💡Alright, so I have this problem: "How many four-digit perfect squares are palindromes?" Hmm, okay. Let me try to figure this out step by step. First, I need to understand what a four-digit perfect square is and what a palindrome is.A perfect square is a number that's the square of an integer. So, four-digit perfect squares would be numbers like 1000, 1024, 1089, and so on, up to 9999. I think the smallest four-digit number is 1000, and the largest is 9999. So, I need to find all the perfect squares between 1000 and 9999.Now, what's a palindrome? A palindrome is a number that reads the same forwards and backwards. For example, 1221 is a palindrome because if you reverse the digits, you still get 1221. Similarly, 1331 is also a palindrome. So, for a four-digit number to be a palindrome, the first digit must be the same as the last digit, and the second digit must be the same as the third digit.Okay, so I need to find all four-digit perfect squares that are palindromes. That means I need to check each four-digit perfect square and see if it's a palindrome. But that sounds like a lot of work. Maybe there's a smarter way to do this without checking every single four-digit perfect square.First, let me figure out the range of numbers whose squares are four-digit numbers. The smallest four-digit number is 1000, and the square root of 1000 is approximately 31.62. So, the smallest integer whose square is a four-digit number is 32 because 32 squared is 1024. Similarly, the largest four-digit number is 9999, and the square root of 9999 is approximately 99.99, so the largest integer whose square is a four-digit number is 99 because 99 squared is 9801.So, the four-digit perfect squares are the squares of the integers from 32 to 99. That's 68 numbers in total. Now, I need to find out how many of these squares are palindromes.Since checking all 68 numbers might take some time, maybe I can narrow it down by considering the properties of palindromic numbers. For a four-digit palindrome, the number must have the form ABBA, where A and B are digits. That means the first digit is the same as the last digit, and the second digit is the same as the third digit.So, if I can find squares of numbers between 32 and 99 that have this ABBA structure, those would be the four-digit perfect squares that are palindromes.Let me think about the possible digits A and B. Since it's a four-digit number, A can't be zero. So, A can be from 1 to 9, and B can be from 0 to 9.Now, let's consider the square of a number. The square of a number can end with certain digits only. For example, squares can only end with 0, 1, 4, 5, 6, or 9. So, the last digit of a perfect square must be one of these. Since the number is a palindrome, the first digit must be the same as the last digit. Therefore, the first digit A must also be one of these digits: 0, 1, 4, 5, 6, or 9. But since A can't be zero, A must be 1, 4, 5, 6, or 9.So, A can be 1, 4, 5, 6, or 9. That narrows down the possible palindromic numbers we need to consider.Now, let's think about the structure of the number. It's ABBA, so the number can be expressed as 1000*A + 100*B + 10*B + A, which simplifies to 1001*A + 110*B.So, the number is 1001A + 110B. Since this number must be a perfect square, we can write it as N^2, where N is an integer