 = 128. Position 2: C(2, 1) * 128 = 2 * 128 = 256. Position 3: C(2, 2) * 128 = 1 * 128 = 128. So, positions 1 and 3 are odd, so total subsets where 3 is in an odd position is 128 + 128 = 256. Position 2 is even, so subsets where 3 is in an even position is 256.Wait, that seems interesting. For m = 2, the number of subsets where it's in odd and even positions is equal. For m = 3, same thing: 256 in odd, 256 in even. Maybe this is a pattern.Let me check m = 4. j = 4. The number of subsets where 4 is in position 1: C(3, 0)*2^(10-4)=1*64=64. Position 2: C(3,1)*64=3*64=192. Position 3: C(3,2)*64=3*64=192. Position 4: C(3,3)*64=1*64=64. So, positions 1 and 3 are odd: 64 + 192 = 256. Positions 2 and 4 are even: 192 + 64 = 256. Again, equal.Hmm, so it seems that for m >=2, the number of subsets where m is in an odd position is equal to the number of subsets where it's in an even position, both being 256 for m=2,3,4. Wait, but for m=1, it's different: 512 in odd, 0 in even.Wait, let me verify for m=5. j=5. Subsets where 5 is in position 1: C(4,0)*2^(10-5)=1*32=32. Position 2: C(4,1)*32=4*32=128. Position 3: C(4,2)*32=6*32=192. Position 4: C(4,3)*32=4*32=128. Position 5: C(4,4)*32=1*32=32. So, odd positions: 32 + 192 + 32 = 256. Even positions: 128 + 128 = 256. Again, equal.So, it seems that for m >=2, the number of subsets where m is in an odd position is equal to the number of subsets where it's in an even position, both being 256. For m=1, it's 512 in odd, 0 in even.Wait, but 256 is 2^(10-2) = 256. So, for m >=2, the number of subsets where m is in an odd position is 2^(n-2), where n=10. Similarly, for even positions, it's also 2^(n-2). For m=1, it's 2^(n-1) in odd positions.So, generalizing, for each element m in S:- If m=1, it appears in an odd position in 2^(n-1) subsets, and never in an even position.- If m >=2, it appears in an odd position in 2^(n-2) subsets, and in an even position in 2^(n-2) subsets.Therefore, the total contribution of each element m to the WSUM is:- For m=1: 3 * 2^(n-1) * m- For m >=2: 3 * 2^(n-2) * m + 2 * 2^(n-2) * m = (3 + 2) * 2^(n-2) * m = 5 * 2^(n-2) * mSo, the total WSUM over all subsets is:WSUM_total = 3 * 2^(n-1) * 1 + 5 * 2^(n-2) * (2 + 3 + ... + 10)Wait, let me write it properly.Let n=10.So,WSUM_total = 3 * 2^(10-1) * 1 + 5 * 2^(10-2) * (2 + 3 + ... + 10)Simplify:2^(10-1) = 5122^(10-2) = 256Sum from 2 to 10: Let's compute that.Sum from 1 to 10 is (10*11)/2 = 55So, sum from 2 to 10 is 55 - 1 = 54Therefore,WSUM_total = 3*512*1 + 5*256*54Compute each term:First term: 3*512 = 1536Second term: 5*256 = 1280; 1280*54Compute 1280*54:First, 1280*50 = 64,000Then, 1280*4 = 5,120So, total is 64,000 + 5,120 = 69,120Therefore, WSUM_total = 1536 + 69,120 = 70,656Wait, is that correct? Let me double-check.Wait, 5*256 = 1280, correct.1280*54: Let me compute 1280*54 step by step.54 = 50 + 41280*50 = 64,0001280*4 = 5,12064,000 + 5,120 = 69,120, correct.Then, 3*512 = 1,5361,536 + 69,120 = 70,656Yes, that seems correct.But wait, let me think again. Is the sum from 2 to 10 equal to 54? Yes, because 1+2+...+10=55, so 55-1=54.So, the total WSUM over all subsets is 70,656.But let me think if I missed something. For m=1, it's 3*512*1=1,536. For m=2 to 10, each contributes 5*256*m. So, sum over m=2 to 10 of 5*256*m = 5*256*(2+3+...