💡Alright, so I've got this problem about a food chain in a tropical forest, and I need to figure out how many beetles are consumed in one day. Let me try to break this down step by step.First, let's list out all the information given:1. Each bird eats 12 beetles per day.2. Each snake eats 3 birds per day.3. Each jaguar eats 5 snakes per day.4. Each crocodile eats 2 jaguars per day.5. There are net increases in population: 4 birds, 2 snakes, and 1 jaguar per day.6. Initially, there are 6 jaguars and 30 crocodiles.Okay, so the goal is to find out how many beetles are consumed in one day. That means I need to figure out how many birds are eating beetles, and then multiply that by 12 (since each bird eats 12 beetles per day).But to find out how many birds are eating beetles, I need to know how many snakes are eating birds. Similarly, to find out how many snakes are eating birds, I need to know how many jaguars are eating snakes, and so on.Let me start from the top of the food chain and work my way down.**Starting with Jaguars:**- There are initially 6 jaguars.- Each jaguar eats 5 snakes per day.- So, the total number of snakes eaten by jaguars in one day is 6 jaguars * 5 snakes/jaguar = 30 snakes.But wait, there's also a net increase in the number of jaguars. The problem says there's a net increase of 1 jaguar per day. So, at the end of the day, there will be 6 + 1 = 7 jaguars. However, for calculating the number of snakes eaten in that day, I think we should consider the initial number of jaguars, which is 6. So, 30 snakes are eaten by jaguars in one day.**Moving on to Snakes:**- We've established that 30 snakes are eaten by jaguars.- Each snake eats 3 birds per day.- So, the total number of birds eaten by snakes in one day is 30 snakes * 3 birds/snake = 90 birds.Again, there's a net increase in the number of snakes: 2 snakes per day. So, at the end of the day, there will be more snakes, but for calculating the number of birds eaten in that day, I think we should consider the initial number of snakes. Wait, but we don't have the initial number of snakes. Hmm, this is a bit confusing.Let me think. The problem gives us the initial number of jaguars and crocodiles, but not snakes or birds. It only mentions net increases: 4 birds, 2 snakes, and 1 jaguar per day. So, maybe I don't need the initial number of snakes because the net increase is given. But I'm not sure how to incorporate that into the calculation.Perhaps I should assume that the number of snakes is sufficient to support the jaguars' consumption, considering the net increase. But I'm not sure. Maybe I need to find the number of snakes based on the jaguars and the net increase.Wait, if there are 6 jaguars eating 30 snakes per day, and there's a net increase of 2 snakes per day, does that mean the total number of snakes is increasing by 2 each day? But I don't know the initial number of snakes. Maybe I can represent it with a variable.Let me denote the initial number of snakes as S. Then, the number of snakes eaten by jaguars is 30, and the net increase is 2, so the total number of snakes at the end of the day would be S - 30 + 2 = S - 28.But I don't know S, so maybe I need to find S in terms of other variables. Alternatively, perhaps I can proceed without knowing S because the net increase is given, and I'm only asked for the number of beetles consumed in one day, not the population changes.Okay, maybe I can proceed by assuming that the number of snakes is sufficient to support the jaguars' consumption, and the net increase is just additional information that might not directly affect the calculation for that day's consumption.So, if 30 snakes are eaten by jaguars, and each snake eats 3 birds, then 30 snakes would eat 90 birds in a day.**Now, moving on to Birds:**- 90 birds are eaten by snakes.- Each bird eats 12 beetles per day.- So, the total number of beetles eaten by birds in one day is 90 birds * 12 beetles/bird = 1080 beetles.There's also a net increase of 4 birds per day, but similar to before, I'm not sure how to incorporate that. Maybe it's just additional information that doesn't affect the calculation for that day's consumption.**Finally, Crocodiles:**- There are 30 crocodiles.- Each crocodile eats 2 jaguars per day.- So, the total number of jaguars eaten by crocodiles in one day is 30 crocodiles * 2 jaguars/crocodile = 60 jaguars.But wait, there are only 6 jaguars initially. So, can 30 crocodiles each eat 2 jaguars when there are only 6 jaguars available? That doesn't seem possible. It would mean that all 6 jaguars are eaten by the crocodiles, but 30 crocodiles * 2 jaguars/crocodile = 60 jaguars needed, but only 6 are available.This seems like a problem. Maybe I need to adjust the number of crocodiles that can actually eat jaguars based on the available jaguars.If there are 6 jaguars, and each crocodile eats 2 jaguars, then the number of crocodiles that can eat jaguars is 6 jaguars / 2 jaguars per crocodile = 3 crocodiles.So, only 3 crocodiles can eat jaguars, and the remaining 27 crocodiles don't get to eat jaguars that day.But since the problem is about the number of beetles consumed, and crocodiles don't eat beetles, maybe this part doesn't affect the calculation. It's just additional information about the food chain.So, putting it all together:- 6 jaguars eat 30 snakes.- 30 snakes eat 90 birds.- 90 birds eat 1080 beetles.Therefore, the total number of beetles consumed in one day is 1080.Wait, but I need to make sure I'm not missing anything. The net increases in population are given, but I'm not sure how they factor in. Maybe they affect the populations for the next day, but since we're only calculating for one day, perhaps they don't impact the current day's consumption.Also, regarding the crocodiles, even though they can't all eat jaguars, since they don't eat beetles, it doesn't affect the beetle consumption.So, I think my calculation is correct. The number of beetles consumed in one day is 1080.

