💡Okay, so I need to figure out Benjamin's speed in kilometers per hour. He skated 80 kilometers and was skating for 8 hours. Hmm, speed is generally calculated by dividing the distance traveled by the time taken, right? So, if I take the distance, which is 80 kilometers, and divide it by the time, which is 8 hours, that should give me his speed.Let me write that down: Speed = Distance / Time. Plugging in the numbers, that would be Speed = 80 km / 8 hours. Doing the division, 80 divided by 8 is 10. So, is Benjamin's speed 10 kilometers per hour? That seems straightforward, but let me double-check to make sure I didn't miss anything.Wait, could there be any other factors involved? Like, did Benjamin take any breaks during his skating? The problem doesn't mention anything about breaks, so I guess I can assume he was continuously skating for those 8 hours. Also, is the distance one way or round trip? The problem says he skated 80 kilometers, so I think that's the total distance, not a round trip. So, he went 80 kilometers in 8 hours.Another thing to consider is the units. The distance is in kilometers and the time is in hours, so the speed will naturally be in kilometers per hour, which is what the question is asking for. So, no need to convert units here.Let me visualize this. If Benjamin skated for 8 hours and covered 80 kilometers, then each hour he would have skated 10 kilometers. That makes sense because 10 times 8 is 80. So, his speed is consistently 10 kilometers per hour throughout the 8 hours.I think I've covered all the bases here. The calculation seems simple, but it's always good to verify to avoid any mistakes. Yeah, I'm confident that Benjamin's speed was 10 kilometers per hour.

💡Alright, so I've got this problem here about sets of triangles. It says:Given the sets ( M = {x | x text{ is an isosceles triangle}} ) and ( N = {x | x text{ is a right-angled triangle}} ), then the intersection ( M cap N ) isA. ( {x | x text{ is an isosceles right-angled triangle}} ) B. ( {x | x text{ is an isosceles triangle or a right-angled triangle}} ) C. ( varnothing ) D. ( M )Okay, so I need to figure out what ( M cap N ) is. Let me start by recalling what each set represents.First, set ( M ) consists of all isosceles triangles. An isosceles triangle has at least two sides that are equal in length. That means it could have exactly two equal sides or all three sides equal, which would make it an equilateral triangle. But in general, isosceles triangles have two equal sides and the base being a different length.Set ( N ) consists of all right-angled triangles. A right-angled triangle has one angle that is exactly 90 degrees, and the sides satisfy the Pythagorean theorem: ( a^2 + b^2 = c^2 ), where ( c ) is the hypotenuse.Now, the intersection ( M cap N ) would be the set of triangles that are both isosceles and right-angled. So, I'm looking for triangles that have two equal sides and one right angle.Let me think about what such a triangle would look like. If a triangle is both isosceles and right-angled, it must have two equal sides and a right angle. In a right-angled triangle, the two legs can be equal, making it an isosceles right-angled triangle. This means the two legs are equal, and the angles opposite those legs are each 45 degrees, with the right angle being 90 degrees.So, an isosceles right-angled triangle has angles of 45°, 45°, and 90°, and the sides are in the ratio ( 1 : 1 : sqrt{2} ). This fits both definitions: it's isosceles because two sides are equal, and it's right-angled because it has a 90° angle.Looking back at the options:A. ( {x | x text{ is an isosceles right-angled triangle}} ) B. ( {x | x text{ is an isosceles triangle or a right-angled triangle}} ) C. ( varnothing ) D. ( M )Option A directly describes the set of triangles that are both isosceles and right-angled, which is exactly what the intersection ( M cap N ) should be. Option B describes the union of the two sets, not the intersection. Option C suggests there's no overlap, which isn't true because isosceles right-angled triangles do exist. Option D says the intersection is just set M, which isn't correct because not all isosceles triangles are right-angled.Therefore, the correct answer should be A.

