🧠:<think>Okay, so I've got this math problem here. It says that |2a + b| and √(3b + 12) are opposite in sign. Then, I need to find the square root of 2a - 3b and solve the equation ax² + 4b - 2 = 0 in terms of x. Hmm, let me try to figure this out step by step.First, let's understand what it means for |2a + b| and √(3b + 12) to be opposite in sign. Absolute value, |2a + b|, is always non-negative because absolute value can't be negative. Similarly, a square root, √(3b + 12), is also always non-negative because you can't have a negative square root in real numbers. So, if they're opposite in sign, one has to be positive and the other negative. But wait, both of these expressions are non-negative. The only way they can be opposite in sign is if one of them is zero. Because zero is neither positive nor negative. So, if one is zero, the other can be positive or zero, but since they need to be opposite, the other must be positive or negative. But since both expressions are non-negative, the only possibility is that one is zero, and the other is positive. So, let's consider two cases:1. |2a + b| = 0 and √(3b + 12) > 02. |2a + b| > 0 and √(3b + 12) = 0Let's start with the first case: |2a + b| = 0. This implies that 2a + b = 0. And since √(3b + 12) > 0, the expression inside the square root must be positive. So, 3b + 12 > 0. Let's solve these two equations.From 2a + b = 0, we can express b in terms of a: b = -2a. Now, substitute b into the inequality 3b + 12 > 0:3(-2a) + 12 > 0-6a + 12 > 0-6a > -12Divide both sides by -6, remembering to flip the inequality sign:a < 2So, in this case, a is less than 2, and b = -2a.Now, let's look at the second case: |2a + b| > 0 and √(3b + 12) = 0. For the square root to be zero, the expression inside must be zero:3b + 12 = 03b = -12b = -4And since |2a + b| > 0, that means 2a + b ≠ 0. Let's check if 2a + (-4) ≠ 0:2a - 4 ≠ 02a ≠ 4a ≠ 2So, in this case, b is -4, and a is not equal to 2.Wait a minute, so we have two possible scenarios:1. a < 2 and b = -2a2. b = -4 and a ≠ 2But the problem doesn't specify which case it is, so do we have to consider both? Or is there something I'm missing?Let me think. Since |2a + b| and √(3b + 12) are opposite in sign, and we've established that one must be zero and the other positive. So, maybe both cases are possible? But then, for part (1), finding the square root of 2a - 3b, it might depend on the values of a and b. Similarly, for part (2), solving the equation ax² + 4b - 2 = 0, again, it depends on a and b.But the problem doesn't specify any additional constraints on a and b, so maybe both cases are valid. Therefore, perhaps I need to explore both possibilities.Let's start with the first case: a < 2 and b = -2a.So, from b = -2a, let's compute 2a - 3b:2a - 3b = 2a - 3(-2a) = 2a + 6a = 8aSo, the square root of 2a - 3b is √(8a). But since a < 2, and we don't know the exact value of a, this expression remains as √(8a). However, for the square root to be real, 8a must be non-negative. Since a is a real number, and a < 2, 8a can be negative if a is negative. Wait, but if a is negative, then 8a is negative, and we can't take the square root of a negative number in real numbers. So, 8a must be greater than or equal to zero. Hence, a ≥ 0.Therefore, in this case, a is between 0 and 2 (0 ≤ a < 2), and b = -2a.So, for part (1), √(2a - 3b) = √(8a). Since a is between 0 and 2, √(8a) is a real number.For part (2), solving ax² + 4b - 2 = 0, let's substitute