between 32 and 99.So, N^2 = 1001A + 110B.Hmm, maybe I can factor this expression. Let's see:1001A + 110B = 11*(91A + 10B).So, N^2 = 11*(91A + 10B).This implies that 11 divides N^2, which means that 11 must divide N because 11 is a prime number. Therefore, N must be a multiple of 11.So, N = 11*K, where K is an integer.Since N is between 32 and 99, K must be between 3 and 9 because 11*3=33 and 11*9=99.So, K can be 3, 4, 5, 6, 7, 8, or 9.Therefore, N can be 33, 44, 55, 66, 77, 88, or 99.Okay, so now I only need to check the squares of these multiples of 11 between 33 and 99.Let me list them:33^2 = 108944^2 = 193655^2 = 302566^2 = 435677^2 = 592988^2 = 774499^2 = 9801Now, let's check each of these to see if they are palindromes.Starting with 33^2 = 1089. Is 1089 a palindrome? Let's see: the first digit is 1, the last digit is 9. Since 1 ≠ 9, it's not a palindrome.Next, 44^2 = 1936. First digit 1, last digit 6. 1 ≠ 6, not a palindrome.55^2 = 3025. First digit 3, last digit 5. 3 ≠ 5, not a palindrome.66^2 = 4356. First digit 4, last digit 6. 4 ≠ 6, not a palindrome.77^2 = 5929. First digit 5, last digit 9. 5 ≠ 9, not a palindrome.88^2 = 7744. First digit 7, last digit 4. 7 ≠ 4, not a palindrome.99^2 = 9801. First digit 9, last digit 1. 9 ≠ 1, not a palindrome.Hmm, none of these squares of multiples of 11 are palindromes. That's interesting.Wait a minute, did I miss any numbers? I thought N had to be a multiple of 11 because N^2 is divisible by 11, which is prime. So, N must be divisible by 11. Therefore, I think I covered all possible candidates.But just to be thorough, let me check if there are any four-digit perfect squares that are palindromes outside of these multiples of 11. Maybe I made a mistake in assuming that N must be a multiple of 11.Wait, let's think again. If N^2 is divisible by 11, then N must be divisible by 11 because 11 is prime. So, N must be a multiple of 11. Therefore, I didn't miss any numbers.But just to double-check, let me consider a few other four-digit perfect squares and see if any of them are palindromes.For example, 32^2 = 1024. Is 1024 a palindrome? 1 vs. 4, no.34^2 = 1156. 1 vs. 6, no.35^2 = 1225. 1 vs. 5, no.36^2 = 1296. 1 vs. 6, no.37^2 = 1369. 1 vs. 9, no.38^2 = 1444. 1 vs. 4, no.39^2 = 1521. 1 vs. 1, okay, first and last digits are the same. Now, check the second and third digits: 5 vs. 2. 5 ≠ 2, so not a palindrome.40^2 = 1600. 1 vs. 0, no.41^2 = 1681. 1 vs. 1, same. Now, second digit 6 vs. third digit 8. 6 ≠ 8, not a palindrome.42^2 = 1764. 1 vs. 4, no.43^2 = 1849. 1 vs. 9, no.44^2 = 1936. Already checked, not a palindrome.45^2 = 2025. 2 vs. 5, no.46^2 = 2116. 2 vs. 6, no.47^2 = 2209. 2 vs. 9, no.48^2 = 2304. 2 vs. 4, no.49^2 = 2401. 2 vs. 1, no.50^2 = 2500. 2 vs. 0, no.51^2 = 2601. 2 vs. 1, no.52^2 = 2704. 2 vs. 4, no.53^2 = 2809. 2 vs. 9, no.54^2 = 2916. 2 vs. 6, no.55^2 = 3025. Checked, not a palindrome.56^2 = 3136. 3 vs. 6, no.57^2 = 3249. 3 vs. 9, no.58^2 = 3364. 3 vs. 4, no.59^2 = 3481. 3 vs. 1, no.60^2 = 3600. 3 vs. 0, no.61^2 = 3721. 3 vs. 1, no.62^2 = 3844. 3 vs. 4, no.63^2 = 3969. 3 vs. 9, no.64^2 = 4096. 4 vs. 6, no.65^2 = 4225. 4 vs. 5, no.66^2 = 4356. Checked, not a palindrome.67^2 = 4489. 4 vs. 9, no.68^2 = 4624. 4 vs. 4, same. Now, second digit 6 vs. third digit 2. 6 ≠ 2, not a palindrome.69^2 = 4761. 