💡Okay, so I have this problem about a sequence {a_n} where a₁ = 1, a₂ = 6, and the point (a_n - a_{n-1}, a_{n+1}) lies on the graph of the function f(x) = 4x. There are three parts to this problem, and I need to tackle them one by one. Let me start with part (1).**Problem (1): Prove that the sequence {a_{n+1} - 2a_n} is a geometric sequence, and find the general formula for the sequence {a_n}.**Alright, so first, I need to understand what it means for the point (a_n - a_{n-1}, a_{n+1}) to lie on the graph of f(x) = 4x. That means if I plug in x = a_n - a_{n-1}, the function f(x) should give me y = a_{n+1}. So, mathematically, that translates to:a_{n+1} = 4(a_n - a_{n-1})So, this is a recurrence relation. Let me write that down:a_{n+1} = 4a_n - 4a_{n-1}Hmm, okay. Now, the problem wants me to prove that the sequence {a_{n+1} - 2a_n} is a geometric sequence. Let me denote b_n = a_{n+1} - 2a_n. So, I need to show that {b_n} is a geometric sequence.Let me compute b_{n} and b_{n+1} to see if there's a common ratio.First, b_n = a_{n+1} - 2a_nFrom the recurrence relation, a_{n+1} = 4a_n - 4a_{n-1}, so substituting that in:b_n = (4a_n - 4a_{n-1}) - 2a_n = 2a_n - 4a_{n-1} = 2(a_n - 2a_{n-1})Wait, that's interesting. So, b_n = 2(a_n - 2a_{n-1})But notice that a_n - 2a_{n-1} is similar to b_{n-1}:b_{n-1} = a_n - 2a_{n-1}So, substituting back, we get:b_n = 2b_{n-1}Ah! So, each term b_n is twice the previous term b_{n-1}. That means {b_n} is a geometric sequence with common ratio 2.Now, to confirm, let's compute the first term of {b_n}. Since b_n = a_{n+1} - 2a_n, let's compute b₁:b₁ = a₂ - 2a₁ = 6 - 2*1 = 6 - 2 = 4So, the first term is 4, and the common ratio is 2. Therefore, the general term for {b_n} is:b_n = 4 * 2^{n-1} = 2^{2} * 2^{n-1} = 2^{n+1}So, b_n = 2^{n+1}But b_n = a_{n+1} - 2a_n, so:a_{n+1} - 2a_n = 2^{n+1}This is another recurrence relation. Let me write it as:a_{n+1} = 2a_n + 2^{n+1}Hmm, okay. So, this is a nonhomogeneous linear recurrence relation. To solve this, I can use the method of solving linear recurrences.First, let's write the homogeneous part:a_{n+1} - 2a_n = 0The characteristic equation is r - 2 = 0, so r = 2. Therefore, the general solution to the homogeneous equation is:a_n^{(h)} = C * 2^nNow, we need a particular solution to the nonhomogeneous equation. The nonhomogeneous term is 2^{n+1}, which is similar to the homogeneous solution. So, I need to multiply by n to find a particular solution.Let me assume a particular solution of the form:a_n^{(p)} = D * n * 2^nLet's substitute this into the recurrence relation:a_{n+1}^{(p)} = 2a_n^{(p)} + 2^{n+1}Compute a_{n+1}^{(p)}:a_{n+1}^{(p)} = D*(n+1)*2^{n+1}Compute 2a_n^{(p)}:2a_n^{(p)} = 2*D*n*2^n = D*n*2^{n+1}So, substituting into the equation:D*(n+1)*2^{n+1} = D*n*2^{n+1} + 2^{n+1}Divide both sides by 2^{n+1}:D*(n + 1) = D*n + 1Simplify:D*n + D = D*n + 1Subtract D*n from both sides:D = 1So, D = 1. Therefore, the particular solution is:a_n^{(p)} = n * 2^nTherefore, the general solution is:a_n = a_n^{(h)} + a_n^{(p)} = C * 2^n + n * 2^n = (C + n) * 2^nNow, apply the initial condition to find C. Let's use n = 1:a₁ = (C + 1) * 2^1 = 2(C + 1) = 1So, 2(C + 1) = 1 => C + 1 = 1/2 => C = -1/2Therefore, the general term is:a_n = (-1/2 + n) * 2^n = (n - 1/2) * 2^nAlternatively, we can write this as:a_n = (2n - 1) * 2^{n - 1}Let me verify this formula with the given initial terms.For n = 1:a₁ = (2*1 - 1) * 2^{0} = (2 - 1)*1 = 1*1 = 1 ✓For n = 2:a₂ = (2*2 - 1) * 2^{1} = (4 - 1)*2 = 3*2 = 6 ✓Good, it matches the given terms. So, part (1) is done.