b = -2a:a x² + 4(-2a) - 2 = 0a x² - 8a - 2 = 0Let's factor out 'a' from the first two terms:a(x² - 8) - 2 = 0But this doesn't seem to help much. Maybe rearrange the equation:a x² = 8a + 2Then, if a ≠ 0, we can divide both sides by a:x² = 8 + 2/aSo, x = ±√(8 + 2/a)But since a is between 0 and 2, and a ≠ 0, this is valid.Now, let's look at the second case: b = -4 and a ≠ 2.So, let's compute 2a - 3b:2a - 3(-4) = 2a + 12So, the square root of 2a - 3b is √(2a + 12)Since b = -4, and we have no constraints on a except a ≠ 2, we need to ensure that 2a + 12 is non-negative for the square root to be real. So, 2a + 12 ≥ 0 ⇒ a ≥ -6So, as long as a ≥ -6, this is valid.For part (2), solving ax² + 4b - 2 = 0, substitute b = -4:a x² + 4(-4) - 2 = 0a x² - 16 - 2 = 0a x² - 18 = 0So, a x² = 18If a ≠ 0, then x² = 18/aTherefore, x = ±√(18/a)But since a ≠ 2, and a can be any real number except 2, as long as a > 0, this is valid. If a is negative, then x² would be negative, which is not possible in real numbers, so we must have a > 0.So, in this case, a > 0 and a ≠ 2, and x = ±√(18/a)Putting it all together, we have two scenarios:1. When a < 2 and b = -2a: - √(2a - 3b) = √(8a) - Solutions for x: x = ±√(8 + 2/a)2. When b = -4 and a ≠ 2: - √(2a - 3b) = √(2a + 12) - Solutions for x: x = ±√(18/a)But wait, the problem states that |2a + b| and √(3b + 12) are opposite in sign. So, in the first case, |2a + b| = 0 and √(3b + 12) > 0. In the second case, |2a + b| > 0 and √(3b + 12) = 0. So, both cases are possible, and thus, both sets of solutions are valid depending on the values of a and b.However, the problem asks us to find the square root of 2a - 3b and solve the equation in terms of x. So, it seems like we need to express the answers in terms of a and b, but since a and b are related through the initial condition, perhaps we can find specific values.Wait, maybe I misinterpreted the initial condition. Let me go back.The problem says that |2a + b| and √(3b + 12) are opposite in sign. So, they have opposite signs. Since |2a + b| is always non-negative, and √(3b + 12) is also always non-negative, the only way they can be opposite in sign is if one is zero and the other is positive. Because if both are non-negative, to be opposite in sign, one must be positive and the other negative, but since they can't be negative, the only possibility is one is zero, and the other is positive.Therefore, we have two possibilities:1. |2a + b| = 0 and √(3b + 12) > 02. |2a + b| > 0 and √(3b + 12) = 0In the first case:|2a + b| = 0 ⇒ 2a + b = 0 ⇒ b = -2aAnd √(3b + 12) > 0 ⇒ 3b + 12 > 0 ⇒ 3(-2a) + 12 > 0 ⇒ -6a + 12 > 0 ⇒ -6a > -12 ⇒ a < 2So, in this case, a < 2, b = -2aIn the second case:|2a + b| > 0 ⇒ 2a + b ≠ 0And √(3b + 12) = 0 ⇒ 3b + 12 = 0 ⇒ b = -4So, in this case, b = -4 and 2a + (-4) ≠ 0 ⇒ 2a ≠ 4 ⇒ a ≠ 2Therefore, we have two distinct cases:Case 1:- b = -2a- a < 2Case 2:- b = -4- a ≠ 2Now, for each case, we can compute the required values.Starting with Case 1:Compute 2a - 3b:2a - 3b = 2a - 3(-2a) = 2a + 6a = 8aTherefore, √(2a - 3b) = √(8a)Since a < 2, and for √(8a) to be real, 8a ≥ 0 ⇒ a ≥ 0So, in this case, 0 ≤ a < 2, and √(2a - 3b) = √(8a)Now, for the equation ax² + 4b - 2 = 0:Substitute b = -2a:a x² + 4(-2a) - 2 = 0 ⇒ a x² - 8a - 2 = 0Let's solve for x²:a x² = 8a + 2 ⇒ x² = (8a + 2)/aAssuming a ≠ 0, which it isn't because a ≥ 0 and a < 2, and a = 0 would make |2a + b| = 0, but then b = 0, which conflicts with b = -2a, but if a = 0, b = 0, but then √(3b + 12) = √12 > 0, which is valid. Wait, if a = 0, then 2a + b = 0 ⇒ b = 0, but then 3b + 12 = 12 ⇒ √12 > 0, so that's okay. So, a can be 0 as well.But if a = 0, the equation becomes 0 x² + 4b - 2 = 0 ⇒ 4b - 2 = 0 ⇒ b = 0.5But wait, in Case 1, b = -2a, so if a = 0, b = 0. But from the equation, b would have to be 0.5. That's a contradiction. So, a cannot be 0 in this case.Therefore, a > 0 in Case 1.So, x² = (8a + 2)/a = 8 + 2/aThus, x = ±√(8 + 2/a)So, for Case 1, the solutions are:√(2a - 3b) = √(8a)andx = ±√(8 + 2/a)Now, moving to Case 2:Compute 2a - 3b:2a - 3(-4) = 2a + 12Therefore, √(2a - 3b) = √(2a + 12)Since b = -4, and a ≠ 2, we need to ensure that 2a + 12 ≥ 0 ⇒ a ≥ -6So, a ≥ -6, a ≠ 2Now, for the equation ax² + 4b - 2 = 0:Substitute b = -4:a x² + 4(-4) - 2 = 0 ⇒ a x² - 16 - 2 = 0 ⇒ a x² - 18 = 0 ⇒ a x² = 18 ⇒ x² = 18/aAssuming a ≠ 0, which it isn't because a ≠ 2 and if a = 0, then the equation becomes 0 x² - 18 = 0 ⇒ -18 = 0, which is impossible. So, a ≠ 0 is already satisfied.Thus, x = ±√(18/a)So, for Case 2, the solutions are:√(2a - 3b) = √(2a + 12)andx = ±√(18/a)But the problem doesn't specify which case to consider, so perhaps we need to consider both possibilities. However, since the problem asks to "find the square root of 2a - 3b" and "solve the equation ax² + 4b - 2 = 0 in terms of x", it might be expecting specific numerical answers rather than expressions in terms of a. This suggests that there might be specific values for a and b.Wait, maybe I need to find a and b such that both conditions are satisfied, and then compute the required expressions. Let me see.From the initial condition, either Case 1 or Case 2 must hold. So, we have two separate scenarios.But perhaps the problem expects us to consider both cases and provide the answers accordingly.Alternatively, maybe there's a unique solution for a and b. Let me check.If I consider both cases:In Case 1: b = -2a, a < 2In Case 2: b = -4, a ≠ 2So, unless a = -2, which would make b = 4 in Case 1, but in Case 2, b is fixed at -4.Wait, maybe there's a unique solution when combining both cases. Let me see.If we suppose that both cases hold simultaneously, which would require:From Case 1: b = -2aFrom Case 2: b = -4Therefore, -2a = -4 ⇒ a = 2But in Case 1, a < 2, and in Case 2, a ≠ 2. So, a = 2 is excluded from both cases. Therefore, there is no overlap between the two cases.Therefore, we have two distinct possibilities for a and b, leading to two different sets of solutions for parts (1) and (2).But the problem doesn't specify any further constraints, so perhaps we need to present both possibilities.However, looking back at the problem statement, it says "Given that |2a + b| and √(3b + 12) are opposite in sign."This implies that one is positive and the other is negative. But since both are non-negative, the only way is one is zero and the other is positive.Therefore, we have two cases:1. |2a + b| = 0 and √(3b + 12) > 02. |2a + b| > 0 and √(3b + 12) = 0So, let's solve for a and b in each case.Case 1:|2a + b| = 0 ⇒ 2a + b = 0 ⇒ b = -2a√(3b + 12) > 0 ⇒ 3b + 12 > 0 ⇒ 3(-2a) + 12 > 0 ⇒ -6a + 12 > 0 ⇒ -6a > -12 ⇒ a < 2So, in this case, a < 2 and b = -2aCase 2:|2a + b| > 0 ⇒ 2a + b ≠ 0√(3b + 12) = 0 ⇒ 3b + 12 = 0 ⇒ b = -4So, in this case, b = -4 and 2a + (-4) ≠ 0 ⇒ 2a ≠ 4 ⇒ a ≠ 2Therefore, we have two possible sets of solutions for a and b:1. b = -2a, a < 22. b = -4, a ≠ 2Now, let's compute the required expressions for each case.Case 1:Compute 2a - 3b:2a - 3b = 2a - 3(-2a) = 2a + 6a = 8aTherefore, √(2a - 3b) = √(8a)Since a < 2 and a ≥ 0 (because √(8a) must be real), 0 ≤ a < 2Now, solve ax² + 4b - 2 = 0:Substitute b = -2a:a x² + 4(-2a) - 2 = 0 ⇒ a x² - 8a - 2 = 0Let's solve for x²:a x² = 8a + 2 ⇒ x² = (8a + 2)/a = 8 + 2/aThus, x = ±√(8 + 2/a)Case 2:Compute 2a - 3b:2a - 3(-4) = 2a + 12Therefore, √(2a - 3b) = √(2a + 12)Since b = -4, and a ≠ 2, we need 2a + 12 ≥ 0 ⇒ a ≥ -6Now, solve ax² + 4b - 2 = 0:Substitute b = -4:a x² + 4(-4) - 2 = 0 ⇒ a x² - 16 - 2 = 0 ⇒ a x² - 18 = 0 ⇒ a x² = 18 ⇒ x² = 18/aThus, x = ±√(18/a)So, summarizing:For Case 1 (b = -2a, a < 2):1. √(2a - 3b) = √(8a)2. Solutions for x: x = ±√(8 + 2/a)For Case 2 (b = -4, a ≠ 2):1. √(2a - 3b) = √(2a + 12)2. Solutions for x: x = ±√(18/a)But the problem doesn't specify which case to consider, so perhaps we need to present both possibilities.Alternatively, maybe there's a unique solution when considering both cases together, but as we saw earlier, a = 2 is excluded from both cases, so there's no overlap.Therefore, the answers depend on the values of a and b, which are either in Case 1 or Case 2.But the problem asks to "find the square root of 2a - 3b" and "solve the equation ax² + 4b - 2 = 0 in terms of x". It doesn't specify to express in terms of a, so perhaps it's expecting numerical values.Wait, but without specific values for a and b, we can't compute numerical answers. Therefore, maybe the problem expects expressions in terms of a, as we derived earlier.Alternatively, perhaps the problem assumes that both cases are valid, and we need to provide both possibilities.But looking back, the problem states that |2a + b| and √(3b + 12) are opposite in sign. Since both are non-negative, one must be zero and the other positive. Therefore, we have two distinct scenarios, each leading to different expressions for a and b, and thus different answers for parts (1) and (2).Therefore, the answers are:(1) √(2a - 3b) is either √(8a) or √(2a + 12), depending on whether we're in Case 1 or Case 2.(2) The solutions for x are either ±√(8 + 2/a) or ±√(18/a), again depending on the case.But the problem doesn't specify which case to use, so perhaps we need to consider both possibilities.Alternatively, maybe there's a unique solution when considering both cases together, but as we saw earlier, a = 2 is excluded, so there's no overlap.Therefore, the answers are as above.But let's check if there's a specific solution.Suppose we consider both cases together:From Case 1: b = -2aFrom Case 2: b = -4Setting them equal: -2a = -4 ⇒ a = 2But in Case 1, a < 2, and in Case 2, a ≠ 2. Therefore, a = 2 is not a valid solution in either case.Therefore, there's no overlap, and the two cases are distinct.Thus, the answers are as derived above.Therefore, the square root of 2a - 3b is either √(8a) or √(2a + 12), and the solutions for x are either ±√(8 + 2/a) or ±√(18/a), depending on whether we're in Case 1 or Case 2.But since the problem doesn't specify which case, perhaps we need to present both possibilities.Alternatively, maybe the problem expects us to consider that both cases are possible and provide both sets of answers.But given the way the problem is phrased, it's likely that there's a unique solution, which means I might have made a mistake in interpreting the cases.Wait, let's think differently. The problem says that |2a + b| and √(3b + 12) are opposite in sign. Since both are non-negative, the only