💡Okay, so I have this problem about expanding (1 + 2x)^n, and I need to find two things: first, which term's binomial coefficient is the largest, and second, which term's coefficient is the largest. The problem also mentions that the sum of the coefficients of the first three terms is 201. Hmm, let me try to break this down step by step.Starting with the expansion of (1 + 2x)^n. I remember that the binomial expansion of (a + b)^n is the sum from k=0 to n of (n choose k) * a^(n-k) * b^k. In this case, a is 1 and b is 2x. So, the expansion will be the sum from k=0 to n of (n choose k) * 1^(n-k) * (2x)^k. Simplifying that, since 1 to any power is just 1, it becomes the sum from k=0 to n of (n choose k) * 2^k * x^k.Now, the problem says that the sum of the coefficients of the first three terms is 201. Let me figure out what those first three terms are. The first term is when k=0: (n choose 0)*2^0*x^0 = 1. The second term is when k=1: (n choose 1)*2^1*x^1 = 2n x. The third term is when k=2: (n choose 2)*2^2*x^2 = 4*(n(n-1)/2) x^2 = 2n(n-1) x^2.But wait, the problem mentions the sum of the coefficients of the first three terms. So, I need to consider the coefficients without the x terms. That means for the first term, the coefficient is 1. For the second term, the coefficient is 2n. For the third term, the coefficient is 2n(n-1). So, adding these together: 1 + 2n + 2n(n - 1) = 201.Let me write that equation down:1 + 2n + 2n(n - 1) = 201.Simplify this equation step by step. First, expand the 2n(n - 1):2n(n - 1) = 2n^2 - 2n.So, substituting back into the equation:1 + 2n + 2n^2 - 2n = 201.Wait, the +2n and -2n cancel each other out:1 + 2n^2 = 201.Subtract 1 from both sides:2n^2 = 200.Divide both sides by 2:n^2 = 100.Take the square root of both sides:n = 10.Since n is a positive integer, n = 10.Alright, so n is 10. Now, moving on to the first question: (1) Find which term's binomial coefficient is the largest.I remember that in the binomial expansion of (a + b)^n, the binomial coefficients (n choose k) are largest around the middle term. Specifically, when n is even, the largest coefficient is at k = n/2, and when n is odd, the largest coefficients are at k = (n - 1)/2 and k = (n + 1)/2.Since n = 10, which is even, the largest binomial coefficient is at k = 10/2 = 5. So, the 6th term (since we start counting from k=0) has the largest binomial coefficient.Wait, let me verify that. The binomial coefficients for n=10 are symmetric, so (10 choose 0) = 1, (10 choose 1) = 10, (10 choose 2) = 45, (10 choose 3) = 120, (10 choose 4) = 210, (10 choose 5) = 252, and then they start decreasing: (10 choose 6) = 210, etc. So yes, the largest binomial coefficient is at k=5, which is the 6th term.So, for part (1), the answer is the 6th term.Now, moving on to part (2): Find which term's coefficient is the largest.Hmm, this is slightly different. The coefficient of each term isn't just the binomial coefficient, but also includes the factor from the expansion, which in this case is 2^k. So, the coefficient of the (k+1)th term is (10 choose k) * 2^k.I need to find the value of k where this coefficient is maximized. I remember that to find the maximum term in a binomial expansion with coefficients multiplied by some factor, we can use the ratio test. Specifically, we can look at the ratio of consecutive coefficients and find where this ratio changes from greater than 1 to less than 1. The term where this happens is the maximum.So, let's denote T_k = (10 choose k) * 2^k.We can compute the ratio T_{k+1}/T_k:T_{k+1}/T_k = [(10 choose (k+1)) * 2^{k+1}] / [(10 choose k) * 2^k] = [ (10 - k)/(k + 1) ) ] * 2.Simplify that:= [2*(10 - k)] / (k + 1).We need to find when this ratio is greater than 1 and when it's less than 1.Set the ratio equal to 1:[2*(10 - k)] / (k + 1) = 1.Multiply both sides by (k + 1):2*(10 - k) = k + 1.Expand:20 - 2k = k + 1.Bring all terms to one side:20 - 2k - k - 1 = 0 => 19 - 3k = 0.Solve for k:3k = 19 => k = 19/3 ≈ 6.333.So, when k < 19/3 ≈ 6.333, the ratio T_{k+1}/T_k > 1, meaning the coefficients are increasing. When k > 19/3, the ratio < 1, meaning the coefficients are decreasing. Therefore, the maximum coefficient occurs around k = 6.333.But since k must be an integer, we check k=6 and k=7.Compute T_6 and T_7.First, T_6 = (10 choose 6) * 2^6.