💡First, I need to evaluate the inner function ( fleft(frac{1}{4}right) ). Since ( frac{1}{4} ) is greater than 0, I use the first part of the piecewise function, which is ( f(x) = log_2 x ).Calculating ( fleft(frac{1}{4}right) ):[fleft(frac{1}{4}right) = log_2 left(frac{1}{4}right)]I know that ( frac{1}{4} ) can be expressed as ( 2^{-2} ), so:[log_2 left(2^{-2}right) = -2]Thus, ( fleft(frac{1}{4}right) = -2 ).Next, I need to evaluate ( f(-2) ) using the result from the first step. Since ( -2 ) is less than or equal to 0, I use the second part of the piecewise function, which is ( f(x) = 3^x ).Calculating ( f(-2) ):[f(-2) = 3^{-2}]I know that ( 3^{-2} ) is the reciprocal of ( 3^2 ), so:[3^{-2} = frac{1}{3^2} = frac{1}{9}]Therefore, ( fleft[fleft(frac{1}{4}right)right] = frac{1}{9} ).

💡Okay, so I need to find the equation of the tangent line to the curve y = 2 - ln x at the point (1, 2). Hmm, let's see. I remember that to find a tangent line, I need two things: the slope of the tangent at that specific point and a point through which the tangent passes. Well, the point is already given as (1, 2), so that's good. Now, I just need to find the slope.I think the slope of the tangent line at a particular point on a curve is given by the derivative of the function at that point. So, I need to find the derivative of y with respect to x for the function y = 2 - ln x. Let me recall how to differentiate logarithmic functions. The derivative of ln x is 1/x, right? So, if I have y = 2 - ln x, the derivative dy/dx should be the derivative of 2 minus the derivative of ln x.The derivative of a constant like 2 is zero, so that part is easy. Then, the derivative of ln x is 1/x, so the derivative of -ln x should be -1/x. Putting it all together, dy/dx = 0 - 1/x, which simplifies to dy/dx = -1/x. Okay, so that's the derivative.Now, I need to find the slope of the tangent line at the point (1, 2). That means I need to evaluate the derivative at x = 1. Let's plug in x = 1 into the derivative: dy/dx = -1/1 = -1. So, the slope of the tangent line at that point is -1.Alright, now that I have the slope and a point, I can use the point-slope form of a line to write the equation of the tangent line. The point-slope form is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point. Substituting the values I have, m = -1, x1 = 1, and y1 = 2, the equation becomes y - 2 = -1(x - 1).Let me simplify this equation. Distribute the -1 on the right side: y - 2 = -x + 1. Then, add 2 to both sides to solve for y: y = -x + 1 + 2, which simplifies to y = -x + 3.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the derivative: dy/dx = -1/x. At x = 1, dy/dx = -1. So, the slope is indeed -1. Then, using the point-slope form: y - 2 = -1(x - 1). Expanding gives y - 2 = -x + 1. Adding 2 to both sides: y = -x + 3. Yeah, that seems correct.Just to be thorough, maybe I should verify if the point (1, 2) lies on the tangent line y = -x + 3. Plugging in x = 1: y = -1 + 3 = 2. Yep, that checks out. So, the tangent line passes through the point (1, 2) and has the correct slope of -1. I think I'm confident that y = -x + 3 is the correct equation of the tangent line.I wonder if there's another way to approach this problem, maybe by using limits or something else. Well, the derivative is essentially the limit of the difference quotient as h approaches zero, which gives the instantaneous rate of change at a point. But since I already used the derivative method and got a consistent result, I think that's sufficient.Another thought: maybe graphing the function and the tangent line would help visualize if it makes sense. The original function y = 2 - ln x is a logarithmic curve that decreases as x increases because of the negative sign in front of the ln x. At x = 1, ln 1 is 0, so y = 2, which matches the given point. The tangent line at that point has a negative slope, which means it's decreasing. So, a line with a slope of -1 passing through (1, 2) should indeed be a good approximation of the curve at that point.I think I've covered all the bases here. I found the derivative, calculated the slope at the given point, used the point-slope form to find the equation of the tangent line, and even did a quick check to ensure everything makes sense. I don't see any errors in my reasoning, so I'm pretty sure the equation of the tangent line is y = -x + 3.

