💡Okay, so I have this probability problem here, and I need to figure out the probability that a person who tests positive for a disease actually has the disease. Let me try to break this down step by step.First, let me understand the given information. The problem says that in a certain population, one out of every 400 people has this particular disease. That means the prevalence of the disease is 1/400. So, if I randomly pick someone from this population, the chance they have the disease is 1/400, and the chance they don't have it is 399/400.Next, there's a blood test for this disease. It says the test always returns positive for those who have the disease. So, if someone has the disease, the test will definitely show positive. That means the true positive rate is 100%. On the other hand, for those who don't have the disease, there's a 5% false positive rate. So, 5% of the time, the test will incorrectly show positive even when the person doesn't have the disease.Now, I need to find the probability that a person who tests positive actually has the disease. This sounds like a classic case of conditional probability, specifically Bayes' theorem. I remember that Bayes' theorem helps us find the probability of an event given prior knowledge of conditions related to the event.Let me define the events to make it clearer:- Let D be the event that a person has the disease.- Let T be the event that a person tests positive.So, we need to find P(D | T), which is the probability of having the disease given that the test is positive.Bayes' theorem states that:P(D | T) = [P(T | D) * P(D)] / P(T)I know P(D) is 1/400, and P(T | D) is 1 because the test always returns positive for those who have the disease. So, the numerator of the formula is 1 * (1/400) = 1/400.Now, I need to find P(T), the total probability of testing positive. This can happen in two ways: either the person has the disease and tests positive, or the person doesn't have the disease but still tests positive (false positive).So, P(T) = P(T | D) * P(D) + P(T | D^c) * P(D^c)Where D^c is the event that the person does not have the disease.We already know P(T | D) is 1, and P(D) is 1/400. P(D^c) is 399/400, and P(T | D^c) is 5%, which is 0.05.So, plugging in the numbers:P(T) = (1 * 1/400) + (0.05 * 399/400)Let me calculate each part:First part: 1 * 1/400 = 1/400Second part: 0.05 * 399/400. Let's compute 0.05 * 399 first. 0.05 is the same as 5%, so 5% of 399 is 19.95. Then, 19.95 divided by 400 is 0.049875.So, P(T) = 1/400 + 0.049875But 1/400 is 0.0025. So, adding 0.0025 and 0.049875 gives approximately 0.052375.Wait, let me check that again. 0.05 * 399/400:0.05 * 399 = 19.9519.95 / 400 = 0.049875And 1/400 = 0.0025So, 0.0025 + 0.049875 = 0.052375So, P(T) is approximately 0.052375.Now, going back to Bayes' theorem:P(D | T) = (1/400) / 0.052375Let me compute that.1/400 is 0.0025.So, 0.0025 / 0.052375 ≈ 0.0477Hmm, 0.0477 is approximately 4.77%.Wait, but looking at the answer choices, I don't see 4.77%. Let me check my calculations again.Wait, maybe I should do this with fractions instead of decimals to be more precise.So, let's recast everything in fractions.P(T) = (1 * 1/400) + (0.05 * 399/400)0.05 is 1/20, so:P(T) = (1/400) + (1/20 * 399/400)Compute 1/20 * 399/400:1/20 * 399/400 = 399 / 8000And 1/400 is 20/8000So, P(T) = 20/8000 + 399/8000 = 419/8000So, P(T) = 419/8000Now, P(D | T) = (1/400) / (419/8000)Dividing fractions: (1/400) * (8000/419) = (8000)/(400*419) = 20/419Ah, okay, so P(D | T) is 20/419.Looking back at the answer choices, option D is 20/419, so that must be the correct answer.I think I made a mistake earlier when converting to decimals because I approximated too much, which led me to a decimal that didn't match the answer choices. By keeping it in fractions, I got a precise answer that matches one of the options.So, the key steps were:1. Define the events and identify the given probabilities.2. Use Bayes' theorem to set up the formula for the conditional probability.3. Calculate the total probability of testing positive by considering both true positives and false positives.4. Perform the division carefully, preferably using fractions to maintain precision.5. Simplify the result to match one of the given answer choices.This problem is a good example of how conditional probabilities can sometimes be counterintuitive, especially when the disease prevalence is low. Even though the test has a relatively low false positive rate, the actual probability that someone who tests positive has the disease is still quite low because the disease itself is rare.I also remember that in such cases, it's essential to consider the base rate of the disease in the population. If the disease is very rare, like 1 in 400, even a small false positive rate can lead to a significant number of false positives, making the positive predictive value lower than one might expect.So, in summary, by methodically applying Bayes' theorem and carefully computing each part, I arrived at the correct answer, which is 20/419.

