💡Okay, so I have this sequence problem with three parts. Let me try to tackle each part step by step. Starting with part (1): If a₁, m, a₂ form an arithmetic sequence, find the value of m.Alright, arithmetic sequence means that the difference between consecutive terms is constant. So, if a₁, m, a₂ are in arithmetic sequence, then the difference between m and a₁ should be the same as the difference between a₂ and m. First, I know that a₁ is given as 1. Then, a₂ is calculated using the recursive formula: a₂ = (1/8)a₁² + m. Since a₁ is 1, that simplifies to a₂ = (1/8)(1) + m = 1/8 + m.Now, since a₁, m, a₂ are in arithmetic sequence, the difference between m and a₁ should equal the difference between a₂ and m. So, m - a₁ = a₂ - m. Plugging in the values, that becomes m - 1 = (1/8 + m) - m. Simplifying the right side, it's 1/8 + m - m, which is just 1/8. So, m - 1 = 1/8. Solving for m, I add 1 to both sides: m = 1 + 1/8 = 9/8. Wait, let me double-check that. If a₁ is 1, m is 9/8, then a₂ is 1/8 + 9/8 = 10/8 = 5/4. Now, checking the arithmetic sequence: 1, 9/8, 5/4. The difference between 9/8 and 1 is 1/8, and the difference between 5/4 and 9/8 is 5/4 - 9/8 = 10/8 - 9/8 = 1/8. Yep, that works. So, m is 9/8.Moving on to part (2): If m = 0, find the general term aₙ of the sequence {aₙ}.Alright, so m is 0, so the recursive formula becomes aₙ₊₁ = (1/8)aₙ². Starting with a₁ = 1.Hmm, this seems like a quadratic recurrence relation. Maybe I can find a pattern or find a way to express it in terms of powers.Let me compute the first few terms to see if I can spot a pattern.a₁ = 1a₂ = (1/8)(1)² = 1/8a₃ = (1/8)(1/8)² = (1/8)(1/64) = 1/512a₄ = (1/8)(1/512)² = (1/8)(1/262144) = 1/2097152Hmm, that's getting really small quickly. It seems like each term is 1 over 8 raised to some power. Let me see:a₁ = 1 = 8⁰a₂ = 1/8 = 8⁻¹a₃ = 1/512 = 8⁻³Wait, 8³ is 512, so 1/512 is 8⁻³.a₄ = 1/2097152. Let's see, 8⁶ is 262144, so 8⁷ is 2097152. So, a₄ = 8⁻⁷.Wait, so the exponents are 0, -1, -3, -7,... That seems like each exponent is doubling and subtracting 1? Let me check:From a₁ to a₂: exponent goes from 0 to -1. That's subtracting 1.From a₂ to a₃: exponent goes from -1 to -3. That's subtracting 2.From a₃ to a₄: exponent goes from -3 to -7. That's subtracting 4.Wait, so the amount subtracted is doubling each time: 1, 2, 4,... So, the exponent is 0 - 1 - 2 - 4 - 8 - ... which is like - (2ⁿ - 1). Wait, let's see:At n=1: exponent is 0n=2: exponent is -1 = -(2¹ - 1)n=3: exponent is -3 = -(2² - 1)n=4: exponent is -7 = -(2³ - 1)Ah! So, for aₙ, the exponent is -(2ⁿ⁻¹ - 1). Therefore, aₙ = 8^{-(2ⁿ⁻¹ - 1)} = 8^{1 - 2ⁿ⁻¹}.Wait, let me verify:For n=1: 8^{1 - 2⁰} = 8^{1 - 1} = 8⁰ = 1. Correct.n=2: 8^{1 - 2¹} = 8^{-1} = 1/8. Correct.n=3: 8^{1 - 2²} = 8^{-3} = 1/512. Correct.n=4: 8^{1 - 2³} = 8^{-7} = 1/2097152. Correct.So, the general term is aₙ = 8^{1 - 2ⁿ⁻¹}. Alternatively, that can be written as 8^{(1 - 2^{n-1})}.Alternatively, since 8 is 2³, we can write it as 2^{3(1 - 2^{n-1})} = 2^{3 - 3*2^{n-1}}.But 8^{1 - 2^{n-1}} is simpler. So, I think that's the general term.Now, part (3): If for any positive integer n, aₙ < 4, find the maximum value of m.Okay, so we need to find the maximum m such that all terms of the sequence are less than 4.Given the recursive formula aₙ₊₁ = (1/8)aₙ² + m, starting from a₁ = 1.We need to ensure that aₙ < 4 for all n. So, we need to find the maximum m such that this condition holds.First, let's think about the behavior of the sequence. If m is too large, the terms might grow beyond 4. So, we need to find the threshold m where the sequence just stays below 4.Perhaps we can consider the fixed point of the recurrence. A fixed point is a value x such that x = (1/8)x² + m.Solving for x: (1/8)x² - x + m = 0.Multiply both sides by 8: x² - 8x + 8m = 0.Solutions: x = [8 ± sqrt(64 - 32m)] / 2 = 4 ± sqrt(16 - 8m).For real solutions, the discriminant must be non-negative: 16 - 8m ≥ 0 ⇒ m ≤ 2.So, if m > 2, there are no real fixed points, which might mean the sequence could diverge.But we need to ensure that aₙ < 4 for all n. Let's see.If m = 2, then the fixed point equation becomes x = (1/8)x² + 2.So, x² - 8x + 16 = 0 ⇒ (x - 4)² = 0 ⇒ x = 4.So, 4 is the fixed point when m = 2.Now, let's check if starting from a₁ = 1, the sequence stays below 4 when m = 2.Compute a₂ = (1/8)(1)² + 2 = 1/8 + 2 = 17/8 ≈ 2.125 < 4.a₃ = (1/8)(17/8)² + 2 = (1/8)(289/64) + 2 = 289/512 + 2 ≈ 0.564 + 2 = 2.564 < 4.a₄ = (1/8)(2.564)² + 2 ≈ (1/8)(6.573) + 2 ≈ 0.821 + 2 = 2.821 < 4.a₅ ≈ (1/8)(2.821)² + 2 ≈ (1/8)(7.958) + 2 ≈ 0.995 + 2 = 2.995 < 4.a₆ ≈ (1/8)(2.995)² + 2 ≈ (1/8)(8.97) + 2 ≈ 1.121 + 2 = 3.121 < 4.a₇ ≈ (1/8)(3.121)² + 2 ≈ (1/8)(9.74) + 2 ≈ 1.2175 + 2 = 3.2175 < 4.a₈ ≈ (1/8)(3.2175)² + 2 ≈ (1/8)(10.35) + 2 ≈ 1.29375 + 2 = 3.29375 < 4.a₉ ≈ (1/8)(3.29375)² + 2 ≈ (1/8)(10.84) + 2 ≈ 1.355 + 2 = 3.355 < 4.a₁₀ ≈ (1/8)(3.355)² + 2 ≈ (1/8)(11.26) + 2 ≈ 1.4075 + 2 = 3.4075 < 4.It seems like the sequence is approaching 4 but never reaching it when m = 2. So, m = 2 is a candidate.But we need to ensure that for m > 2, the sequence will eventually exceed 4.Suppose m = 2 + ε for some ε > 0.Then, a₂ = (1/8)(1)² + 2 + ε = 1/8 + 2 + ε = 17/8 + ε ≈ 2.125 + ε.a₃ = (1/8)(17/8 + ε)² + 2 + ε.Let's approximate:(17/8 + ε)² ≈ (17/8)² + 2*(17/8)*ε = 289/64 + (34/8)ε = 289/64 + 4.25ε.So, a₃ ≈ (1/8)(289/64 + 4.25ε) + 2 + ε ≈ (289/512) + (4.25/8)ε + 2 + ε ≈ 0.564 + 0.53125ε + 2 + ε ≈ 2.564 + 1.53125ε.Similarly, a₄ ≈ (1/8)(2.564 + 1.53125ε)² + 2 + ε.Expanding the square: ≈ (2.564)² + 2*2.564*1.53125ε ≈ 6.573 + 7.859ε.So, a₄ ≈ (1/8)(6.573 + 7.859ε) + 2 + ε ≈ 0.8216 + 0.9824ε + 2 + ε ≈ 2.8216 + 1.9824ε.Continuing this, each term seems to be increasing more rapidly as ε increases. So, eventually, the terms will exceed 4.Therefore, m = 2 is the maximum value where the sequence stays below 4 for all n.Wait, but let me think again. If m = 2, the fixed point is 4, but the sequence approaches 4 asymptotically. So, does it ever reach 4? No, because each term is less than 4. So, for m = 2, aₙ approaches 4 but never exceeds it. Therefore, m = 2 is the maximum value.If m were larger than 2, say m = 2.1, then the fixed point equation x = (1/8)x² + 2.1 would have no real solutions because the discriminant would be negative. That means the sequence might diverge to infinity, but we need to confirm.Alternatively, even if there's no fixed point, the sequence could still approach a limit, but in this case, since the function f(x) = (1/8)x² + m is increasing for x > 0, and if m > 2, f(x) grows faster, so the sequence might not converge and instead increase without bound.But we need to ensure that aₙ < 4 for all n. So, if m > 2, even if the sequence doesn't diverge immediately, it might eventually exceed 4.To confirm, let's try m = 2.1.a₁ = 1a₂ = (1/8)(1) + 2.1 = 0.125 + 2.1 = 2.225a₃ = (1/8)(2.225)² + 2.1 ≈ (1/8)(4.9506) + 2.1 ≈ 0.6188 + 2.1 ≈ 2.7188a₄ ≈ (1/8)(2.7188)² + 2.1 ≈ (1/8)(7.392) + 2.1 ≈ 0.924 + 2.1 ≈ 3.024a₅ ≈ (1/8)(3.024)² + 2.1 ≈ (1/8)(9.144) + 2.1 ≈ 1.143 + 2.1 ≈ 3.243a₆ ≈ (1/8)(3.243)² + 2.1 ≈ (1/8)(10.516) + 2.1 ≈ 1.3145 + 2.1 ≈ 3.4145a₇ ≈ (1/8)(3.4145)² + 2.1 ≈ (1/8)(11.656) + 2.1 ≈ 1.457 + 2.1 ≈ 3.557a₈ ≈ (1/8)(3.557)² + 2.1 ≈ (1/8)(12.656) + 2.1 ≈ 1.582 + 2.1 ≈ 3.682a₉ ≈ (1/8)(3.682)² + 2.1 ≈ (1/8)(13.556) + 2.1 ≈ 1.6945 + 2.1 ≈ 3.7945a₁₀ ≈ (1/8)(3.7945)² + 2.1 ≈ (1/8)(14.403) + 2.1 ≈ 1.8004 + 2.1 ≈ 3.9004a₁₁ ≈ (1/8)(3.9004)² + 2.1 ≈ (1/8)(15.213) + 2.1 ≈ 1.9016 + 2.1 ≈ 4.0016Oh, look at that! At n=11, a₁₁ ≈ 4.0016, which is just over 4. So, for m=2.1, the sequence exceeds 4 at n=11. Therefore, m=2.1 is too big.Similarly, if we try m=2, the sequence approaches 4 but never exceeds it. So, m=2 is indeed the maximum value where aₙ < 4 for all n.Therefore, the maximum value of m is 2.