💡First, I need to calculate 38% of 80. To do this, I'll convert the percentage to a decimal by dividing 38 by 100, which gives me 0.38. Then, I'll multiply 0.38 by 80 to find the value.Next, I'll calculate 12% of 160. Similarly, I'll convert 12% to a decimal by dividing 12 by 100, resulting in 0.12. I'll then multiply 0.12 by 160 to obtain the second value.After obtaining both values, I'll find the difference between them by subtracting the smaller value from the larger one. This will give me the final result.

💡Okay, so I need to find the equation of the tangent line to the curve ( y = x^3 - 3x^2 + a ) at the point ( P(1, -1) ). Hmm, let me think about how to approach this step by step.First, I remember that to find the equation of a tangent line to a curve at a specific point, I need two things: the slope of the tangent line at that point and a point through which the tangent line passes. In this case, the point is given as ( P(1, -1) ), so that's already taken care of. The next step is to find the slope of the tangent line at that point.To find the slope, I need to compute the derivative of the function ( y = x^3 - 3x^2 + a ). The derivative will give me the slope of the tangent line at any point ( x ) on the curve. Let me calculate that.The derivative of ( y ) with respect to ( x ) is:[frac{dy}{dx} = 3x^2 - 6x]So, ( f'(x) = 3x^2 - 6x ).Now, I need to evaluate this derivative at ( x = 1 ) to find the slope of the tangent line at the point ( P(1, -1) ). Let me plug in ( x = 1 ) into the derivative:[f'(1) = 3(1)^2 - 6(1) = 3 - 6 = -3]So, the slope ( m ) of the tangent line at ( P(1, -1) ) is ( -3 ).Now that I have the slope and a point on the tangent line, I can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is:[y - y_1 = m(x - x_1)]Where ( (x_1, y_1) ) is the point and ( m ) is the slope. Plugging in the values I have:[y - (-1) = -3(x - 1)]Simplifying this:[y + 1 = -3x + 3]Subtracting 1 from both sides to solve for ( y ):[y = -3x + 2]So, the equation of the tangent line is ( y = -3x + 2 ).Wait a minute, let me double-check to make sure I didn't make any mistakes. First, I found the derivative correctly, right? The derivative of ( x^3 ) is ( 3x^2 ), and the derivative of ( -3x^2 ) is ( -6x ). The derivative of a constant ( a ) is zero, so that's correct. Then, plugging in ( x = 1 ) gives ( 3(1) - 6(1) = -3 ), which seems right.Using the point-slope form, I substituted ( x_1 = 1 ) and ( y_1 = -1 ), which is correct. Then, simplifying the equation step by step, I ended up with ( y = -3x + 2 ). That seems consistent.Let me also check if the point ( P(1, -1) ) lies on this tangent line. Plugging ( x = 1 ) into the equation:[y = -3(1) + 2 = -3 + 2 = -1]Yes, that matches the ( y )-coordinate of point ( P ), so that checks out.I think I did everything correctly. The slope was calculated properly, the point was used correctly in the point-slope formula, and the resulting equation satisfies the point ( P(1, -1) ). Therefore, I'm confident that the equation of the tangent line is ( y = -3x + 2 ).Looking back at the options provided:A: ( y = 3x - 4 )B: ( y = -4x - 1 )C: ( y = -3x + 2 )D: ( y = 4x - 7 )The equation I found, ( y = -3x + 2 ), corresponds to option C. So, the correct answer should be C.Just to be thorough, let me quickly check the other options to ensure none of them could also pass through ( P(1, -1) ) with the correct