💡To determine the sum of all positive integer values of ( k ) for which the quadratic equation ( x^2 - kx + 18 = 0 ) has integer solutions, we can use Vieta's formulas. 1. **Vieta's Formulas**: For a quadratic equation ( x^2 - kx + 18 = 0 ) with roots ( alpha ) and ( beta ): - The sum of the roots is ( alpha + beta = k ). - The product of the roots is ( alpha beta = 18 ).2. **Finding Integer Pairs**: We need to find all pairs of integers ( (alpha, beta) ) such that their product is 18. These pairs are: - ( (1, 18) ) - ( (-1, -18) ) - ( (2, 9) ) - ( (-2, -9) ) - ( (3, 6) ) - ( (-3, -6) ) - ( (6, 3) ) - ( (-6, -3) ) - ( (9, 2) ) - ( (-9, -2) ) - ( (18, 1) ) - ( (-18, -1) )3. **Calculating ( k )**: For each pair, calculate ( k = alpha + beta ) and consider only the positive values: - ( 1 + 18 = 19 ) - ( -1 + (-18) = -19 ) (not positive) - ( 2 + 9 = 11 ) - ( -2 + (-9) = -11 ) (not positive) - ( 3 + 6 = 9 ) - ( -3 + (-6) = -9 ) (not positive) - ( 6 + 3 = 9 ) - ( -6 + (-3) = -9 ) (not positive) - ( 9 + 2 = 11 ) - ( -9 + (-2) = -11 ) (not positive) - ( 18 + 1 = 19 ) - ( -18 + (-1) = -19 ) (not positive)4. **Distinct Positive Values of ( k )**: The distinct positive values of ( k ) are 19, 11, and 9.5. **Summing the Values**: Add these distinct positive values together: [ 19 + 11 + 9 = 39 ]Therefore, the sum of all such values of ( k ) is (boxed{39}).

💡First, I recognize that the problem involves compound interest, which means the interest is calculated and added to the principal multiple times over the investment period. The formula for compound interest is:[ A = P left(1 + frac{r}{n}right)^{nt} ]Where:- ( A ) is the amount of money accumulated after ( t ) years, including interest.- ( P ) is the principal amount (the initial amount of money).- ( r ) is the annual interest rate (in decimal).- ( n ) is the number of times that interest is compounded per year.- ( t ) is the time the money is invested for, in years.Given:- ( A = 121.00000000000001 )- ( P = 100 )- ( n = 2 ) (since the interest is compounded semiannually)- ( t = 1 ) yearPlugging these values into the formula:[ 121 = 100 left(1 + frac{r}{2}right)^{2 times 1} ]Simplifying the equation:[ 1.21 = left(1 + frac{r}{2}right)^2 ]Taking the square root of both sides to solve for ( r ):[ sqrt{1.21} = 1 + frac{r}{2} ][ 1.1 = 1 + frac{r}{2} ]Subtracting 1 from both sides:[ 0.1 = frac{r}{2} ]Multiplying both sides by 2:[ 0.2 = r ]Converting the decimal to a percentage:[ r = 20% ]Therefore, the annual interest rate of the account is 20%.

