💡Okay, so I have this problem here about two vectors, a and b. The problem says that the magnitude of vector a is √2, the magnitude of vector b is 2, and when you add vectors a and b together, the resulting vector is perpendicular to vector a. I need to find the angle between vector a and vector b. Hmm, okay, let me try to break this down step by step.First, I remember that if two vectors are perpendicular, their dot product is zero. So, since (a + b) is perpendicular to a, that means the dot product of (a + b) and a should be zero. Let me write that down:(a + b) · a = 0Alright, now I need to expand this dot product. I think the dot product distributes over addition, so:a · a + b · a = 0Which simplifies to:|a|² + a · b = 0I know that |a| is √2, so |a|² is (√2)², which is 2. So now I have:2 + a · b = 0So, a · b must be -2. Okay, that's the dot product of a and b. Now, I also remember that the dot product can be expressed in terms of the magnitudes of the vectors and the cosine of the angle between them. The formula is:a · b = |a| |b| cosθWhere θ is the angle between vectors a and b. I know |a| is √2 and |b| is 2, so plugging those in:a · b = √2 * 2 * cosθWhich simplifies to:a · b = 2√2 cosθBut earlier, I found that a · b is -2. So, setting them equal:2√2 cosθ = -2Now, I need to solve for cosθ. Let's divide both sides by 2√2:cosθ = -2 / (2√2)Simplify the numerator and denominator:cosθ = -1 / √2Hmm, okay, so cosθ is -1/√2. I need to find the angle θ where the cosine is -1/√2. I remember that cos(45°) is √2/2, which is the same as 1/√2. But here it's negative, so the angle must be in a different quadrant. Since cosine is negative in the second and third quadrants, but angles between vectors are typically considered between 0 and 180 degrees, so it must be in the second quadrant.What angle has a cosine of -1/√2? I think it's 135 degrees, which is 3π/4 radians. Let me confirm that:cos(135°) = cos(180° - 45°) = -cos(45°) = -√2/2 = -1/√2Yes, that's correct. So θ is 135 degrees or 3π/4 radians.Wait, let me just make sure I didn't make any mistakes in my calculations. Starting from the beginning:1. (a + b) is perpendicular to a, so their dot product is zero.2. Expanded the dot product to get |a|² + a · b = 0.3. Plugged in |a| = √2, so |a|² = 2.4. Therefore, a · b = -2.5. Expressed a · b as |a||b|cosθ, which is √2 * 2 * cosθ = 2√2 cosθ.6. Set that equal to -2, so 2√2 cosθ = -2.7. Divided both sides by 2√2 to get cosθ = -1/√2.8. Recognized that θ must be 135 degrees or 3π/4 radians.Everything seems to check out. I don't think I made any calculation errors. So, the angle between vectors a and b is 135 degrees or 3π/4 radians.Just to visualize it, if vector a is pointing in some direction, and vector b is such that when you add them, the resultant vector is perpendicular to a. That means vector b must be pointing in a direction that, when added to a, the resultant vector is at a right angle to a. This makes sense with the angle being greater than 90 degrees, as the vectors are pointing in somewhat opposite directions to create that perpendicular resultant vector.I think that's all. I can't think of any other steps or considerations needed here. The key was recognizing the perpendicularity condition and using the dot product formula appropriately. It's a good reminder of how dot products relate to angles between vectors.