4 vs. 1, no.70^2 = 4900. 4 vs. 0, no.71^2 = 5041. 5 vs. 1, no.72^2 = 5184. 5 vs. 4, no.73^2 = 5329. 5 vs. 9, no.74^2 = 5476. 5 vs. 6, no.75^2 = 5625. 5 vs. 5, same. Now, second digit 6 vs. third digit 2. 6 ≠ 2, not a palindrome.76^2 = 5776. 5 vs. 6, no.77^2 = 5929. Checked, not a palindrome.78^2 = 6084. 6 vs. 4, no.79^2 = 6241. 6 vs. 1, no.80^2 = 6400. 6 vs. 0, no.81^2 = 6561. 6 vs. 1, no.82^2 = 6724. 6 vs. 4, no.83^2 = 6889. 6 vs. 9, no.84^2 = 7056. 7 vs. 6, no.85^2 = 7225. 7 vs. 5, no.86^2 = 7396. 7 vs. 6, no.87^2 = 7569. 7 vs. 9, no.88^2 = 7744. Checked, not a palindrome.89^2 = 7921. 7 vs. 1, no.90^2 = 8100. 8 vs. 0, no.91^2 = 8281. 8 vs. 1, no.92^2 = 8464. 8 vs. 4, no.93^2 = 8649. 8 vs. 9, no.94^2 = 8836. 8 vs. 6, no.95^2 = 9025. 9 vs. 5, no.96^2 = 9216. 9 vs. 6, no.97^2 = 9409. 9 vs. 9, same. Now, second digit 4 vs. third digit 0. 4 ≠ 0, not a palindrome.98^2 = 9604. 9 vs. 4, no.99^2 = 9801. Checked, not a palindrome.Okay, so after checking all four-digit perfect squares, none of them are palindromes. That's surprising. I thought there might be at least one, but apparently not.Wait, let me think again. Did I make a mistake in my initial assumption that N must be a multiple of 11? Because I concluded that N^2 is divisible by 11, so N must be divisible by 11. But is that necessarily true?Yes, because 11 is a prime number. If a prime divides a square, it must divide the base. So, if 11 divides N^2, then 11 divides N. So, my initial assumption was correct. Therefore, I didn't miss any numbers by only checking multiples of 11.But just to be absolutely sure, let me think about the structure of the palindrome. It's ABBA, which is 1001A + 110B. So, it's divisible by 11 because 1001 is 11*91 and 110 is 11*10. So, the entire number is divisible by 11. Therefore, its square root must also be divisible by 11, which is why N must be a multiple of 11.So, I think my reasoning is sound. Therefore, there are no four-digit perfect squares that are palindromes.But wait, I recall that 7744 is a square and a palindrome. Wait, is that true?Let me check: 88^2 = 7744. Is 7744 a palindrome? Yes, because reversing the digits gives 4477, which is not the same as 7744. Wait, no, that's not correct. Wait, 7744 reversed is 4477, which is different. So, 7744 is not a palindrome.Wait, maybe I confused it with another number. Let me think. Is there a four-digit palindrome that's a perfect square?I think 2112 is a palindrome, but is it a perfect square? Let me check: 46^2 = 2116, which is close but not 2112. 45^2 = 2025, 46^2 = 2116, 47^2 = 2209. So, 2112 is not a perfect square.What about 6776? Is that a perfect square? Let's see: 82^2 = 6724, 83^2 = 6889. So, 6776 is not a perfect square.How about 8118? 90^2 = 8100, 91^2 = 8281. So, 8118 is not a perfect square.Wait, maybe I'm overcomplicating this. Since I've already checked all four-digit perfect squares and none of them are palindromes, I think the answer is zero.But just to make sure, let me think about the possible palindromic squares. I know that in two digits, 121 is a palindrome and a perfect square (11^2). In three digits, 121, 484, and 676 are palindromic squares. But in four digits, I don't recall any.So, based on my thorough checking, I think the number of four-digit perfect squares that are palindromes is zero.

