💡Okay, so I have this sequence problem here. Let me read it again to make sure I understand. It says that a₁ is 1, a₂ is 2, and then for n+2, the term is defined as (2aₙaₙ₊₁)/(aₙ + aₙ₊₁). And the question is asking about the limit of the sequence {aₙ} as n approaches infinity. The options are: A. Does not exist, B. Equals 1, C. Equals 3/2, D. Equals 2.Hmm, okay. So, first, let's write down what we know. The first term is 1, the second term is 2, and each subsequent term is defined by this formula involving the previous two terms. The formula looks a bit complicated, but maybe I can simplify it or find a pattern.Let me try computing the first few terms to see if I can spot a pattern. So, a₁ is 1, a₂ is 2. Let's compute a₃.a₃ = (2a₁a₂)/(a₁ + a₂) = (2*1*2)/(1 + 2) = 4/3 ≈ 1.333...Okay, a₃ is 4/3. Now, let's compute a₄.a₄ = (2a₂a₃)/(a₂ + a₃) = (2*2*(4/3))/(2 + 4/3) = (16/3)/(10/3) = (16/3)*(3/10) = 16/10 = 8/5 = 1.6Alright, a₄ is 8/5. Let's do a₅.a₅ = (2a₃a₄)/(a₃ + a₄) = (2*(4/3)*(8/5))/(4/3 + 8/5) = (64/15)/( (20 + 24)/15 ) = (64/15)/(44/15) = 64/44 = 16/11 ≈ 1.4545...Hmm, a₅ is 16/11. Let's do a₆.a₆ = (2a₄a₅)/(a₄ + a₅) = (2*(8/5)*(16/11))/(8/5 + 16/11) = (256/55)/( (88 + 80)/55 ) = (256/55)/(168/55) = 256/168 = 32/21 ≈ 1.5238...Okay, a₆ is 32/21. Let's keep going to a₇.a₇ = (2a₅a₆)/(a₅ + a₆) = (2*(16/11)*(32/21))/(16/11 + 32/21) = (1024/231)/( (336 + 352)/231 ) = (1024/231)/(688/231) = 1024/688 = 64/43 ≈ 1.4884...Hmm, a₇ is 64/43. Let's compute a₈.a₈ = (2a₆a₇)/(a₆ + a₇) = (2*(32/21)*(64/43))/(32/21 + 64/43) = (4096/903)/( (1384 + 1344)/903 ) = (4096/903)/(2728/903) = 4096/2728 ≈ 1.5018...So, a₈ is approximately 1.5018. Let's do a₉.a₉ = (2a₇a₈)/(a₇ + a₈) ≈ (2*(1.4884)*(1.5018))/(1.4884 + 1.5018) ≈ (2*2.237)/(2.9902) ≈ 4.474/2.9902 ≈ 1.496...So, a₉ is approximately 1.496. Hmm, so it seems like the sequence is oscillating around 1.5, getting closer each time. So, maybe the limit is 3/2, which is 1.5.But wait, let me think more carefully. Maybe I can find a pattern or a way to solve this recurrence relation.Looking at the recurrence relation: aₙ₊₂ = (2aₙaₙ₊₁)/(aₙ + aₙ₊₁). Hmm, this looks familiar. It's similar to the harmonic mean of aₙ and aₙ₊₁. Because the harmonic mean of two numbers x and y is 2xy/(x + y). So, each term is the harmonic mean of the two previous terms.Interesting. So, the sequence is defined by taking harmonic means of consecutive terms. I wonder if there's a known behavior for such sequences.I recall that if you take means repeatedly, sometimes the sequence converges to a particular value. For example, if you take arithmetic means, it might converge to the average of the initial terms. But harmonic mean is different.Alternatively, maybe I can transform the sequence to make it easier to handle. Since harmonic mean is involved, perhaps taking reciprocals would help. Let me try that.Let me define bₙ = 1/aₙ. Then, let's see what the recurrence becomes.Given that aₙ₊₂ = (2aₙaₙ₊₁)/(aₙ + aₙ₊₁), taking reciprocals:1/aₙ₊₂ = (aₙ + aₙ₊₁)/(2aₙaₙ₊₁) = (1/(2aₙ) + 1/(2aₙ₊₁)) = (bₙ + bₙ₊₁)/2.So, the recurrence for bₙ is:bₙ₊₂ = (bₙ + bₙ₊₁)/2.Ah, that's a much simpler recurrence! So, the sequence {bₙ} is defined by the average of the two previous terms. That is, each term is the average of the two before it.This is a linear recurrence relation, and I think it's a type of second-order linear recurrence. Maybe I can solve it using characteristic equations or find a pattern.Given that bₙ₊₂ = (bₙ + bₙ₊₁)/2, let's write the characteristic equation.Assume a solution of the form bₙ = r^n. Plugging into the recurrence:r^{n+2} = (r^n + r^{n+1})/2Divide both sides by r^n:r^2 = (1 + r)/2Multiply both sides by 2:2r^2 = 1 + rBring all terms to one side:2r^2 - r - 1 = 0Now, solve this quadratic equation:r = [1 ± sqrt(1 + 8)] / 4 = [1 ± 3]/4So, the roots are:r = (1 + 3)/4 = 4/4 = 1andr = (1 - 3)/4 = (-2)/4 = -1/2Therefore, the general solution for bₙ is:bₙ = A*(1)^n + B*(-1/2)^n = A + B*(-1/2)^nWhere A and B are constants determined by the initial conditions.Now, let's find A and B using the initial terms.Given that a₁ = 1, so b₁ = 1/a₁ = 1.Similarly, a₂ = 2, so b₂ = 1/a₂ = 1/2.So, we have:For n = 1: b₁ = A + B*(-1/2)^1 = A - B/2 = 1For n = 2: b₂ = A + B*(-1/2)^2 = A + B/4 = 1/2So, we have a system of two equations:1. A - (B/2) = 12. A + (B/4) = 1/2Let me solve this system.Subtract equation 2 from equation 1:(A - B/2) - (A + B/4) = 1 - 1/2Simplify:A - B/2 - A - B/4 = 1/2Combine like terms:(-B/2 - B/4) = 1/2Convert to common denominator:(-2B/4 - B/4) = 1/2So:(-3B)/4 = 1/2Multiply both sides by 4:-3B = 2Divide by -3:B = -2/3Now, substitute B back into equation 1:A - (-2/3)/2 = 1Simplify:A + (1/3) = 1Therefore:A = 1 - 1/3 = 2/3So, the general solution for bₙ is:bₙ = 2/3 + (-2/3)*(-1/2)^nSimplify:bₙ = 2/3 + (2/3)*(1/2)^nBecause (-2/3)*(-1/2)^n = (2/3)*(1/2)^n when n is even, and negative when n is odd. But since we have (-1/2)^n, it alternates sign. However, since we're dealing with the limit as n approaches infinity, the term (1/2)^n will go to zero regardless of the sign.Therefore, as n approaches infinity, the term (2/3)*(1/2)^n approaches zero, so:lim_{n→∞} bₙ = 2/3 + 0 = 2/3But remember, bₙ = 1/aₙ, so:lim_{n→∞} aₙ = 1 / (lim_{n→∞} bₙ) = 1 / (2/3) = 3/2So, the limit of the sequence {aₙ} is 3/2.Let me double-check my steps to make sure I didn't make a mistake.1. Defined bₙ = 1/aₙ, transformed the recurrence correctly into bₙ₊₂ = (bₙ + bₙ₊₁)/2.2. Solved the characteristic equation, found roots 1 and -1/2.3. Wrote the general solution as bₙ = A + B*(-1/2)^n.4. Applied initial conditions b₁ = 1 and b₂ = 1/2 to solve for A and B.5. Found A = 2/3 and B = -2/3.6. Expressed bₙ as 2/3 + (2/3)*(1/2)^n, considering the absolute value since the limit is concerned.7. Took the limit as n approaches infinity, which gives 2/3.8. Converted back to aₙ by taking reciprocal, resulting in 3/2.Everything seems to check out. The computations for the first few terms also suggested oscillation around 1.5, which supports the conclusion that the limit is 3/2.Therefore, the answer should be option C, which is 3/2.**Final Answer**boxed{C}

💡First, I recognize that the diagonal of a square divides it into two congruent 45-45-90 triangles. In such triangles, the legs are equal, and the hypotenuse is (sqrt{2}) times the length of each leg.Given the side length of the square is (100sqrt{3}) cm, this value represents the length of each leg of the 45-45-90 triangle formed by the diagonal.To find the length of the diagonal (the hypotenuse), I multiply the side length by (sqrt{2}):[text{Diagonal} = 100sqrt{3} times sqrt{2} = 100sqrt{6} text{ cm}]

💡Okay, so I have this problem here with two sets, A and B, and I need to find their intersection. Let me try to figure this out step by step.First, let's look at set A. It's defined as all real numbers x such that x squared minus x minus 2 is less than zero. So, A is {x | x² - x - 2 < 0}. Hmm, that looks like a quadratic inequality. I remember that to solve quadratic inequalities, I need to find the roots of the quadratic equation first and then determine the intervals where the inequality holds.Let me write down the quadratic equation: x² - x - 2 = 0. To solve this, I can factor it or use the quadratic formula. Let me try factoring first. I need two numbers that multiply to -2 and add up to -1. Hmm, 1 and -2? Yes, because 1 * (-2) = -2 and 1 + (-2) = -1. So, the equation factors as (x + 1)(x - 2) = 0. Therefore, the roots are x = -1 and x = 2.Now, to solve the inequality x² - x - 2 < 0, I can use a sign chart. The critical points are x = -1 and x = 2, which divide the number line into three intervals: (-∞, -1), (-1, 2), and (2, ∞). I'll test each interval to see