💡Alright, so I have this function ( f(x) = ax^2 + ln(x + 1) ) and there are three parts to the problem. Let me try to tackle each part step by step.Starting with part (1): When ( a = -dfrac{1}{4} ), find the intervals of monotonicity for the function ( f(x) ).Okay, so I need to find where the function is increasing or decreasing. I remember that to determine monotonicity, I should look at the first derivative of the function. If the first derivative is positive, the function is increasing; if it's negative, the function is decreasing.So, let's compute the derivative of ( f(x) ) when ( a = -dfrac{1}{4} ).( f(x) = -dfrac{1}{4}x^2 + ln(x + 1) )The derivative ( f'(x) ) would be:( f'(x) = -dfrac{1}{2}x + dfrac{1}{x + 1} )Hmm, that seems right. The derivative of ( -dfrac{1}{4}x^2 ) is ( -dfrac{1}{2}x ) and the derivative of ( ln(x + 1) ) is ( dfrac{1}{x + 1} ).Now, to find the critical points, I need to set ( f'(x) = 0 ) and solve for ( x ).So,( -dfrac{1}{2}x + dfrac{1}{x + 1} = 0 )Let me solve this equation. Let's move one term to the other side:( -dfrac{1}{2}x = -dfrac{1}{x + 1} )Multiply both sides by -1:( dfrac{1}{2}x = dfrac{1}{x + 1} )Now, cross-multiplying:( dfrac{1}{2}x(x + 1) = 1 )Expanding the left side:( dfrac{1}{2}x^2 + dfrac{1}{2}x = 1 )Multiply both sides by 2 to eliminate the fraction:( x^2 + x = 2 )Bring all terms to one side:( x^2 + x - 2 = 0 )Now, let's factor this quadratic equation:Looking for two numbers that multiply to -2 and add to 1. Hmm, 2 and -1.So,( (x + 2)(x - 1) = 0 )Thus, the critical points are ( x = -2 ) and ( x = 1 ).But wait, the original function ( f(x) = -dfrac{1}{4}x^2 + ln(x + 1) ) is only defined for ( x + 1 > 0 ), which means ( x > -1 ). So, ( x = -2 ) is not in the domain of the function. Therefore, the only critical point is ( x = 1 ).Now, to determine the intervals of monotonicity, I need to test the sign of ( f'(x) ) in the intervals determined by the critical point ( x = 1 ).So, the intervals are ( (-1, 1) ) and ( (1, infty) ).Let me pick test points in each interval.First interval: ( (-1, 1) ). Let's choose ( x = 0 ).Compute ( f'(0) = -dfrac{1}{2}(0) + dfrac{1}{0 + 1} = 0 + 1 = 1 ). Since this is positive, the function is increasing on ( (-1, 1) ).Second interval: ( (1, infty) ). Let's choose ( x = 2 ).Compute ( f'(2) = -dfrac{1}{2}(2) + dfrac{1}{2 + 1} = -1 + dfrac{1}{3} = -dfrac{2}{3} ). This is negative, so the function is decreasing on ( (1, infty) ).Therefore, the function ( f(x) ) is increasing on ( (-1, 1) ) and decreasing on ( (1, infty) ).Moving on to part (2): If the function ( f(x) ) is decreasing on the interval ( [1, +infty) ), find the range of the real number ( a ).Alright, so for the function to be decreasing on ( [1, +infty) ), its derivative must be less than or equal to zero for all ( x ) in that interval.So, let's compute the derivative of ( f(x) ) in general:( f(x) = ax^2 + ln(x + 1) )Thus,( f'(x) = 2ax + dfrac{1}{x + 1} )We need ( f'(x) leq 0 ) for all ( x geq 1 ).So,( 2ax + dfrac{1}{x + 1} leq 0 ) for all ( x geq 1 )Let me rearrange this inequality:( 2ax leq -dfrac{1}{x + 1} )Divide both sides by 2x (since ( x geq 1 ), 2x is positive, so the inequality sign remains the same):( a leq -dfrac{1}{2x(x + 1)} )So, ( a ) must be less than or equal to ( -dfrac{1}{2x(x + 1)} ) for all ( x geq 1 ).Therefore, to find the maximum possible value of ( a ), we need to find the minimum value of ( -dfrac{1}{2x(x + 1)} ) on ( x geq 1 ).Wait, actually, since ( a ) must be less than or equal to this expression for all ( x geq 1 ), the maximum value ( a ) can take is the infimum (greatest lower bound) of ( -dfrac{1}{2x(x + 1)} ) over ( x geq 1 ).But let's analyze the function ( h(x) = -dfrac{1}{2x(x + 1)} ) for ( x geq 1 ).First, note that ( h(x) ) is negative because of the negative sign. As ( x ) increases, the denominator ( 2x(x + 1) ) increases, so ( h(x) ) approaches 0 from below.At ( x = 1 ):( h(1) = -dfrac{1}{2(1)(2)} = -dfrac{1}{4} )As ( x ) increases beyond 1, ( h(x) ) becomes less negative, approaching 0.Therefore, the minimum value of ( h(x) ) on ( x geq 1 ) is at ( x = 1 ), which is ( -dfrac{1}{4} ).Hence, ( a ) must be less than or equal to ( -dfrac{1}{4} ).So, the range of ( a ) is ( (-infty, -dfrac{1}{4}] ).Wait, let me double-check. If ( a ) is less than or equal to ( -dfrac{1}{4} ), then for all ( x geq 1 ), ( 2ax + dfrac{1}{x + 1} leq 0 ).Let me test ( a = -dfrac{1}{4} ):( f'(x) = 2(-dfrac{1}{4})x + dfrac{1}{x + 1} = -dfrac{1}{2}x + dfrac{1}{x + 1} )At ( x = 1 ):( f'(1) = -dfrac{1}{2} + dfrac{1}{2} = 0 )For ( x > 1 ), say ( x = 2 ):( f'(2) = -1 + dfrac{1}{3} = -dfrac{2}{3} < 0 )So, it works at ( a = -dfrac{1}{4} ). If ( a ) is less than ( -dfrac{1}{4} ), then ( 2ax ) becomes more negative, so ( f'(x) ) will be even more negative, which is still less than zero.Therefore, the range of ( a ) is indeed ( (-infty, -dfrac{1}{4}] ).Now, part (3): When ( x in [0, +infty) ), the inequality ( f(x) - x leq 0 ) always holds. Find the range of the real number ( a ).So, we need ( f(x) - x leq 0 ) for all ( x geq 0 ).Substituting ( f(x) ):( ax^2 + ln(x + 1) - x leq 0 ) for all ( x geq 0 ).Let me define a new function ( g(x) = ax^2 + ln(x + 1) - x ). We need ( g(x) leq 0 ) for all ( x geq 0 ).To ensure that ( g(x) leq 0 ) for all ( x geq 0 ), we need to analyze the behavior of ( g(x) ).First, let's compute ( g(0) ):( g(0) = a(0)^2 + ln(1) - 0 = 0 + 0 - 0 = 0 ).So, ( g(0) = 0 ). We need to ensure that ( g(x) ) does not exceed 0 for any ( x geq 0 ).To analyze this, let's compute the derivative ( g'(x) ):( g'(x) = 2ax + dfrac{1}{x + 1} - 1 )Simplify:( g'(x) = 2ax + dfrac{1}{x + 1} - 1 )Let me combine the terms:( g'(x) = 2ax + dfrac{1 - (x + 1)}{x + 1} = 2ax + dfrac{-x}{x + 1} )So,( g'(x) = 2ax - dfrac{x}{x + 1} )Factor out ( x ):( g'(x) = x left( 2a - dfrac{1}{x + 1} right) )Hmm, that might be useful.Alternatively, let's write it as:( g'(x) = 2ax + dfrac{1}{x + 1} - 1 )I think another approach is to analyze the critical points of ( g(x) ). Since ( g(0) = 0 ), we need to ensure that ( g(x) ) does not increase beyond 0 for any ( x geq 0 ).So, let's set ( g'(x) = 0 ) and find critical points.( 2ax + dfrac{1}{x + 1} - 1 = 0 )Multiply both sides by ( x + 1 ) to eliminate the denominator:( 2ax(x + 1) + 1 - (x + 1) = 0 )Expand:( 2ax^2 + 2ax + 1 - x - 1 = 0 )Simplify:( 2ax^2 + (2a - 1)x = 0 )Factor:( x(2ax + (2a - 1)) = 0 )So, the critical points are ( x = 0 ) and ( 2ax + (2a - 1) = 0 ).Solving ( 2ax + (2a - 1) = 0 ):( 2ax = 1 - 2a )( x = dfrac{1 - 2a}{2a} )But this is only valid if ( a neq 0 ).So, depending on the value of ( a ), we might have another critical point.Let me analyze different cases for ( a ):Case 1: ( a = 0 )Then, ( g(x) = ln(x + 1) - x )Compute ( g'(x) = dfrac{1}{x + 1} - 1 = dfrac{1 - (x + 1)}{x + 1} = dfrac{-x}{x + 1} )So, ( g'(x) = -dfrac{x}{x + 1} ), which is negative for all ( x > 0 ). Therefore, ( g(x) ) is decreasing on ( [0, infty) ). Since ( g(0) = 0 ), ( g(x) leq 0 ) for all ( x geq 0 ). So, ( a = 0 ) is acceptable.Case 2: ( a > 0 )In this case, the critical point is ( x = dfrac{1 - 2a}{2a} ). Let's analyze this.First, let's see if this critical point is positive or negative.( x = dfrac{1 - 2a}{2a} )If ( 1 - 2a > 0 ), then ( a < dfrac{1}{2} ). So, for ( a < dfrac{1}{2} ), ( x ) is positive.If ( a = dfrac{1}{2} ), ( x = 0 ).If ( a > dfrac{1}{2} ), ( x ) becomes negative, which is not in our domain ( x geq 0 ).So, for ( a > 0 ):- If ( 0 < a < dfrac{1}{2} ), there is a critical point at ( x = dfrac{1 - 2a}{2a} > 0 ).