💡Okay, so I have this problem here: "The average of a non-zero real number and its square is 5 times the number. Use a quadratic equation to solve this problem. What is the real number?" Alright, let's try to break this down step by step.First, I need to understand what the problem is asking. It mentions a non-zero real number, which means it's some number that's not zero and is on the number line. Let's call this number "x" for simplicity. The problem says the average of this number and its square is 5 times the number. Hmm, averages involve adding things and dividing by the number of things, right?So, if I have two numbers, their average is (number1 + number2)/2. In this case, the two numbers are "x" and "x squared" (which is x^2). So, the average would be (x + x^2)/2. According to the problem, this average equals 5 times the number, which is 5x. So, I can write this as an equation:(x + x^2)/2 = 5xAlright, now I need to solve this equation for x. Since it's mentioned to use a quadratic equation, I should probably rearrange this equation into the standard quadratic form, which is ax^2 + bx + c = 0.Let me start by getting rid of the denominator to make things simpler. If I multiply both sides of the equation by 2, I'll eliminate the fraction:2 * [(x + x^2)/2] = 2 * 5xThis simplifies to:x + x^2 = 10xNow, I want to get all the terms on one side to set the equation to zero. So, I'll subtract 10x from both sides:x + x^2 - 10x = 0Combining like terms (x - 10x = -9x):x^2 - 9x = 0Okay, so now I have a quadratic equation: x^2 - 9x = 0. To solve this, I can factor it. Let's see, both terms have an x in common, so I can factor out an x:x(x - 9) = 0This gives me two possible solutions:1. x = 02. x - 9 = 0 → x = 9But wait, the problem specifies that the number is non-zero. So, x = 0 is not a valid solution here. That leaves us with x = 9.Let me double-check to make sure this makes sense. If x = 9, then x squared is 81. The average of 9 and 81 is (9 + 81)/2 = 90/2 = 45. And 5 times the number is 5 * 9 = 45. Yep, that checks out.So, the non-zero real number we're looking for is 9.

💡Okay, so I have this problem about an acute triangle ABC. The sides opposite angles A, B, and C are a, b, and c respectively. The given equation is (b² + c² - a²) tan A = √3 bc. Part (1) asks me to find angle A. Hmm, let me think. I remember the Law of Cosines, which says that a² = b² + c² - 2bc cos A. So, if I rearrange that, b² + c² - a² = 2bc cos A. Maybe I can substitute that into the given equation.So, substituting, the left side becomes (2bc cos A) tan A. And that's equal to √3 bc. Let me write that out:2bc cos A * tan A = √3 bcI know that tan A is sin A / cos A, so substituting that in:2bc cos A * (sin A / cos A) = √3 bcThe cos A cancels out:2bc sin A = √3 bcNow, I can divide both sides by bc (assuming bc ≠ 0, which they aren't in a triangle):2 sin A = √3So, sin A = √3 / 2. Hmm, when is sin A equal to √3 / 2? That's at 60 degrees and 120 degrees. But the triangle is acute, so all angles are less than 90 degrees. Therefore, angle A must be 60 degrees.Okay, that seems straightforward. So, part (1) is solved, angle A is 60 degrees.Moving on to part (2). It says if a = 2, find the maximum value of the area S of triangle ABC. First, I know that the area of a triangle can be given by (1/2)ab sin C, but in this case, since we have angle A, maybe I can use (1/2)bc sin A. Since angle A is 60 degrees, sin A is √3 / 2. So, the area S = (1/2) bc (√3 / 2) = (√3 / 4) bc. So, to maximize S, I need to maximize bc.But I also have the given equation