where the quadratic expression is negative.Let's pick a test point in each interval:1. For (-∞, -1), let's choose x = -2: (-2 + 1)(-2 - 2) = (-1)(-4) = 4, which is positive. So, the expression is positive here.2. For (-1, 2), let's choose x = 0: (0 + 1)(0 - 2) = (1)(-2) = -2, which is negative. So, the expression is negative here.3. For (2, ∞), let's choose x = 3: (3 + 1)(3 - 2) = (4)(1) = 4, which is positive. So, the expression is positive here.Since we're looking for where the expression is less than zero, the solution is the interval where it's negative, which is (-1, 2). So, set A is (-1, 2).Alright, now onto set B. It's defined as {y | y = 3^x, x ≤ 0}. So, B is the set of all y such that y is equal to 3 raised to the power of x, where x is less than or equal to zero. Let me think about what this set looks like.Since x is less than or equal to zero, we're dealing with exponents that are zero or negative. Remember that 3^0 is 1, and for negative exponents, 3^(-a) is equal to 1/(3^a). So, as x decreases (becomes more negative), y approaches zero but never actually reaches zero. So, the set B includes all positive real numbers y such that y is greater than 0 and less than or equal to 1. In interval notation, that's (0, 1].Wait, let me double-check that. If x is 0, y is 1. If x is negative, say x = -1, y is 1/3. If x is -2, y is 1/9, and so on. So, as x approaches negative infinity, y approaches zero. So, y is always positive and less than or equal to 1. So, yes, B is (0, 1].Now, I need to find the intersection of sets A and B, which is A ∩ B. Remember, the intersection of two sets is the set of elements that are common to both sets.Set A is (-1, 2), and set B is (0, 1]. So, I need to find the overlap between these two intervals.Let me visualize this on the number line. A spans from -1 to 2, and B spans from just above 0 to 1, including 1. So, the overlapping part would be where both intervals are true. That is, numbers greater than 0 and less than or equal to 1. Because from 0 to 1, both A and B include those numbers. Beyond 1, A continues up to 2, but B stops at 1. Below 0, A goes down to -1, but B doesn't include any numbers less than or equal to 0.Therefore, the intersection A ∩ B is the interval (0, 1]. Let me just make sure I didn't make a mistake here. So, A is (-1, 2), which includes all numbers between -1 and 2. B is (0, 1], which includes all numbers greater than 0 and up to 1. So, the overlap is indeed the numbers greater than 0 and up to 1. So, yes, (0, 1] is correct.Looking back at the options given:A: (-1, 2) B: (-2, 1) C: (-1, 1] D: (0, 1]So, option D is (0, 1], which matches what I found. Therefore, the correct answer should be D.I think that makes sense. I don't see any mistakes in my reasoning. I solved the quadratic inequality correctly, found the correct interval for A, determined the correct interval for B by understanding the exponential function, and then found their intersection by looking at the overlapping parts of the intervals. So, yeah, I'm confident that D is the right answer.**Final Answer**boxed{D}

💡Okay, so I have this problem where each edge of a cube is increased by 60%, and I need to figure out by what percentage the surface area and the volume of the cube increase. Hmm, let's break this down step by step.First, let me recall what a cube is. A cube has all its edges of equal length, right? So, if the original edge length is, say, 's', then all edges are 's'. Now, if each edge is increased by 60%, that means the new edge length will be the original length plus 60% of it. Let me write that down. If the original edge length is 's', then a 60% increase would make the new edge length:New edge length = s + 60% of s = s + 0.6s = 1.6s.Okay, so the new edge length is 1.6 times the original edge length. Got that.Now, moving on to surface area. The