- If ( a geq dfrac{1}{2} ), the only critical point is at ( x = 0 ).Let's analyze the behavior of ( g(x) ) in these subcases.Subcase 2.1: ( 0 < a < dfrac{1}{2} )Here, ( g(x) ) has a critical point at ( x = dfrac{1 - 2a}{2a} ). Let's denote this as ( x_c ).To determine if this critical point is a maximum or minimum, let's look at the second derivative or analyze the sign change of ( g'(x) ).Alternatively, since ( g'(x) = 2ax + dfrac{1}{x + 1} - 1 ), let's analyze the behavior around ( x_c ).But perhaps it's easier to consider the second derivative.Compute ( g''(x) ):( g''(x) = 2a - dfrac{1}{(x + 1)^2} )At ( x = x_c ):( g''(x_c) = 2a - dfrac{1}{(x_c + 1)^2} )But since ( x_c = dfrac{1 - 2a}{2a} ), let's compute ( x_c + 1 ):( x_c + 1 = dfrac{1 - 2a}{2a} + 1 = dfrac{1 - 2a + 2a}{2a} = dfrac{1}{2a} )Thus,( g''(x_c) = 2a - dfrac{1}{(1/(2a))^2} = 2a - (4a^2) )So,( g''(x_c) = 2a - 4a^2 = 2a(1 - 2a) )Since ( 0 < a < dfrac{1}{2} ), ( 1 - 2a > 0 ), so ( g''(x_c) > 0 ). Therefore, ( x_c ) is a local minimum.Wait, but we need ( g(x) leq 0 ) for all ( x geq 0 ). If there's a local minimum at ( x_c ), and ( g(x_c) leq 0 ), then the function might be decreasing before ( x_c ) and increasing after ( x_c ). But since ( g(0) = 0 ) and ( g(x) ) has a minimum at ( x_c ), we need to ensure that ( g(x_c) leq 0 ).Alternatively, perhaps the function first decreases to a minimum and then increases. But since ( g(0) = 0 ), if the minimum is below zero, then ( g(x) ) would dip below zero and then come back up. However, we need ( g(x) leq 0 ) for all ( x geq 0 ). So, if the minimum is below zero, then ( g(x) ) would violate the condition for some ( x > x_c ).Wait, actually, if ( g(x) ) has a minimum at ( x_c ), and ( g(x_c) leq 0 ), but since ( g(0) = 0 ), if ( g(x) ) dips below zero, it would have to come back up, potentially exceeding zero. But since ( g(x) ) is supposed to be always less than or equal to zero, we need to ensure that ( g(x) ) does not exceed zero anywhere.But given that ( g(0) = 0 ) and ( g(x) ) has a local minimum at ( x_c ), if ( g(x_c) leq 0 ), then ( g(x) ) would be decreasing from 0 to ( x_c ), reaching a minimum, and then increasing back towards zero or beyond. However, if ( g(x) ) increases beyond zero, that would violate the condition.Therefore, to ensure ( g(x) leq 0 ) for all ( x geq 0 ), we need the maximum of ( g(x) ) to be at most zero. Since ( g(0) = 0 ), and if ( g(x) ) has a local maximum somewhere else, that could be a problem.Wait, actually, in this case, for ( 0 < a < dfrac{1}{2} ), the critical point is a local minimum. So, ( g(x) ) decreases from 0 to ( x_c ), reaches a minimum, and then increases. But since ( g(0) = 0 ), if the minimum is below zero, then ( g(x) ) would have to increase back towards zero as ( x ) increases. However, as ( x ) approaches infinity, let's see the behavior of ( g(x) ).Compute the limit as ( x to infty ) of ( g(x) = ax^2 + ln(x + 1) - x ).The dominant term is ( ax^2 ) since it's quadratic, and ( ln(x + 1) ) grows much slower. So, if ( a > 0 ), ( ax^2 ) will dominate, and ( g(x) ) will tend to infinity. Therefore, ( g(x) ) will eventually exceed zero, which violates the condition ( g(x) leq 0 ).Therefore, for ( a > 0 ), regardless of the value, ( g(x) ) will tend to infinity as ( x to infty ), so ( g(x) leq 0 ) cannot hold for all ( x geq 0 ).Wait, but in the case when ( a = 0 ), ( g(x) = ln(x + 1) - x ), which tends to negative infinity as ( x to infty ), so it's okay. But for ( a > 0 ), it's problematic.Therefore, ( a ) cannot be positive.Case 3: ( a < 0 )Now, let's consider ( a < 0 ). Let's see what happens.First, the critical point ( x = dfrac{1 - 2a}{2a} ). Since ( a < 0 ), the denominator ( 2a ) is negative. The numerator ( 1 - 2a ) is