from part (1). Let me recall that from the given equation, we had:(b² + c² - a²) tan A = √3 bcWe already found angle A is 60 degrees, so tan A is √3. Let's plug that in:(b² + c² - a²) * √3 = √3 bcDivide both sides by √3:b² + c² - a² = bcSince a = 2, substitute that in:b² + c² - 4 = bcSo, b² + c² - bc = 4I need to maximize bc given that b² + c² - bc = 4. Hmm, how can I approach this? Maybe using some inequality.I remember that for any real numbers, b² + c² ≥ 2bc by the AM-GM inequality. But here, we have b² + c² - bc = 4. So, substituting the inequality:b² + c² - bc ≥ 2bc - bc = bcSo, 4 ≥ bcWait, that's interesting. So, bc ≤ 4. Therefore, the maximum value of bc is 4.But does equality hold? When does b² + c² - bc = 4 and bc = 4? Let's check.If bc = 4, then from b² + c² - bc = 4, we have b² + c² - 4 = 4, so b² + c² = 8.But if bc = 4, then by AM-GM, b² + c² ≥ 2bc = 8. So, equality holds when b = c. So, b = c. Therefore, b = c = 2, since bc = 4.So, when b = c = 2, then a² = b² + c² - 2bc cos A = 4 + 4 - 2*2*2*(1/2) = 8 - 4 = 4, so a = 2, which matches the given a = 2.Therefore, the maximum bc is 4, so the maximum area S is (√3 / 4) * 4 = √3.Wait, let me double-check. If b = c = 2, then the triangle is isoceles with sides b = c = 2 and a = 2. So, it's actually an equilateral triangle? Because all sides are equal. But wait, in an equilateral triangle, all angles are 60 degrees, which is consistent with angle A being 60 degrees. So, that makes sense.So, in that case, the area would indeed be (√3 / 4) * (2)^2 = √3. So, that's correct.Therefore, the maximum area is √3.I think that's solid. I used the Law of Cosines, substituted into the given equation, found angle A, then used the area formula and applied some inequalities to find the maximum bc, leading to the maximum area.**Final Answer**(1) boxed{60^circ}(2) boxed{sqrt{3}}

💡Okay, so I have this problem where I need to find the interval that contains the root of the function f(x) = x² + ln x - 4. The options are A: (0,1), B: (1,2), C: (2,3), and D: (3,4). Hmm, let me think about how to approach this.First, I remember that a root of a function is a value of x where f(x) = 0. So, I need to find an interval where the function crosses the x-axis. To do this, I can use the Intermediate Value Theorem, which says that if a function is continuous on an interval [a, b] and takes on values f(a) and f(b) at each end of the interval, then it also takes on any value between f(a) and f(b). So, if f(a) and f(b) have opposite signs, there must be at least one root between a and b.Alright, so I need to check the value of f(x) at the endpoints of each interval and see where the sign changes. Let's start by evaluating f(x) at each integer point from 0 to 4.First, let's check x=0. But wait, ln(0) is undefined because the natural logarithm of zero is negative infinity. So, f(0) doesn't exist. That means the interval (0,1) might not be valid because the function isn't defined at x=0. Hmm, but maybe the function is defined just to the right of 0? Let me think. As x approaches 0 from the right, ln x approaches negative infinity, so f(x) = x² + ln x - 4 would approach negative infinity. So, near x=0, f(x) is negative. Let me check f(1).f(1) = (1)² + ln(1) - 4 = 1 + 0 - 4 = -3. So, f(1) is -3, which is negative. So, at x=1, the function is still negative. Let's check x=2.f(2) = (2)² + ln(2) - 4 = 4 + ln(2) - 4. Since ln(2) is approximately 0.693, so 4 + 0.693 - 4 = 0.693. So, f(2) is approximately 0.693, which is positive. Okay, so at x=2, the function is positive.So, between x=1 and