answer：Okay, so I have this problem about sequences. It says that {a_n} is a geometric sequence with a1 = 3 and a4 = 24. Then, there's another sequence {a_n + b_n} which is an arithmetic sequence with the first term 4 and a common difference of 1. I need to find the general term formulas for {a_n} and {b_n}, and then calculate the sum of the first n terms for {b_n}.Alright, let's start with the geometric sequence {a_n}. I remember that in a geometric sequence, each term is the previous term multiplied by a common ratio, r. So, the general formula for a geometric sequence is a_n = a1 * r^(n-1). Here, a1 is given as 3, so a_n = 3 * r^(n-1).Now, I need to find the common ratio r. I know that a4 = 24. Using the formula, a4 = 3 * r^(4-1) = 3 * r^3. So, 3 * r^3 = 24. If I divide both sides by 3, I get r^3 = 8. Taking the cube root of both sides, r = 2. So, the common ratio is 2. That means the general term for {a_n} is a_n = 3 * 2^(n-1).Okay, that part seems straightforward. Now, moving on to the arithmetic sequence {a_n + b_n}. An arithmetic sequence has a common difference, which is given as 1, and the first term is 4. The general formula for an arithmetic sequence is c_n = c1 + (n-1)d, where c1 is the first term and d is the common difference. So, in this case, c_n = 4 + (n-1)*1 = n + 3. Therefore, a_n + b_n = n + 3.Since I already have a_n, I can solve for b_n. From a_n + b_n = n + 3, subtracting a_n gives b_n = n + 3 - a_n. Substituting the expression for a_n, which is 3 * 2^(n-1), I get b_n = n + 3 - 3 * 2^(n-1). So, that's the general term for {b_n}.Now, part 2 asks for the sum of the first n terms for {b_n}. To find the sum, I can write it as the sum from k=1 to n of b_k. Since b_k = k + 3 - 3 * 2^(k-1), the sum will be the sum of (k + 3) minus the sum of 3 * 2^(k-1) from k=1 to n.Let's break this down. The sum of (k + 3) from k=1 to n can be split into two separate sums: the sum of k from 1 to n and the sum of 3 from 1 to n. The sum of k from 1 to n is a well-known formula: n(n + 1)/2. The sum of 3 from 1 to n is just 3n. So, combining these, the sum of (k + 3) is n(n + 1)/2 + 3n.Simplifying that, let's get a common denominator for the terms. The first term is n(n + 1)/2, which is (n^2 + n)/2. The second term is 3n, which can be written as 6n/2. So, adding them together, we have (n^2 + n + 6n)/2 = (n^2 + 7n)/2.Now, the other part is the sum of 3 * 2^(k-1) from k=1 to n. This is a geometric series where each term is multiplied by 2. The general formula for the sum of a geometric series is S = a1 * (r^n - 1)/(r - 1), where a1 is the first term, r is the common ratio, and n is the number of terms. Here, the first term when k=1 is 3 * 2^(1-1) = 3 * 1 = 3, and the common ratio is 2. So, the sum is 3 * (2^n - 1)/(2 - 1) = 3*(2^n - 1)/1 = 3*(2^n - 1).Putting it all together, the sum of b_n from k=1 to n is the sum of (k + 3) minus the sum of 3 * 2^(k-1). So, that's (n^2 + 7n)/2 - 3*(2^n - 1). Let's simplify this expression.First, expand the second term: 3*(2^n - 1) = 3*2^n - 3. So, the entire expression becomes (n^2 + 7n)/2 - 3*2^n + 3. To make it look neater, I can write it as (n^2 + 7n)/2 - 3*2^n + 3.Alternatively, I can combine the constant terms: -3*2^n + 3. So, the final expression for the sum is (n^2 + 7n)/2 - 3*2^n + 3.Let me double-check my steps to make sure I didn't make any mistakes. Starting with the geometric sequence, I found r correctly by using a4 = 24. Then, I used the arithmetic sequence formula correctly to express a_n + b_n as n + 3. Subtracting a_n gave me b_n correctly.For the sum, I split the sum into two parts: the sum of (k + 3) and the sum of 3*2^(k-1). I used the correct formulas for both sums, the arithmetic series for (k + 3) and the geometric series for 3*2^(k-1). The algebra seems correct, and I simplified the expression properly.So, I think my final answer for the sum is correct: (n^2 + 7n)/2 - 3*2^n + 3.