**Problem (2): Let the sum of the first n terms of the sequence {a_n} be S_n. Prove that S_n < (n - 1) * 2^{n+1} + 2.**Alright, so S_n = a₁ + a₂ + ... + a_n. We need to show that S_n is less than (n - 1) * 2^{n+1} + 2.First, let's write the expression for S_n using the general term we found:a_n = (2n - 1) * 2^{n - 1}So,S_n = Σ_{k=1}^{n} (2k - 1) * 2^{k - 1}Let me write this out:S_n = 1*2^{0} + 3*2^{1} + 5*2^{2} + ... + (2n - 1)*2^{n - 1}This looks like a sum that can be tackled using the method of differences or generating functions. Let me try the method of differences.Let me denote S = S_n for simplicity.Compute 2S:2S = 1*2^{1} + 3*2^{2} + 5*2^{3} + ... + (2n - 1)*2^{n}Now, subtract S from 2S:2S - S = S = (1*2^{1} + 3*2^{2} + 5*2^{3} + ... + (2n - 1)*2^{n}) - (1*2^{0} + 3*2^{1} + 5*2^{2} + ... + (2n - 1)*2^{n - 1})Let me write this term by term:= [1*2 + 3*2^2 + 5*2^3 + ... + (2n - 1)*2^n] - [1 + 3*2 + 5*2^2 + ... + (2n - 1)*2^{n-1}]Now, let's subtract term by term:- The first term: 1*2 - 1 = 2 - 1 = 1- The second term: 3*2^2 - 3*2 = 12 - 6 = 6- The third term: 5*2^3 - 5*2^2 = 40 - 20 = 20- ...- The k-th term: (2k - 1)*2^k - (2k - 1)*2^{k-1} = (2k - 1)*2^{k-1}(2 - 1) = (2k - 1)*2^{k-1}- ...- The last term: (2n - 1)*2^n - (2n - 1)*2^{n-1} = (2n - 1)*2^{n-1}Wait, actually, when subtracting, each term from the second series cancels out with the previous term in the first series except for the first term of the first series and the last term of the first series.Wait, let me think again. Maybe I should align the terms properly.Let me write both series:First series (2S):1*2^1, 3*2^2, 5*2^3, ..., (2n - 1)*2^nSecond series (S):1*2^0, 3*2^1, 5*2^2, ..., (2n - 1)*2^{n-1}Now, subtracting term by term:- The first term of 2S is 1*2^1, and the first term of S is 1*2^0. So, 1*2^1 - 1*2^0 = 2 - 1 = 1- The second term of 2S is 3*2^2, and the second term of S is 3*2^1. So, 3*2^2 - 3*2^1 = 12 - 6 = 6- The third term of 2S is 5*2^3, and the third term of S is 5*2^2. So, 5*2^3 - 5*2^2 = 40 - 20 = 20- ...- The k-th term of 2S is (2k - 1)*2^k, and the k-th term of S is (2k - 1)*2^{k-1}. So, the difference is (2k - 1)*2^{k} - (2k - 1)*2^{k-1} = (2k - 1)*2^{k-1}(2 - 1) = (2k - 1)*2^{k-1}- ...- The last term of 2S is (2n - 1)*2^n, and there is no corresponding term in S beyond the (2n - 1)*2^{n-1} term. So, the last term in the difference is (2n - 1)*2^nWait, no, actually, when subtracting, the last term of 2S is (2n - 1)*2^n, and the last term of S is (2n - 1)*2^{n-1}. So, the difference is (2n - 1)*2^n - (2n - 1)*2^{n-1} = (2n - 1)*2^{n-1}(2 - 1) = (2n - 1)*2^{n-1}But actually, when subtracting, the terms after the first term in 2S will align with the terms in S starting from the second term. So, actually, the subtraction should be:2S - S = S = [1*2^1 - 1*2^0] + [3*2^2 - 3*2^1] + [5*2^3 - 5*2^2] + ... + [(2n - 1)*2^n - (2n - 1)*2^{n-1}]So, each bracket is (2k - 1)*2^{k} - (2k - 1)*2^{k-1} = (2k - 1)*2^{k-1}(2 - 1) = (2k - 1)*2^{k-1}Wait, but that would mean each term is (2k - 1)*2^{k-1}, but that's the same as the original terms in S. That can't be right because S = 2S - S, which would imply S = S, which is trivial. Hmm, maybe I made a mistake in the subtraction.Wait, let me try a different approach. Let me write 2S and subtract S:2S = 1*2^1 + 3*2^2 + 5*2^3 + ... + (2n - 1)*2^nS = 1*2^0 + 3*2^1 + 