x=2, the function goes from negative at x=1 to positive at x=2. That means, by the Intermediate Value Theorem, there must be a root somewhere between 1 and 2. So, interval B: (1,2) contains a root.Wait, but let me check the other intervals just to make sure. Maybe there are multiple roots?Let's check x=3.f(3) = (3)² + ln(3) - 4 = 9 + ln(3) - 4. ln(3) is approximately 1.0986, so 9 + 1.0986 - 4 = 6.0986. So, f(3) is approximately 6.0986, which is positive.And x=4.f(4) = (4)² + ln(4) - 4 = 16 + ln(4) - 4. ln(4) is approximately 1.386, so 16 + 1.386 - 4 = 13.386. So, f(4) is positive as well.So, from x=2 onwards, the function is positive and increasing. So, it doesn't cross the x-axis again after x=2 because it's just getting more positive. So, the only root is between x=1 and x=2.But wait, let me double-check. Maybe I made a mistake in evaluating f(2). Let me recalculate f(2):f(2) = 2² + ln(2) - 4 = 4 + 0.6931 - 4 = 0.6931. Yeah, that's correct. So, it's positive.And f(1) is -3, which is negative. So, the function crosses from negative to positive between 1 and 2. Therefore, the root is in interval B: (1,2).But wait, the initial problem statement said the correct answer is C: (2,3). Did I make a mistake?Let me check my calculations again.f(1) = 1 + 0 - 4 = -3. Correct.f(2) = 4 + ln(2) - 4 ≈ 0.6931. Correct.f(3) = 9 + ln(3) - 4 ≈ 6.0986. Correct.So, f(1) is negative, f(2) is positive, so the root is between 1 and 2. So, interval B.But the initial answer given was C: (2,3). Maybe I misread the function? Let me check the function again: f(x) = x² + ln x - 4. Yes, that's correct.Wait, maybe the function is f(x) = x³ + ln x - 4? Because if it were x³ instead of x², then f(2) would be 8 + ln(2) - 4 ≈ 4.693, which is still positive, but f(1) is 1 + 0 - 4 = -3. So, still the root would be between 1 and 2.Alternatively, maybe the function is f(x) = x² + ln(x) - 4, which is what I have. Hmm.Wait, maybe I need to consider the behavior of the function more carefully. Let's plot f(x) in my mind.As x approaches 0 from the right, ln x approaches negative infinity, so f(x) approaches negative infinity.At x=1, f(1) = -3.At x=2, f(2) ≈ 0.693.So, the function increases from negative infinity at x approaching 0, reaches -3 at x=1, then increases to positive 0.693 at x=2, and continues increasing beyond that.So, the function crosses the x-axis only once, between x=1 and x=2.Therefore, the root is in interval B: (1,2).But the initial answer was C: (2,3). Maybe the problem was f(x) = x² + ln(x) - 4, but perhaps I misread it as x squared instead of x cubed? Let me check.No, the function is definitely f(x) = x² + ln x - 4.Wait, maybe the problem is f(x) = x² + ln(x - 4). But that would be different. Let me check.No, the function is f(x) = x² + ln x - 4. So, as written.Alternatively, maybe the problem is f(x) = x² + ln(x) - 4, but the interval is (2,3). Let me check f(2) and f(3) again.f(2) ≈ 0.693, positive.f(3) ≈ 6.0986, positive.So, both f(2) and f(3) are positive, so the function doesn't cross the x-axis between 2 and 3. So, the root must be between 1 and 2.Therefore, the correct interval is B: (1,2).But the initial answer given was C: (2,3). Maybe there was a mistake in the initial answer.Alternatively, perhaps I made a mistake in evaluating f(2). Let me calculate f(2) more precisely.f(2) = 2² + ln(2) - 4 = 4 + 0.69314718056 - 4 = 0.69314718056. So, approximately 0.6931, which is positive.So, f(2) is positive, f(1) is negative. So, the root is between 1 and 2.Therefore, the correct answer should be B: (1,2).But the initial answer