💡Alright, so I have this problem about mixing two types of sugar, Type A and Type B, to make what's called assorted sugar. The problem says that if I make 1 kilogram of assorted sugar using 3 parts Type A and 2 parts Type B, it costs 1.32 yuan more than making 1 kilogram using 2 parts Type A and 3 parts Type B. I need to find out how much more expensive 1 kilogram of Type A sugar is compared to 1 kilogram of Type B sugar.Okay, let's break this down. First, I need to understand what it means by "parts." I think it means parts by weight, so if I have 3 parts Type A and 2 parts Type B, the total number of parts is 5. That means in 1 kilogram of assorted sugar, 3/5 of it is Type A and 2/5 is Type B. Similarly, for the other mixture, 2/5 is Type A and 3/5 is Type B.So, let's define some variables to make this clearer. Let's say the cost per kilogram of Type A sugar is 'a' yuan, and the cost per kilogram of Type B sugar is 'b' yuan. Then, the cost of the first mixture (3 parts A and 2 parts B) would be (3/5)a + (2/5)b yuan per kilogram. Similarly, the cost of the second mixture (2 parts A and 3 parts B) would be (2/5)a + (3/5)b yuan per kilogram.According to the problem, the first mixture is 1.32 yuan more expensive than the second mixture. So, if I subtract the cost of the second mixture from the cost of the first mixture, I should get 1.32 yuan. Let's write that equation out:(3/5)a + (2/5)b - [(2/5)a + (3/5)b] = 1.32Okay, let's simplify this equation step by step. First, distribute the negative sign to both terms inside the brackets:(3/5)a + (2/5)b - (2/5)a - (3/5)b = 1.32Now, combine like terms. The terms with 'a' are (3/5)a and -(2/5)a, and the terms with 'b' are (2/5)b and -(3/5)b.For the 'a' terms:(3/5)a - (2/5)a = (1/5)aFor the 'b' terms:(2/5)b - (3/5)b = -(1/5)bSo, putting it all together, we have:(1/5)a - (1/5)b = 1.32Hmm, this looks like we can factor out (1/5):(1/5)(a - b) = 1.32To solve for (a - b), we can multiply both sides of the equation by 5:a - b = 1.32 * 5Let's calculate that:1.32 * 5 = 6.6So, a - b = 6.6This means that Type A sugar is 6.6 yuan more expensive per kilogram than Type B sugar.Wait, let me double-check my steps to make sure I didn't make any mistakes. I defined the variables correctly, set up the equation based on the given cost difference, simplified it step by step, and solved for (a - b). Everything seems to add up. I think I did this correctly.Just to be thorough, let's plug the value back into the original equation to verify:(1/5)(6.6) = 1.32Calculating the left side:1.32 = 1.32Yes, that checks out. So, my answer seems correct.In summary, by setting up the cost equations for both mixtures and using the given cost difference, I was able to determine that Type A sugar is 6.6 yuan more expensive per kilogram than Type B sugar.