5*2^2 + ... + (2n - 1)*2^{n-1}Subtracting S from 2S:2S - S = S = (1*2^1 - 1*2^0) + (3*2^2 - 3*2^1) + (5*2^3 - 5*2^2) + ... + [(2n - 1)*2^n - (2n - 1)*2^{n-1}]So, each term is (2k - 1)*2^{k} - (2k - 1)*2^{k-1} = (2k - 1)*2^{k-1}(2 - 1) = (2k - 1)*2^{k-1}But this is the same as the original terms of S, which would imply S = S, which is not helpful. Hmm, maybe I need to consider the difference differently.Wait, perhaps I should shift the index. Let me write 2S as:2S = 1*2^1 + 3*2^2 + 5*2^3 + ... + (2n - 1)*2^nAnd S as:S = 1*2^0 + 3*2^1 + 5*2^2 + ... + (2n - 1)*2^{n-1}Now, subtract S from 2S:2S - S = S = (1*2^1 - 1*2^0) + (3*2^2 - 3*2^1) + (5*2^3 - 5*2^2) + ... + [(2n - 1)*2^n - (2n - 1)*2^{n-1}]But as I saw earlier, each term is (2k - 1)*2^{k-1}, so:S = Σ_{k=1}^{n} (2k - 1)*2^{k-1} = SWhich is just S = S, which doesn't help. Maybe I need to consider another approach.Alternatively, perhaps I can express S_n in terms of a known sum. Let me recall that:Σ_{k=1}^{n} k*2^{k} = 2 + 2^2 + 3*2^3 + ... + n*2^nBut our sum is Σ_{k=1}^{n} (2k - 1)*2^{k-1} = Σ_{k=1}^{n} (2k*2^{k-1} - 2^{k-1}) = Σ_{k=1}^{n} k*2^{k} - Σ_{k=1}^{n} 2^{k-1}So, let's compute each sum separately.First, compute Σ_{k=1}^{n} k*2^{k}.I recall that the sum Σ_{k=1}^{n} k*r^{k} can be computed using the formula:Σ_{k=1}^{n} k*r^{k} = r(1 - (n+1)r^n + n r^{n+1}) ) / (1 - r)^2For r ≠ 1.In our case, r = 2, so:Σ_{k=1}^{n} k*2^{k} = 2(1 - (n+1)2^n + n*2^{n+1}) ) / (1 - 2)^2Simplify denominator: (1 - 2)^2 = 1So,= 2(1 - (n+1)2^n + n*2^{n+1}) ) / 1= 2 - 2(n+1)2^n + 2n*2^{n+1}Wait, let's compute step by step:First, compute numerator:1 - (n+1)2^n + n*2^{n+1} = 1 - (n+1)2^n + 2n*2^n = 1 + (2n - n -1)2^n = 1 + (n -1)2^nSo, numerator is 1 + (n -1)2^nMultiply by 2:2*(1 + (n -1)2^n) = 2 + 2(n -1)2^n = 2 + (n -1)2^{n+1}So, Σ_{k=1}^{n} k*2^{k} = 2 + (n -1)2^{n+1}Now, compute Σ_{k=1}^{n} 2^{k-1}:This is a geometric series with first term 2^{0} = 1, ratio 2, and n terms.Sum = (2^{n} - 1)/(2 - 1) = 2^{n} - 1Therefore, our original sum S_n = Σ_{k=1}^{n} (2k - 1)*2^{k-1} = Σ_{k=1}^{n} k*2^{k} - Σ_{k=1}^{n} 2^{k-1} = [2 + (n -1)2^{n+1}] - [2^{n} - 1] = 2 + (n -1)2^{n+1} - 2^{n} + 1 = 3 + (n -1)2^{n+1} - 2^{n}Simplify:= 3 + 2^{n+1}(n -1) - 2^{n} = 3 + 2^{n}(2(n -1) -1) = 3 + 2^{n}(2n - 2 -1) = 3 + 2^{n}(2n -3)So, S_n = 3 + (2n -3)2^{n}Now, the problem wants us to prove that S_n < (n -1)2^{n+1} + 2Let me compute (n -1)2^{n+1} + 2:= (n -1)2^{n+1} + 2 = 2(n -1)2^{n} + 2 = 2^{n+1}(n -1) + 2Compare with S_n = 3 + (2n -3)2^{n}Let me write both expressions:S_n = 3 + (2n -3)2^{n}And the upper bound is:(n -1)2^{n+1} + 2 = 2(n -1)2^{n} + 2 = (2n -2)2^{n} + 2So, let's compute the difference between the upper bound and S_n:Upper bound - S_n = [(2n -2)2^{n} + 2] - [3 + (2n -3)2^{n}] = (2n -2)2^{n} - (2n -3)2^{n} + 2 -3 = [ (2n -2 -2n +3) ]2^{n} -1 = (1)2^{n} -1 = 2^{n} -1Since 2^{n} -1 is always positive for n ≥1, we have:Upper bound - S_n = 2^{n} -1 >0 => S_n < (n -1)2^{n+1} +2Therefore, the inequality holds. So, part (2) is proved.**Problem (3): Let C_n = 3^n - λ*(-1)^n*(a_n)/(n - 1/2) (n ∈ ℕ*, λ is a non-zero real number). If for any n ∈ ℕ*, C_{n+1} > C_n always holds, find the range of values for the real number λ.