was C: (2,3). Maybe the problem was different? Or perhaps I misread the function.Wait, let me check the function again: f(x) = x² + ln x - 4. Yes.Alternatively, maybe the function is f(x) = x² + ln(x) - 4, but the interval is (2,3). Let me check f(2) and f(3) again.f(2) ≈ 0.693, positive.f(3) ≈ 6.0986, positive.So, no sign change between 2 and 3. Therefore, no root in (2,3).Therefore, the correct interval is (1,2), which is option B.So, maybe the initial answer was incorrect. Or perhaps I misread the function.Alternatively, maybe the function is f(x) = x² + ln(x) - 4, but the root is actually in (2,3). Let me check f(2) and f(3) again.Wait, f(2) is approximately 0.693, which is positive, and f(3) is approximately 6.0986, which is positive. So, no sign change.Wait, unless I made a mistake in calculating f(2). Let me do it step by step.f(2) = 2² + ln(2) - 4 = 4 + ln(2) - 4 = ln(2) ≈ 0.6931. Correct.f(1) = 1 + 0 - 4 = -3. Correct.So, the function goes from negative at x=1 to positive at x=2, so the root is in (1,2).Therefore, the correct answer is B: (1,2).But the initial answer was C: (2,3). Maybe the problem was different, or perhaps I misread it.Alternatively, maybe the function is f(x) = x² + ln(x) - 4, but the root is actually in (2,3). Let me check f(2) and f(3) again.Wait, f(2) is positive, f(3) is positive. So, no root in (2,3). Therefore, the root must be in (1,2).So, I think the initial answer was incorrect. The correct interval is B: (1,2).But to be thorough, let me check f(1.5) to see where the root is.f(1.5) = (1.5)² + ln(1.5) - 4 = 2.25 + ln(1.5) - 4.ln(1.5) ≈ 0.4055.So, f(1.5) ≈ 2.25 + 0.4055 - 4 ≈ 2.6555 - 4 ≈ -1.3445. So, f(1.5) is negative.So, at x=1.5, f(x) is still negative. So, the function goes from negative at x=1.5 to positive at x=2. Therefore, the root is between 1.5 and 2.So, within interval (1,2), the root is between 1.5 and 2.Therefore, the correct interval is (1,2), which is option B.So, I think the initial answer was incorrect. The correct answer should be B: (1,2).

💡Okay, so I have this problem about sequences. There's a sequence {a_n} that follows the rule a_n - a_{n+1} = a_{n+1}a_n for all natural numbers n. Then there's another sequence {b_n} where each term is the reciprocal of a_n, meaning b_n = 1/a_n. I'm told that the sum of the first ten terms of {b_n} is 65, and I need to find a general formula for a_n.Alright, let's break this down step by step. First, I need to understand the relationship given for {a_n}. The equation is a_n - a_{n+1} = a_{n+1}a_n. Hmm, that looks a bit complicated, but maybe I can manipulate it to find a recursive formula or something more manageable.Let me try to rearrange the equation. Starting with a_n - a_{n+1} = a_{n+1}a_n. If I move a_{n+1} to the other side, I get a_n = a_{n+1} + a_{n+1}a_n. Hmm, that doesn't seem immediately helpful. Maybe I can factor out a_{n+1} on the right side. So, a_n = a_{n+1}(1 + a_n). Hmm, still not too helpful.Wait, maybe I can divide both sides by a_n a_{n+1} to simplify. Let's see:Starting from a_n - a_{n+1} = a_{n+1}a_n.Divide both sides by a_n a_{n+1}:(a_n - a_{n+1}) / (a_n a_{n+1}) = 1.Breaking that fraction into two terms:a_n / (a_n a_{n+1}) - a_{n+1} / (a_n a_{n+1}) = 1.Simplify each term:1/a_{n+1} - 1/a_n = 1.Oh! That's interesting. So, 1/a_{n+1} - 1/a_n = 1. But wait, since b_n = 1/a_n, this equation becomes b_{n+1} - b_n = 1.So, the difference between consecutive terms of {b_n} is 1. That means {b_n} is an arithmetic sequence with a common difference of 1. That's a key insight.Now, since {b_n} is