💡Alright, so I've got this problem about geometric and arithmetic sequences. It's divided into three parts, and I need to figure out each step by step. Let me start with part (1).**Problem (1):** Find the general formulas for the geometric sequence {a_n} and the arithmetic sequence {b_n}.Given:- a₁ = b₃ = 3- a₂ = b₉- a₃ = b₂₇First, let me recall what a geometric sequence is. A geometric sequence has a common ratio between consecutive terms. So, if a₁ is the first term, then a₂ = a₁ * q, a₃ = a₁ * q², and so on, where q is the common ratio.Similarly, an arithmetic sequence has a common difference between consecutive terms. If b₁ is the first term and d is the common difference, then bₙ = b₁ + (n-1)d.Given that a₁ = b₃ = 3. So, let's write that down:a₁ = 3b₃ = 3Since b₃ is the third term of the arithmetic sequence, it can be expressed as:b₃ = b₁ + 2d = 3So, we have one equation: b₁ + 2d = 3.Next, we know that a₂ = b₉. Let's express both sides.a₂ is the second term of the geometric sequence, so:a₂ = a₁ * q = 3qb₉ is the ninth term of the arithmetic sequence:b₉ = b₁ + 8dSo, setting them equal:3q = b₁ + 8dSimilarly, a₃ = b₂₇.a₃ = a₁ * q² = 3q²b₂₇ = b₁ + 26dSo,3q² = b₁ + 26dNow, we have three equations:1) b₁ + 2d = 32) 3q = b₁ + 8d3) 3q² = b₁ + 26dOur unknowns are b₁, d, and q. Let's try to solve these equations.From equation 1, we can express b₁ in terms of d:b₁ = 3 - 2dNow, substitute this into equations 2 and 3.Equation 2 becomes:3q = (3 - 2d) + 8d = 3 + 6dSo,3q = 3 + 6dDivide both sides by 3:q = 1 + 2dEquation 3 becomes:3q² = (3 - 2d) + 26d = 3 + 24dSo,3q² = 3 + 24dDivide both sides by 3:q² = 1 + 8dNow, from equation 2, we have q = 1 + 2d. Let's substitute this into equation 3.q² = (1 + 2d)² = 1 + 4d + 4d²But from equation 3, q² = 1 + 8dSo,1 + 4d + 4d² = 1 + 8dSubtract 1 from both sides:4d + 4d² = 8dSubtract 8d from both sides:4d² - 4d = 0Factor:4d(d - 1) = 0So, d = 0 or d = 1But the problem states that the common difference d is non-zero, so d = 1.Now, substitute d = 1 back into equation 1:b₁ + 2(1) = 3 => b₁ = 3 - 2 = 1So, b₁ = 1, d = 1.Now, from equation 2, q = 1 + 2d = 1 + 2(1) = 3So, q = 3.Therefore, the general formulas are:For the geometric sequence {a_n}:a_n = a₁ * q^{n-1} = 3 * 3^{n-1} = 3^nFor the arithmetic sequence {b_n}:b_n = b₁ + (n - 1)d = 1 + (n - 1)(1) = nSo, b_n = nLet me double-check these results.Given a₁ = 3, a₂ = 3*3 = 9, a₃ = 27b₃ = 3, which is correct because b_n = n, so b₃ = 3.a₂ = 9, which should equal b₉. Since b₉ = 9, that's correct.a₃ = 27, which should equal b₂₇. Since b₂₇ = 27, that's correct.So, part (1) seems solved.**Problem (2):** Let S_n be the sum of the first n terms of {a_n}, and T_n be the sum of the first n terms of {b_n}. Prove that T_n / S_n ≤ 1/3.First, let's write expressions for S_n and T_n.For the geometric series {a_n}, the sum S_n is:S_n = a₁ + a₂ + ... + a_n = 3 + 9 + 27 + ... + 3^nThe formula for the sum of a geometric series is S_n = a₁*(q^n - 1)/(q - 1). Here, a₁ = 3, q = 3.So,S_n = 3*(3^n - 1)/(3 - 1) = (3^{n+1} - 3)/2For the arithmetic series {b_n}, the