**Alright, so we have C_n defined as:C_n = 3^n - λ*(-1)^n*(a_n)/(n - 1/2)We need to find the range of λ such that C_{n+1} > C_n for all n ∈ ℕ*.First, let's express C_{n+1} - C_n >0.Compute C_{n+1} - C_n:= [3^{n+1} - λ*(-1)^{n+1}*(a_{n+1})/(n +1 -1/2)] - [3^n - λ*(-1)^n*(a_n)/(n -1/2)]Simplify:= 3^{n+1} - 3^n - λ*(-1)^{n+1}*(a_{n+1})/(n +1/2) + λ*(-1)^n*(a_n)/(n -1/2)Factor out 3^n:= 3^n(3 -1) + λ*(-1)^n [ (a_{n+1})/(n +1/2) + (a_n)/(n -1/2) ]Wait, let me check the signs:- λ*(-1)^{n+1} = λ*(-1)^n*(-1) = -λ*(-1)^nSo, the expression becomes:= 3^{n+1} - 3^n - λ*(-1)^{n+1}*(a_{n+1})/(n +1/2) + λ*(-1)^n*(a_n)/(n -1/2)= 3^n(3 -1) + λ*(-1)^n [ (a_n)/(n -1/2) + (a_{n+1})/(n +1/2) ]Wait, no:Wait, let's re-express:C_{n+1} - C_n = 3^{n+1} - 3^n - λ*(-1)^{n+1}*(a_{n+1})/(n +1/2) + λ*(-1)^n*(a_n)/(n -1/2)= 3^n(3 -1) + λ*(-1)^n [ (a_n)/(n -1/2) + (a_{n+1})/(n +1/2) ]Wait, that doesn't seem right. Let me compute term by term:First term: 3^{n+1} - 3^n = 3^n(3 -1) = 2*3^nSecond term: -λ*(-1)^{n+1}*(a_{n+1})/(n +1/2) = λ*(-1)^n*(a_{n+1})/(n +1/2)Third term: +λ*(-1)^n*(a_n)/(n -1/2)So, combining the second and third terms:λ*(-1)^n [ (a_{n+1})/(n +1/2) + (a_n)/(n -1/2) ]Therefore, overall:C_{n+1} - C_n = 2*3^n + λ*(-1)^n [ (a_{n+1})/(n +1/2) + (a_n)/(n -1/2) ]We need this to be greater than 0 for all n ∈ ℕ*.So,2*3^n + λ*(-1)^n [ (a_{n+1})/(n +1/2) + (a_n)/(n -1/2) ] > 0Let me denote:D_n = [ (a_{n+1})/(n +1/2) + (a_n)/(n -1/2) ]So, the inequality becomes:2*3^n + λ*(-1)^n D_n > 0We need this to hold for all n ≥1.Let me compute D_n:First, recall that a_n = (2n -1)2^{n-1}So,a_n = (2n -1)2^{n-1}Similarly,a_{n+1} = (2(n+1) -1)2^{(n+1)-1} = (2n +1)2^nSo, compute D_n:D_n = [ (2n +1)2^n / (n +1/2) ) ] + [ (2n -1)2^{n-1} / (n -1/2) ) ]Simplify each term:First term:(2n +1)2^n / (n +1/2) = (2n +1)2^n / ( (2n +1)/2 ) = (2n +1)2^n * 2/(2n +1) ) = 2^{n+1}Second term:(2n -1)2^{n-1} / (n -1/2) = (2n -1)2^{n-1} / ( (2n -1)/2 ) = (2n -1)2^{n-1} * 2/(2n -1) ) = 2^{n}Therefore, D_n = 2^{n+1} + 2^{n} = 2^{n}(2 +1) = 3*2^{n}So, D_n = 3*2^{n}Therefore, the inequality becomes:2*3^n + λ*(-1)^n *3*2^{n} > 0Factor out 3^n:= 3^n [2 + λ*(-1)^n *3*(2/3)^n ] >0Wait, let me factor 3^n:Wait, 2*3^n + λ*(-1)^n *3*2^{n} = 3^n*2 + 3*2^{n}λ*(-1)^nFactor out 3^n:= 3^n [2 + λ*(-1)^n * (2/3)^n *3 ]Wait, 3*2^{n} = 3*(2/3)^n *3^nSo,= 3^n [2 + λ*(-1)^n *3*(2/3)^n ]So,= 3^n [2 + 3λ*(-1)^n*(2/3)^n ]We need this to be greater than 0 for all n.Since 3^n >0 for all n, the inequality reduces to:2 + 3λ*(-1)^n*(2/3)^n >0So,2 + 3λ*(-1)^n*(2/3)^n >0Let me write this as:2 + 3λ*(-1)^n*(2/3)^n >0Let me denote r_n = (-1)^n*(2/3)^nSo, the inequality becomes:2 + 3λ r_n >0We need this to hold for all n ∈ ℕ*.So, for each n, 2 + 3λ r_n >0Let me analyze r_n:r_n = (-1)^n*(2/3)^n = [ (-2/3) ]^nSo, r_n is positive when n is even, and negative when n is odd.Let me consider two cases: n even and n odd.**Case 1: n is even**Let n = 2k, k ∈ ℕ*Then, r_n = [ (-2/3) ]^{2k} = (4/9)^k >0So, the inequality becomes:2 + 3λ*(4/9)^k >0Since (4/9)^k >0, and λ is a real number, we need:2 + 3λ*(4/9)^k >0To ensure this for all k, we need to consider the minimal value of the expression.As k increases, (4/9)^k decreases towards 0. So, the minimal value occurs as k approaches infinity, but since k is finite, the minimal value is when k is as large as possible, but since k can be any positive integer, the minimal value is approached as k→∞, which is 2 + 0 =2 >0. So, for even n, the inequality holds for any λ, because 2 + positive term is always greater than 0.Wait, but actually, for n even, r_n is positive, so 3λ r_n is positive if λ >0, and negative if λ <0.Wait, let me correct that.Wait, for n even, r_n is positive, so 3λ r_n is positive if λ >0, and negative if λ <0.So, the inequality is:2 + 3λ r_n >0If λ >0, then 3λ r_n >0, so 2 + positive >0, which is always true.If λ <0, then 3λ r_n <0, so we need:2 + 3λ r_n >0 => 3λ r_n > -2 => λ > (-2)/(3 r_n)But since r_n = (4/9)^k >0, and λ <0, we have:λ > (-2)/(3*(4/9)^k )But since λ is negative, and (-2)/(3*(4/9)^k ) is negative, we need:λ > (-2)/(3*(4/9)^k )But as k increases, (4/9)^k decreases, so (-2)/(3*(4/9)^k ) becomes more negative. Therefore, the most restrictive condition is when k is minimal, i.e., k=1 (n=2):For n=2, k=1:r_n = (4/9)^1 =4/9So,2 + 3λ*(4/9) >0 => 2 + (4/3)λ >0 => (4/3)λ > -2 => λ > (-2)*(3/4) = -3/2So, for n even, to satisfy the inequality for all even n, we must have λ > -3/2But since λ is negative in this case, the condition is λ > -3/2**Case 2: n is odd**Let n = 2k +1, k ∈ ℕ*Then, r_n = [ (-2/3) ]^{2k+1} = (-2/3)*(4/9)^k <0So, the inequality becomes:2 + 3λ*(-2/3)*(4/9)^k >0Simplify:2 - 2λ*(4/9)^k >0So,2 > 2λ*(4/9)^kDivide both sides by 2:1 > λ*(4/9)^kSince (4/9)^k >0, and λ is a real number, we need:λ < 1/( (4/9)^k )But since n is odd, k can be 0,1,2,...Wait, for n=1, k=0:r_n = (-2/3)^1 = -2/3So, the inequality becomes:2 + 3λ*(-2/3) >0 => 2 - 2λ >0 => -2λ > -2 => λ <1For n=3, k=1:r_n = (-2/3)^3 = -8/27So,2 + 3λ*(-8/27) >0 => 2 - (8/9)λ >0 => (8/9)λ <2 => λ < (2)*(9/8) = 9/4But wait, for n=3, the condition is λ <9/4, but for n=1, it's λ <1. Since we need the inequality to hold for all n, including n=1, the most restrictive condition is λ <1.Similarly, for higher odd n, the condition becomes less restrictive because (4/9)^k decreases, so 1/( (4/9)^k ) increases, meaning λ can be larger. But since we need the inequality to hold for all n, including n=1, the most restrictive condition is λ <1.Therefore, combining both cases:For even n: λ > -3/2For odd n: λ <1Additionally, we need to consider the sign of λ.Wait, in the case of even n, we considered λ <0, but in reality, λ can be positive or negative. Let me re-examine.Wait, in the even case, when n is even, r_n is positive. So, the term 3λ r_n is positive if λ >0, and negative if λ <0.But in the inequality 2 + 3λ r_n >0, regardless of λ's sign, we need to ensure that the expression is positive.Wait, perhaps I should consider the two cases separately:Case 1: λ >0Then, for even n, 3λ r_n >0, so 2 + positive >0, which is always true.For odd n, 3λ r_n <0, so we need 2 + negative >0 => 2 > |3λ r_n|But r_n for odd n is negative, so |r_n| = | (-2/3)^n | = (2/3)^nSo,2 > 3λ*(2/3)^nWhich is:λ < 2 / [3*(2/3)^n ] = 2*(3/2)^n /3 = (2/3)*(3/2)^n = (3/2)^{n-1}But this must hold for all n odd.The minimal value of (3/2)^{n-1} occurs when n is minimal, i.e., n=1:For n=1:λ < (3/2)^{0}=1For n=3:λ < (3/2)^{2}=9/4For n=5:λ < (3/2)^4=81/16So, the most restrictive condition is λ <1.Case 2: λ <0For even n, r_n is positive, so 3λ r_n <0, so we need:2 + 3λ r_n >0 => 2 > -3λ r_n => since λ <0, -3λ >0, so:2 > -3λ r_n => -3λ r_n <2 => since r_n >0,-3λ < 2 / r_n => since λ <0, multiply both sides by -1 (reverse inequality):3λ > -2 / r_nBut r_n = (4/9)^k for n=2k, so:3λ > -2 / (4/9)^k => 3λ > -2*(9/4)^kSince λ <0, we can write:λ > -2*(9/4)^k /3 = - (2/3)*(9/4)^kWe need this to hold for all k ≥1 (since n=2k, k≥1).The minimal value of - (2/3)*(9/4)^k occurs as k increases, since (9/4)^k grows exponentially. Therefore, the most restrictive condition is when k=1 (n=2):λ > - (2/3)*(9/4) = - (2/3)*(9/4) = - (18/12) = -3/2So, for λ <0, we must