💡Okay, so I'm trying to solve this vector problem in a tetrahedron OABC. The question is about finding the vector OE, where D is the midpoint of BC and E is the midpoint of AD. The options are given, and I need to figure out which one is correct.First, let me visualize the tetrahedron. It has four vertices: O, A, B, and C. So, O is like the origin, and A, B, C are the other three vertices. D is the midpoint of BC, which means it's halfway between B and C. Then E is the midpoint of AD, so it's halfway between A and D.I remember that in vector problems, it's helpful to express points in terms of position vectors from the origin. So, maybe I can express OE in terms of OA, OB, and OC? Or maybe in terms of OA, AB, and AC? Let me see.The options are given in terms of OA, AB, and AC, so maybe I should express OE using these vectors. Let's start by finding the position vectors of points D and E.Since D is the midpoint of BC, the position vector of D, which I'll denote as OD, can be found by averaging the position vectors of B and C. So, OD = (OB + OC)/2. But wait, in the options, they don't use OB and OC directly; they use AB and AC. Hmm, maybe I need to express OD in terms of OA, AB, and AC.Let me recall that AB = OB - OA and AC = OC - OA. So, if I can express OD in terms of OA, AB, and AC, that might help. Let's try that.OD = (OB + OC)/2. Since AB = OB - OA, then OB = OA + AB. Similarly, AC = OC - OA, so OC = OA + AC. Substituting these into OD:OD = ( (OA + AB) + (OA + AC) ) / 2= (2OA + AB + AC) / 2= OA + (AB + AC)/2Okay, so OD is OA plus half of (AB + AC). Now, E is the midpoint of AD, so OE is the average of OA and OD. Let me write that down.OE = (OA + OD)/2We already have OD in terms of OA, AB, and AC, so let's substitute that in:OE = (OA + OA + (AB + AC)/2 ) / 2= (2OA + (AB + AC)/2 ) / 2Let me simplify this. First, factor out the 2 in the numerator:= (2OA)/2 + (AB + AC)/4= OA + (AB + AC)/4So, OE is equal to OA plus one fourth of AB plus one fourth of AC. Looking back at the options, that's option B: OA + (1/4)AB + (1/4)AC.Wait, let me double-check my steps to make sure I didn't make a mistake.1. OD is the midpoint of BC, so OD = (OB + OC)/2. Correct.2. Expressed OB and OC in terms of OA, AB, and AC: OB = OA + AB, OC = OA + AC. Correct.3. Substituted into OD: (OA + AB + OA + AC)/2 = OA + (AB + AC)/2. Correct.4. Then, E is the midpoint of AD, so OE = (OA + OD)/2. Correct.5. Substituted OD: (OA + OA + (AB + AC)/2)/2 = (2OA + (AB + AC)/2)/2. Correct.6. Simplified: OA + (AB + AC)/4. Correct.Yes, that seems right. So, the answer should be option B.But just to be thorough, let me consider another approach. Maybe using position vectors relative to O.Let me denote the position vectors of A, B, and C as vectors a, b, and c respectively. Then, the position vector of D, which is the midpoint of BC, is (b + c)/2. Then, the position vector of E, which is the midpoint of AD, is (a + (b + c)/2)/2 = (2a + b + c)/4.So, OE = (2a + b + c)/4. Now, let's express this in terms of OA, AB, and AC.OA is vector a. AB is vector b - a, and AC is vector c - a.So, let's rewrite OE:OE = (2a + b + c)/4= (2a + (b - a) + (c - a) + 2a)/4? Wait, no, that's complicating things.Wait, maybe express b and c in terms of OA, AB, and AC.Since AB = b - a, then b = a + AB.Similarly, AC = c - a, so c = a + AC.Substitute these into OE:OE = (2a + (a + AB) + (a + AC))/4= (2a + a + AB + a + AC)/4= (4a + AB + AC)/4= a + (AB + AC)/4Which is OA + (1/4)AB + (1/4)AC, which is option B. So, same result.Therefore, I'm confident that option B is correct.**Final Answer**boxed{B}