+10) = 5*256*54 = 69,120. So, total is 1,536 + 69,120 = 70,656.Yes, that seems correct.Alternatively, I can think of it as:Total WSUM = sum_{A subset of S} WSUM(A) = sum_{A subset of S} [3*(sum of odd-indexed elements) + 2*(sum of even-indexed elements)]= 3*sum_{A subset of S} (sum of odd-indexed elements) + 2*sum_{A subset of S} (sum of even-indexed elements)Then, interchanging the order of summation:= 3*sum_{m in S} m * (number of subsets where m is in an odd position) + 2*sum_{m in S} m * (number of subsets where m is in an even position)Which is exactly what I did earlier.So, the total WSUM is 70,656.But wait, let me check if the number of subsets where m is in an odd position is indeed 2^(n-2) for m >=2.For m=2, j=2. The number of subsets where m is in an odd position is sum_{k=0}^{floor((j-1)/2)} C(j-1, 2k) * 2^(n - j)}.Wait, no, earlier I thought it's 2^(n-2), but actually, for m=2, j=2, the number of subsets where m is in an odd position is C(1,0)*2^(8) + C(1,2)*2^(8). But C(1,2)=0, so it's just 256. Similarly, for m=3, j=3, the number of subsets where m is in an odd position is C(2,0)*2^(7) + C(2,2)*2^(7) = 128 + 128 = 256. So, yes, for m >=2, the number of subsets where m is in an odd position is 256, which is 2^(10-2)=256.Similarly, the number of subsets where m is in an even position is also 256.Therefore, my earlier calculation is correct.So, the final answer is 70,656.But wait, let me express it in terms of powers of 2 and see if it can be simplified.70,656 = 70,656But 2^16 = 65,53670,656 - 65,536 = 5,1205,120 = 5 * 1,024 = 5 * 2^10So, 70,656 = 2^16 + 5*2^10But 2^16 = 65,5365*2^10 = 5,12065,536 + 5,120 = 70,656Alternatively, 70,656 = 2^10*(2^6 + 5) = 1,024*(64 + 5) = 1,024*69 = 70,656Yes, 1,024*69 = 70,656So, another way to write it is 2^10 * 69 = 70,656But 69 is 64 + 5, which is 2^6 + 5.Alternatively, 70,656 = 2^10 * (2^6 + 5) = 2^10 * 69But I don't think it's necessary to write it in a different form unless required.So, the final answer is 70,656.Wait, but let me check the calculation again because sometimes when dealing with exponents, it's easy to make a mistake.Compute 5*256*54:First, 256*54:256*50 = 12,800256*4 = 1,024So, 12,800 + 1,024 = 13,824Then, 5*13,824 = 69,120Yes, correct.Then, 3*512 = 1,5361,536 + 69,120 = 70,656Yes, correct.So, the sum of WSUMs over all subsets of S is 70,656.But wait, let me think if there's another way to approach this problem, maybe using generating functions or something else, to confirm the result.Alternatively, consider that each element m contributes to the WSUM based on its position in the subsets. For m=1, it's always in the first position when included, so it's multiplied by 3. For m>=2, it can be in either an odd or even position, each contributing 3 or 2 respectively, and each with equal probability across subsets.Wait, but the number of subsets where m is in an odd position is equal to the number where it's in an even position, which is 2^(n-2). So, the expected contribution of m is (3 + 2)/2 * m * number of subsets where m is included.But actually, the total contribution is 3*(number of subsets where m is in odd position) + 2*(number of subsets where m is in even position). Since these numbers are equal for m >=2, it's (3 + 2)/2 * (number of subsets where m is included) * m.Wait, but the number of subsets where m is included is 2^(n-1). Because for each element, it can be included or not, so total subsets including m is 2^(n-1).But for m >=2, the number of subsets where m is in an odd position is 2^(n-2), and same for even. So, the total contribution is 3*2^(n-2) + 2*2^(n-2) = 5*2^(n-2) per m.Which is the same as before.So, for m=1, it's 3*2^(n-1), and for m>=2, it's 5*2^(n-2)*m.So, the total WSUM is 3*2^(n-1) + 5*2^(n-2)*(sum from m=2 to 10 of m)Which is what I computed earlier.So, yes, 70,656 is the correct answer.I think I've verified it sufficiently. So, the sum of WSUMs over all subsets of S is 70,656.