way they can be opposite in sign is if one is zero and the other is positive. Therefore, we have two possibilities:1. |2a + b| = 0 and √(3b + 12) > 02. |2a + b| > 0 and √(3b + 12) = 0So, in the first case:|2a + b| = 0 ⇒ 2a + b = 0 ⇒ b = -2aAnd √(3b + 12) > 0 ⇒ 3b + 12 > 0 ⇒ 3(-2a) + 12 > 0 ⇒ -6a + 12 > 0 ⇒ a < 2So, b = -2a, a < 2In the second case:√(3b + 12) = 0 ⇒ 3b + 12 = 0 ⇒ b = -4And |2a + b| > 0 ⇒ 2a + b ≠ 0 ⇒ 2a - 4 ≠ 0 ⇒ a ≠ 2So, b = -4, a ≠ 2Therefore, we have two distinct sets of solutions for a and b.Now, let's compute the required expressions for each case.Case 1:Compute 2a - 3b:2a - 3b = 2a - 3(-2a) = 2a + 6a = 8aTherefore, √(2a - 3b) = √(8a)Since a < 2, and a ≥ 0 (because √(8a) must be real), 0 ≤ a < 2Now, solve ax² + 4b - 2 = 0:Substitute b = -2a:a x² + 4(-2a) - 2 = 0 ⇒ a x² - 8a - 2 = 0Let's solve for x²:a x² = 8a + 2 ⇒ x² = (8a + 2)/a = 8 + 2/aThus, x = ±√(8 + 2/a)Case 2:Compute 2a - 3b:2a - 3(-4) = 2a + 12Therefore, √(2a - 3b) = √(2a + 12)Since b = -4, and a ≠ 2, we need 2a + 12 ≥ 0 ⇒ a ≥ -6Now, solve ax² + 4b - 2 = 0:Substitute b = -4:a x² + 4(-4) - 2 = 0 ⇒ a x² - 16 - 2 = 0 ⇒ a x² - 18 = 0 ⇒ a x² = 18 ⇒ x² = 18/aThus, x = ±√(18/a)So, in conclusion, depending on the case, the answers are:For Case 1 (b = -2a, a < 2):1. √(2a - 3b) = √(8a)2. x = ±√(8 + 2/a)For Case 2 (b = -4, a ≠ 2):1. √(2a - 3b) = √(2a + 12)2. x = ±√(18/a)Since the problem doesn't specify which case to use, we need to present both possibilities.However, the problem might expect a specific numerical answer, implying that there's a unique solution. Let me check if there's a way to determine a and b uniquely.Wait, if we consider both cases together, we have:From Case 1: b = -2a, a < 2From Case 2: b = -4, a ≠ 2If we set b = -2a equal to b = -4, we get a = 2, but in Case 1, a < 2, and in Case 2, a ≠ 2. Therefore, a = 2 is not a valid solution in either case, meaning there's no overlap. Thus, the two cases are distinct, and we can't determine unique values for a and b without additional information.Therefore, the answers must be expressed in terms of a, as we've done above.So, the final answers are:(1) The square root of 2a - 3b is either √(8a) or √(2a + 12), depending on whether |2a + b| is zero or √(3b + 12) is zero.(2) The solutions for x are either ±√(8 + 2/a) or ±√(18/a), corresponding to the respective cases.However, since the problem asks to "find the square root of 2a - 3b" and "solve the equation...", it's likely expecting expressions rather than multiple possibilities. Therefore, the answers are:(1) √(8a) or √(2a + 12)(2) x = ±√(8 + 2/a) or x = ±√(18/a)But to present them neatly, perhaps we can box both possibilities.Alternatively, if the problem expects a single answer, it might be that a and b are uniquely determined, but as we've seen, there are two distinct cases. Therefore, unless there's a constraint I'm missing, both cases are valid.In conclusion, the answers are as above.</think>

🧠:<think>Alright, so I've got this sequence problem here, and it's divided into three parts. Let me try to tackle each part step by step. First, the sequence is defined by ( a_1 = 1 ) and ( a_{n+1} = 1 - frac{1}{4a_n} ) for ( n in mathbb{N}^* ). That means it's a recursively defined sequence, and I need to work with this definition.Part (1): Let ( b_n = frac{2}{2a_n - 1} ). I need to prove that ( {b_n} ) is an arithmetic sequence. Hmm, okay. An arithmetic sequence has a common difference, so I need to show that ( b_{n+1} - b_n ) is constant for all ( n ). Let me compute ( b_{n+1} ) in terms of ( a_{n+1} ) and then relate it back to ( a_n ).Given ( a_{n+1} = 1 - frac{1}{4a_n} ), I can plug this into the expression for ( b_{n+1} ):( b_{n+1} = frac{2}{2a_{n+1} - 1} = frac{2}{2left(1 - frac{1}{4a_n}right) - 1} ).Let me simplify the denominator:( 2left(1 - frac{1}{4a_n}right) - 1 = 2 - frac{2}{4a_n} - 1 = 1 - frac{1}{2a_n} ).So, ( b_{n+1} = frac{2}{1 - frac{1}{2a_n}} ).Hmm, maybe I can write this differently. Let me factor out the denominator:( 1 - frac{1}{2a_n} = frac{2a_n - 1}{2a_n} ), so the reciprocal is ( frac{2a_n}{2a_n - 1} ).Thus, ( b_{n+1} = 2 times frac{2a_n}{2a_n - 1} = frac{4a_n}{2a_n - 1} ).Wait, but ( b_n = frac{2}{2a_n - 1} ). So, ( frac{4a_n}{2a_n - 1} = 2a_n times frac{2}{2a_n - 1} = 2a_n times b_n ).But this seems a bit tangled. Maybe I can express ( a_n ) in terms of ( b_n ). From ( b_n = frac{2}{2a_n - 1} ), we can solve for ( a_n ):Multiply both sides by ( 2a_n - 1 ): ( b_n(2a_n - 1) = 2 ).So, ( 2b_n a_n - b_n = 2 ).Then, ( 2b_n a_n = 2 + b_n ), so ( a_n = frac{2 + b_n}{2b_n} ).Let me plug this into the expression for ( b_{n+1} ):( b_{n+1} = frac{4a_n}{2a_n - 1} ).Substituting ( a_n = frac{2 + b_n}{2b_n} ):First, compute ( 4a_n = 4 times frac{2 + b_n}{2b_n} = frac{8 + 4b_n}{2b_n} = frac{4 + 2b_n}{b_n} ).Next, compute ( 2a_n - 1 = 2 times frac{2 + b_n}{2b_n} - 1 = frac{2(2 + b_n)}{2b_n} - 1 = frac{2 + b_n}{b_n} - 1 = frac{2 + b_n - b_n}{b_n} = frac{2}{b_n} ).So, ( b_{n+1} = frac{frac{4 + 2b_n}{b_n}}{frac{2}{b_n}} = frac{4 + 2b_n}{2} = 2 + b_n ).Ah, that simplifies nicely! Therefore, ( b_{n+1} = b_n + 2 ), which means that each term is 2 more than the previous term. So, ( {b_n} ) is indeed an arithmetic sequence with a common difference of 2.Given that ( a_1 = 1 ), let's compute ( b_1 ):( b_1 = frac{2}{2a_1 - 1} = frac{2}{2 times 1 - 1} = frac{2}{1} = 2 ).Therefore, the sequence ( {b_n} ) starts at 2 and increases by 2 each time, so ( b_n = 2n ).Part (2): Now, we have ( c_n = 6^n + (-1)^{n-1} lambda cdot 2^{b_n} ). We need to determine if there exists a ( lambda ) such that ( c_{n+1} > c_n ) for all ( n in mathbb{N}^+ ). If such a ( lambda ) exists, we need to find its range.First, let's express ( c_{n+1} ) and ( c_n ):( c_{n+1} = 6^{n+1} + (-1)^{(n+1)-1} lambda cdot 2^{b_{n+1}} = 6^{n+1} + (-1)^n lambda cdot 2^{b_{n+1}} ).Similarly, ( c_n = 6^n + (-1)^{n-1} lambda cdot 2^{b_n} ).We need ( c_{n+1} > c_n ), so subtract ( c_n ) from both sides:( c_{n+1} - c_n = 6^{n+1} - 6^n + (-1)^n lambda cdot 2^{b_{n+1}} - (-1)^{n-1} lambda cdot 2^{b_n} > 0 ).Simplify each part:First, ( 6^{n+1} - 6^n = 6^n(6 - 1) = 5 times 6^n ).Second, the lambda terms:( (-1)^n lambda cdot 2^{b_{n+1}} - (-1)^{n-1} lambda cdot 2^{b_n} = (-1)^n lambda cdot 2^{b_{n+1}} + (-1)^n lambda cdot 2^{b_n} ) because ( -(-1)^{n-1} = (-1)^n ).So, factor out ( (-1)^n lambda ):( (-1)^n lambda (2^{b_{n+1}} + 2^{b_n}) ).But wait, ( b_{n+1} = b_n + 2 ) since it's an arithmetic sequence with common difference 2. So, ( 2^{b_{n+1}} = 2^{b_n + 2} = 4 times 2^{b_n} ).Thus, the lambda term becomes:( (-1)^n lambda (4 times 2^{b_n} + 2^{b_n}) = (-1)^n lambda times 5 times 2^{b_n} ).So, putting it all together:( c_{n+1} - c_n = 5 times 6^n + (-1)^n times 5 times lambda times 2^{b_n} > 0 ).We can factor out the 5:( 5 left(6^n + (-1)^n lambda 2^{b_n}right) > 0 ).Since 5 is positive, we can divide both sides by 5:( 6^n + (-1)^n lambda 2^{b_n} > 0 ).So, the condition