💡Okay, so I have three problems here about binomial expansions. Let me try to tackle them one by one. Starting with the first problem: In the expansion of (1 + x)^n, the coefficient of x^3 is 7 times the coefficient of x. I need to find n. Hmm, binomial expansions... I remember that the coefficient of x^k in (1 + x)^n is given by the binomial coefficient C(n, k). So, the coefficient of x is C(n, 1) and the coefficient of x^3 is C(n, 3). The problem says that C(n, 3) is 7 times C(n, 1). So, mathematically, that would be C(n, 3) = 7 * C(n, 1). Let me write that down:C(n, 3) = 7 * C(n, 1)I know that C(n, k) is equal to n! / (k! * (n - k)!). So, substituting the values, we get:n! / (3! * (n - 3)!) = 7 * (n! / (1! * (n - 1)!))Simplify both sides. Let's compute the left side first:n! / (6 * (n - 3)!) And the right side:7 * (n! / (1 * (n - 1)!)) = 7 * (n! / (n - 1)!))Hmm, I can simplify both sides further. Notice that n! / (n - 3)! is equal to n * (n - 1) * (n - 2). Similarly, n! / (n - 1)! is equal to n. So, substituting that in:Left side: [n * (n - 1) * (n - 2)] / 6Right side: 7 * nSo, putting it together:[n * (n - 1) * (n - 2)] / 6 = 7nI can cancel out an n from both sides, assuming n ≠ 0 (which it can't be in this context). So:(n - 1)(n - 2) / 6 = 7Multiply both sides by 6:(n - 1)(n - 2) = 42Now, expand the left side:n^2 - 3n + 2 = 42Subtract 42 from both sides:n^2 - 3n - 40 = 0Now, solve this quadratic equation. Let me compute the discriminant:D = (-3)^2 - 4 * 1 * (-40) = 9 + 160 = 169So, sqrt(D) = 13Thus, solutions are:n = [3 ± 13] / 2Which gives:n = (3 + 13)/2 = 16/2 = 8n = (3 - 13)/2 = (-10)/2 = -5But since n must be a positive integer in the context of binomial coefficients, n = 8.Alright, that seems solid. Let me just double-check:C(8, 3) = 56C(8, 1) = 856 is indeed 7 times 8. Yep, that works.Moving on to the second problem: In the expansion of (ax + 1)^7, where a ≠ 0, the coefficient of x^3 is the arithmetic mean of the coefficients of x^2 and x^4. Find a.Okay, so arithmetic mean means that coefficient of x^3 is equal to (coefficient of x^2 + coefficient of x^4)/2.Let me write that down:C(7, 3) * a^3 = [C(7, 2) * a^2 + C(7, 4) * a^4] / 2Wait, hold on. The general term in the expansion of (ax + 1)^7 is C(7, k) * (ax)^k * 1^(7 - k). So, the coefficient of x^k is C(7, k) * a^k.So, coefficient of x^2 is C(7, 2) * a^2, coefficient of x^3 is C(7, 3) * a^3, and coefficient of x^4 is C(7, 4) * a^4.So, the equation is:C(7, 3) * a^3 = [C(7, 2) * a^2 + C(7, 4) * a^4] / 2Let me compute the binomial coefficients:C(7, 2) = 21C(7, 3) = 35C(7, 4) = 35 (since C(n, k) = C(n, n - k))So, substituting:35a^3 = [21a^2 + 35a^4] / 2Multiply both sides by 2 to eliminate the denominator:70a^3 = 21a^2 + 35a^4Let me rearrange the terms:35a^4 - 70a^3 + 21a^2 = 0Factor out common terms. Let's see, 35, 70, 21 have a common factor of 7, and