💡Okay, so I have this inequality: ( x^2 - 2x - 2 - m < 0 ), and I need to find the range of values for ( m ) such that there exists some ( x ) in the interval [2, 4] where this inequality holds. Hmm, let's see how to approach this.First, maybe I can rewrite the inequality to make it a bit clearer. If I move ( m ) to the other side, it becomes ( x^2 - 2x - 2 < m ). So, this means that ( m ) has to be greater than ( x^2 - 2x - 2 ) for some ( x ) between 2 and 4.Now, if I think about this, ( m ) just needs to be larger than the minimum value of the quadratic function ( f(x) = x^2 - 2x - 2 ) on the interval [2, 4]. Because if ( m ) is larger than the minimum, there will definitely be some ( x ) where ( f(x) < m ).Wait, is that right? Let me double-check. If ( m ) is greater than the minimum value of ( f(x) ) on [2, 4], then there exists at least one ( x ) where ( f(x) < m ). Yeah, that makes sense because the function reaches its minimum somewhere, and if ( m ) is just a bit larger than that, it will be above the function at that point.So, I need to find the minimum value of ( f(x) = x^2 - 2x - 2 ) on the interval [2, 4]. To do that, I can analyze the function. It's a quadratic, so it has a parabola shape. Since the coefficient of ( x^2 ) is positive, it opens upwards, meaning the vertex is the minimum point.The vertex of a quadratic ( ax^2 + bx + c ) is at ( x = -frac{b}{2a} ). Here, ( a = 1 ) and ( b = -2 ), so the vertex is at ( x = -frac{-2}{2*1} = 1 ). Hmm, but 1 is not in our interval [2, 4]. That means the minimum on [2, 4] isn't at the vertex but somewhere else.Since the parabola opens upwards and the vertex is at x=1, which is to the left of our interval, the function is increasing on the interval [2, 4]. So, the minimum value on [2, 4] should be at x=2, and the maximum at x=4.Let me calculate ( f(2) ) and ( f(4) ) to confirm. For ( x = 2 ):( f(2) = (2)^2 - 2*(2) - 2 = 4 - 4 - 2 = -2 ).For ( x = 4 ):( f(4) = (4)^2 - 2*(4) - 2 = 16 - 8 - 2 = 6 ).So, the function increases from -2 at x=2 to 6 at x=4. Therefore, the minimum value on [2, 4] is -2, and the maximum is 6.Going back to the inequality ( x^2 - 2x - 2 < m ), since the minimum value of the left side is -2, for there to exist an ( x ) in [2, 4] such that the inequality holds, ( m ) must be greater than -2. If ( m ) were less than or equal to -2, then ( x^2 - 2x - 2 ) would never be less than ( m ) because the smallest it gets is -2.Wait, let me think again. If ( m ) is greater than -2, then there exists some ( x ) where ( f(x) < m ). Specifically, near x=2, where ( f(x) ) is -2, if ( m ) is just a bit larger, say -1, then at x=2, ( f(x) = -2 < -1 ), so the inequality holds. Similarly, for any ( m > -2 ), there will be some x near 2 where the inequality is satisfied.But what if ( m ) is exactly -2? Then, at x=2, ( f(x) = -2 ), so the inequality becomes ( -2 < -2 ), which is false. So, ( m ) has to be strictly greater than -2.Therefore, the range of ( m ) is all real numbers greater than -2. In interval notation, that's ( (-2, +infty) ).I think that makes sense. The key was realizing that the function is increasing on [2, 4], so the minimum is at x=2, and ( m ) just needs to be larger than that minimum for the inequality to hold somewhere in the interval.**Final Answer**The range of values for ( m ) is boxed{(-2, +infty)}.

💡Okay, so I have this problem here where I'm given a function f(x) defined as f₁(x) = x/(1+x). Then, for n ≥ 2, fₙ(x) is defined recursively as fₙ₋₁(f(x)). I need to find the sum f(1) + f(2) + … + f(n) + f₁(1) + f₂(1) + … + fₙ(1). Hmm, that sounds a bit complicated, but let me try to break it down step by step.First, let me make sure I understand the notation. The function fₙ(x) is defined recursively, meaning each subsequent function is built from the previous one. So f₂(x) would be f₁(f(x)), f₃(x) would be f₂(f(x)), and so on. Got it.Now, the problem is asking for two separate sums: one is the sum of f(k) from k=1 to n, and the other is the sum of fₖ(1) from k=1 to n. Then, we need to add these two sums together. Let me write this out:Total Sum = [f(1) + f(2) + … + f(n)] + [f₁(1) + f₂(1) + … + fₙ(1)]So, I need to compute each of these parts separately and then add them up.Let me start with the first part: f(1) + f(2) + … + f(n). Since f(x) = x/(1+x), each term f(k) is just k/(1+k). So, this sum becomes:Sum₁ = 1/2 + 2/3 + 3/4 + … + n/(n+1)Okay, that's straightforward. Now, what about the second sum: f₁(1) + f₂(1) + … + fₙ(1). Let's compute each term here.We know that f₁(x) = x/(1+x), so f₁(1) = 1/2. Now, f₂(x) is f₁(f(x)). Let me compute f₂(1):f₂(1) = f₁(f(1)) = f₁(1/2) = (1/2)/(1 + 1/2) = (1/2)/(3/2) = 1/3Similarly, f₃(1) would be f₂(f(1)) = f₂(1/2). Let me compute that:f₂(1/2) = f₁(f(1/2)) = f₁( (1/2)/(1 + 1/2) ) = f₁(1/3) = (1/3)/(1 + 1/3) = (1/3)/(4/3) = 1/4Wait a minute, I see a pattern here. Let me test it with f₄(1):f₄(1) = f₃(f(1)) = f₃(1/2) = f₂(f(1/2)) = f₂(1/3) = f₁(f(1/3)) = f₁(1/4) = (1/4)/(1 + 1/4) = 1/5So, it seems like fₖ(1) = 1/(1 + k). Let me verify this for k=1,2,3,4:- f₁(1) = 1/2 = 1/(1+1)- f₂(1) = 1/3 = 1/(1+2)- f₃(1) = 1/4 = 1/(1+3)- f₄(1) = 1/5 = 1/(1+4)Yes, that's consistent. So, in general, fₖ(1) = 1/(1 + k). Therefore, the second sum is:Sum₂ = 1/2 + 1/3 + 1/4 + … + 1/(n + 1)So, now I have both sums:Sum₁ = 1/2 + 2/3 + 3/4 + … + n/(n + 1)Sum₂ = 1/2 + 1/3 + 1/4 + … + 1/(n + 1)Now, I need to compute Total Sum = Sum₁ + Sum₂.Let me write out both sums together:Total Sum = [1/2 + 2/3 + 3/4 + … + n/(n + 1)] + [1/2 + 1/3 + 1/4 + … + 1/(n + 1)]Hmm, let me see if I can combine these terms. Let's pair the terms from Sum₁ and Sum₂:First term: 1/2 (from Sum₁) + 1/2 (from Sum₂) = 1Second term: 2/3 (from Sum₁) + 1/3 (from Sum₂) = 1Third term: 3/4 (from Sum₁) + 1/4 (from Sum₂) = 1...n-th term: n/(n + 1) (from Sum₁) + 1/(n + 1) (from Sum₂) = 1Wait, so each pair adds up to 1. How many such pairs are there?Looking at Sum₁, it has n terms: from k=1 to k=n, each term is k/(k + 1). Similarly, Sum₂ also has n terms: from k=1 to k=n, each term is 1/(k + 1). So, when I add them together, I have n pairs, each adding up to 1.Therefore, Total Sum = 1 + 1 + 1 + … + 1 (n times) = nWow, that's neat! So, the total sum is just n.Let me double-check this with a small value of n to make sure I didn't make a mistake.Let's take n = 1:Sum₁ = f(1) = 1/2Sum₂ = f₁(1) = 1/2Total Sum = 1/2 + 1/2 = 1, which is equal to n=1. Okay, that works.How about n=2:Sum₁ = f(1) + f(2) = 1/2 + 2/3 = 7/6Sum₂ = f₁(1) + f₂(1) = 1/2 + 1/3 = 5/6Total Sum = 7/6 + 5/6 = 12/6 = 2, which is equal to n=2. Perfect.Another test with n=3:Sum₁ = 1/2 + 2/3 + 3/4 = (6/12 + 8/12 + 9/12) = 23/12Sum₂ = 1/2 + 1/3 + 1/4 = (6/12 + 4/12 + 3/12) = 13/12Total Sum = 23/12 + 13/12 = 36/12 = 3, which is equal to n=3. Great, that works too.So, it seems my reasoning is correct. Each pair of terms from Sum₁ and Sum₂ adds up to 1, and since there are n such pairs, the total sum is n.Therefore, the answer should be n.**Final Answer**The value of the given expression is boxed{n}.

💡Okay, so I have this problem here that I need to prove. It says that for any natural number ( n ), the sum over all sequences ( i_1 < i_2 < cdots < i_k ) taken from the set ( {1, 2, ldots, n} ) with ( k = 1, 2, ldots, n ) of the reciprocals of the products ( frac{1}{i_1 i_2 cdots i_k} ) equals ( n ). Hmm, that sounds a bit complicated, but let me try to break it down. So, for each ( k ) from 1 to ( n ), I need to consider all possible combinations of ( k ) distinct numbers from 1 to ( n ), multiply their reciprocals, and then sum all those up. The claim is that this total sum is just ( n ). Let me start by understanding what exactly is being summed. For each ( k ), we're looking at all subsets of size ( k ) from the set ( {1, 2, ldots, n} ). For each such subset, we take the product of its elements and then take the reciprocal of that product. Finally, we sum all these reciprocals over all possible subsets for each ( k ) and then add them all together. Wait, but the problem statement says the summation is over all sequences ( i_1 < i_2 < cdots < i_k ) with ( k = 1, 2, ldots, n ). So, actually, it's considering all non-empty subsets of ( {1, 2, ldots, n} ), since ( k ) ranges from 1 to ( n ). Each subset contributes the reciprocal of the product of its elements. So, another way to phrase this is: if I take all non-empty subsets of ( {1, 2, ldots, n} ), compute the reciprocal of the product of each subset's elements, and sum them all up, the total should be ( n ). Let me test this with small values of ( n ) to see if it holds, which might give me some intuition.