sum T_n is:T_n = b₁ + b₂ + ... + b_n = 1 + 2 + 3 + ... + nThe formula for the sum of an arithmetic series is T_n = n*(b₁ + b_n)/2. Here, b₁ = 1, b_n = n.So,T_n = n*(1 + n)/2 = n(n + 1)/2So, now, we need to compute T_n / S_n:T_n / S_n = [n(n + 1)/2] / [(3^{n+1} - 3)/2] = [n(n + 1)] / [3^{n+1} - 3]We need to prove that this ratio is ≤ 1/3 for all n.So, we need to show:[n(n + 1)] / [3^{n+1} - 3] ≤ 1/3Multiply both sides by [3^{n+1} - 3], which is positive since 3^{n+1} > 3 for n ≥ 1.So,n(n + 1) ≤ (3^{n+1} - 3)/3 = 3^n - 1Thus, the inequality reduces to:n(n + 1) ≤ 3^n - 1We need to show that for all positive integers n, n(n + 1) ≤ 3^n - 1.Let me test this for small n:n=1: 1*2=2 ≤ 3^1 -1=2. So, 2 ≤ 2, which holds.n=2: 2*3=6 ≤ 3^2 -1=8. 6 ≤ 8, holds.n=3: 3*4=12 ≤ 3^3 -1=26. 12 ≤26, holds.n=4: 4*5=20 ≤ 81 -1=80. 20 ≤80, holds.n=5:5*6=30 ≤243 -1=242. 30 ≤242, holds.It seems to hold for these values. Now, let's try to prove it in general.We can use mathematical induction.**Base case:** n=1, as above, 2 ≤ 2, holds.**Inductive step:** Assume that for some k ≥1, k(k + 1) ≤ 3^k -1.We need to show that (k +1)(k +2) ≤ 3^{k+1} -1.Let's compute 3^{k+1} -1 = 3*3^k -1.From the inductive hypothesis, 3^k ≥ k(k +1) +1.So, 3^{k+1} -1 = 3*3^k -1 ≥ 3*(k(k +1) +1) -1 = 3k(k +1) +3 -1 = 3k(k +1) +2Now, we need to show that (k +1)(k +2) ≤ 3k(k +1) +2Let me compute the difference:3k(k +1) +2 - (k +1)(k +2) = 3k(k +1) +2 - [k^2 + 3k +2] = 3k² +3k +2 -k² -3k -2 = 2k²Since k ≥1, 2k² ≥0, so 3k(k +1) +2 ≥ (k +1)(k +2)Thus, 3^{k+1} -1 ≥ (k +1)(k +2)Therefore, by induction, the inequality holds for all n ≥1.Hence, T_n / S_n ≤1/3.Alternatively, another approach is to consider the function f(n) = T_n / S_n and show it's decreasing.Compute f(n) = [n(n +1)] / [3^{n+1} -3]Compute f(n+1) - f(n):= [(n+1)(n+2)] / [3^{n+2} -3] - [n(n +1)] / [3^{n+1} -3]Let me factor out common terms:= (n+1)[(n+2)/(3^{n+2} -3) - n/(3^{n+1} -3)]Let me compute the difference inside the brackets:(n+2)/(3^{n+2} -3) - n/(3^{n+1} -3)= (n+2)/(3*(3^{n+1} -1)) - n/(3^{n+1} -3)= (n+2)/(3*(3^{n+1} -1)) - n/(3^{n+1} -3)Let me denote A = 3^{n+1} -1, then 3^{n+2} -3 = 3*(3^{n+1} -1) = 3ASo, the expression becomes:(n+2)/(3A) - n/(A - 2)Wait, that might complicate things. Alternatively, let's compute a common denominator.Let me write both fractions with denominator (3^{n+2} -3)(3^{n+1} -3):= (n+2)(3^{n+1} -3) - n(3^{n+2} -3) / [(3^{n+2} -3)(3^{n+1} -3)]Simplify numerator:= (n+2)(3^{n+1} -3) - n*3^{n+2} + 3n= (n+2)3^{n+1} -3(n+2) -n*3^{n+2} +3n= (n+2)3^{n+1} -3n -6 -n*3^{n+2} +3n= (n+2)3^{n+1} -n*3^{n+2} -6Factor 3^{n+1}:= 3^{n+1}[n +2 -3n] -6= 3^{n+1}(-2n +2) -6= -2(n -1)3^{n+1} -6So, numerator is negative because 3^{n+1} is positive, and multiplied by -2(n -1), which is negative for n ≥1.Thus, the entire expression f(n+1) - f(n) is negative, meaning f(n) is decreasing.Since f(n) is decreasing, the maximum value occurs at n=1, which is f(1)= [1*2]/[3^2 -3] = 2/(9 -3)=2/6=1/3.Therefore, f(n) ≤1/3 for all n ≥1.Hence, T_n / S_n ≤1/3.