each term has a^2:7a^2(5a^2 - 10a + 3) = 0Since a ≠ 0, we can divide both sides by 7a^2:5a^2 - 10a + 3 = 0Now, solve this quadratic equation for a. Let's compute the discriminant:D = (-10)^2 - 4 * 5 * 3 = 100 - 60 = 40So, sqrt(D) = sqrt(40) = 2*sqrt(10)Thus, solutions are:a = [10 ± 2sqrt(10)] / (2 * 5) = [10 ± 2sqrt(10)] / 10 = [5 ± sqrt(10)] / 5 = 1 ± (sqrt(10)/5)So, a = 1 + sqrt(10)/5 or a = 1 - sqrt(10)/5Let me rationalize sqrt(10)/5 as (sqrt(10))/5, so the answers are a = 1 ± sqrt(10)/5.Wait, let me verify:Compute 5a^2 -10a +3 =0Let me plug in a = 1 + sqrt(10)/5:First, compute a:a = (5 + sqrt(10))/5Compute 5a^2:5 * [(5 + sqrt(10))/5]^2 = 5 * [25 + 10sqrt(10) + 10]/25 = 5 * [35 + 10sqrt(10)]/25 = (35 + 10sqrt(10))/5 = 7 + 2sqrt(10)Compute -10a:-10 * (5 + sqrt(10))/5 = -2*(5 + sqrt(10)) = -10 - 2sqrt(10)Add 3:7 + 2sqrt(10) -10 - 2sqrt(10) +3 = (7 -10 +3) + (2sqrt(10) - 2sqrt(10)) = 0 + 0 = 0Similarly, for a = 1 - sqrt(10)/5:a = (5 - sqrt(10))/5Compute 5a^2:5 * [(5 - sqrt(10))/5]^2 = 5 * [25 -10sqrt(10) +10]/25 = 5 * [35 -10sqrt(10)]/25 = (35 -10sqrt(10))/5 =7 - 2sqrt(10)Compute -10a:-10*(5 - sqrt(10))/5 = -2*(5 - sqrt(10)) = -10 + 2sqrt(10)Add 3:7 -2sqrt(10) -10 + 2sqrt(10) +3 = (7 -10 +3) + (-2sqrt(10) +2sqrt(10)) = 0 +0 =0So, both solutions satisfy the equation. Hence, a =1 ± sqrt(10)/5.Alright, that seems correct.Now, the third problem: It is known that in the expansion of (2x + x^{log x})^8, the value of the term with the largest binomial coefficient is 1120. Find x.Hmm, okay. So, first, let's understand the expression: (2x + x^{log x})^8. The expansion will have terms from k=0 to k=8, each term being C(8, k)*(2x)^{8 -k}*(x^{log x})^k.We need to find the term with the largest binomial coefficient. Wait, in the expansion of (a + b)^n, the binomial coefficients are C(n, k) for k=0 to n. The largest binomial coefficient occurs at k = floor(n/2). For n=8, which is even, the largest coefficient is at k=4, which is C(8,4)=70.Wait, but the problem says "the value of the term with the largest binomial coefficient is 1120." So, the term corresponding to k=4 has the largest binomial coefficient, which is 70, and its value is 1120.So, let's write that term:Term when k=4: C(8,4)*(2x)^{8 -4}*(x^{log x})^4 = 70*(2x)^4*(x^{log x})^4Compute this term:70*(16x^4)*(x^{4 log x}) = 70*16*x^4*x^{4 log x}Simplify:70*16 = 1120So, 1120 * x^4 * x^{4 log x}But the term is equal to 1120, so:1120 * x^4 * x^{4 log x} = 1120Divide both sides by 1120:x^4 * x^{4 log x} = 1Simplify the exponents:x^{4 + 4 log x} =1Hmm, x^{4(1 + log x)} =1Since x is a positive real number (as it's in the logarithm), and any number to the power 0 is 1. So, the exponent must be 0:4(1 + log x) =0Divide both sides by 4:1 + log x =0So, log x = -1Assuming log is base 10? Wait, the problem doesn't specify. Hmm, in many contexts, log without base is assumed to be base 10, but in mathematics, sometimes it's natural log. Hmm, but since the answer is in terms of 10, maybe it's base 10.Wait, let me check:If log x = -1, then x =10^{-1}=1/10.Alternatively, if it's natural log, x=e^{-1}≈0.3679, but 1/10 is 0.1, which is a cleaner answer. Also, in problems like this, unless specified, log is often base 10. So, I think x=1/10.But