have λ > -3/2Therefore, combining both cases:If λ >0, then λ <1If λ <0, then λ > -3/2So, overall, the range of λ is:-3/2 < λ <1But λ ≠0, as given.Therefore, the range is (-3/2, 0) ∪ (0,1)Let me verify this with n=1 and n=2.For n=1:C₁ =3^1 - λ*(-1)^1*(a₁)/(1 -1/2)=3 - λ*(-1)*(1)/(1/2)=3 + 2λC₂=3^2 - λ*(-1)^2*(a₂)/(2 -1/2)=9 - λ*(1)*(6)/(3/2)=9 - λ*(6)*(2/3)=9 -4λWe need C₂ > C₁:9 -4λ >3 +2λ =>9 -3 >4λ +2λ =>6>6λ =>λ <1Which matches our condition.For n=2:C₂=9 -4λC₃=3^3 - λ*(-1)^3*(a₃)/(3 -1/2)=27 - λ*(-1)*( (2*3 -1)2^{3-1} )/(5/2)=27 - λ*(-1)*(5*4)/(5/2)=27 - λ*(-1)*(20)/(5/2)=27 - λ*(-1)*(8)=27 +8λWe need C₃ >C₂:27 +8λ >9 -4λ =>27 -9 > -4λ -8λ =>18 >-12λ =>Divide both sides by -12 (inequality sign flips):-18/12 <λ => -3/2 <λWhich matches our condition.Therefore, the range of λ is (-3/2,0) ∪ (0,1)

💡First, let's define the number of marbles Ringo has as ( r ) and the number of marbles Paul has as ( p ).According to the problem:- Ringo has 5 marbles left when he tries to fill bags with 8 marbles each. This can be expressed as: [ r equiv 5 pmod{8} ]- Paul has 6 marbles left when he tries to fill bags with 8 marbles each. This can be expressed as: [ p equiv 6 pmod{8} ]When they combine their marbles, the total number of marbles they have together is ( r + p ). To find the number of marbles left over after filling as many bags as possible with 8 marbles each, we add the two congruences:[r + p equiv 5 + 6 pmod{8}][r + p equiv 11 pmod{8}]Simplifying ( 11 mod 8 ):[11 equiv 3 pmod{8}]This means that when Ringo and Paul combine their marbles, they will have 3 marbles left over after filling as many bags of 8 marbles as possible.

💡Alright, so I have this inequality to prove: if (a, b, c > 0), then[frac{a^{3}}{a^{2}+ab+b^{2}} + frac{b^{3}}{b^{2}+bc+c^{2}} + frac{c^{3}}{c^{2}+ac+a^{2}} geq frac{a+b+c}{3}.]Hmm, okay. I remember that inequalities like this often involve techniques like the AM-GM inequality or Cauchy-Schwarz. Let me think about how to approach this.First, looking at each term on the left-hand side, they have a similar structure: (frac{x^3}{x^2 + xy + y^2}). Maybe I can find a way to simplify or bound each of these terms.Wait, I recall that sometimes with fractions like this, you can decompose them or find a relationship between the numerator and denominator. Let me try to manipulate one of these terms.Take (frac{a^3}{a^2 + ab + b^2}). Maybe I can write this as:[frac{a^3}{a^2 + ab + b^2} = a - frac{ab(a + b)}{a^2 + ab + b^2}.]Is that right? Let me check:[a - frac{ab(a + b)}{a^2 + ab + b^2} = frac{a(a^2 + ab + b^2) - ab(a + b)}{a^2 + ab + b^2} = frac{a^3 + a^2b + ab^2 - a^2b - ab^2}{a^2 + ab + b^2} = frac{a^3}{a^2 + ab + b^2}.]Yes, that works. So, each term can be written as (x - frac{xy(x + y)}{x^2 + xy + y^2}).So, applying this to all three terms, the left-hand side becomes:[left(a - frac{ab(a + b)}{a^2 + ab + b^2}right) + left(b - frac{bc(b + c)}{b^2 + bc + c^2}right) + left(c - frac{ac(a + c)}{c^2 + ac + a^2}right).]Simplifying, that's:[a + b + c - left(frac{ab(a + b)}{a^2 + ab + b^2} + frac{bc(b + c)}{b^2 + bc + c^2} + frac{ac(a + c)}{c^2 + ac + a^2}right).]So, the original inequality becomes:[a + b + c - left(frac{ab(a + b)}{a^2 + ab + b^2} + frac{bc(b + c)}{b^2 + bc + c^2} + frac{ac(a + c)}{c^2 + ac + a^2}right) geq frac{a + b + c}{3}.]Subtracting (a + b + c) from both sides:[- left(frac{ab(a + b)}{a^2 + ab + b^2} + frac{bc(b + c)}{b^2 + bc + c^2} + frac{ac(a + c)}{c^2 + ac + a^2}right) geq -frac{2(a + b + c)}{3}.]Multiplying both sides by -1 (which reverses the inequality):[frac{ab(a + b)}{a^2 + ab + b^2} + frac{bc(b + c)}{b^2 + bc + c^2} + frac{ac(a + c)}{c^2 + ac + a^2} leq frac{2(a + b + c)}{3}.]Okay, so now I need to show that the sum of these fractions is less than or equal to (frac{2(a + b + c)}{3}).Looking at each term individually, maybe I can find an upper bound for (frac{ab(a + b)}{a^2 + ab + b^2}).Let me consider (frac{ab(a + b)}{a^2 + ab + b^2}). I wonder if I can relate this to something simpler, like (frac{a + b}{3}).Wait, I remember that (a^2 + ab + b^2 geq 3ab) by the AM-GM inequality. Let me verify that:The AM-GM inequality states that for non-negative numbers, the arithmetic mean is greater than or equal to the geometric mean. For three terms, it would be:[frac{a^2 + ab + b^2}{3} geq sqrt[3]{a^2 cdot ab cdot b^2} = sqrt[3]{a^3b^3} = ab.]Multiplying both sides by 3:[a^2 + ab + b^2 geq 3ab.]Yes, that's correct. So, since (a^2 + ab + b^2 geq 3ab), we can write:[frac{ab(a + b)}{a^2 + ab + b^2} leq frac{ab(a + b)}{3ab} = frac{a + b}{3}.]Ah, that's useful! So, each term (frac{ab(a + b)}{a^2 + ab + b^2}) is less than or equal to (frac{a + b}{3}).Similarly, applying this to the other terms:[frac{bc(b + c)}{b^2 + bc + c^2} leq frac{b + c}{3},][frac{ac(a + c)}{c^2 + ac + a^2} leq frac{a + c}{3}.]Adding these inequalities together:[frac{ab(a + b)}{a^2 + ab + b^2} + frac{bc(b + c)}{b^2 + bc + c^2} + frac{ac(a + c)}{c^2 + ac + a^2} leq frac{a + b}{3} + frac{b + c}{3} + frac{a + c}{3}.]Simplifying the right-hand side:[frac{a + b + b + c + a + c}{3} = frac{2a + 2b + 2c}{3} = frac{2(a + b + c)}{3}.]So, we've shown that:[frac{ab(a + b)}{a^2 + ab + b^2} + frac{bc(b + c)}{b^2 + bc + c^2} + frac{ac(a + c)}{c^2 + ac + a^2} leq frac{2(a + b + c)}{3}.]Going back to our earlier expression, this means:[a + b + c - left(frac{ab(a + b)}{a^2 + ab + b^2} + frac{bc(b + c)}{b^2 + bc + c^2} + frac{ac(a + c)}{c^2 + ac + a^2}right) geq a + b + c - frac{2(a + b + c)}{3}.]Simplifying the right-hand side:[a + b + c - frac{2(a + b + c)}{3} = frac{3(a + b + c) - 2(a + b + c)}{3} = frac{a + b + c}{3}.]Therefore, we've shown that:[frac{a^{3}}{a^{2}+ab+b^{2}} + frac{b^{3}}{b^{2}+bc+c^{2}} + frac{c^{3}}{c^{2}+ac+a^{2}} geq frac{a + b + c}{3}.]Which is exactly what we needed to prove. So, the inequality holds true for all positive real numbers (a, b, c).**Final Answer**The inequality is proven, and the result is boxed{frac{a+b+c}{3}}.

💡First, I need to identify the first ten prime numbers. These are 2, 3, 5, 7, 11, 13, 17, 19, 23, and 29.Next, I'll calculate the total number of possible pairs when drawing two primes without replacement. This can be done using the combination formula C(10, 2), which equals 45.To find the probability that the sum of two primes is also a prime number, I need to determine how many such pairs exist. Since the sum of two odd primes will always be even (and greater than 2), it cannot be a prime number. Therefore, one of the primes in the pair must be 2, the only even prime.I'll then check the sums of 2 with each of the other primes:- 2 + 3 = 5 (prime)- 2 + 5 = 7 (prime)- 2 + 11 = 13 (prime)- 2 + 17 = 19 (prime)- 2 + 23 = 25 (not prime)- 2 + 29 = 31 (prime)There are 6 valid pairs where the sum is a prime number.Finally, the probability is the number of successful pairs divided by the total number of pairs, which is 6/45. Simplifying this fraction gives 2/15.