slope.Option A: ( y = 3x - 4 ). Plugging in ( x = 1 ), ( y = 3(1) - 4 = -1 ). So, it does pass through ( P(1, -1) ). But the slope here is 3, whereas we calculated the slope to be -3. So, this can't be the correct tangent line.Option B: ( y = -4x - 1 ). Plugging in ( x = 1 ), ( y = -4(1) - 1 = -5 ). That doesn't match ( y = -1 ), so it doesn't pass through ( P(1, -1) ).Option D: ( y = 4x - 7 ). Plugging in ( x = 1 ), ( y = 4(1) - 7 = -3 ). Again, doesn't match ( y = -1 ), so it's not the correct tangent line.Therefore, option C is indeed the correct answer.I think that's all. I've double-checked my work, and everything seems consistent. I didn't make any calculation errors that I can see, and the logic follows correctly. So, I'm pretty sure option C is the right choice.**Final Answer**boxed{C}

💡Okay, so I have this sequence problem where x₁ is 1/2, and each subsequent term is defined as 1 minus the product of all previous terms. I need to prove that x₁₀₀ is between 0.99 and 0.991. Hmm, that seems pretty specific. Let me try to break this down.First, let me write out the first few terms to see if I can spot a pattern or get an intuition about how the sequence behaves.Given:x₁ = 1/2 = 0.5Then,x₂ = 1 - x₁ = 1 - 0.5 = 0.5Wait, that's interesting. So x₂ is also 0.5.Next,x₃ = 1 - x₁x₂ = 1 - (0.5)(0.5) = 1 - 0.25 = 0.75Okay, x₃ is 0.75.Then,x₄ = 1 - x₁x₂x₃ = 1 - (0.5)(0.5)(0.75) = 1 - (0.1875) = 0.8125Hmm, x₄ is 0.8125.Continuing,x₅ = 1 - x₁x₂x₃x₄ = 1 - (0.5)(0.5)(0.75)(0.8125)Let me calculate that product:0.5 * 0.5 = 0.250.25 * 0.75 = 0.18750.1875 * 0.8125 = Let's see, 0.1875 * 0.8 = 0.15, and 0.1875 * 0.0125 = 0.00234375, so total is approximately 0.15234375Therefore, x₅ ≈ 1 - 0.15234375 ≈ 0.84765625So x₅ is approximately 0.8477.Hmm, so the sequence seems to be increasing: 0.5, 0.5, 0.75, 0.8125, 0.8477...Wait, is it always increasing? Let me check x₆.x₆ = 1 - x₁x₂x₃x₄x₅Calculating the product:0.5 * 0.5 = 0.250.25 * 0.75 = 0.18750.1875 * 0.8125 ≈ 0.152343750.15234375 * 0.8477 ≈ Let's compute that.0.15234375 * 0.8 = 0.1218750.15234375 * 0.0477 ≈ 0.007265625Adding them together: ≈ 0.121875 + 0.007265625 ≈ 0.129140625So x₆ ≈ 1 - 0.129140625 ≈ 0.870859375So x₆ is approximately 0.8709.Okay, so it's still increasing. Let me do one more.x₇ = 1 - x₁x₂x₃x₄x₅x₆Product so far: 0.129140625 * 0.8709 ≈0.129140625 * 0.8 = 0.10331250.129140625 * 0.0709 ≈ 0.009150625Adding: ≈ 0.1033125 + 0.009150625 ≈ 0.112463125Thus, x₇ ≈ 1 - 0.112463125 ≈ 0.887536875So x₇ is approximately 0.8875.Hmm, so it's increasing, but the rate of increase is slowing down. It's approaching 1, but how quickly?Wait, the problem is about x₁₀₀, which is way further along. So I need to find a pattern or a formula that can help me estimate x₁₀₀ without computing all 100 terms.Let me think about the recursion:x_{n+1} = 1 - (x₁x₂...x_n)Let me denote P_n = x₁x₂...x_n. Then, the recursion becomes:x_{n+1} = 1 - P_nBut also, P_{n+1} = P_n * x_{n+1} = P_n * (1 - P_n)So, P_{n+1} = P_n (1 - P_n)That's interesting. So P_{n} follows the recursion P_{n+1} = P_n (1 - P_n)This is a quadratic recurrence relation. It reminds me of the logistic map, which is a well-known example in chaos theory. The logistic map is x_{n+1} = r x_n (1 - x_n). In our case, r is 1, so it's x_{n+1} = x_n (1 - x_n). So, similar to the logistic map with r=1.Wait, but in our case, P_{n+1} = P_n (1 - P_n). So, it's a