answer：First, I need to determine the total amount Kanul had initially. He spent 3000 on raw materials and 1000 on machinery. Additionally, he spent 30% of his total amount as cash.Let’s denote the total amount as ( T ).The total amount spent is the sum of the amounts spent on raw materials, machinery, and the 30% of the total amount ( T ).So, the equation representing this situation is:[3000 , (text{raw materials}) + 1000 , (text{machinery}) + 0.30T , (text{30% of total amount}) = T]Now, let's solve for ( T ):[3000 + 1000 + 0.30T = T][4000 + 0.30T = T][0.30T = T - 4000][0.30T = 1T - 4000][0.30T - 1T = -4000][-0.70T = -4000]Now, divide both sides by -0.70 to solve for ( T ):[T = frac{-4000}{-0.70}][T = 5714.29]So, the total amount Kanul had as cash with him was approximately 5714.29.

answer：Okay, so I have this problem with 10 springs, each originally 0.5 meters long, and each has a spring constant of 200 N/m. A mass of 2 kg is hung on each spring, and they're all hung in a series. I need to find the total length of the resulting chain, and I can neglect the mass of the springs themselves.Alright, let's break this down. First, I know that when you hang a mass on a spring, it stretches due to the force of gravity pulling down on the mass. The amount it stretches depends on the spring constant and the force applied. Hooke's Law tells me that the force exerted by the spring is equal to the spring constant multiplied by the extension of the spring: F = kΔx.But in this case, it's not just one spring; it's 10 springs connected in series, each with a mass hanging on them. So, the top spring has to support all 10 masses, the next one supports 9 masses, and so on, until the bottom spring, which only supports its own mass. That means each spring experiences a different force, and therefore, each will extend by a different amount.Let me write this out step by step.1. **Understanding the forces on each spring:** - The first spring (topmost) has to support all 10 masses. So, the force on it is 10 times the weight of one mass. - The second spring supports 9 masses, so the force is 9 times the weight. - This pattern continues until the tenth spring, which only supports 1 mass.2. **Calculating the force on each spring:** - The weight of one mass is given by W = m * g, where m = 2 kg and g = 9.8 m/s². - So, W = 2 kg * 9.8 m/s² = 19.6 N. - Therefore, the force on the first spring is 10 * 19.6 N = 196 N. - The force on the second spring is 9 * 19.6 N = 176.4 N. - Continuing this way, the forces on the springs are 196 N, 176.4 N, 156.8 N, ..., down to 19.6 N.3. **Calculating the extension of each spring:** - Using Hooke's Law, Δx = F / k, where k is the spring constant (200 N/m). - For the first spring: Δx₁ = 196 N / 200 N/m = 0.98 m. - For the second spring: Δx₂ = 176.4 N / 200 N/m = 0.882 m. - Similarly, for the third spring: Δx₃ = 156.8 N / 200 N/m = 0.784 m. - Continuing this pattern, the extensions decrease by 0.196 m each time.4. **Summing up the extensions:** - Now, I need to add up all these extensions to find the total extension of the chain. - The extensions are: 0.98 m, 0.882 m, 0.784 m, ..., 0.196 m. - This is an arithmetic series where the first term a₁ = 0.98 m, the last term a₁₀ = 0.196 m, and the number of terms n = 10. - The sum of an arithmetic series is given by S = n/2 * (a₁ + a₁₀). - So, S = 10/2 * (0.98 + 0.196) = 5 * 1.176 = 5.88 m.5. **Calculating the total length of the chain:** - Each spring is originally 0.5 m long, and there are 10 springs, so the total natural length is 10 * 0.5 m = 5 m. - Adding the total extension of 5.88 m to the natural length gives the total length of the chain: 5 m + 5.88 m = 10.88 m.Wait a minute, that seems a bit long. Let me double-check my calculations.- The weight per mass is correct: 2 kg * 9.8 = 19.6 N.- The forces on each spring are correctly calculated as multiples of 19.6 N.- The extensions using Hooke's Law seem right: each force divided by 200 N/m.- The arithmetic series sum: I think I might have made a mistake here. Let me recalculate the sum.Actually, the extensions are: 0.98, 0.882, 0.784, 0.686, 0.588, 0.49, 0.392, 0.294, 0.196, and 0.098 m.Wait, that's 10 terms. Let me list them properly:1. 0.98 m2. 0.882 m3. 0.784 m4. 0.686 m5. 0.588 m6. 0.49 m7. 0.392 m8. 0.294 m9. 0.196 m10. 0.098 mNow, adding these up:0.98 + 0.882 = 1.8621.862 + 0.784 = 2.6462.646 + 0.686 = 3.3323.332 + 0.588 = 3.923.92 + 0.49 = 4.414.41 + 0.392 = 4.8024.802 + 0.294 = 5.0965.096 + 0.196 = 5.2925.292 + 0.098 = 5.39 mAh, I see. Earlier, I incorrectly added 0.98 and 0.196 to get 1.176 and multiplied by 5, but that's not the correct way to sum the series because the series isn't symmetric around the middle term. Instead, I should have added each term sequentially.So, the total extension is 5.39 m, not 5.88 m. Therefore, the total length of the chain is the natural length plus the total extension: 5 m + 5.39 m = 10.39 m.That makes more sense. So, the final length of the chain is 10.39 meters.