💡Okay, so I have this problem where there's an n-degree polynomial p(x), and for each integer k from 0 to n, p(k) equals k divided by (k+1). I need to find p(n+1). Hmm, okay, let's see how to approach this.First, I remember that polynomials can be determined uniquely if we know their degree and their values at certain points. Since p(x) is an n-degree polynomial and we have n+1 points (from k=0 to k=n), that should uniquely define p(x). But how do I use this information to find p(n+1)?Maybe I can construct another polynomial that relates to p(x). Let me think... If I consider (x+1)p(x) - x, that might be useful. Let's call this new polynomial Q(x). So, Q(x) = (x+1)p(x) - x.Now, let's evaluate Q(x) at the points k = 0, 1, 2, ..., n. For each k, p(k) = k/(k+1). So, plugging into Q(k):Q(k) = (k+1)*p(k) - k = (k+1)*(k/(k+1)) - k = k - k = 0.So, Q(k) = 0 for all k from 0 to n. That means Q(x) has roots at x = 0, 1, 2, ..., n. Since Q(x) is a polynomial of degree n+1 (because p(x) is degree n and multiplied by (x+1), which is degree 1), it can be written as Q(x) = a*x*(x-1)*(x-2)*...*(x-n), where a is some constant.Now, I need to find the value of a. To do that, I can choose a value of x that isn't a root, so maybe x = -1. Let's plug x = -1 into Q(x):Q(-1) = (-1 + 1)*p(-1) - (-1) = 0*p(-1) + 1 = 1.On the other hand, using the product form:Q(-1) = a*(-1)*(-1 - 1)*(-1 - 2)*...*(-1 - n).Let's simplify this. Each term in the product is (-1 - k) where k goes from 0 to n. So, for each k, (-1 - k) = -(k + 1). Therefore, the entire product becomes:(-1)^{n+1} * (1*2*3*...*(n+1)) = (-1)^{n+1}*(n+1)!.So, Q(-1) = a*(-1)^{n+1}*(n+1)! = 1.Solving for a:a = 1 / [(-1)^{n+1}*(n+1)!] = (-1)^{n+1} / (n+1)!.Okay, so now we have Q(x) expressed as:Q(x) = [(-1)^{n+1} / (n+1)!] * x*(x - 1)*(x - 2)*...*(x - n).But remember, Q(x) = (x + 1)p(x) - x. So, we can solve for p(x):(x + 1)p(x) = Q(x) + x = [(-1)^{n+1} / (n+1)!] * x*(x - 1)*(x - 2)*...*(x - n) + x.Therefore, p(x) = [Q(x) + x] / (x + 1).Now, we need to find p(n+1). Let's plug x = n+1 into the expression:p(n+1) = [Q(n+1) + (n+1)] / (n+2).First, let's compute Q(n+1):Q(n+1) = [(-1)^{n+1} / (n+1)!] * (n+1)*(n+1 - 1)*(n+1 - 2)*...*(n+1 - n).Simplifying the product inside:(n+1)*(n)*(n-1)*...*1 = (n+1)!.So, Q(n+1) = [(-1)^{n+1} / (n+1)!] * (n+1)! = (-1)^{n+1}.Therefore, p(n+1) = [(-1)^{n+1} + (n+1)] / (n+2).Hmm, let's see. Depending on whether n is even or odd, (-1)^{n+1} will be either -1 or 1.If n is even, then n+1 is odd, so (-1)^{n+1} = -1.If n is odd, then n+1 is even, so (-1)^{n+1} = 1.So, let's break it down:Case 1: n is even.p(n+1) = [-1 + (n+1)] / (n+2) = (n) / (n+2).Case 2: n is odd.p(n+1) = [1 + (n+1)] / (n+2) = (n+2) / (n+2) = 1.So, putting it all together, p(n+1) is 1 if n is odd, and n/(n+2) if n is even.I think that makes sense. Let me double-check with a small n, say n=1.If n=1, then p(x) is a linear polynomial. We have p(0)=0/1=0, p(1)=1/2.So, p(x) is a line passing through (0,0) and (1,1/2). The equation is p(x) = (1/2)x.Then p(2) should be (1/2)*2 = 1. Since n=1 is odd, according to our result, p(n+1)=1. That matches.Another