**Problem (3):** Define c_n as:c_n = { a_{(n+1)/2} / (b_n +2) if n is odd, -a_{n/2} / [(b_n -1)(b_n +1)] if n is even }Find the sum from i=1 to 2n of c_i.First, let's write down c_i for both cases.Given that {a_n} is 3^n, and {b_n} is n.So, for odd i:c_i = a_{(i+1)/2} / (b_i +2) = 3^{(i+1)/2} / (i +2)For even i:c_i = -a_{i/2} / [(b_i -1)(b_i +1)] = -3^{i/2} / [(i -1)(i +1)]So, c_i alternates between these two expressions depending on whether i is odd or even.We need to compute the sum from i=1 to 2n of c_i.Let me separate the sum into odd and even terms.Number of terms: 2n.Number of odd terms: n (i=1,3,5,...,2n-1)Number of even terms: n (i=2,4,6,...,2n)So, sum = sum_{k=1}^n c_{2k-1} + sum_{k=1}^n c_{2k}Compute each sum separately.First, compute sum_{k=1}^n c_{2k-1}:c_{2k-1} = 3^{(2k -1 +1)/2} / (2k -1 +2) = 3^{k} / (2k +1)So, sum_{k=1}^n c_{2k-1} = sum_{k=1}^n [3^k / (2k +1)]Similarly, compute sum_{k=1}^n c_{2k}:c_{2k} = -3^{2k /2} / [(2k -1)(2k +1)] = -3^{k} / [(2k -1)(2k +1)]So, sum_{k=1}^n c_{2k} = -sum_{k=1}^n [3^k / ((2k -1)(2k +1))]Therefore, total sum:sum_{i=1}^{2n} c_i = sum_{k=1}^n [3^k / (2k +1)] - sum_{k=1}^n [3^k / ((2k -1)(2k +1))]Let me combine these two sums:= sum_{k=1}^n [3^k / (2k +1) - 3^k / ((2k -1)(2k +1))]Factor out 3^k / (2k +1):= sum_{k=1}^n [3^k / (2k +1) * (1 - 1/(2k -1))]Simplify inside the brackets:1 - 1/(2k -1) = (2k -1 -1)/(2k -1) = (2k -2)/(2k -1) = 2(k -1)/(2k -1)So,= sum_{k=1}^n [3^k / (2k +1) * 2(k -1)/(2k -1)]= 2 sum_{k=1}^n [3^k (k -1) / ((2k +1)(2k -1))]Hmm, this seems a bit complicated. Maybe there's a telescoping approach.Let me look at the individual terms:For each k, the term is:3^k / (2k +1) - 3^k / [(2k -1)(2k +1)]Let me write this as:3^k [1/(2k +1) - 1/((2k -1)(2k +1))]= 3^k [ (2k -1) -1 ] / [(2k -1)(2k +1)]= 3^k [2k -2] / [(2k -1)(2k +1)]= 3^k * 2(k -1) / [(2k -1)(2k +1)]So, the term is 2*3^k*(k -1)/[(2k -1)(2k +1)]Hmm, not sure if this telescopes directly. Maybe we can express this as a difference of two terms.Let me think about whether 3^k*(k -1) can be expressed in terms of 3^{k-1} or something.Alternatively, perhaps we can write this as a telescoping series by considering the difference between consecutive terms.Let me consider the expression:Let me define d_k = 3^{k} / (2k -1)Then, let's compute d_{k} - d_{k+1}:= 3^{k} / (2k -1) - 3^{k+1} / (2k +1)= 3^{k} [1/(2k -1) - 3/(2k +1)]= 3^{k} [ (2k +1 - 3(2k -1)) / ((2k -1)(2k +1)) ]= 3^{k} [ (2k +1 -6k +3) / ((2k -1)(2k +1)) ]= 3^{k} [ (-4k +4) / ((2k -1)(2k +1)) ]= 3^{k} * (-4)(k -1) / ((2k -1)(2k +1))= -4 * [3^{k} (k -1) / ((2k -1)(2k +1))]But our term is 2*3^k*(k -1)/[(2k -1)(2k +1)]So, we have:2*3^k*(k -1)/[(2k -1)(2k +1)] = (-1/2)(d_k - d_{k+1})Therefore,sum_{k=1}^n [2*3^k*(k -1)/((2k -1)(2k +1))] = (-1/2) sum_{k=1}^n (d_k - d_{k+1})This is a telescoping series:= (-1/2)[d_1 - d_{n+1}]Compute d_1 = 3^1 / (2*1 -1) = 3 /1 =3d_{n+1}=3^{n+1}/(2(n+1)-1)=3^{n+1}/(2n +1)Therefore,sum = (-1/2)[3 - 3^{n+1}/(2n +1)] = (-1/2)*3 + (1/2)*3^{n+1}/(2n +1) = -3/2 + 3^{n+1}/(2*(2n +1))So, the total sum is:sum_{i=1}^{2n} c_i = -3/2 + 3^{n+1}/(4n +2)Alternatively, factor out 1/2:= (3^{n+1}/(4n +2)) - 3/2So, the final expression is:sum_{i=1}^{2n} c_i = (3^{n+1})/(4n +2) - 3/2Let me verify this with n=1.For n=1, sum from i=1 to 2 of c_i.Compute c₁ and c₂.c₁ = a_{(1+1)/2}/(b₁ +2) = a₁/(1 +2)=3/3=1c₂ = -a_{2/2}/[(b₂ -1)(b₂ +1)] = -a₁/(1*3)= -3/3= -1So, sum =1 + (-1)=0Now, plug n=1 into the formula:(3^{2})/(4*1 +2) -3/2=9/6 -3/2=3/2 -3/2=0. Correct.Another test with n=2.Compute sum from i=1 to 4 of c_i.c₁=1, c₂=-1, c₃= a_{(3+1)/2}/(3 +2)=a₂/5=9/5=1.8c₄= -a_{4/2}/[(4 -1)(4 +1)]= -a₂/(3*5)= -9/15= -0.6Sum=1 -1 +1.8 -0.6=1.2Now, plug n=2 into the formula:(3^{3})/(4*2 +2) -3/2=27/10 -3/2=2.7 -1.5=1.2. Correct.Another test with n=3.Compute sum from i=1 to 6 of c_i.c₁=1, c₂=-1, c₃=9/5=1.8, c₄=-9/15=-0.6, c₅=a_{(5+1)/2}/(5 +2)=a₃/7=27/7≈3.857, c₆=-a_{6/2}/[(6 -1)(6 +1)]= -a₃/(5*7)= -27/35≈-0.771Sum≈1 -1 +1.8 -0.6 +3.857 -0.771≈1 -1=0; 0 +1.8=1.8; 1.8 -0.6=1.2; 1.2 +3.857≈5.057; 5.057 -0.771≈4.286Now, plug n=3 into the formula:(3^{4})/(4*3 +2) -3/2=81/14 -3/2≈5.7857 -1.5≈4.2857. Correct.So, the formula seems to hold.**Final Answer**(1) The general formulas are boxed{a_n = 3^n} and boxed{b_n = n}.(2) It is proven that boxed{dfrac{T_n}{S_n} leq dfrac{1}{3}}.(3) The sum is boxed{dfrac{3^{n+1}}{4n + 2} - dfrac{3}{2}}.

💡First, I need to calculate 35% of 180. This can be done by multiplying 180 by 0.35, which gives me 63.Next, I need to find one-third of this value. Dividing 63 by 3 results in 21.To find one-third less than 63, I subtract 21 from 63, which equals 42.Now, I subtract this result from the original number, 180, to find the remaining value: 180 minus 42 equals 138.Then, I calculate three-fifths of this remaining value. Multiplying 138 by 3/5 gives me 82.8.Finally, I raise this value to the power of 2: 82.8 squared equals 6857.84.

💡First, I need to determine the total number of students in the class, which is 25.Next, I'll calculate the total marks obtained by the 3 students who scored 95 marks each:3 students * 95 marks = 285 marks.Then, I'll calculate the total marks obtained by the 3 students who scored zero:3 students * 0 marks = 0 marks.After that, I'll find out how many students are left:25 total students - 3 students with 95 marks - 3 students with 0 marks = 19 students.These 19 students have an average score of 45 marks. So, their total marks are:19 students * 45 marks = 855 marks.Now, I'll add up all the marks to get the total marks for the whole class:285 marks + 0 marks + 855 marks = 1140 marks.Finally, I'll calculate the average marks for the whole class by dividing the total marks by the number of students:1140 marks / 25 students = 45.6 marks.Therefore, the average marks for the whole class is 45.6.