answer：Okay, so I need to find the equation of the circle that's symmetric to the given circle C with respect to the line l. Hmm, let me start by recalling what it means for a circle to be symmetric with respect to a line. I think it means that if you reflect the original circle over the line, you get the symmetric circle. So, the symmetric circle should have the same radius as the original circle, but its center should be the reflection of the original center over the line.Alright, so first, I need to find the center and radius of the original circle C. The equation given is x² + y² - 6x + 8y + 24 = 0. I remember that the general form of a circle is x² + y² + Dx + Ey + F = 0, and the center is at (-D/2, -E/2) with radius sqrt((D/2)² + (E/2)² - F). Let me apply that here.So, for this equation, D is -6, E is 8, and F is 24. Therefore, the center should be at (-(-6)/2, -8/2) which is (3, -4). The radius would be sqrt(((-6)/2)² + (8/2)² - 24). Calculating that: (-6/2) is -3, squared is 9; (8/2) is 4, squared is 16. So, 9 + 16 is 25, minus 24 is 1. So, the radius is sqrt(1) which is 1. Got it, so the original circle has center (3, -4) and radius 1.Now, I need to find the reflection of this center (3, -4) over the line l: x - 3y - 5 = 0. I think the process involves finding the point (a, b) such that the line l is the perpendicular bisector of the segment joining (3, -4) and (a, b). So, two things: the midpoint of (3, -4) and (a, b) lies on line l, and the line segment connecting (3, -4) and (a, b) is perpendicular to line l.First, let me find the midpoint. The midpoint M would be ((3 + a)/2, (-4 + b)/2). Since M lies on line l, it must satisfy the equation x - 3y - 5 = 0. So, plugging in the midpoint coordinates:( (3 + a)/2 ) - 3*( (-4 + b)/2 ) - 5 = 0Let me simplify this equation:Multiply through by 2 to eliminate denominators:(3 + a) - 3*(-4 + b) - 10 = 0Simplify each term:3 + a + 12 - 3b - 10 = 0Combine like terms:(3 + 12 - 10) + a - 3b = 0That's 5 + a - 3b = 0So, equation 1: a - 3b = -5Now, the second condition is that the line segment connecting (3, -4) and (a, b) is perpendicular to line l. The slope of line l is given by rearranging the equation x - 3y - 5 = 0 to slope-intercept form:x - 5 = 3y => y = (1/3)x - 5/3So, the slope of line l is 1/3. Therefore, the slope of the perpendicular line should be the negative reciprocal, which is -3.So, the slope of the segment connecting (3, -4) and (a, b) is (b - (-4))/(a - 3) = (b + 4)/(a - 3). This slope should be equal to -3.So, equation 2: (b + 4)/(a - 3) = -3Let me write that as:b + 4 = -3(a - 3)Simplify:b + 4 = -3a + 9Bring all terms to one side:b + 4 + 3a - 9 = 0 => 3a + b - 5 = 0So, equation 2: 3a + b = 5Now, I have two equations:1) a - 3b = -52) 3a + b = 5I need to solve this system of equations for a and b.Let me use substitution or elimination. Let's try elimination.From equation 1: a = 3b - 5Plugging this into equation 2:3*(3b - 5) + b = 5Simplify:9b - 15 + b = 5Combine like terms:10b - 15 = 5Add 15 to both sides:10b = 20Divide by 10:b = 2Now, plug b = 2 back into equation 1:a - 3*2 = -5 => a - 6 = -5 => a = 1So, the reflected center is at (1, 2). Therefore, the symmetric circle has center (1, 2) and the same radius 1. So, its equation should be (x - 1)² + (y - 2)² = 1.Looking at the options, that's option B: (x - 1)² + (y - 2)² = 1.Wait, let me double-check my calculations to make sure I didn't make a mistake.First, midpoint:( (3 + 1)/2, (-4 + 2)/2 ) = (2, -1)Does this lie on line l: x - 3y -5 = 0?Plug in: 2 - 3*(-1) -5 = 2 + 3 -5 = 0. Yes, that works.Slope of the segment: (2 - (-4))/(1 - 3) = 6/(-2) = -3, which is correct because it's perpendicular to line l with slope 1/3.So, everything checks out. The symmetric circle is indeed (x - 1)² + (y - 2)² = 1, which is option B.**Final Answer**boxed{B}