(10 choose 6) = 210, and 2^6 = 64. So, T_6 = 210 * 64 = 13,440.T_7 = (10 choose 7) * 2^7.(10 choose 7) = 120, and 2^7 = 128. So, T_7 = 120 * 128 = 15,360.Wait, that's larger than T_6. Hmm, but according to the ratio test, the maximum should be around k=6.333, so between 6 and 7.But let's check T_7 and T_8.Wait, T_8 = (10 choose 8) * 2^8 = 45 * 256 = 11,520.So, T_7 is larger than both T_6 and T_8, so T_7 is the maximum.Wait, but according to the ratio, when k=6, the ratio T_7/T_6 = [2*(10 - 6)] / (6 + 1) = (2*4)/7 ≈ 8/7 ≈ 1.142 > 1, so T_7 > T_6.Similarly, when k=7, T_8/T_7 = [2*(10 - 7)] / (7 + 1) = (2*3)/8 = 6/8 = 0.75 < 1, so T_8 < T_7.Therefore, the maximum coefficient is at k=7, which is the 8th term.Wait, but let me double-check my calculations.Compute T_6: (10 choose 6) = 210, 2^6 = 64, so 210*64 = 13,440.T_7: (10 choose 7) = 120, 2^7 = 128, so 120*128 = 15,360.T_8: (10 choose 8) = 45, 2^8 = 256, so 45*256 = 11,520.Yes, T_7 is indeed larger than both T_6 and T_8, so the maximum coefficient is at k=7, which is the 8th term.Wait, but earlier I thought the ratio test suggested k≈6.333, so between 6 and 7, but since k must be integer, we check k=6 and k=7, and found that T_7 is larger. So, the maximum is at k=7, which is the 8th term.Therefore, for part (2), the answer is the 8th term.Wait, but let me make sure I didn't make any calculation errors.Compute T_6: 210 * 64.210 * 60 = 12,600.210 * 4 = 840.Total: 12,600 + 840 = 13,440. Correct.T_7: 120 * 128.120 * 100 = 12,000.120 * 28 = 3,360.Total: 12,000 + 3,360 = 15,360. Correct.T_8: 45 * 256.45 * 200 = 9,000.45 * 56 = 2,520.Total: 9,000 + 2,520 = 11,520. Correct.Yes, so T_7 is indeed the largest.Therefore, the answers are:(1) The 6th term has the largest binomial coefficient.(2) The 8th term has the largest coefficient.I think that's it. Let me just recap to make sure I didn't miss anything.Given (1 + 2x)^10, the sum of the first three coefficients is 201, which led us to n=10.For the binomial coefficients, the largest is at k=5 (6th term).For the coefficients including the 2^k factor, the largest is at k=7 (8th term).Yes, that seems consistent.**Final Answer**(1) The binomial coefficient of the boxed{6text{th}} term is the largest. (2) The coefficient of the boxed{8text{th}} term is the largest.

💡Okay, so I have this function f(x) = x³ + sin x + 1, and I know that f(a) = 2. I need to find the value of f(-a). Hmm, let's see. First, I should probably write down what f(a) is. So, f(a) = a³ + sin a + 1 = 2. That means if I plug in a into the function, I get 2. Now, I need to find f(-a). Let me write that out too: f(-a) = (-a)³ + sin(-a) + 1.Wait, I remember that when you have a negative exponent, like (-a)³, it's the same as -a³ because the negative sign gets cubed too. So, (-a)³ = -a³. Okay, that's straightforward.Now, what about sin(-a)? I think sine is an odd function, which means sin(-x) = -sin x. So, sin(-a) should be equal to -sin a. That makes sense because sine is symmetric about the origin. So, sin(-a) = -sin a.Putting that back into the expression for f(-a), we have f(-a) = -a³ + (-sin a) + 1. Simplifying that, it becomes f(-a) = -a³ - sin a + 1.Hmm, I notice that in f(a), I have a³ + sin a + 1 = 2. Maybe I can use this to substitute into the expression for f(-a). Let me see.From f(a) = 2, I can rearrange it to find a³ + sin a. So, a³ + sin a = 2 - 1 = 1. Therefore, a³ + sin a = 1.Now, looking back at f(-a), which is -a³ - sin a + 1, I can factor out the negative sign from the first two terms. So, f(-a) = -(a³ + sin a) + 1.But wait, I already know that a³ + sin a = 1. So, substituting that in, f(-a) = -(1) + 1 = -1 + 1 = 0.Oh, so f(-a) equals 0. That seems right. Let me double-check my steps to make sure I didn't make any mistakes.1. I started with f(a) = a³ + sin a + 1 = 2.2. Then, I found f(-a) = (-a)³ + sin(-a) + 1 = -a³ - sin a + 1.3. I recognized that a³ + sin a = 1 from the first equation.4. Substituted that into f(-a) to get -(1) + 1 = 0.Everything seems to check out. I don't think I made any errors in the algebra or the properties of the functions involved. So, I'm pretty confident that f(-a) is indeed 0.