arithmetic, I can write its general term as b_n = b_1 + (n - 1)d, where d is the common difference. Here, d = 1, so b_n = b_1 + (n - 1)*1 = b_1 + n - 1.But I also know that the sum of the first ten terms of {b_n} is 65. The sum of an arithmetic sequence can be calculated using the formula S_n = n/2 * (2a_1 + (n - 1)d). In this case, n = 10, S_10 = 65, a_1 = b_1, and d = 1.Plugging into the formula:65 = 10/2 * (2b_1 + (10 - 1)*1)Simplify:65 = 5 * (2b_1 + 9)Divide both sides by 5:13 = 2b_1 + 9Subtract 9 from both sides:4 = 2b_1Divide both sides by 2:b_1 = 2So, the first term of {b_n} is 2. Therefore, the general term is:b_n = 2 + (n - 1)*1 = n + 1But remember, b_n = 1/a_n, so:1/a_n = n + 1Taking reciprocals:a_n = 1/(n + 1)So, the general term of the sequence {a_n} is 1 divided by (n + 1).Let me double-check my steps to make sure I didn't make any mistakes. Starting from the given recurrence relation, I manipulated it to find a relationship between b_{n+1} and b_n, which turned out to be an arithmetic sequence. Then, using the sum formula for an arithmetic sequence, I found the first term, and hence the general term. Everything seems to check out.I think that's the solution. The sequence {a_n} is given by a_n = 1/(n + 1).**Final Answer**The general term of the sequence is boxed{a_n = dfrac{1}{n + 1}}.

💡Okay, so I have this problem where I need to find the value of e over a, given several ratios involving a, b, c, d, and e. Let me try to figure this out step by step. First, let me write down all the given ratios to keep track:1. a/b = 52. b/c = 1/43. c/d = 74. d/e = 1/2I need to find e/a. Hmm, so I have these ratios connecting a to b, b to c, c to d, and d to e. It seems like a chain of ratios. Maybe I can link them together somehow to get from a to e.Let me think about how ratios work. If I have a/b = 5, that means a is 5 times b. Similarly, b/c = 1/4 means b is 1/4 of c, or c is 4 times b. Then c/d = 7 means c is 7 times d, so d is c divided by 7. Finally, d/e = 1/2 means d is half of e, so e is twice d.Maybe I can express all variables in terms of one variable and then find e in terms of a. Let's try that.Starting with a/b = 5, so a = 5b.Then, b/c = 1/4, so b = (1/4)c, which means c = 4b.Next, c/d = 7, so c = 7d, which implies d = c/7.But since c = 4b, then d = (4b)/7.Then, d/e = 1/2, so d = (1/2)e, which means e = 2d.Substituting d from above, e = 2*(4b/7) = 8b/7.Now, I have e expressed in terms of b: e = (8/7)b.But I also have a expressed in terms of b: a = 5b.So, to find e/a, I can substitute both expressions:e/a = (8/7)b / (5b).Wait, the b's cancel out, right? So e/a = (8/7)/5 = 8/(7*5) = 8/35.Hmm, that seems straightforward. Let me double-check my steps to make sure I didn't make a mistake.1. a = 5b2. c = 4b3. d = c/7 = (4b)/74. e = 2d = 2*(4b/7) = 8b/75. e/a = (8b/7)/(5b) = 8/35Yes, that looks correct. Each step follows logically from the previous one, and the b terms cancel out as expected. So, e/a should be 8/35.Alternatively, I could have multiplied all the given ratios together to get a chain from a to e. Let me try that approach as a verification.Multiplying the four ratios:(a/b) * (b/c) * (c/d) * (d/e) = 5 * (1/4) * 7 * (1/2)Simplifying the left side:(a/b) * (b/c) = a/c(a/c) * (c/d) = a/d(a/d) * (d/e) = a/eSo, the left side simplifies to a/e.On the right side:5 * (1/4) = 5/45/4 * 7 = 35/435/4 * (1/2) = 35/8So, a/e = 35/8Therefore, e/a is the reciprocal, which is 8/35.This matches my earlier result. So, both methods give the same answer, which is reassuring.I think I'm confident now that e/a is 8/35.