answer：Okay, so I need to prove that if the limit of the ratio of consecutive terms ( frac{a_{n+1}}{a_n} ) as ( n ) approaches infinity is equal to some constant ( a ), then the limit of the nth root of ( a_n ) as ( n ) approaches infinity is also equal to ( a ). Hmm, that sounds a bit abstract, but let me try to break it down step by step.First, I remember that when dealing with sequences and their limits, especially involving ratios and roots, there are some standard theorems or properties that might be useful here. Maybe something related to the root test or the ratio test from calculus? Wait, the root test is about the nth root of the absolute value of the terms in a series, and the ratio test is about the limit of the ratio of consecutive terms. So, this problem seems to connect these two concepts.Let me write down what I know:Given:[lim_{n to infty} frac{a_{n+1}}{a_n} = a]Need to prove:[lim_{n to infty} sqrt[n]{a_n} = a]Okay, so I need to relate the ratio of consecutive terms to the nth root of the term itself. Maybe I can express ( a_n ) in terms of the previous terms and then take the nth root?Let me think about how ( a_n ) can be expressed. If I have ( a_{n+1} = a_n cdot frac{a_{n+1}}{a_n} ), then recursively, ( a_n ) can be written as the product of all previous ratios. So, starting from some initial term ( a_1 ), we can write:[a_n = a_1 cdot frac{a_2}{a_1} cdot frac{a_3}{a_2} cdot ldots cdot frac{a_n}{a_{n-1}}]Oh, that's a telescoping product! All the intermediate terms cancel out, leaving:[a_n = a_1 cdot prod_{k=1}^{n-1} frac{a_{k+1}}{a_k}]So, ( a_n ) is the product of the initial term ( a_1 ) and all the ratios ( frac{a_{k+1}}{a_k} ) from ( k = 1 ) to ( k = n-1 ). That makes sense.Now, if I take the nth root of ( a_n ), I get:[sqrt[n]{a_n} = sqrt[n]{a_1 cdot prod_{k=1}^{n-1} frac{a_{k+1}}{a_k}}]Since the nth root of a product is the product of the nth roots, this can be rewritten as:[sqrt[n]{a_n} = sqrt[n]{a_1} cdot prod_{k=1}^{n-1} sqrt[n]{frac{a_{k+1}}{a_k}}]Hmm, that's a bit complicated. Maybe there's a better way to approach this. I recall that if the limit of the ratio ( frac{a_{n+1}}{a_n} ) is ( a ), then for large ( n ), ( frac{a_{n+1}}{a_n} ) is approximately ( a ). So, maybe ( a_n ) behaves roughly like ( a_1 cdot a^{n-1} ) for large ( n ).If that's the case, then:[sqrt[n]{a_n} approx sqrt[n]{a_1 cdot a^{n-1}} = sqrt[n]{a_1} cdot a^{frac{n-1}{n}} = sqrt[n]{a_1} cdot a^{1 - frac{1}{n}}]As ( n ) approaches infinity, ( sqrt[n]{a_1} ) approaches 1 because any constant raised to the power of ( frac{1}{n} ) tends to 1 as ( n ) becomes large. Similarly, ( a^{1 - frac{1}{n}} ) approaches ( a ) because ( frac{1}{n} ) approaches 0. Therefore, the entire expression approaches ( a ).But wait, this is more of an intuitive argument rather than a rigorous proof. I need to make this more precise.Let me consider taking the natural logarithm of ( sqrt[n]{a_n} ) to simplify the