test: n=2.p(x) is a quadratic polynomial with p(0)=0, p(1)=1/2, p(2)=2/3.Let's find p(3). According to our formula, since n=2 is even, p(3)=2/(2+2)=2/4=1/2.Let me compute p(x). Let's set up the system:p(0)=0: a*0^2 + b*0 + c = 0 => c=0.p(1)=1/2: a + b = 1/2.p(2)=2/3: 4a + 2b = 2/3.From the first equation: a + b = 1/2.From the second: 4a + 2b = 2/3.Multiply the first equation by 2: 2a + 2b = 1.Subtract from the second: (4a + 2b) - (2a + 2b) = 2/3 - 1 => 2a = -1/3 => a = -1/6.Then b = 1/2 - a = 1/2 + 1/6 = 2/3.So, p(x) = (-1/6)x^2 + (2/3)x.Compute p(3):p(3) = (-1/6)*9 + (2/3)*3 = (-3/2) + 2 = 1/2.Which matches our formula. So, it seems correct.Alright, so I think the conclusion is that p(n+1) is 1 when n is odd and n/(n+2) when n is even.

💡Okay, so I have this problem in polar coordinates. It says that there's a line passing through two points, A and B, and this line intersects a curve C at exactly one point. I need to find the value of r. Hmm, let me try to break this down step by step.First, let me understand the points A and B. Point A is given as (√3, 2π/3) and point B is (3, π/2). These are in polar coordinates, which means they are represented as (ρ, θ), where ρ is the radius and θ is the angle. I think I need to convert these points into Cartesian coordinates to make it easier to work with lines and curves.Alright, to convert from polar to Cartesian coordinates, I remember the formulas:x = ρ * cos(θ)y = ρ * sin(θ)So, let's convert point A first. For point A, ρ is √3 and θ is 2π/3.Calculating x-coordinate:x_A = √3 * cos(2π/3)I know that cos(2π/3) is equal to cos(120°), which is -1/2. So,x_A = √3 * (-1/2) = -√3/2Now, the y-coordinate:y_A = √3 * sin(2π/3)Sin(2π/3) is sin(120°), which is √3/2. So,y_A = √3 * (√3/2) = (3)/2So, point A in Cartesian coordinates is (-√3/2, 3/2).Now, let's convert point B. Point B is (3, π/2).Calculating x-coordinate:x_B = 3 * cos(π/2)Cos(π/2) is 0, so x_B = 0Calculating y-coordinate:y_B = 3 * sin(π/2)Sin(π/2) is 1, so y_B = 3Therefore, point B in Cartesian coordinates is (0, 3).Okay, so now I have points A(-√3/2, 3/2) and B(0, 3). I need to find the equation of the line passing through these two points.To find the equation of a line, I can use the slope-intercept form, which is y = mx + b, where m is the slope and b is the y-intercept.First, let's find the slope (m). The formula for slope between two points (x1, y1) and (x2, y2) is:m = (y2 - y1)/(x2 - x1)Plugging in the coordinates of points A and B:m = (3 - 3/2)/(0 - (-√3/2)) = (3/2)/(√3/2)Simplify the numerator and denominator:(3/2) divided by (√3/2) is equal to (3/2) * (2/√3) = 3/√3 = √3So, the slope m is √3.Now, to find the y-intercept (b), I can use one of the points. Let's use point B(0, 3). Plugging into y = mx + b:3 = √3 * 0 + bSo, b = 3Therefore, the equation of the line AB is y = √3 x + 3.Alright, now I need to consider the curve C, which is given by ρ = 2r sinθ. The problem states that this curve has exactly one common point with the line AB. So, this means that the line is tangent to the curve C.First, let me convert the polar equation of curve C into Cartesian coordinates to