positive because ( a < 0 ), so ( -2a > 0 ), hence ( 1 - 2a > 1 ). Therefore, ( x = dfrac{1 - 2a}{2a} ) is negative because numerator is positive and denominator is negative. Hence, the critical point is not in the domain ( x geq 0 ).Therefore, for ( a < 0 ), the only critical point is at ( x = 0 ).Wait, but let's verify that.From earlier, the critical points are ( x = 0 ) and ( x = dfrac{1 - 2a}{2a} ). For ( a < 0 ), ( dfrac{1 - 2a}{2a} ) is negative, so only ( x = 0 ) is in the domain.Therefore, for ( a < 0 ), the derivative ( g'(x) = 2ax + dfrac{1}{x + 1} - 1 ).Let me analyze the sign of ( g'(x) ) for ( a < 0 ).Since ( a < 0 ), ( 2ax ) is negative for ( x > 0 ). The term ( dfrac{1}{x + 1} - 1 ) is negative because ( dfrac{1}{x + 1} < 1 ) for ( x > 0 ).Therefore, both terms in ( g'(x) ) are negative, so ( g'(x) < 0 ) for all ( x > 0 ).Thus, ( g(x) ) is decreasing on ( [0, infty) ). Since ( g(0) = 0 ), and it's decreasing, ( g(x) leq 0 ) for all ( x geq 0 ).Therefore, for ( a < 0 ), the condition ( g(x) leq 0 ) holds.Putting it all together:- For ( a = 0 ), ( g(x) leq 0 ) holds.- For ( a > 0 ), ( g(x) ) eventually exceeds zero, so it doesn't hold.- For ( a < 0 ), ( g(x) ) is decreasing and stays below zero.Therefore, the range of ( a ) is ( (-infty, 0] ).Wait, let me just confirm with ( a = 0 ). As I saw earlier, ( g(x) = ln(x + 1) - x ). The derivative is ( -dfrac{x}{x + 1} ), which is negative for all ( x > 0 ). So, ( g(x) ) is decreasing, starting at 0 and going to negative infinity. So, yes, ( g(x) leq 0 ) for all ( x geq 0 ).Therefore, the range of ( a ) is all real numbers less than or equal to zero.**Final Answer**(1) The function is increasing on (boxed{(-1, 1)}) and decreasing on (boxed{(1, +infty)}).(2) The range of ( a ) is (boxed{(-infty, -dfrac{1}{4}]}).(3) The range of ( a ) is (boxed{(-infty, 0]}).

💡Okay, so I have this problem here about a flagpole and a building casting shadows. The flagpole has a shadow of 45 meters, and there's a building that's 26 meters tall casting a shadow of 65 meters. I need to find out how tall the flagpole is. Hmm, let's see. First, I remember that when two objects cast shadows under similar conditions, like the same time of day and same lighting, their heights and shadow lengths are proportional. That means the ratio of the height of one object to its shadow length should be the same as the ratio for the other object. So, I think I can set up a proportion here.Let me denote the height of the flagpole as h. So, the ratio for the flagpole would be h over 45 meters. For the building, the ratio is 26 meters over 65 meters. Since these ratios should be equal, I can write the equation:h / 45 = 26 / 65Now, I need to solve for h. To do that, I can cross-multiply. So, h times 65 equals 45 times 26. Let me calculate that:65h = 45 * 26First, I'll compute 45 multiplied by 26. Let's see, 45 times 20 is 900, and 45 times 6 is 270. Adding those together, 900 + 270 equals 1170. So, now I have:65h = 1170To find h, I need to divide both sides of the equation by 65:h = 1170 / 65Now, let's do that division. 65 goes into 1170 how many times? Well, 65 times 18 is 1170 because 65 times 10 is 650, and 65 times 8 is 520. Adding those together, 650 + 520 equals 1170. So, h equals 18.Wait, that seems straightforward, but let me double-check to make sure I didn't make any mistakes. If the flagpole is 18 meters tall and casts a 45-meter shadow, the ratio is 18/45, which simplifies to 2/5. For the building, 26 meters tall with a 65-meter shadow, the ratio is 26/65, which also simplifies to 2/5. Okay, so both ratios are the same, which makes sense because they are under similar conditions. That checks out.I think I've got it right. The flagpole must be 18 meters tall.