surface area of a cube is calculated by the formula 6 times the square of the edge length because a cube has 6 faces, each being a square with area s². So, the original surface area (SA₁) is:SA₁ = 6s².With the new edge length, the new surface area (SA₂) would be:SA₂ = 6*(1.6s)².Let me compute that. First, square the new edge length:(1.6s)² = (1.6)² * s² = 2.56s².So, the new surface area is:SA₂ = 6 * 2.56s² = 15.36s².Now, to find the percentage increase in surface area, I need to compare the increase to the original surface area. The increase in surface area is:Increase in SA = SA₂ - SA₁ = 15.36s² - 6s² = 9.36s².To find the percentage increase:Percentage increase in SA = (Increase in SA / Original SA) * 100% = (9.36s² / 6s²) * 100%.Simplifying that:(9.36 / 6) * 100% = 1.56 * 100% = 156%.Alright, so the surface area increases by 156%. That seems straightforward.Now, onto the volume. The volume of a cube is given by the edge length cubed. So, the original volume (V₁) is:V₁ = s³.With the new edge length, the new volume (V₂) is:V₂ = (1.6s)³.Let me compute that:(1.6)³ = 1.6 * 1.6 * 1.6.First, 1.6 * 1.6 = 2.56.Then, 2.56 * 1.6. Let me calculate that:2.56 * 1.6:- 2 * 1.6 = 3.2- 0.56 * 1.6 = 0.896Adding them together: 3.2 + 0.896 = 4.096.So, (1.6s)³ = 4.096s³.Therefore, the new volume is:V₂ = 4.096s³.Now, the increase in volume is:Increase in V = V₂ - V₁ = 4.096s³ - s³ = 3.096s³.To find the percentage increase in volume:Percentage increase in V = (Increase in V / Original V) * 100% = (3.096s³ / s³) * 100%.Simplifying that:3.096 * 100% = 309.6%.So, the volume increases by 309.6%.Wait, let me double-check my calculations to make sure I didn't make any mistakes.For the surface area:- Original SA: 6s².- New edge: 1.6s.- New SA: 6*(1.6s)² = 6*2.56s² = 15.36s².- Increase: 15.36s² - 6s² = 9.36s².- Percentage: (9.36/6)*100 = 156%. That seems correct.For the volume:- Original V: s³.- New edge: 1.6s.- New V: (1.6s)³ = 4.096s³.- Increase: 4.096s³ - s³ = 3.096s³.- Percentage: (3.096/1)s³ *100 = 309.6%. That also seems correct.Hmm, just to be thorough, let me think about the percentage increases in terms of scaling factors.When you scale a linear dimension by a factor, the surface area scales by the square of that factor, and the volume scales by the cube of that factor.In this case, the scaling factor is 1.6 (60% increase).So, surface area scaling factor: (1.6)² = 2.56, which is a 156% increase (since 2.56 - 1 = 1.56, which is 156%).Volume scaling factor: (1.6)³ = 4.096, which is a 309.6% increase (since 4.096 - 1 = 3.096, which is 309.6%).Yes, that matches my earlier calculations. So, I'm confident that the surface area increases by 156% and the volume increases by 309.6%.Just to make sure I didn't confuse percentage increase with the scaling factor, let me clarify:- Scaling factor for surface area: 2.56, which means the new surface area is 256% of the original. Therefore, the increase is 256% - 100% = 156%.- Scaling factor for volume: 4.096, which means the new volume is 409.6% of the original. Therefore, the increase is 409.6% - 100% = 309.6%.Yes, that makes sense. So, the percentage increase is the scaling factor minus 1, multiplied by 100%.Another way to think about it is:If something increases by 60%, it becomes 160% of its original size. So, for surface area, which depends on two dimensions, the increase would be (1.6)^2 = 2.56, which is 256% of the original, so a 156% increase.Similarly, for volume, which depends on three dimensions, the increase would be (1.6)^3 = 4.096, which is 409.6% of the original, so a 309.6% increase.I think that's solid. I don't see any mistakes in my reasoning. The calculations check out both directly and through the scaling factors. So, I'm confident in the results.**Final Answer**The surface area increases by boxed{156%} and the volume increases by boxed{309.6%}.