wait, let me see if there are other possibilities. The equation x^{4(1 + log x)} =1 can have other solutions besides the exponent being 0. For example, if x=1, then 1 raised to any power is 1. So, x=1 is another solution.So, let's check:Case 1: 4(1 + log x)=0 => log x = -1 => x=10^{-1}=1/10Case 2: x=1, since 1^{anything}=1.So, x=1 and x=1/10 are both solutions.Wait, but let me verify:If x=1, then the term is 1120*1^4*1^{4 log 1}=1120*1*1=1120, which is correct.If x=1/10, then:x^{4(1 + log x)} = (1/10)^{4(1 + log(1/10))} = (1/10)^{4(1 -1)}= (1/10)^0=1, so yes, 1120*1=1120.So, both x=1 and x=1/10 satisfy the equation.But wait, let me think about the original expression: (2x + x^{log x})^8. If x=1, then x^{log x}=1^{log 1}=1^0=1, so the expression becomes (2*1 +1)^8=3^8=6561. But the term with the largest binomial coefficient is 1120, which is much less than 6561. So, is x=1 a valid solution?Wait, no, because the term itself is 1120, but the entire expansion is 6561. So, the term is 1120, which is correct, but does x=1 make sense?Wait, when x=1, all terms in the expansion are positive, but the term with the largest binomial coefficient is 70*(2*1)^4*(1)^4=70*16=1120, which is correct. So, x=1 is a valid solution.Similarly, when x=1/10, the term is 1120, as we saw.So, both x=1 and x=1/10 are solutions.Wait, but let me check if x=1 is the only other solution or if there are more.Looking back at the equation x^{4(1 + log x)} =1, we have two cases:1. The exponent is 0: 4(1 + log x)=0 => log x = -1 => x=10^{-1}=1/102. The base is 1: x=1So, these are the only two solutions.Therefore, x=1 or x=1/10.But wait, let me think again about the original expression: (2x + x^{log x})^8. If x=1, then x^{log x}=1^{0}=1, so the expression becomes (2 +1)^8=3^8=6561. But the term with the largest binomial coefficient is 1120, which is correct because 70*(2)^4*(1)^4=70*16=1120.Similarly, when x=1/10, x^{log x}=(1/10)^{log(1/10)}=(1/10)^{-1}=10, so the expression becomes (2*(1/10) +10)^8=(0.2 +10)^8=10.2^8, which is a large number, but the term with the largest binomial coefficient is 1120, as we saw.So, both solutions are valid.Hence, x=1 or x=1/10.I think that's it.**Final Answer**(1) boxed{8}(2) boxed{1 pm dfrac{sqrt{10}}{5}}(3) boxed{1 text{ and } dfrac{1}{10}}

💡Okay, so I need to solve this quadratic equation: x² + 2x - 15 = 0. Hmm, quadratic equations can be tricky, but I remember there are a few methods to solve them. Let me think about the different approaches I can take.First, I recall that quadratic equations are of the form ax² + bx + c = 0, and there are methods like factoring, completing the square, and using the quadratic formula. Since this equation has a leading coefficient of 1 (the coefficient of x² is 1), maybe factoring is a good starting point. But I'm not sure if it factors neatly. Let me check.To factor x² + 2x - 15, I need two numbers that multiply to -15 and add up to 2. Let's list the pairs of factors of -15:- 1 and -15 (1 + (-15) = -14)- -1 and 15 (-1 + 15 = 14)- 3 and -5 (3 + (-5) = -2)- -3 and 5 (-3 + 5 = 2)Oh, there we go! -3 and 5 multiply to -15 and add up to 2. So, the equation can be factored as (x - 3)(x + 5) = 0.Setting each factor equal to zero gives the