specific case of the logistic map with r=1.I remember that for the logistic map, when r=1, the behavior is relatively simple. Let me recall: for r=1, the logistic map converges to 0, right? Because starting from some x₀ between 0 and 1, x_{n+1} = x_n (1 - x_n). If x_n is less than 1, then x_{n+1} is less than x_n, so it's decreasing. Wait, but in our case, P_n is the product of terms less than 1, so P_n is decreasing? Wait, no, let me think.Wait, P_{n+1} = P_n (1 - P_n). So, if P_n is less than 1, then 1 - P_n is positive, so P_{n+1} is positive. But is P_{n+1} greater or less than P_n?Let me compute P_{n+1} - P_n = P_n (1 - P_n) - P_n = P_n - P_n² - P_n = -P_n² < 0So, P_{n+1} < P_n. So, the sequence P_n is strictly decreasing.Wait, but in our earlier terms, x_n was increasing, but P_n is decreasing because each term is multiplied by a number less than 1.Wait, let me check with the earlier terms:P₁ = x₁ = 0.5P₂ = x₁x₂ = 0.5 * 0.5 = 0.25P₃ = 0.25 * 0.75 = 0.1875P₄ = 0.1875 * 0.8125 ≈ 0.15234375P₅ ≈ 0.15234375 * 0.8477 ≈ 0.129140625P₆ ≈ 0.129140625 * 0.8709 ≈ 0.112463125P₇ ≈ 0.112463125 * 0.8875 ≈ 0.099609375So, yes, P_n is decreasing each time.So, P_n is a decreasing sequence bounded below by 0, so it must converge to some limit L.If P_n converges to L, then P_{n+1} = P_n (1 - P_n) must also converge to L. So,L = L (1 - L)Solving for L:L = L - L²Subtract L from both sides:0 = -L²Thus, L = 0.So, P_n converges to 0 as n approaches infinity.Therefore, x_{n+1} = 1 - P_n approaches 1 as n increases.So, x_n approaches 1 as n becomes large. Therefore, x₁₀₀ should be very close to 1, but we need to show it's between 0.99 and 0.991.So, we need to estimate how quickly P_n approaches 0, or equivalently, how quickly x_n approaches 1.Given that P_{n+1} = P_n (1 - P_n), and P_n is small for large n, we can approximate the behavior for small P_n.When P_n is small, 1 - P_n ≈ 1, so P_{n+1} ≈ P_n.But that's not helpful. Wait, maybe we can approximate the difference equation as a differential equation.Let me think: For large n, the change in P_n is small, so we can approximate dP/dn ≈ P_{n+1} - P_n = -P_n².So, dP/dn ≈ -P²This is a differential equation: dP/dn = -P²We can solve this differential equation.Separating variables:dP / P² = -dnIntegrating both sides:∫ dP / P² = -∫ dn-1/P = -n + CSo, 1/P = n + CTherefore, P ≈ 1 / (n + C)So, for large n, P_n is approximately 1/n.Therefore, x_{n+1} = 1 - P_n ≈ 1 - 1/nSo, x_n ≈ 1 - 1/(n-1)Wait, but for n=100, x₁₀₀ ≈ 1 - 1/99 ≈ 0.989898...But the problem states that x₁₀₀ is between 0.99 and 0.991, which is approximately 0.99 to 0.991. So, 0.9899 is less than 0.99, which contradicts the lower bound. Hmm, so maybe my approximation is too rough.Wait, perhaps the approximation P_n ≈ 1/n is not precise enough. Maybe we can get a better approximation.Let me consider the recurrence relation P_{n+1} = P_n (1 - P_n)We can write this as:1/P_{n+1} = 1/(P_n (1 - P_n)) = 1/P_n * 1/(1 - P_n) ≈ (1/P_n)(1 + P_n) for small P_nSo,1/P_{n+1} ≈ 1/P_n + 1Therefore, 1/P_{n+1} - 1/P_n ≈ 1This suggests that 1/P_n behaves like a sequence increasing by approximately 1 each time, so 1/P_n ≈ n + CWhich is consistent with our earlier differential equation approximation.But perhaps we can get a better approximation by considering higher-order terms.Let me write:1/P_{n+1} = 1/(P_n (1 - P_n)) = 1/P_n * (1 + P_n + P_n² + P_n³ + ... )Since P_n is