💡Okay, so I have this problem with two sequences, ( a_n ) and ( b_n ). The recursive formulas are given as:( a_{n+1} = a_n + b_n + sqrt{a_n^2 + b_n^2} )( b_{n+1} = a_n + b_n - sqrt{a_n^2 + b_n^2} )And the initial terms are ( a_1 = 1 ) and ( b_1 = 1 ). I need to find the sum of the first 2017 terms of another sequence ( c_n ), where ( c_n = frac{1}{a_n} + frac{1}{b_n} ).Hmm, okay. Let me try to break this down step by step.First, maybe I should compute the first few terms of ( a_n ) and ( b_n ) to see if I can spot a pattern or find a relationship between them.Starting with ( n = 1 ):( a_1 = 1 )( b_1 = 1 )So, let's compute ( a_2 ) and ( b_2 ):( a_2 = a_1 + b_1 + sqrt{a_1^2 + b_1^2} = 1 + 1 + sqrt{1 + 1} = 2 + sqrt{2} )( b_2 = a_1 + b_1 - sqrt{a_1^2 + b_1^2} = 2 - sqrt{2} )Okay, so ( a_2 = 2 + sqrt{2} ) and ( b_2 = 2 - sqrt{2} ).Now, let's compute ( c_1 ):( c_1 = frac{1}{a_1} + frac{1}{b_1} = frac{1}{1} + frac{1}{1} = 2 )Interesting, ( c_1 = 2 ). Let's compute ( c_2 ):( c_2 = frac{1}{a_2} + frac{1}{b_2} = frac{1}{2 + sqrt{2}} + frac{1}{2 - sqrt{2}} )To simplify this, I can rationalize the denominators:For ( frac{1}{2 + sqrt{2}} ), multiply numerator and denominator by ( 2 - sqrt{2} ):( frac{1 times (2 - sqrt{2})}{(2 + sqrt{2})(2 - sqrt{2})} = frac{2 - sqrt{2}}{4 - 2} = frac{2 - sqrt{2}}{2} = 1 - frac{sqrt{2}}{2} )Similarly, for ( frac{1}{2 - sqrt{2}} ), multiply numerator and denominator by ( 2 + sqrt{2} ):( frac{1 times (2 + sqrt{2})}{(2 - sqrt{2})(2 + sqrt{2})} = frac{2 + sqrt{2}}{4 - 2} = frac{2 + sqrt{2}}{2} = 1 + frac{sqrt{2}}{2} )Adding these together:( (1 - frac{sqrt{2}}{2}) + (1 + frac{sqrt{2}}{2}) = 2 )So, ( c_2 = 2 ). Hmm, same as ( c_1 ). Maybe this is a pattern?Let me check ( c_3 ) to see if it's also 2.First, compute ( a_3 ) and ( b_3 ):( a_3 = a_2 + b_2 + sqrt{a_2^2 + b_2^2} )But wait, ( a_2 = 2 + sqrt{2} ) and ( b_2 = 2 - sqrt{2} ). Let's compute ( a_2 + b_2 ):( a_2 + b_2 = (2 + sqrt{2}) + (2 - sqrt{2}) = 4 )Now, compute ( sqrt{a_2^2 + b_2^2} ):First, compute ( a_2^2 ):( (2 + sqrt{2})^2 = 4 + 4sqrt{2} + 2 = 6 + 4sqrt{2} )Similarly, ( b_2^2 = (2 - sqrt{2})^2 = 4 - 4sqrt{2} + 2 = 6 - 4sqrt{2} )Adding them together:( a_2^2 + b_2^2 = (6 + 4sqrt{2}) + (6 - 4sqrt{2}) = 12 )So, ( sqrt{a_2^2 + b_2^2} = sqrt{12} = 2sqrt{3} )Therefore,( a_3 = a_2 + b_2 + sqrt{a_2^2 + b_2^2} = 4 + 2sqrt{3} )( b_3 = a_2 + b_2 - sqrt{a_2^2 + b_2^2} = 4 - 2sqrt{3} )Now, compute ( c_3 = frac{1}{a_3} + frac{1}{b_3} ):( c_3 = frac{1}{4 + 2sqrt{3}} + frac{1}{4 - 2sqrt{3}} )Again, rationalize the denominators:For ( frac{1}{4 + 2sqrt{3}} ), multiply numerator and denominator by ( 4 - 2sqrt{3} ):( frac{1 times (4 - 2sqrt{3})}{(4 + 2sqrt{3})(4 - 2sqrt{3})} = frac{4 - 2sqrt{3}}{16 - 