better understand its shape.Starting with ρ = 2r sinθ.I know that in polar coordinates, ρ = √(x² + y²) and sinθ = y/ρ. So, substituting these into the equation:ρ = 2r * (y/ρ)Multiply both sides by ρ:ρ² = 2r yBut ρ² is x² + y², so:x² + y² = 2r yLet me rearrange this equation to recognize the shape. I can complete the square for the y-term:x² + y² - 2r y = 0x² + (y² - 2r y + r²) = r²x² + (y - r)² = r²Ah, so this is the equation of a circle with center at (0, r) and radius r. So, curve C is a circle centered at (0, r) with radius r.Now, the line AB is y = √3 x + 3, and it is tangent to this circle. Since the line is tangent to the circle, the distance from the center of the circle to the line must be equal to the radius of the circle.The formula for the distance from a point (h, k) to the line Ax + By + C = 0 is:Distance = |A h + B k + C| / √(A² + B²)First, let's write the equation of line AB in the standard form Ax + By + C = 0.Starting with y = √3 x + 3, subtract y from both sides:√3 x - y + 3 = 0So, A = √3, B = -1, C = 3.The center of the circle is (0, r). Plugging into the distance formula:Distance = |√3 * 0 + (-1) * r + 3| / √((√3)² + (-1)²)Simplify numerator and denominator:Numerator: |0 - r + 3| = |3 - r|Denominator: √(3 + 1) = √4 = 2So, the distance is |3 - r| / 2.Since the line is tangent to the circle, this distance must equal the radius of the circle, which is r.Therefore, we have the equation:|3 - r| / 2 = rThis is an absolute value equation, so we need to consider two cases.Case 1: 3 - r ≥ 0, which implies r ≤ 3.In this case, |3 - r| = 3 - r, so the equation becomes:(3 - r)/2 = rMultiply both sides by 2:3 - r = 2rAdd r to both sides:3 = 3rDivide both sides by 3:r = 1Case 2: 3 - r < 0, which implies r > 3.In this case, |3 - r| = r - 3, so the equation becomes:(r - 3)/2 = rMultiply both sides by 2:r - 3 = 2rSubtract r from both sides:-3 = rBut r is given as a positive real number (r > 0), so r = -3 is not acceptable.Therefore, the only valid solution is r = 1.Let me just double-check my steps to make sure I didn't make a mistake.1. Converted points A and B correctly to Cartesian coordinates. Yes, seems right.2. Calculated the slope correctly: (3 - 3/2)/(0 - (-√3/2)) = (3/2)/(√3/2) = √3. That seems correct.3. Found the equation of the line as y = √3 x + 3. Plugging in point B (0,3) checks out.4. Converted the polar equation ρ = 2r sinθ to Cartesian coordinates: x² + y² = 2r y, then completed the square to get x² + (y - r)² = r². That looks correct, it's a circle with center (0, r) and radius r.5. Calculated the distance from center (0, r) to the line √3 x - y + 3 = 0: |√3*0 - 1*r + 3| / √( (√3)^2 + (-1)^2 ) = |3 - r| / 2. That seems right.6. Set the distance equal to radius r: |3 - r| / 2 = r. Solved for both cases: Case 1: r = 1, which is valid since r ≤ 3 and r > 0. Case 2: r = -3, which is invalid because r must be positive.Therefore, r = 1 is the correct solution.I think I covered all the steps and checked each part. I don't see any mistakes in my reasoning. So, I'm confident that r = 1 is the correct answer.**Final Answer**The value of ( r ) is boxed{1}.