small, we can approximate:1/P_{n+1} ≈ 1/P_n + 1 + P_n + P_n² + ...But since P_n is small, higher powers are negligible, so:1/P_{n+1} ≈ 1/P_n + 1 + P_nTherefore,1/P_{n+1} - 1/P_n ≈ 1 + P_nBut 1/P_{n+1} - 1/P_n ≈ 1 + P_nIf we let Q_n = 1/P_n, then:Q_{n+1} - Q_n ≈ 1 + 1/Q_nSo,Q_{n+1} ≈ Q_n + 1 + 1/Q_nThis is a better approximation.Now, for large n, Q_n is large, so 1/Q_n is small. So, we can approximate:Q_{n+1} ≈ Q_n + 1 + 1/Q_nThis suggests that Q_n grows a bit faster than linearly.Let me try to model this.Assume Q_n ≈ n + c + d ln n + ...Let me plug this into the recurrence:Q_{n+1} ≈ (n + 1) + c + d ln(n + 1) ≈ n + 1 + c + d ln n + d/(n)On the other hand,Q_{n+1} ≈ Q_n + 1 + 1/Q_n ≈ (n + c + d ln n) + 1 + 1/(n + c + d ln n)≈ n + c + d ln n + 1 + 1/nSo, equating both expressions:n + 1 + c + d ln n + d/(n) ≈ n + c + d ln n + 1 + 1/nSimplify:Left side: n + 1 + c + d ln n + d/nRight side: n + c + d ln n + 1 + 1/nSo, equating:d/n ≈ 1/nThus, d = 1Therefore, Q_n ≈ n + c + ln nSo, Q_n ≈ n + c + ln nTherefore, P_n ≈ 1/(n + c + ln n)So, for large n, P_n is approximately 1/(n + ln n + c)Therefore, x_{n+1} = 1 - P_n ≈ 1 - 1/(n + ln n + c)So, x_n ≈ 1 - 1/(n - 1 + ln(n - 1) + c)But we need to find c.Wait, maybe we can use the initial terms to estimate c.Wait, but for n=1, P₁=0.5, so Q₁=2.For n=2, P₂=0.25, Q₂=4.For n=3, P₃=0.1875, Q₃≈5.333...Wait, let's compute Q_n for n=1,2,3,4,5,6,7:n=1: Q₁=2n=2: Q₂=4n=3: Q₃≈5.333...n=4: Q₄≈6.571...Wait, let me compute Q₄:P₄≈0.15234375, so Q₄≈6.57142857Similarly, Q₅≈1/0.129140625≈7.746Q₆≈1/0.112463125≈8.892Q₇≈1/0.099609375≈10.04So, Q_n seems to be increasing by roughly 1 each time, but slightly more.Wait, let's see:From n=1 to n=2: Q increases by 2 (from 2 to 4)From n=2 to n=3: Q increases by ~1.333 (from 4 to ~5.333)From n=3 to n=4: Q increases by ~1.238 (from ~5.333 to ~6.571)From n=4 to n=5: Q increases by ~1.175 (from ~6.571 to ~7.746)From n=5 to n=6: Q increases by ~1.146 (from ~7.746 to ~8.892)From n=6 to n=7: Q increases by ~1.148 (from ~8.892 to ~10.04)So, the increments are decreasing, approaching 1.So, for large n, Q_{n+1} - Q_n ≈1 + 1/Q_n≈1 + 0=1, since Q_n is large.Therefore, for large n, Q_n≈n + c + ln nBut let's see if we can get a better approximation.Assume Q_n = n + c + ln n + dThen,Q_{n+1} = (n + 1) + c + ln(n + 1) + d≈ n + 1 + c + ln n + 1/n + dOn the other hand,Q_{n+1} ≈ Q_n + 1 + 1/Q_n ≈ (n + c + ln n + d) + 1 + 1/(n + c + ln n + d)≈ n + c + ln n + d + 1 + 1/nSo, equate:n + 1 + c + ln n + 1/n + d ≈ n + c + ln n + d + 1 + 1/nSimplify:Left side: n + 1 + c + ln n + 1/n + dRight side: n + c + ln n + d + 1 + 1/nThey are equal, so our assumption holds.Therefore, Q_n ≈ n + c + ln nBut we need to find c.Wait, let's use the initial terms to estimate c.From n=1, Q₁=2=1 + c + ln 1=1 + c +0=1 + cThus, c=1So, Q_n≈n +1 + ln nTherefore, P_n≈1/(n +1 + ln n)So, x_{n+1}=1 - P_n≈1 -1/(n +1 + ln n)Therefore, x_n≈1 -1/(n -1 +1 + ln(n -1))=1 -1/(n + ln(n -1))So, for n=100,x₁₀₀≈1 -1/(100 + ln 99)Compute ln 99:ln 99≈4.5951So,x₁₀₀≈1 -1/(100 +4.5951)=1 -1/104.5951≈1 -0.00956≈0.99044So, approximately 0.99044, which is between 0.99 and 0.991.But the problem asks to prove that 0.99 <x₁₀₀<0.991.So, my approximation gives x₁₀₀≈0.9904, which is within the desired bounds.But I need to make this rigorous.So, perhaps I can show that P_n is slightly less than 1/(n + ln n + c), so x_{n}=1 - P_n is slightly more than 1 -1/(n + ln n + c)Alternatively, maybe I can bound P_n.Wait, from the recurrence Q_{n+1}=1/P_{n+1}=1/(P_n(1 - P_n))=1/P_n +1/(1 - P_n)But 1/(1 - P_n)=1 + P_n + P_n² +...So,Q_{n+1}=Q_n +1 + P_n + P_n² +...Since P_n is small, P_n² is negligible, so Q_{n+1}≈Q_n +1 + P_nBut P_n=1/Q_nSo,Q_{n+1}≈Q_n +1 +1/Q_nThus,Q_{n+1} - Q_n≈1 +1/Q_nThis is a better approximation.So, summing from n=1 to N,Q_{N+1} - Q₁≈Σ_{n=1}^N (1 +1/Q_n)=N + Σ_{n=1}^N 1/Q_nBut Q_n≈n +1 + ln n, so 1/Q_n≈1/(n +1 + ln n)Thus,Σ_{n=1}^N 1/Q_n≈Σ_{n=1}^N 1/(n +1 + ln n)This sum is approximately ln N, since it's similar to the harmonic series.Wait, but actually, Σ_{n=1}^N 1/(n + c)≈ln N + γ, where γ is Euler-Mascheroni constant.But in our case, c=1 + ln n, which complicates things.Wait, perhaps I can approximate Σ_{n=1}^N 1/(n +1 + ln n)≈Σ_{n=1}^N 1/(n +1)≈ln(N +1)But actually, 1/(n +1 + ln n) <1/(n +1), so Σ_{n=1}^N 1/(n +1 + ln n) < Σ_{n=1}^N 1/(n +1)=H_{N+1} -1≈ln(N+1)+γ -1Similarly, 1/(n +1 + ln n) >1/(n +1 + ln N) for n <NWait, maybe it's too vague.Alternatively, perhaps I can use integral approximation.Σ_{n=1}^N 1/(n +1 + ln n)≈∫_{1}^{N} 1/(x +1 + ln x) dxBut this integral is complicated.Alternatively, maybe I can use the approximation Q_n≈n +1 + ln n + cBut I already used that.Alternatively, perhaps I can use the relation Q_{n+1}=Q_n +1 +1/Q_nSo, for large n, Q_{n}≈n +1 + ln n + cThen, Q_{n+1}≈(n +1) +1 + ln(n +1) +c≈n +2 + ln n + ln(1 +1/n) +c≈n +2 + ln n +1/n +cBut also,Q_{n+1}=Q_n +1 +1/Q_n≈(n +1 + ln n +c) +1 +1/(n +1 + ln n +c)≈n +2 + ln n +c +1/(n +1 + ln n +c)Comparing both expressions:n +2 + ln n +c +1/(n +1 + ln n +c)≈n +2 + ln n +c +1/nSo,1/(n +1 + ln n +c)≈1/nThus,n +1 + ln n +c≈nWhich implies 1 + ln n +c≈0, which is not possible.Hmm, maybe my initial assumption is missing something.Alternatively, perhaps I can use the relation Q_{n+1}=Q_n +1 +1/Q_nSo, for large n, Q_{n}≈n +1 + ln nThen,Q_{n+1}=Q_n +1 +1/Q_n≈(n +1 + ln n) +1 +1/(n +1 + ln n)≈n +2 + ln n +1/(n +1 + ln n)But also,Q_{n+1}= (n +1) +1 + ln(n +1)=n +2 + ln(n +1)=n +2 + ln n + ln(1 +1/n)≈n +2 + ln n +1/nSo, equating both expressions:n +2 + ln n +1/(n +1 + ln n)≈n +2 + ln n +1/nThus,1/(n +1 + ln n)≈1/nWhich implies n +1 + ln n≈n, which is not true.So, my approximation is missing something.Alternatively, perhaps I need to include another term in the expansion.Let me assume Q_n≈n + a + b ln n + cThen,Q_{n+1}=Q_n +1 +1/Q_n≈(n + a + b ln n +c) +1 +1/(n + a + b ln n +c)≈n + a + b ln n +c +1 +1/nOn the other hand,Q_{n+1}= (n +1) + a + b ln(n +1) +c≈n +1 +a +b(ln n +1/n) +c≈n +1 +a +b ln n +b/n +cSo, equate:n + a + b ln n +c +1 +1/n≈n +1 +a +b ln n +b/n +cSimplify:Left side: n + a + b ln n +c +1 +1/nRight side: n +1 +a +b ln n +b/n +cThus,Left side: n + (a +1) + b ln n +c +1/nRight side: n + (1 +a) + b ln n +c +b/nSo, equating:1/n≈b/nThus, b=1Therefore, Q_n≈n +a + ln n +cBut we still need to find a and c.Wait, let's use the initial terms.From n=1:Q₁=2=1 +a + ln 1 +c=1 +a +0 +c=1 +a +cFrom n=2:Q₂=4=2 +a + ln 2 +c=2 +a +0.6931 +cFrom n=3:Q₃≈5.333=3 +a + ln 3 +c≈3 +a +1.0986 +cSo, we have three equations:1) 1 +a +c=2 => a +c=12) 2 +a +0.6931 +c=4 => a +c=4 -2 -0.6931=1.3069But from equation 1, a +c=1, which contradicts equation 2.So, this suggests that the approximation Q_n≈n +a + ln n +c is not accurate enough.Alternatively, perhaps I need to include another term, like Q_n≈n +a + ln n +b ln ln n +cBut this is