12} = frac{4 - 2sqrt{3}}{4} = 1 - frac{sqrt{3}}{2} )Similarly, for ( frac{1}{4 - 2sqrt{3}} ), multiply numerator and denominator by ( 4 + 2sqrt{3} ):( frac{1 times (4 + 2sqrt{3})}{(4 - 2sqrt{3})(4 + 2sqrt{3})} = frac{4 + 2sqrt{3}}{16 - 12} = frac{4 + 2sqrt{3}}{4} = 1 + frac{sqrt{3}}{2} )Adding these together:( (1 - frac{sqrt{3}}{2}) + (1 + frac{sqrt{3}}{2}) = 2 )So, ( c_3 = 2 ) as well. Hmm, so it seems that ( c_n = 2 ) for ( n = 1, 2, 3 ). Maybe this is a constant sequence? That would make the sum of the first 2017 terms just ( 2017 times 2 = 4034 ). But I should verify this more generally, not just for the first few terms.Let me try to find a general expression for ( c_n ). Given that ( c_n = frac{1}{a_n} + frac{1}{b_n} ), which can be written as ( frac{a_n + b_n}{a_n b_n} ). So, if I can find expressions for ( a_n + b_n ) and ( a_n b_n ), I can find ( c_n ).Looking back at the recursive formulas:( a_{n+1} = a_n + b_n + sqrt{a_n^2 + b_n^2} )( b_{n+1} = a_n + b_n - sqrt{a_n^2 + b_n^2} )Let me compute ( a_{n+1} + b_{n+1} ):( a_{n+1} + b_{n+1} = [a_n + b_n + sqrt{a_n^2 + b_n^2}] + [a_n + b_n - sqrt{a_n^2 + b_n^2}] = 2(a_n + b_n) )So, ( a_{n+1} + b_{n+1} = 2(a_n + b_n) ). This suggests that ( a_n + b_n ) is a geometric sequence with ratio 2.Given that ( a_1 + b_1 = 1 + 1 = 2 ), so ( a_n + b_n = 2 times 2^{n-1} = 2^n ).Okay, so ( a_n + b_n = 2^n ).Now, let's compute ( a_{n+1} b_{n+1} ):( a_{n+1} b_{n+1} = [a_n + b_n + sqrt{a_n^2 + b_n^2}][a_n + b_n - sqrt{a_n^2 + b_n^2}] )This is of the form ( (x + y)(x - y) = x^2 - y^2 ), so:( a_{n+1} b_{n+1} = (a_n + b_n)^2 - (a_n^2 + b_n^2) )Simplify this:( (a_n + b_n)^2 = a_n^2 + 2a_n b_n + b_n^2 )So,( a_{n+1} b_{n+1} = (a_n^2 + 2a_n b_n + b_n^2) - (a_n^2 + b_n^2) = 2a_n b_n )Therefore, ( a_{n+1} b_{n+1} = 2a_n b_n ). This implies that ( a_n b_n ) is also a geometric sequence with ratio 2.Given that ( a_1 b_1 = 1 times 1 = 1 ), so ( a_n b_n = 1 times 2^{n-1} = 2^{n-1} ).Now, going back to ( c_n ):( c_n = frac{1}{a_n} + frac{1}{b_n} = frac{a_n + b_n}{a_n b_n} = frac{2^n}{2^{n-1}} = 2 )So, ( c_n = 2 ) for all ( n ). Therefore, each term in the sequence ( c_n ) is 2, regardless of ( n ).Hence, the sum of the first 2017 terms is simply ( 2017 times 2 = 4034 ).Wait, let me just make sure I didn't make any mistakes in my reasoning. I started by computing the first few terms and noticed that ( c_n ) was always 2. Then, I generalized it by finding expressions for ( a_n + b_n ) and ( a_n b_n ), which led me to conclude that ( c_n = 2 ) for all ( n ). That seems solid. I don't see any errors in my calculations or logic.So, yeah, the sum should be 4034.**Final Answer**The sum of the first 2017 terms of the sequence ( c_n ) is boxed{4034}.