💡Okay, so I've got this problem where Malcolm writes a positive integer on a piece of paper. Then he doubles that integer and subtracts 1, writing the result down. After that, he doubles the second integer and adds 1, writing that third result down too. The catch is that all three numbers he writes down have to be prime. I need to figure out all possible values for the first integer.Alright, let's start by breaking down the problem step by step. Let me denote the first integer as ( n ). According to the problem, the second number Malcolm writes is ( 2n - 1 ), and the third number is ( 2(2n - 1) + 1 ). Let me simplify that third expression:( 2(2n - 1) + 1 = 4n - 2 + 1 = 4n - 1 ).So, we have three numbers: ( n ), ( 2n - 1 ), and ( 4n - 1 ). All of these need to be prime numbers.Since ( n ) is a positive integer and a prime, let's consider the possible values of ( n ). Primes are numbers greater than 1 that have no positive divisors other than 1 and themselves. The smallest primes are 2, 3, 5, 7, 11, and so on.Let me start by testing the smallest primes to see if they satisfy all the conditions.First, let's try ( n = 2 ):- The second number is ( 2(2) - 1 = 4 - 1 = 3 ), which is prime.- The third number is ( 4(2) - 1 = 8 - 1 = 7 ), which is also prime.So, ( n = 2 ) works.Next, let's try ( n = 3 ):- The second number is ( 2(3) - 1 = 6 - 1 = 5 ), which is prime.- The third number is ( 4(3) - 1 = 12 - 1 = 11 ), which is prime as well.So, ( n = 3 ) also works.Now, let's try ( n = 5 ):- The second number is ( 2(5) - 1 = 10 - 1 = 9 ). Wait, 9 is not a prime number because it's divisible by 3. So, ( n = 5 ) doesn't work.Moving on to ( n = 7 ):- The second number is ( 2(7) - 1 = 14 - 1 = 13 ), which is prime.- The third number is ( 4(7) - 1 = 28 - 1 = 27 ). 27 is not prime since it's divisible by 3. So, ( n = 7 ) doesn't work either.Let's check ( n = 11 ):- The second number is ( 2(11) - 1 = 22 - 1 = 21 ). 21 is not prime; it's divisible by 3 and 7. So, ( n = 11 ) is out.Hmm, it seems like as ( n ) increases, the second and third numbers might not be prime. Maybe I should look for a pattern or a mathematical reason why only small primes work.Let me think about the expressions ( 2n - 1 ) and ( 4n - 1 ). If ( n ) is a prime greater than 3, it can be expressed in the form ( 6k pm 1 ) because all primes greater than 3 are of that form. Let me test this idea.Suppose ( n = 6k + 1 ):- Then, ( 2n - 1 = 2(6k + 1) - 1 = 12k + 2 - 1 = 12k + 1 ).- And ( 4n - 1 = 4(6k + 1) - 1 = 24k + 4 - 1 = 24k + 3 ).Now, ( 24k + 3 ) can be factored as ( 3(8k + 1) ), which means it's divisible by 3. Since it's greater than 3, it can't be prime. So, ( n = 6k + 1 ) doesn't work.Similarly, suppose ( n = 6k - 1 ):- Then, ( 2n - 1 = 2(6k - 1) - 1 = 12k - 2 - 1 = 12k - 3 ).- And ( 4n - 1 = 4(6k - 1) - 1 = 24k - 4 - 1 = 24k - 5 ).Looking at ( 12k - 3 ), this can be factored as ( 3(4k - 1) ), so it's divisible by 3 and greater than 3, hence not prime. Therefore, ( n = 6k - 1 ) also doesn't work.This suggests that primes of the form ( 6k pm 1 ) don't satisfy the conditions beyond the smallest ones. So, the only candidates left are ( n = 2 ) and ( n = 3 ), which we've already confirmed work.Just to be thorough, let me check ( n = 1 ). Wait, 1 isn't considered a prime number, so that's out of the question.What about ( n = 4 )? It's not prime, so we can skip it. ( n = 6 ) isn't prime either. ( n = 8 ) is not prime, and so on. It seems like beyond 3, the next integers either aren't prime or don't satisfy the conditions for the second and third numbers.Therefore, after testing the small primes and analyzing the structure of the expressions, the only possible values for ( n ) that make all three numbers prime are 2 and 3.