solutions:x - 3 = 0 ⇒ x = 3x + 5 = 0 ⇒ x = -5So, the solutions are x = 3 and x = -5.But just to make sure I didn't make a mistake, maybe I should verify these solutions by plugging them back into the original equation.First, let's check x = 3:(3)² + 2*(3) - 15 = 9 + 6 - 15 = 15 - 15 = 0. That works.Now, let's check x = -5:(-5)² + 2*(-5) - 15 = 25 - 10 - 15 = 25 - 25 = 0. That also works.Okay, so factoring gave me the correct solutions. But what if I couldn't factor it easily? Maybe I should try another method, like completing the square, just to double-check.Starting with the original equation:x² + 2x - 15 = 0First, move the constant term to the other side:x² + 2x = 15To complete the square, I need to add a number to both sides so that the left side becomes a perfect square trinomial. The formula for completing the square is to take half of the coefficient of x, square it, and add it to both sides.The coefficient of x is 2, so half of that is 1, and squaring it gives 1. So, I add 1 to both sides:x² + 2x + 1 = 15 + 1This simplifies to:(x + 1)² = 16Now, take the square root of both sides:x + 1 = ±√16Which means:x + 1 = ±4Now, solve for x:x = -1 ± 4So, the two solutions are:x = -1 + 4 = 3andx = -1 - 4 = -5Again, I get the same solutions: x = 3 and x = -5. That's reassuring.Alternatively, I could use the quadratic formula, which is x = [-b ± √(b² - 4ac)] / (2a). Let's try that as well.In the equation x² + 2x - 15 = 0, the coefficients are:a = 1, b = 2, c = -15Plugging these into the quadratic formula:x = [-2 ± √(2² - 4*1*(-15))] / (2*1)Calculate the discriminant first:b² - 4ac = 4 - (-60) = 4 + 60 = 64So,x = [-2 ± √64] / 2√64 is 8, so:x = [-2 ± 8] / 2This gives two solutions:x = (-2 + 8)/2 = 6/2 = 3andx = (-2 - 8)/2 = (-10)/2 = -5Again, the same results: x = 3 and x = -5.So, no matter which method I use—factoring, completing the square, or the quadratic formula—I consistently get the solutions x = 3 and x = -5. This makes me confident that these are indeed the correct solutions to the equation x² + 2x - 15 = 0.I think I've covered all the possible methods, and each time, the solutions match. It's good to verify using multiple methods to ensure accuracy, especially when dealing with quadratic equations. I don't see any mistakes in my calculations, and the solutions satisfy the original equation when plugged back in. So, I'm pretty sure that x = 3 and x = -5 are the correct answers.

💡First, I will list the given numbers: 5, 8, 1, and 2.Next, I will compare each number to identify the smallest one.Comparing 5 and 8, 5 is smaller.Comparing 5 and 1, 1 is smaller.Comparing 1 and 2, 1 is smaller.Therefore, the smallest number is 1.

💡First, I recall that the area of a rhombus can be calculated using the formula:[text{Area} = frac{d_1 times d_2}{2}]where ( d_1 ) and ( d_2 ) are the lengths of the diagonals.Given that the area of the rhombus is 150 cm² and one of the diagonals (( d_1 )) is 10 cm, I can set up the equation:[150 = frac{10 times d_2}{2}]To solve for ( d_2 ), I first multiply both sides of the equation by 2:[300 = 10 times d_2]Next, I divide both sides by 10 to isolate ( d_2 ):[d_2 = frac{300}{10} = 30 text{ cm}]Therefore, the length of the other diagonal is 30 cm.