**Case 1: ( n = 1 )**- The only subset is ( {1} ).- The sum is ( frac{1}{1} = 1 ), which equals ( n = 1 ). So, it works.**Case 2: ( n = 2 )**- Subsets: ( {1}, {2}, {1, 2} ).- Their reciprocals: ( frac{1}{1} = 1 ), ( frac{1}{2} = 0.5 ), ( frac{1}{1 times 2} = 0.5 ).- Sum: ( 1 + 0.5 + 0.5 = 2 ), which equals ( n = 2 ). Good.**Case 3: ( n = 3 )**- Subsets: ( {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3} ).- Their reciprocals: ( 1, 0.5, frac{1}{3} approx 0.333 ), ( 0.5, approx 0.333, approx 0.166 ), ( approx 0.166 ).- Sum: ( 1 + 0.5 + 0.333 + 0.5 + 0.333 + 0.166 + 0.166 approx 3 ). So, it works here too.Okay, so it seems to hold for small ( n ). Now, how can I generalize this?I remember that generating functions are often useful in combinatorics for dealing with sums over subsets. Maybe I can construct a generating function where the coefficients correspond to these reciprocals.Let me consider the product:[prod_{i=1}^{n} left(1 + frac{1}{i}xright)]If I expand this product, the coefficient of ( x^k ) will be the sum of all reciprocals of products of ( k ) distinct elements from ( {1, 2, ldots, n} ). So, if I set ( x = 1 ), the sum of all coefficients will be the total sum we're trying to compute.Let me write that out:[prod_{i=1}^{n} left(1 + frac{1}{i}right)]Wait, if I set ( x = 1 ), each term becomes ( 1 + frac{1}{i} ). Let's compute this product.Compute ( prod_{i=1}^{n} left(1 + frac{1}{i}right) ):- For ( i = 1 ): ( 1 + 1 = 2 )- For ( i = 2 ): ( 1 + frac{1}{2} = frac{3}{2} )- For ( i = 3 ): ( 1 + frac{1}{3} = frac{4}{3} )- ...- For ( i = n ): ( 1 + frac{1}{n} = frac{n+1}{n} )So, the product becomes:[2 times frac{3}{2} times frac{4}{3} times cdots times frac{n+1}{n}]I notice that this is a telescoping product. Each numerator cancels with the denominator of the next term:- The 2 in the numerator cancels with the 2 in the denominator of the next term.- The 3 in the numerator cancels with the 3 in the denominator of the following term.- This continues all the way up to ( n ).So, after cancellation, we're left with:[frac{n+1}{1} = n + 1]But wait, the generating function evaluated at ( x = 1 ) gives ( n + 1 ). However, the sum we're interested in is the sum of all coefficients except the constant term (which corresponds to the empty subset). The constant term is 1, so the sum of all reciprocals is ( (n + 1) - 1 = n ).That's exactly what we wanted to prove! So, by considering the generating function and evaluating it at ( x = 1 ), we found that the sum of all reciprocals of products of non-empty subsets is ( n ).Let me just recap to make sure I didn't skip any steps:1. I considered the generating function ( prod_{i=1}^{n} left(1 + frac{1}{i}xright) ).2. By expanding this product, the coefficients of ( x^k ) give the sum of reciprocals for each subset size ( k ).3. Evaluating the generating function at ( x = 1 ) gives the total sum of all coefficients, which includes the empty subset (which contributes 1).4. Subtracting 1 gives the sum over all non-empty subsets, which equals ( n ).This seems solid. I tested it with small ( n ) and it worked, and the telescoping product argument makes sense. I don't see any flaws in this reasoning. So, I think this is a valid proof.**Final Answer**The sum is equal to ( boxed{n} ).

💡First, I need to determine how many eggs were used in the afternoon by subtracting the number of eggs used in the morning from the total number of eggs used throughout the day.The total number of eggs used during the day is 1339, and the number of eggs used in the morning is 816.By subtracting 816 from 1339, I can find out how many eggs were used in the afternoon.So, 1339 minus 816 equals 523.Therefore, the number of eggs used in the afternoon is 523.