💡To find the largest three-digit integer congruent to (12 pmod{15}), I start by expressing such integers in the form (15n + 12), where (n) is an integer.Next, I set up the inequality to ensure the number remains a three-digit integer:[15n + 12 < 1000]Subtracting 12 from both sides gives:[15n < 988]Dividing both sides by 15:[n < frac{988}{15} approx 65.87]Since (n) must be an integer, the largest possible value for (n) is 65.Finally, substituting (n = 65) into the expression:[15 cdot 65 + 12 = 987]Therefore, the largest three-digit integer congruent to 12 modulo 15 is 987.

💡To find the smallest three-digit integer ( n ) that satisfies the given conditions, we start by translating the problem into congruences:1. If Alicia buys 7 more candies, the total will be a multiple of 9: [ n + 7 equiv 0 pmod{9} implies n equiv 2 pmod{9} ]2. If Alicia loses 9 candies, the total will be a multiple of 6: [ n - 9 equiv 0 pmod{6} implies n equiv 3 pmod{6} ]Next, we solve these congruences using the Chinese Remainder Theorem. We express ( n ) from the first congruence:[n = 9k + 2]Substituting this into the second congruence:[9k + 2 equiv 3 pmod{6} implies 9k equiv 1 pmod{6}]Since ( 9 equiv 3 pmod{6} ), we simplify to:[3k equiv 1 pmod{6}]Solving for ( k ), we find:[k equiv 5 pmod{6} implies k = 6m + 5]Substituting back into the expression for ( n ):[n = 9(6m + 5) + 2 = 54m + 47]To find the smallest three-digit ( n ), we solve:[54m + 47 geq 100 implies m geq 1]For ( m = 1 ):[n = 54(1) + 47 = 101]Verification:- ( 101 + 7 = 108 ) (a multiple of 9)- ( 101 - 9 = 92 ) (a multiple of 6)Thus, the smallest three-digit number ( n ) is:[boxed{101}]

💡Alright, so I need to figure out how to write the number 360 in what's called "base prime." Honestly, I'm not entirely sure what that means, but I think it has something to do with prime numbers. Maybe it's related to prime factorization? Let me think.Okay, prime factorization is breaking down a number into its prime number components. For example, 360 can be broken down into primes like 2, 3, and 5. Let me try that. So, starting with 360, I can divide it by 2 first.360 divided by 2 is 180. Okay, that's straightforward. Now, 180 can be divided by 2 again, giving me 90. Dividing 90 by 2 once more gives me 45. Hmm, 45 isn't divisible by 2, so I'll move to the next prime number, which is 3.45 divided by 3 is 15. Then, 15 divided by 3 is 5. Finally, 5 is a prime number itself, so I stop there. So, putting it all together, the prime factors of 360 are 2^3 * 3^2 * 5^1.Wait, but the question is about "base prime." I'm not sure if that means something different. Maybe it's about representing the number in a base that's related to primes? I'm a bit confused. Let me see if I can find an example or something.I recall that in positional numeral systems, like base 10 or base 2, each digit represents a power of the base. Maybe "base prime" refers to using primes as the base? But that doesn't quite make sense because bases are usually single numbers, not multiple primes.Perhaps "base prime" is referring to the exponents in the prime factorization? So, if 360 is 2^3 * 3^2 * 5^1, then maybe the exponents 3, 2, and 1 are the "base prime" representation? But then what about the primes themselves? Do they play a role in the base?Alternatively, maybe "base prime" means expressing the number in a base that's a prime number, like base 2, base 3, or base 5. But 360 in base 2 would be a binary number, which is different from what I did earlier with prime factors.I'm getting more confused. Let me try to clarify. If I think of "base prime" as a way to represent numbers using their prime factors, then perhaps it's about listing the exponents of the primes in order. So, starting from the smallest prime, which is 2, then 3, then 5, and so on.Given that, for 360, the exponents are 3 for 2, 2 for 3, and 1 for 5. If I list these exponents in order, it would be 3, 2, 1. But how does that translate into a number? Maybe concatenating them? So, 321? But that doesn't seem right because 321 is just another number, not necessarily related to 360 in a meaningful way.Wait, maybe I need to consider the next prime after 5, which is 7, and since 7 doesn't divide 360, its exponent is 0. So, including that, the exponents would be 3, 2, 1, 0. If I concatenate those, it would be 3210. Hmm, that seems like a possible answer, but I'm not entirely sure.Let me check if this makes sense. If I take the exponents of the primes in order and write them down, starting from 2, then 3, then 5, then 7, and so on, and if a prime doesn't divide the number, its exponent is 0. So, for 360, it's 2^3 * 3^2 * 5^1 * 7^0 * 11^0 * ..., and so on. But writing all those zeros would be tedious, so maybe we just stop at the last non-zero