💡Alright, so I have this problem here where I need to prove that the integral of f(x) squared from 0 to 1 is at least n squared. The function f is continuous on [0,1], and it satisfies these conditions where the integral of x^k times f(x) from 0 to 1 is equal to 1 for every k from 0 up to n-1. Hmm, okay.First, I remember that when dealing with integrals like this, especially with multiple conditions, it might be related to something like the Cauchy-Schwarz inequality or maybe orthogonal functions. I'm not entirely sure, but let me think.So, the function f has these integral conditions with powers of x. That makes me think of moments in probability or something like that. Maybe I can represent f in terms of some basis functions that are orthogonal with respect to the weight function 1 on [0,1]. Wait, that sounds like Legendre polynomials, right? Legendre polynomials are orthogonal on the interval [-1,1], but maybe there's an analogous set on [0,1].Alternatively, maybe I can use the concept of orthonormal basis functions. If I can express f as a combination of these basis functions, then the integral of f squared would just be the sum of the squares of the coefficients. That might be useful because the given conditions could relate to the inner products of f with these basis functions.Let me recall that for a function space, if I have an orthonormal basis, then any function can be written as a linear combination of these basis functions, and the norm squared is just the sum of the squares of the coefficients. So, if I can find such a basis where the given conditions correspond to the coefficients, then I can maybe bound the norm.Wait, but the given conditions are integrals of x^k f(x). So, these are like inner products of f with the monomials x^k. But monomials aren't orthogonal on [0,1], so that might complicate things. Maybe I need to use some kind of orthogonalization process, like Gram-Schmidt, to create orthogonal polynomials from the monomials.Yes, that sounds right. If I can construct an orthonormal set of polynomials up to degree n-1, then f can be expressed in terms of these polynomials, and the given conditions would relate to the coefficients in that expansion.Let me try to outline the steps:1. Construct an orthonormal basis of polynomials {p_0, p_1, ..., p_{n-1}} on [0,1] with respect to the inner product <g,h> = ∫₀¹ g(x)h(x) dx.2. Express f(x) as a linear combination of these basis polynomials: f(x) = a_0 p_0(x) + a_1 p_1(x) + ... + a_{n-1} p_{n-1}(x) + g(x), where g(x) is orthogonal to all p_i(x).3. Use the given conditions ∫₀¹ x^k f(x) dx = 1 for k = 0,1,...,n-1 to find relations between the coefficients a_i.4. Compute the integral of f(x)^2 and show that it's at least n².Wait, but constructing the orthonormal basis might be complicated. Maybe there's a simpler approach using the fact that the monomials x^k are linearly independent and form a basis for the space of polynomials of degree less than n.Alternatively, perhaps using the concept of dual basis or something related to linear algebra. Since we have n conditions, maybe f is related to the dual space, and the integral of f squared can be minimized subject to those conditions.Oh, right! This seems like a problem that can be approached using the method of Lagrange multipliers. We can think of minimizing ∫₀¹ f(x)^2 dx subject to the constraints ∫₀¹ x^k f(x) dx = 1 for k = 0,1,...,n-1.Yes, that sounds promising. So, setting up a functional with Lagrange multipliers for each constraint.Let me recall that in calculus of variations, to minimize a functional subject to constraints, we introduce Lagrange multipliers. So, the functional to minimize would be:J[f] = ∫₀¹ f(x)^2 dx - Σ_{k=0}^{n-1} λ_k (∫₀¹ x^k f(x) dx - 1)Taking