💡Okay, so I have this problem where I need to find the sum of the reciprocals of all positive integers that only have prime factors 2, 3, 5, and 7. That set is called B. Each number in B can be written as 2^a * 3^b * 5^c * 7^d where a, b, c, d are non-negative integers. I need to find the sum of 1/n for all n in B and express it as a fraction p/q where p and q are coprime, then find p+q.Hmm, I remember that when dealing with sums of reciprocals of numbers with specific prime factors, it's related to the Riemann zeta function or Euler products. But maybe I can break it down into simpler parts.Each number n in B is a product of powers of 2, 3, 5, and 7. So, the reciprocal 1/n can be expressed as 1/(2^a * 3^b * 5^c * 7^d). Since the exponents a, b, c, d are independent, I can separate the sum into four separate geometric series.Let me write that out:Sum_{n in B} 1/n = Sum_{a=0}^∞ Sum_{b=0}^∞ Sum_{c=0}^∞ Sum_{d=0}^∞ 1/(2^a * 3^b * 5^c * 7^d)Because multiplication distributes over addition, this can be factored into the product of four separate sums:= (Sum_{a=0}^∞ 1/2^a) * (Sum_{b=0}^∞ 1/3^b) * (Sum_{c=0}^∞ 1/5^c) * (Sum_{d=0}^∞ 1/7^d)Okay, so each of these is a geometric series. I remember that the sum of a geometric series Sum_{k=0}^∞ r^k is 1/(1 - r) when |r| < 1.Let me compute each sum individually.First, Sum_{a=0}^∞ 1/2^a. Here, r = 1/2, so the sum is 1/(1 - 1/2) = 1/(1/2) = 2.Next, Sum_{b=0}^∞ 1/3^b. Here, r = 1/3, so the sum is 1/(1 - 1/3) = 1/(2/3) = 3/2.Then, Sum_{c=0}^∞ 1/5^c. Here, r = 1/5, so the sum is 1/(1 - 1/5) = 1/(4/5) = 5/4.Lastly, Sum_{d=0}^∞ 1/7^d. Here, r = 1/7, so the sum is 1/(1 - 1/7) = 1/(6/7) = 7/6.Now, I need to multiply all these sums together:2 * (3/2) * (5/4) * (7/6)Let me compute this step by step.First, 2 * (3/2) = (2 * 3)/2 = 3.Then, 3 * (5/4) = (3 * 5)/4 = 15/4.Next, 15/4 * (7/6) = (15 * 7)/(4 * 6) = 105/24.Wait, 105 and 24 can be simplified. Let me see if they have a common factor. 105 is 5 * 21, which is 5 * 3 * 7. 24 is 8 * 3, which is 2^3 * 3. So, the common factor is 3.Divide numerator and denominator by 3:105 ÷ 3 = 3524 ÷ 3 = 8So, 105/24 simplifies to 35/8.Thus, the total sum is 35/8. Now, I need to express this as p/q where p and q are coprime. 35 and 8 have no common factors other than 1, so p = 35 and q = 8.Therefore, p + q = 35 + 8 = 43.So, the answer should be 43, which corresponds to option D.**Final Answer**boxed{D}