💡Alright, so I have this equation to solve: (frac{x}{2x - 1} + frac{2}{1 - 2x} = 3). Hmm, it's a fractional equation, which means I need to get rid of the denominators to make it easier to solve. I remember that to eliminate denominators, I can multiply both sides of the equation by the least common denominator (LCD). First, let me identify the denominators here. They are (2x - 1) and (1 - 2x). Wait a minute, (1 - 2x) looks like the negative of (2x - 1). So, (1 - 2x = -(2x - 1)). That might be useful later on.Okay, so the denominators are (2x - 1) and (-(2x - 1)). The LCD would be (2x - 1) because it's the same as the other denominator except for the negative sign. But since one denominator is negative, I need to be careful with the signs when I multiply through.Let me write down the equation again:[frac{x}{2x - 1} + frac{2}{1 - 2x} = 3]Since (1 - 2x = -(2x - 1)), I can rewrite the second fraction as:[frac{2}{1 - 2x} = frac{2}{-(2x - 1)} = -frac{2}{2x - 1}]So now, the equation becomes:[frac{x}{2x - 1} - frac{2}{2x - 1} = 3]Great, now both fractions have the same denominator, (2x - 1). I can combine them:[frac{x - 2}{2x - 1} = 3]Now, to eliminate the denominator, I'll multiply both sides of the equation by (2x - 1):[(x - 2) = 3(2x - 1)]Okay, so after eliminating the denominators, the equation simplifies to (x - 2 = 3(2x - 1)). Let me check the options given:A: (x + 2 = 3)B: (x - 2 = 3)C: (x + 2 = 3(2x - 1))D: (x - 2 = 3(2x - 1))Looks like option D matches what I got: (x - 2 = 3(2x - 1)). So, I think D is the correct answer.But just to be thorough, let me solve the equation further to make sure I didn't make any mistakes.Starting with:[x - 2 = 3(2x - 1)]First, expand the right side:[x - 2 = 6x - 3]Now, let's get all the x terms on one side and constants on the other. Subtract (x) from both sides:[-2 = 5x - 3]Then, add 3 to both sides:[1 = 5x]So, (x = frac{1}{5}).Let me check if this solution makes sense by plugging it back into the original equation:[frac{frac{1}{5}}{2(frac{1}{5}) - 1} + frac{2}{1 - 2(frac{1}{5})} = 3]Simplify the denominators:First denominator: (2(frac{1}{5}) - 1 = frac{2}{5} - 1 = -frac{3}{5})Second denominator: (1 - 2(frac{1}{5}) = 1 - frac{2}{5} = frac{3}{5})So, the equation becomes:[frac{frac{1}{5}}{-frac{3}{5}} + frac{2}{frac{3}{5}} = 3]Simplify the fractions:First term: (frac{1}{5} div -frac{3}{5} = -frac{1}{3})Second term: (2 div frac{3}{5} = frac{10}{3})So, adding them together:[-frac{1}{3} + frac{10}{3} = frac{9}{3} = 3]Which matches the right side of the equation. So, (x = frac{1}{5}) is indeed the correct solution.Therefore, going back to the original question, after eliminating the denominators, the equation becomes (x - 2 = 3(2x - 1)), which is option D. So, I'm confident that D is the correct answer.

💡First, I need to determine the total distance the boy runs. Since the field is a square with each side measuring 35 meters, the perimeter of the field is 4 times the length of one side.Next, I'll calculate the perimeter:[text{Perimeter} = 4 times 35, text{m} = 140, text{m}]The boy takes 56 seconds to run this distance. To find his speed in meters per second (m/s), I'll use the formula:[text{Speed} = frac{text{Distance}}{text{Time}} = frac{140, text{m}}{56, text{s}} = 2.5, text{m/s}]To convert this speed to kilometers per hour (km/h), I'll use the conversion factor:[1, text{m/s} = 3.6, text{km/h}][text{Speed in km/h} = 2.5, text{m/s} times 3.6 = 9, text{km/h}]

💡Okay, so I've got these two equations here: 2x minus y equals 5, and x plus 2y equals 5. I need to find the value of x. Hmm, I remember from class that when you have two equations with two variables, you can solve them using either substitution or elimination. I think elimination might be easier here because the coefficients look manageable.First, let me write down the equations again to make sure I have them right:1. 2x - y = 52. x + 2y = 5Alright, so I need to eliminate one of the variables. Let's see, if I can get the coefficients of y to be the same or opposites, I can add or subtract the equations to eliminate y. Looking at the first equation, the coefficient of y is -1, and in the second equation, it's 2. If I multiply the first equation by 2, then the coefficient of y will be -2, which is the opposite of 2 in the second equation. That should work.So, multiplying the first equation by 2:2*(2x - y) = 2*5Which gives:4x - 2y = 10Now, I have the modified first equation:4x - 2y = 10And the second equation remains:x + 2y = 5Now, if I add these two equations together, the y terms should cancel out:(4x - 2y) + (x + 2y) = 10 + 5Simplifying the left side:4x + x - 2y + 2y = 5x + 0y = 5xAnd the right side:10 + 5 = 15So, 5x = 15To solve for x, I divide both sides by 5:x = 15 / 5x = 3Wait, let me double-check that. If x is 3, then I can plug it back into one of the original equations to find y and make sure everything makes sense.Let's use the second equation: x + 2y = 5Plugging in x = 3:3 + 2y = 5Subtract 3 from both sides:2y = 2Divide by 2:y = 1Now, let's check these values in the first equation to make sure:2x - y = 52*3 - 1 = 6 - 1 = 5Yes, that works out. So, x = 3 and y = 1 satisfy both equations. Therefore, the value of x is indeed 3.