the variation of J[f] with respect to f and setting it to zero would give the Euler-Lagrange equation.So, δJ = 2 ∫₀¹ f(x) δf(x) dx - Σ_{k=0}^{n-1} λ_k ∫₀¹ x^k δf(x) dx = 0This implies that 2 f(x) - Σ_{k=0}^{n-1} λ_k x^k = 0, so f(x) = (1/2) Σ_{k=0}^{n-1} λ_k x^k.So, f is a polynomial of degree at most n-1.Now, we can substitute this back into the constraints to solve for the λ_k.So, for each k from 0 to n-1:∫₀¹ x^k f(x) dx = ∫₀¹ x^k (1/2 Σ_{m=0}^{n-1} λ_m x^m) dx = (1/2) Σ_{m=0}^{n-1} λ_m ∫₀¹ x^{k+m} dx = 1So, (1/2) Σ_{m=0}^{n-1} λ_m / (k + m + 1) = 1This gives us a system of n equations:Σ_{m=0}^{n-1} λ_m / (k + m + 1) = 2 for k = 0,1,...,n-1This is a linear system in the variables λ_0, λ_1, ..., λ_{n-1}.Let me denote A as the matrix where A_{k,m} = 1 / (k + m + 1), for k,m = 0,1,...,n-1.Then, the system is A λ = 2 e, where e is the vector of ones.So, to solve for λ, we need to invert the matrix A.But what is the matrix A? It's a Hilbert matrix, which is known to be invertible and has a specific inverse.I remember that the Hilbert matrix has entries A_{i,j} = 1/(i + j + 1), and its inverse has integer entries. The inverse of the Hilbert matrix is known and can be expressed in terms of binomial coefficients.Specifically, the inverse of the Hilbert matrix H has entries:H^{-1}_{i,j} = (-1)^{i+j} (i + j) binom{n + i - 1}{n - j - 1} binom{n + j - 1}{n - i - 1} binom{i + j}{i}Wait, that seems complicated. Maybe I should look for a simpler expression or see if there's a pattern.Alternatively, perhaps instead of computing the inverse explicitly, I can use properties of the Hilbert matrix to find the solution.But maybe there's a smarter way. Since f is expressed as a linear combination of x^k, and we're dealing with integrals, perhaps we can relate this to orthogonal polynomials.Wait, going back to the beginning, if f is a polynomial of degree n-1, then it can be expressed as f(x) = Σ_{k=0}^{n-1} c_k x^k.Then, the constraints are ∫₀¹ x^m f(x) dx = 1 for m = 0,1,...,n-1.So, writing this out:Σ_{k=0}^{n-1} c_k ∫₀¹ x^{m + k} dx = 1Which is:Σ_{k=0}^{n-1} c_k / (m + k + 1) = 1This is the same system as before, so f is determined by solving this system.Once we have f, we can compute ∫₀¹ f(x)^2 dx, which is Σ_{k=0}^{n-1} Σ_{l=0}^{n-1} c_k c_l ∫₀¹ x^{k + l} dx = Σ_{k=0}^{n-1} Σ_{l=0}^{n-1} c_k c_l / (k + l + 1)But this seems messy. Maybe instead of computing it directly, I can use the fact that the minimum of ∫ f^2 dx subject to the given constraints is achieved by f being the orthogonal projection onto the space spanned by the monomials x^k.Wait, that might be the key. The minimum of ∫ f^2 dx over all f satisfying the constraints is equal to the square of the norm of the vector of ones in the dual space, scaled appropriately.Alternatively, using the Cauchy-Schwarz inequality in the space of functions.Wait, let me think differently. Suppose I consider the space of polynomials of degree less than n, which is n-dimensional. The function f is in this space, and the constraints are n linearly independent conditions.Therefore, f is uniquely determined by these conditions, and the integral of f squared is the squared norm of f in this space.Now, to find the minimum of ∫ f^2 dx subject to the constraints ∫ x^k f dx = 1, we can use the method of Lagrange multipliers as I did before, leading to f being a linear combination of the monomials.But perhaps there's a more straightforward way using the properties of Hilbert spaces.Wait, another approach: consider the function f and the monomials x^k as vectors in L^2([0,1]). Then, the given conditions are that the inner product of f with each x^k is 1.We need to find the minimum of ||f||^2 subject to <f, x^k> = 1 for k = 0,1,...,n-1.This is a constrained optimization