💡Alright, so I have this problem here where I'm given two sets, A and B. Set A is {2, 0, 1, 6}, and set B is defined as all real numbers k such that k squared minus 2 is in A, and k minus 2 is not in A. I need to find the product of all elements in set B. The options are 48, 8 times the square root of 3, 96, and 192. Hmm, okay, let's break this down step by step.First, let me make sure I understand the problem correctly. Set A is straightforward—it's just four numbers: 2, 0, 1, and 6. Set B is a bit more complex because it's defined based on some conditions involving k. Specifically, for a number k to be in set B, two things must be true:1. When I take k, square it, and subtract 2, the result should be an element of set A.2. When I take k and subtract 2, the result should **not** be an element of set A.So, my task is to find all real numbers k that satisfy both of these conditions and then multiply them all together to get the product. That product should be one of the given options.Let me start by tackling the first condition: k² - 2 ∈ A. Since A is {2, 0, 1, 6}, this means that k² - 2 must be equal to one of these numbers. So, I can set up equations for each element in A and solve for k.Let's list out the equations:1. k² - 2 = 22. k² - 2 = 03. k² - 2 = 14. k² - 2 = 6I'll solve each of these equations one by one.**First Equation: k² - 2 = 2**Adding 2 to both sides:k² = 4Taking the square root of both sides:k = ±2So, potential solutions here are k = 2 and k = -2.But wait, I need to check the second condition for these values of k. The second condition is that k - 2 ∉ A.Let's check k = 2:k - 2 = 2 - 2 = 0But 0 is in set A, right? So, k = 2 doesn't satisfy the second condition because k - 2 is in A. Therefore, k = 2 is **not** included in set B.Now, let's check k = -2:k - 2 = -2 - 2 = -4Is -4 in set A? Set A is {2, 0, 1, 6}, so no, -4 is not in A. Therefore, k = -2 satisfies both conditions and is included in set B.So, from the first equation, only k = -2 is valid.**Second Equation: k² - 2 = 0**Adding 2 to both sides:k² = 2Taking the square root of both sides:k = ±√2So, potential solutions are k = √2 and k = -√2.Again, I need to check the second condition for these values.First, k = √2:k - 2 = √2 - 2Is √2 - 2 in set A? Let's see, set A is {2, 0, 1, 6}. √2 is approximately 1.414, so √2 - 2 is approximately -0.586. That's not in A. So, k = √2 is valid.Next, k = -√2:k - 2 = -√2 - 2That's approximately -1.414 - 2 = -3.414, which is also not in A. So, k = -√2 is also valid.Therefore, both k = √2 and k = -√2 are included in set B.**Third Equation: k² - 2 = 1**Adding 2 to both sides:k² = 3Taking the square root of both sides:k = ±√3So, potential solutions are k = √3 and k = -√3.Checking the second condition:First, k = √3:k - 2 = √3 - 2 ≈ 1.732 - 2 ≈ -0.268Not in A, so valid.Next, k = -√3:k - 2 = -√3 - 2 ≈ -1.732 - 2 ≈ -3.732Also not in A, so valid.Thus, both k = √3 and k = -√3 are included in set B.**Fourth Equation: k² - 2 = 6**Adding 2 to both sides:k² = 8Taking the square root of both sides:k = ±2√2So, potential solutions are k = 2√2 and k = -2√2.Checking the second condition:First, k = 2√2:k - 2 = 2√2 - 2 ≈ 2.828 - 2 ≈ 0.828Is 0.828 in set A? Set A is {2, 0, 1, 6}. 0.828 is not in A, so k = 2√2 is valid.Next, k = -2√2:k - 2 = -2√2 - 2 ≈ -2.828 - 2 ≈ -4.828Not in A, so valid.Therefore, both k = 2√2 and k = -2√2 are included in set B.