💡I need to find the value of (sin 600^circ). First, I remember that angles in trigonometry can be more than (360^circ), and these are called coterminal angles. Coterminal angles share the same terminal side on the unit circle, which means they have the same trigonometric function values. To simplify (600^circ), I'll subtract (360^circ) until I get an angle between (0^circ) and (360^circ). Subtracting (360^circ) once from (600^circ) gives:[600^circ - 360^circ = 240^circ]So, (sin 600^circ = sin 240^circ).Next, I'll determine the reference angle for (240^circ). The reference angle is the acute angle formed by the terminal side and the x-axis. Since (240^circ) is in the third quadrant, I subtract (180^circ) from (240^circ):[240^circ - 180^circ = 60^circ]The reference angle is (60^circ).In the third quadrant, both sine and cosine values are negative. I know that (sin 60^circ = frac{sqrt{3}}{2}). Therefore, in the third quadrant:[sin 240^circ = -frac{sqrt{3}}{2}]Thus, (sin 600^circ = -frac{sqrt{3}}{2}).

💡Okay, so I need to find the smallest natural number ( n ) such that both ( 2n ) is a perfect square and ( 15n ) is a perfect cube. Hmm, let's break this down step by step.First, let's think about what it means for a number to be a perfect square or a perfect cube. A perfect square has all its prime factors with even exponents, and a perfect cube has all its prime factors with exponents that are multiples of three.So, let's factorize ( n ) into its prime factors. Let's say ( n = 2^a times 3^b times 5^c times ldots ), where ( a, b, c, ldots ) are non-negative integers. Now, ( 2n = 2^{a+1} times 3^b times 5^c times ldots ). For this to be a perfect square, all the exponents in the prime factorization must be even. That means:- ( a + 1 ) must be even,- ( b ) must be even,- ( c ) must be even,- and so on for any other prime factors.Similarly, ( 15n = 3^{b+1} times 5^{c+1} times 2^a times ldots ). For this to be a perfect cube, all the exponents must be multiples of three. So:- ( a ) must be a multiple of three,- ( b + 1 ) must be a multiple of three,- ( c + 1 ) must be a multiple of three,- and so on for any other prime factors.Okay, so let's write down these conditions:1. From ( 2n ) being a perfect square: - ( a + 1 ) is even ⇒ ( a ) is odd. - ( b ) is even. - ( c ) is even.2. From ( 15n ) being a perfect cube: - ( a ) is a multiple of three. - ( b + 1 ) is a multiple of three. - ( c + 1 ) is a multiple of three.So, we need to find the smallest ( n ), which means we need to find the smallest exponents ( a, b, c ) that satisfy all these conditions.Let's start with ( a ):- ( a ) must be odd and a multiple of three. The smallest such number is 3.Next, ( b ):- ( b ) must be even, and ( b + 1 ) must be a multiple of three. Let's find the smallest ( b ) that satisfies this. - If ( b = 2 ), then ( b + 1 = 3 ), which is a multiple of three. Perfect.Now, ( c ):- ( c ) must be even, and ( c + 1 ) must be a multiple of three. Let's find the smallest ( c ) that satisfies this. - If ( c = 2 ), then ( c + 1 = 3 ), which is a multiple of three. Perfect.So, putting it all together, ( a = 3 ), ( b = 2 ), ( c = 2 ). Therefore, the smallest ( n ) is:( n = 2^3 times 3^2 times 5^2 )Calculating that:- ( 2^3 = 8 )- ( 3^2 = 9 )- ( 5^2 = 25 )Multiplying these together:( 8 times 9 = 72 )( 72 times 25 = 1800 )So, ( n = 1800 ).Let me double-check to make sure this works:- ( 2n = 2 times 1800 = 3600 ). Is 3600 a perfect square? Yes, because ( 60^2 = 3600 ).- ( 15n = 15 times 1800 = 27000 ). Is 27000 a perfect cube? Yes, because ( 30^3 = 27000 ).Looks good! So, the smallest interesting number is 1800.**Final Answer**The smallest interesting number is boxed{1800}.