getting too complicated.Alternatively, maybe I can use the relation Q_{n+1}=Q_n +1 +1/Q_n and try to bound Q_n.Note that Q_{n+1}=Q_n +1 +1/Q_nSo, Q_{n+1} > Q_n +1Thus, Q_n > Q₁ + (n -1)*1=2 +n -1=n +1Similarly, Q_{n+1}=Q_n +1 +1/Q_n < Q_n +1 +1/(n +1)Because Q_n >n +1Thus,Q_{n+1} < Q_n +1 +1/(n +1)So, summing from n=1 to N,Q_{N+1} < Q₁ + Σ_{n=1}^N [1 +1/(n +1)] =2 +N + Σ_{n=1}^N 1/(n +1)=2 +N + (H_{N+1} -1)Where H_{N+1} is the (N+1)th harmonic number.But H_{N+1}≈ln(N+1) + γ +1/(2(N+1)) -...So,Q_{N+1} <2 +N + ln(N+1) + γ -1=1 +N + ln(N+1) + γThus,Q_n <n + ln n + γ +1Therefore,P_n=1/Q_n >1/(n + ln n + γ +1)Similarly, from the lower bound Q_n >n +1Thus,P_n=1/Q_n <1/(n +1)Therefore,1/(n + ln n + γ +1) < P_n <1/(n +1)Thus,1 -1/(n +1) <x_{n+1}=1 - P_n <1 -1/(n + ln n + γ +1)So, for n=99,x₁₀₀=1 - P_{99} <1 -1/(99 + ln 99 + γ +1)=1 -1/(100 + ln 99 + γ)Compute ln 99≈4.5951, γ≈0.5772Thus,1/(100 +4.5951 +0.5772)=1/(105.1723)≈0.009506Thus,x₁₀₀ <1 -0.009506≈0.990494Similarly,x₁₀₀=1 - P_{99} >1 -1/(99 + ln 99 + γ +1)=1 -1/(100 + ln 99 + γ)≈1 -0.009506≈0.990494Wait, but that gives the same value.Wait, no, actually, the lower bound is:x_{n+1}=1 - P_n >1 -1/(n + ln n + γ +1)So, for n=99,x₁₀₀ >1 -1/(99 + ln 99 + γ +1)=1 -1/(100 + ln 99 + γ)≈1 -0.009506≈0.990494And the upper bound is:x_{n+1}=1 - P_n <1 -1/(n +1)=1 -1/100=0.99Wait, that can't be, because 0.990494 <0.99 is false.Wait, no, actually, the upper bound is x_{n+1}=1 - P_n <1 -1/(n +1)So, for n=99,x₁₀₀=1 - P_{99} <1 -1/100=0.99But wait, our earlier approximation gave x₁₀₀≈0.9904, which is greater than 0.99, but the upper bound is 0.99, which contradicts.This suggests that my bounding is not tight enough.Wait, perhaps I made a mistake in the inequalities.Let me re-examine.We have:Q_{n+1}=Q_n +1 +1/Q_nSo, Q_{n+1} > Q_n +1Thus, Q_n > Q₁ + (n -1)*1=2 +n -1=n +1Therefore, P_n=1/Q_n <1/(n +1)Thus, x_{n+1}=1 - P_n >1 -1/(n +1)So, for n=99,x₁₀₀=1 - P_{99} >1 -1/100=0.99Similarly, from the upper bound:Q_{n+1}=Q_n +1 +1/Q_n < Q_n +1 +1/(n +1)Thus,Q_{n+1} < Q_n +1 +1/(n +1)Summing from n=1 to N,Q_{N+1} < Q₁ + Σ_{n=1}^N [1 +1/(n +1)]=2 +N + Σ_{n=1}^N 1/(n +1)=2 +N + (H_{N+1} -1)H_{N+1}≈ln(N+1) + γ +1/(2(N+1)) -...Thus,Q_{N+1} <1 +N + ln(N+1) + γ +1/(2(N+1))Therefore,Q_n <n + ln n + γ +1 +1/(2n)Thus,P_n=1/Q_n >1/(n + ln n + γ +1 +1/(2n))Therefore,x_{n+1}=1 - P_n <1 -1/(n + ln n + γ +1 +1/(2n))So, for n=99,x₁₀₀=1 - P_{99} <1 -1/(99 + ln 99 + γ +1 +1/(2*99))=1 -1/(100 + ln 99 + γ +1/198)Compute:ln 99≈4.5951γ≈0.57721/198≈0.00505Thus,Denominator≈100 +4.5951 +0.5772 +0.00505≈105.17735Thus,1/105.17735≈0.009506Thus,x₁₀₀ <1 -0.009506≈0.990494Therefore, combining both bounds:0.99 <x₁₀₀<0.990494But the problem states that x₁₀₀ is between 0.99 and 0.991, which is approximately 0.99 to 0.991.But according to this, x₁₀₀ <0.990494, which is less than 0.991, so the upper bound holds.But the lower bound is 0.99, which is less than x₁₀₀.Wait, but according to our earlier approximation, x₁₀₀≈0.9904, which is between 0.99 and 0.991.But according to the bounds, x₁₀₀ <0.990494, which is less than 0.991, and x₁₀₀ >0.99.So, 0.99 <x₁₀₀<0.990494<0.991Thus, 0.99 <x₁₀₀<0.991Therefore, the desired inequality holds.But to make this rigorous, I need to show that:1 -1/(n +1) <x_{n+1}=1 - P_n <1 -1/(n + ln n + γ +1)For n=99,1 -1/100=0.99 <x₁₀₀<1 -1/(100 + ln 99 + γ +1)≈0.990494<0.991Thus, 0.99 <x₁₀₀<0.991Therefore, the proof is complete.