💡Okay, so I have this problem where there's a sequence {a_n} defined by a recursion formula: a_n = 3a_{n-1} + 3^n - 1 for n ≥ 2, and the initial term a_1 is 5. The question is asking me to find a real number λ such that when I take (a_n + λ)/3^n, this new sequence forms an arithmetic sequence. Alright, let me break this down. First, I need to recall what an arithmetic sequence is. An arithmetic sequence is one where the difference between consecutive terms is constant. So, if I have a sequence {b_n}, it's arithmetic if b_n - b_{n-1} is the same for all n. Alternatively, another way to think about it is that the average of two consecutive terms is equal to the term in between, so 2b_n = b_{n-1} + b_{n+1} for all n. Maybe that's a property I can use here.So, let me define b_n as (a_n + λ)/3^n. The goal is to find λ such that {b_n} is arithmetic. Since {b_n} is arithmetic, it must satisfy the condition 2b_n = b_{n-1} + b_{n+1} for all n ≥ 2. But before jumping into that, maybe I can first compute the first few terms of {a_n} to get a sense of what's going on. Let's see:Given a_1 = 5.Then, a_2 = 3a_1 + 3^2 - 1 = 3*5 + 9 - 1 = 15 + 9 - 1 = 23.a_3 = 3a_2 + 3^3 - 1 = 3*23 + 27 - 1 = 69 + 27 - 1 = 95.a_4 = 3a_3 + 3^4 - 1 = 3*95 + 81 - 1 = 285 + 81 - 1 = 365.Hmm, okay, so the terms are growing pretty quickly. Now, if I compute b_n for n=1,2,3,4, maybe I can see a pattern.Compute b_1 = (a_1 + λ)/3^1 = (5 + λ)/3.b_2 = (a_2 + λ)/3^2 = (23 + λ)/9.b_3 = (a_3 + λ)/3^3 = (95 + λ)/27.b_4 = (a_4 + λ)/3^4 = (365 + λ)/81.Now, for {b_n} to be arithmetic, the difference between consecutive terms should be constant. So, let's compute the differences:b_2 - b_1 = (23 + λ)/9 - (5 + λ)/3 = (23 + λ - 15 - 3λ)/9 = (8 - 2λ)/9.Similarly, b_3 - b_2 = (95 + λ)/27 - (23 + λ)/9 = (95 + λ - 69 - 3λ)/27 = (26 - 2λ)/27.And b_4 - b_3 = (365 + λ)/81 - (95 + λ)/27 = (365 + λ - 285 - 3λ)/81 = (80 - 2λ)/81.For {b_n} to be arithmetic, these differences should be equal. So, set b_2 - b_1 = b_3 - b_2 and solve for λ.So, (8 - 2λ)/9 = (26 - 2λ)/27.Multiply both sides by 27 to eliminate denominators:3*(8 - 2λ) = 26 - 2λ24 - 6λ = 26 - 2λBring all terms to one side:24 - 6λ - 26 + 2λ = 0-2 - 4λ = 0-4λ = 2λ = -2/4 = -1/2.Wait, so λ is -1/2. Let me check if this works for the next difference as well.Compute b_3 - b_2 with λ = -1/2:(26 - 2*(-1/2))/27 = (26 +1)/27 = 27/27 = 1.Similarly, b_4 - b_3 should also be 1.Compute (80 - 2*(-1/2))/81 = (80 +1)/81 = 81/81 = 1.Okay, so the differences are consistent. So, λ = -1/2.Alternatively, maybe I can approach this using the recursion relation. Since {b_n} is arithmetic, it has a constant difference, say d. So, b_n = b_1 + (n-1)d.But since b_n = (a_n + λ)/3^n, that would mean a_n = 3^n b_n - λ.So, substituting into the recursion:a_n = 3a_{n-1} + 3^n - 1So, 3^n b_n - λ = 3*(3^{n-1} b_{n-1} - λ) + 3^n - 1Simplify the right-hand side:3*(3^{n-1} b_{n-1} - λ) = 3^n b_{n-1} - 3λSo, 3^n b_n - λ = 3^n b_{n-1} - 3λ + 3^n - 1Bring all terms to the left:3^n b_n - λ - 3^n b_{n-1} + 3λ - 3^n + 1 = 0Factor out 3^n:3^n (b_n - b_{n-1} - 1) + ( -λ + 3λ + 1 ) = 0Simplify:3^n (b_n - b_{n-1} - 1) + (2λ + 1) = 0But since {b_n} is arithmetic, b_n - b_{n-1} = d, a constant. So, substitute:3^n (d - 1) + (2λ + 1) = 0This equation must hold for all n ≥ 2. The only way this can be true is if the coefficient of 3^n is zero, and the constant term is also zero.So, set the coefficient of 3^n to zero:d - 1 = 0 ⇒ d = 1And the constant term:2λ + 1 = 0 ⇒ λ = -1/2So, this confirms the earlier result. Therefore, λ is indeed -1/2.I think that's solid. I approached it both by computing the differences and by substituting into the recursion, and both methods led me to λ = -1/2. So, I feel confident that's the correct answer.**Final Answer**The real number λ is boxed{-dfrac{1}{2}}.