expression. The natural logarithm is a continuous function, so if I can find the limit of the logarithm, I can exponentiate the result to get the original limit.Let:[L = lim_{n to infty} sqrt[n]{a_n}]Taking the natural logarithm of both sides:[ln L = lim_{n to infty} frac{1}{n} ln a_n]So, if I can find ( lim_{n to infty} frac{1}{n} ln a_n ), I can find ( L ).Now, from earlier, we have:[a_n = a_1 cdot prod_{k=1}^{n-1} frac{a_{k+1}}{a_k}]Taking the natural logarithm of both sides:[ln a_n = ln a_1 + sum_{k=1}^{n-1} ln left( frac{a_{k+1}}{a_k} right )]So,[frac{1}{n} ln a_n = frac{ln a_1}{n} + frac{1}{n} sum_{k=1}^{n-1} ln left( frac{a_{k+1}}{a_k} right )]As ( n ) approaches infinity, ( frac{ln a_1}{n} ) approaches 0 because ( ln a_1 ) is a constant. Therefore, we have:[ln L = lim_{n to infty} frac{1}{n} sum_{k=1}^{n-1} ln left( frac{a_{k+1}}{a_k} right )]Now, this looks like the average of the terms ( ln left( frac{a_{k+1}}{a_k} right ) ) from ( k = 1 ) to ( k = n-1 ). If the limit of ( frac{a_{k+1}}{a_k} ) as ( k ) approaches infinity is ( a ), then the limit of ( ln left( frac{a_{k+1}}{a_k} right ) ) is ( ln a ).But does the average of these terms approach ( ln a ) as ( n ) becomes large? I think so, because if the terms ( ln left( frac{a_{k+1}}{a_k} right ) ) approach ( ln a ), then their average should also approach ( ln a ). This is similar to the idea that if a sequence converges to a limit, then the average of the first ( n ) terms also converges to the same limit.To make this more precise, let's consider that for large ( k ), ( ln left( frac{a_{k+1}}{a_k} right ) ) is close to ( ln a ). So, for sufficiently large ( k ), say ( k geq N ), ( ln left( frac{a_{k+1}}{a_k} right ) ) is within ( epsilon ) of ( ln a ).Therefore, the sum ( sum_{k=1}^{n-1} ln left( frac{a_{k+1}}{a_k} right ) ) can be split into two parts: the first ( N ) terms and the remaining ( n - N - 1 ) terms. The first part is a constant with respect to ( n ), and the second part is approximately ( (n - N - 1) ln a ).So, as ( n ) becomes very large, the average ( frac{1}{n} sum_{k=1}^{n-1} ln left( frac{a_{k+1}}{a_k} right ) ) is approximately:[frac{1}{n} left( C + (n - N - 1) ln a right ) approx frac{C}{n} + ln a - frac{N + 1}{n} ln a]As ( n to infty ), both ( frac{C}{n} ) and ( frac{N + 1}{n} ln a ) approach 0, leaving ( ln a ). Therefore,[ln L = ln a implies L = a]So, we've shown that:[lim_{n to infty} sqrt[n]{a_n} = a]This completes the proof.**Final Answer**boxed{a}

answer：First, I need to determine how many times the prime number 7 appears in the prime factorization of the product of integers from 1 to 50, which is 50 factorial (50!).To count the number of times 7 appears as a factor in the numbers from 1 to 50, I will divide 50 by 7. This gives me the number of multiples of 7 within that range.Next, I will take the quotient from the previous division and divide it by 7 again. This step accounts for numbers that are multiples of higher powers of 7, such as 49 (which is 7 squared).I will continue this process until the quotient is less than 7, ensuring that I have accounted for all possible factors of 7 in the prime factorization.Finally, I will sum up all the quotients obtained from each division to find the total number of times 7 appears as a factor in 50!. This total will give me the greatest integer ( z ) for which ( 7^z ) is a factor of ( m ).