problem in a Hilbert space. The minimum is achieved when f is the orthogonal projection of some vector onto the space spanned by the x^k's.But more formally, the minimum of ||f||^2 subject to <f, e_k> = 1 for k=1,...,n, where {e_k} are the monomials, is given by the reciprocal of the maximum eigenvalue of the Gram matrix of the constraints.Wait, perhaps using the concept of the reproducing kernel Hilbert space. The minimum norm solution is given by the kernel evaluated at certain points.Alternatively, maybe using the fact that the minimum is achieved when f is orthogonal to the space orthogonal to the constraints.Wait, I'm getting a bit tangled here. Let me go back to the Lagrange multipliers approach.We have f(x) = Σ_{k=0}^{n-1} λ_k x^k / 2, as derived earlier.Then, the integral of f squared is:∫₀¹ (Σ_{k=0}^{n-1} λ_k x^k / 2)^2 dx = (1/4) Σ_{k=0}^{n-1} Σ_{l=0}^{n-1} λ_k λ_l ∫₀¹ x^{k + l} dx = (1/4) Σ_{k=0}^{n-1} Σ_{l=0}^{n-1} λ_k λ_l / (k + l + 1)But we also have the constraints:Σ_{m=0}^{n-1} λ_m / (k + m + 1) = 2 for each k.So, if I denote the matrix A as before, with A_{k,m} = 1/(k + m + 1), then the constraints are A λ = 2 e, and the integral of f squared is (1/4) λ^T A λ.So, we need to find λ such that A λ = 2 e, and then compute (1/4) λ^T A λ.But since A is invertible, we can write λ = 2 A^{-1} e.Therefore, the integral becomes (1/4) (2 A^{-1} e)^T A (2 A^{-1} e) = (1/4) * 4 e^T A^{-1} A A^{-1} e = e^T A^{-1} e.Wait, no, let's compute it step by step.λ = 2 A^{-1} eThen, λ^T A λ = (2 e^T A^{-1}) A (2 A^{-1} e) = 4 e^T A^{-1} A A^{-1} e = 4 e^T A^{-1} eTherefore, the integral is (1/4) * 4 e^T A^{-1} e = e^T A^{-1} e.So, ∫₀¹ f(x)^2 dx = e^T A^{-1} e.Now, we need to compute e^T A^{-1} e, where A is the Hilbert matrix of size n x n.I recall that for the Hilbert matrix, the sum e^T A^{-1} e is equal to n^2. Wait, is that true?Let me check for small n.For n=1, A is [1], so A^{-1} is [1], and e^T A^{-1} e = 1, which is 1^2.For n=2, A is:[1, 1/2][1/2, 1/3]The inverse of A is:[3, -2][-2, 4]So, e^T A^{-1} e = [1,1] [3, -2; -2, 4] [1;1] = (3 - 2) + (-2 + 4) = 1 + 2 = 3, which is not 2^2=4. Hmm, that contradicts my thought.Wait, maybe I made a mistake in computing the inverse.Wait, let's compute A^{-1} for n=2.Given A = [[1, 1/2], [1/2, 1/3]]The determinant is (1)(1/3) - (1/2)^2 = 1/3 - 1/4 = 1/12So, A^{-1} = (1/det) * [[1/3, -1/2], [-1/2, 1]]So, A^{-1} = 12 * [[1/3, -1/2], [-1/2, 1]] = [[4, -6], [-6, 12]]Then, e^T A^{-1} e = [1,1] [4, -6; -6, 12] [1;1] = (4 -6) + (-6 +12) = (-2) + (6) = 4, which is 2^2.Ah, okay, so for n=2, it works.Similarly, for n=1, it's 1.For n=3, let's see.A is:[1, 1/2, 1/3][1/2, 1/3, 1/4][1/3, 1/4, 1/5]Computing A^{-1} is more involved, but I recall that for the Hilbert matrix, the sum e^T A^{-1} e is n^2.Yes, in fact, it's a known result that for the Hilbert matrix H_n, the sum of the elements of H_n^{-1} is n^2.Therefore, e^T A^{-1} e = n^2.Hence, ∫₀¹ f(x)^2 dx = n^2.But wait, in the problem statement, we are to prove that ∫₀¹ f(x)^2 dx ≥ n^2.But according to this, the minimum is exactly n^2, achieved when f is the unique function satisfying the constraints, which is the polynomial we found.Therefore, for any function f satisfying the given conditions, ∫₀¹ f(x)^2 dx is at least n^2, with equality when f is the polynomial solution found via Lagrange multipliers.So, putting it all together, the minimum value is n^2, hence the integral is at least n^2.I think that makes sense. The key was realizing that the problem reduces to minimizing the L^2 norm subject to the given constraints, which leads to solving a system involving the Hilbert matrix, and knowing that the sum e^T A^{-1} e for the Hilbert matrix is n^2.Therefore, the integral of f squared is at least n squared.