💡Okay, so I have this problem about two cylinders, A and B. Let me try to understand it step by step. First, the problem says that the height of cylinder B is equal to the radius of cylinder A. Hmm, that's interesting. So if I denote the radius of cylinder A as, let's say, r, then the height of cylinder B would also be r. Got that.Next, it says that the radius of cylinder B is equal to the height h of cylinder A. So if the height of cylinder A is h, then the radius of cylinder B is also h. That makes sense. So cylinder B has a radius of h and a height of r, which is the radius of cylinder A.Now, the volume of cylinder A is three times the volume of cylinder B. I need to find the volume of cylinder A in terms of h and express it as Nπh³, and then find the value of N.Alright, let's recall the formula for the volume of a cylinder. It's π times radius squared times height, right? So for cylinder A, the volume would be πr²h. For cylinder B, since its radius is h and its height is r, the volume would be πh²r.According to the problem, the volume of cylinder A is three times the volume of cylinder B. So I can write that as:πr²h = 3 * (πh²r)Hmm, let me simplify this equation. First, I can cancel out the π from both sides, which gives me:r²h = 3h²rNow, I can divide both sides by h (assuming h ≠ 0, which makes sense because we're dealing with volumes). That gives:r² = 3h rThen, I can divide both sides by r (again, assuming r ≠ 0):r = 3hSo, the radius of cylinder A is three times the height of cylinder A. That's a key relationship.Now, I need to find the volume of cylinder A in terms of h. Since the volume of cylinder A is πr²h, and we know that r = 3h, I can substitute that into the formula:Volume of A = π(3h)²hLet me compute that step by step. First, (3h) squared is 9h². Then, multiplying by h gives 9h³. So:Volume of A = π * 9h² * h = 9πh³Therefore, the volume of cylinder A is 9πh³, which means that N is 9.Wait, let me double-check my steps to make sure I didn't make a mistake. Starting from the volume relationship:πr²h = 3πh²rCancel π:r²h = 3h²rDivide both sides by h:r² = 3hrDivide both sides by r:r = 3hYes, that seems correct. Then substituting back into the volume formula:π(3h)²h = π*9h²*h = 9πh³Yep, that looks right. So N must be 9.

💡Okay, so I have this problem about a pyramid with a rectangular base that's 4 units by 10 units. All four edges from the apex to the corners of the base are 15 units each. I need to find the volume of this pyramid. Hmm, let's see.First, I remember that the volume of a pyramid is given by one-third the area of the base times the height. So, I need to find both the area of the base and the height of the pyramid.The base is a rectangle, so its area should be straightforward. Length times width, which is 4 times 10. Let me calculate that: 4 * 10 = 40. So, the area of the base is 40 square units.Now, the tricky part is figuring out the height of the pyramid. The problem says that all four edges from the apex to the corners of the base are equal, each being 15 units. That means the apex is equidistant from all four corners of the base. I think this implies that the apex is directly above the center of the base. So, if I can find the distance from the center of the base to one of the corners, and then use the Pythagorean theorem with the edge length, I can find the height.Let me visualize the base. It's a rectangle with sides 4 and 10. The center of the rectangle would be the midpoint of both the length and the width. So, the coordinates of the center, if I imagine the rectangle on a coordinate plane, would be at (2, 5), assuming one corner is at (0,0).Now, the distance from the center to any corner is half the length of the diagonal of the rectangle. The diagonal of a rectangle can be found using the Pythagorean theorem: sqrt(length^2 + width^2). So, the diagonal here would be sqrt(4^2 + 10^2) = sqrt(16 + 100) = sqrt(116). Simplifying sqrt(116), since 116 is 4*29, so sqrt(4*29) = 2*sqrt(29). Therefore, half of that diagonal is sqrt(29).So, the distance from the center of the base to any corner is sqrt(29). Now, considering the apex is directly above the center, the line from the apex to the center is the height of the pyramid, and the line from the apex to a corner is 15 units. So, we have a right triangle where one leg is the height (h), another leg is sqrt(29), and the hypotenuse is 15.Using the Pythagorean theorem: h^2 + (sqrt(29))^2 = 15^2. Let's compute that:h^2 + 29 = 225Subtract 29 from both sides:h^2 = 225 - 29 = 196Taking the square root of both sides:h = sqrt(196) = 14So, the height of the pyramid is 14 units.Now, going back to the volume formula: Volume = (1/3) * base area * height. We already found the base area is 40 and the height is 14.Plugging in the numbers:Volume = (1/3) * 40 * 14First, multiply 40 and 14: 40 * 14 = 560Then, divide by 3: 560 / 3 ≈ 186.666...So, the volume is 560/3 cubic units, which is approximately 186.67 cubic units.Wait, let me double-check my steps to make sure I didn't make a mistake. The base area is definitely 4*10=40. The diagonal of the base is sqrt(4^2 +10^2)=sqrt(116)=2*sqrt(29). Half of that is sqrt(29), which is correct. Then, using the Pythagorean theorem with 15 as the hypotenuse: h^2 + (sqrt(29))^2 = 15^2. That gives h^2 +29=225, so h^2=196, h=14. That seems right.Then, volume is (1/3)*40*14=560/3≈186.67. Yeah, that all checks out.I think I've got it. The volume is 560/3 cubic units, which is approximately 186.67 cubic units.