**So, compiling all the valid k values we found:**From the first equation: k = -2From the second equation: k = √2, k = -√2From the third equation: k = √3, k = -√3From the fourth equation: k = 2√2, k = -2√2Therefore, set B is:B = { -2, -√2, √2, -√3, √3, -2√2, 2√2 }Wait, hold on, let me list them all:-2, √2, -√2, √3, -√3, 2√2, -2√2So, that's seven elements in total.Now, I need to find the product of all these elements.Hmm, okay, so the product is:(-2) × (√2) × (-√2) × (√3) × (-√3) × (2√2) × (-2√2)That's a lot of terms. Let me see if I can pair them up or simplify step by step.First, let's note that multiplication is commutative and associative, so I can rearrange and group terms as needed.Let me group the constants and the radicals separately.First, the constants:-2, and then in the radicals, we have √2, -√2, √3, -√3, 2√2, -2√2.Wait, actually, all terms are either constants or radicals. Let me see:Wait, actually, all terms are either constants or multiples of radicals. So, perhaps I can group the constants and the radicals separately.But actually, let's see:Looking at each term:1. (-2)2. (√2)3. (-√2)4. (√3)5. (-√3)6. (2√2)7. (-2√2)So, let's pair terms that are negatives of each other or can be multiplied to simplify.Let me pair term 2 and term 3: (√2) × (-√2) = - (√2 × √2) = -2Similarly, pair term 4 and term 5: (√3) × (-√3) = - (√3 × √3) = -3Pair term 6 and term 7: (2√2) × (-2√2) = - (2√2 × 2√2) = - (4 × 2) = -8So now, we have:(-2) × (-2) × (-3) × (-8)Wait, let me make sure:After pairing:(√2 × -√2) = -2(√3 × -√3) = -3(2√2 × -2√2) = -8And the remaining term is (-2)So, the product becomes:(-2) × (-2) × (-3) × (-8)Wait, no, hold on. Let's recount:Original terms:1. (-2)2. (√2)3. (-√2)4. (√3)5. (-√3)6. (2√2)7. (-2√2)After pairing 2 & 3: (√2 × -√2) = -2After pairing 4 & 5: (√3 × -√3) = -3After pairing 6 & 7: (2√2 × -2√2) = -8So, now, the product is:(-2) × (-2) × (-3) × (-8)Wait, but hold on, that would be four terms:1. (-2) from the first pairing2. (-3) from the second pairing3. (-8) from the third pairing4. The remaining term is (-2)Wait, no, actually, the first term is (-2), and then the pairings give us three more terms: (-2), (-3), (-8). So, total four terms.So, the product is:(-2) × (-2) × (-3) × (-8)Let me compute this step by step.First, multiply (-2) × (-2) = 4Then, multiply 4 × (-3) = -12Then, multiply -12 × (-8) = 96So, the product is 96.Wait, but let me double-check because sometimes when dealing with multiple negative signs, it's easy to make a mistake.Alternatively, let's compute the product step by step without pairing:Start with (-2) × (√2) × (-√2) × (√3) × (-√3) × (2√2) × (-2√2)First, multiply (-2) × (√2) = -2√2Then, multiply by (-√2): (-2√2) × (-√2) = 2 × (√2 × √2) = 2 × 2 = 4Next, multiply by (√3): 4 × √3 = 4√3Then, multiply by (-√3): 4√3 × (-√3) = -4 × (√3 × √3) = -4 × 3 = -12Next, multiply by (2√2): -12 × 2√2 = -24√2Then, multiply by (-2√2): -24√2 × (-2√2) = 48 × (√2 × √2) = 48 × 2 = 96So, same result: 96.Therefore, the product of all elements in set B is 96, which corresponds to option C.But just to make sure I didn't miss any elements or make any calculation errors, let me recount:Set B has the following elements:-2, √2, -√2, √3, -√3, 2√2, -2√2That's seven elements. When I paired them, I paired three pairs and had one element left, which was (-2). Wait, no, actually, when I paired the radicals, I had three pairs, each of which gave me a constant, and then the remaining term was (-2). So, in total, four constants: (-2), (-2), (-3), (-8). Multiplying them together gives 96.Alternatively, when I multiplied step by step, I also got 96.Therefore, I'm confident that the product is 96, which is option C.**Final Answer**boxed{C}