exponent.In that case, for 360, the exponents are 3, 2, 1, and then 0 for 7. So, writing them together, it would be 3210. That seems plausible. But I'm still not 100% certain if this is the correct interpretation of "base prime."Alternatively, maybe "base prime" refers to expressing the number in a base that's a prime number, like base 5 or base 7, but that doesn't seem to align with the initial thought process about prime factorization.I think I need to go back to the original problem and see if there's any additional context or clues. The user simply asked, "What is 360 written in base prime?" Without more information, it's a bit ambiguous. However, considering the mention of prime factorization in the initial response, it's likely that "base prime" refers to the exponents of the primes in the factorization.So, to summarize, if I break down 360 into its prime factors: 2^3 * 3^2 * 5^1. Listing the exponents in order of the primes (2, 3, 5, 7, ...), we get 3, 2, 1, 0, ... Since 7 doesn't divide 360, its exponent is 0. Therefore, the "base prime" representation would be 3210.But I'm still a bit unsure if this is the correct interpretation. Maybe I should look for another way to represent 360 in a prime-related base. For example, if I consider base 2, base 3, or base 5, how would 360 be represented in those bases?Let's try converting 360 to base 2. To do that, I divide 360 by 2 repeatedly and record the remainders.360 ÷ 2 = 180, remainder 0180 ÷ 2 = 90, remainder 090 ÷ 2 = 45, remainder 045 ÷ 2 = 22, remainder 122 ÷ 2 = 11, remainder 011 ÷ 2 = 5, remainder 15 ÷ 2 = 2, remainder 12 ÷ 2 = 1, remainder 01 ÷ 2 = 0, remainder 1Reading the remainders from bottom to top, 360 in base 2 is 101101000.That's interesting, but I don't think that's what the question is asking for, since it's about prime bases, not converting to a prime base.Similarly, converting 360 to base 3:360 ÷ 3 = 120, remainder 0120 ÷ 3 = 40, remainder 040 ÷ 3 = 13, remainder 113 ÷ 3 = 4, remainder 14 ÷ 3 = 1, remainder 11 ÷ 3 = 0, remainder 1So, 360 in base 3 is 111100.Again, that's a different representation, but I don't think it's what the question is asking for.Similarly, base 5:360 ÷ 5 = 72, remainder 072 ÷ 5 = 14, remainder 214 ÷ 5 = 2, remainder 42 ÷ 5 = 0, remainder 2So, 360 in base 5 is 2420.Still, I don't think that's the "base prime" representation.Going back to the initial thought, if "base prime" refers to the exponents in the prime factorization, then 360 being 2^3 * 3^2 * 5^1 * 7^0 would translate to exponents 3, 2, 1, 0. Concatenating these gives 3210.But to be thorough, let's consider if there's another interpretation. Maybe "base prime" refers to a numbering system where each digit represents a prime number, and the position represents the power. For example, the first digit represents 2^something, the second digit represents 3^something, and so on.If that's the case, then 360 being 2^3 * 3^2 * 5^1 would be represented as 3 (for 2^3), 2 (for 3^2), 1 (for 5^1), and 0 for the next prime, which is 7. So, again, it would be 3210.This seems consistent with the initial thought. Therefore, I think the "base prime" representation of 360 is 3210.However, to ensure I'm not missing something, let me consider if there's a different way to interpret "base prime." Maybe it's a base where the digits are prime numbers instead of digits 0-9. But that would complicate things because primes are infinite and not easily mapped to digits.Alternatively, maybe "base prime" refers to a base that's a prime number, but as I saw earlier, converting 360 to base 2, 3, or 5 gives different results, and none of them seem to align with the prime factorization exponents.Given all this, I think the most plausible interpretation is that "base prime" refers to the exponents in the prime factorization, listed in order of the primes. Therefore, 360 in base prime is 3210.

💡First, I need to determine the number of seeds that germinated in the first plot. Since 25% of 300 seeds germinated, I calculate 0.25 multiplied by 300, which equals 75 seeds.Next, I find the total number of seeds planted in both plots by adding 300 and 200, resulting in 500 seeds.The problem states that 31% of the total seeds germinated. To find the total number of germinated seeds, I calculate 0.31 multiplied by 500, which equals 155 seeds.Subtracting the germinated seeds from the first plot (75) from the total germinated seeds (155) gives me the number of seeds that germinated in the second plot: 155 minus 75 equals 80 seeds.Finally, to find the percentage of seeds that germinated in the second plot, I set up the equation (x/100) multiplied by 200 equals 80. Solving for x, I divide 80 by 2, which gives me 40. Therefore, 40% of the seeds in the second plot germinated.