simplifies to:( 6^n + (-1)^n lambda 2^{b_n} > 0 ) for all ( n ).Let me analyze this inequality for even and odd ( n ).First, note that ( b_n = 2n ), as established in part (1). So, ( 2^{b_n} = 2^{2n} = 4^n ).So, the inequality becomes:( 6^n + (-1)^n lambda 4^n > 0 ).Let me write this as:( 6^n + (-1)^n lambda 4^n > 0 ).We can factor out ( 4^n ):( 4^n left( left(frac{6}{4}right)^n + (-1)^n lambda right) > 0 ).Since ( 4^n > 0 ) for all ( n ), the sign of the entire expression depends on the term in the brackets:( left(frac{6}{4}right)^n + (-1)^n lambda > 0 ).Simplify ( frac{6}{4} ) to ( frac{3}{2} ):( left(frac{3}{2}right)^n + (-1)^n lambda > 0 ).So, the inequality is:( left(frac{3}{2}right)^n + (-1)^n lambda > 0 ) for all ( n ).Let me consider cases when ( n ) is even and when ( n ) is odd.Case 1: ( n ) is even.Let ( n = 2k ) for some integer ( k geq 1 ). Then, ( (-1)^n = 1 ). So, the inequality becomes:( left(frac{3}{2}right)^{2k} + lambda > 0 ).Which is:( left(frac{9}{4}right)^k + lambda > 0 ).Since ( left(frac{9}{4}right)^k ) is positive and increases as ( k ) increases, the smallest value occurs when ( k = 1 ), which is ( frac{9}{4} ).Thus, for even ( n ), the inequality becomes:( frac{9}{4} + lambda > 0 Rightarrow lambda > -frac{9}{4} ).Case 2: ( n ) is odd.Let ( n = 2k - 1 ) for some integer ( k geq 1 ). Then, ( (-1)^n = -1 ). So, the inequality becomes:( left(frac{3}{2}right)^{2k - 1} - lambda > 0 ).Which is:( left(frac{3}{2}right)^{2k - 1} > lambda ).Since ( left(frac{3}{2}right)^{2k - 1} ) is positive and increases as ( k ) increases, the smallest upper bound for ( lambda ) occurs when ( k = 1 ), which is ( frac{3}{2} ).Thus, for odd ( n ), the inequality becomes:( frac{3}{2} > lambda ).So, combining both cases, we have:( -frac{9}{4} < lambda < frac{3}{2} ).Therefore, such a ( lambda ) exists, and it must lie in the interval ( left( -frac{9}{4}, frac{3}{2} right) ).Part (3): We need to prove that for all positive integers ( n ),( frac{1}{b_1(b_1 + 1)} + frac{1}{b_2(b_2 + 1)} + ldots + frac{1}{b_n(b_n + 1)} < frac{13}{42} ).First, from part (1), we know ( b_n = 2n ). So, each term in the sum is:( frac{1}{2n(2n + 1)} ).Let me write out the first few terms to see if I can find a telescoping pattern or some way to bound the sum.The general term is ( frac{1}{2n(2n + 1)} ). Let me try partial fractions:( frac{1}{2n(2n + 1)} = frac{A}{2n} + frac{B}{2n + 1} ).Multiplying both sides by ( 2n(2n + 1) ):( 1 = A(2n + 1) + B(2n) ).Expanding:( 1 = (2A + 2B)n + A ).This must hold for all ( n ), so coefficients must match:- Coefficient of ( n ): ( 2A + 2B = 0 Rightarrow A = -B ).- Constant term: ( A = 1 ).Thus, ( A = 1 ), so ( B = -1 ).Therefore,( frac{1}{2n(2n + 1)} = frac{1}{2n} - frac{1}{2n + 1} ).So, the sum becomes:( sum_{k=1}^n left( frac{1}{2k} - frac{1}{2k + 1} right ) ).Let me write this out:( left( frac{1}{2} - frac{1}{3} right ) + left( frac{1}{4} - frac{1}{5} right ) + ldots + left( frac{1}{2n} - frac{1}{2n + 1} right ) ).Hmm, this doesn't telescope in the usual sense, but maybe I can compare it to another series or bound it.Alternatively, notice that each term ( frac{1}{2k} - frac{1}{2k + 1} = frac{1}{2k(2k + 1)} ), which is positive and decreasing.But I need a bound on the sum. Let me see if I can express this sum in terms of harmonic numbers or something else.Wait, another approach: Let me consider the difference between consecutive terms. Maybe I can find an upper bound for each term and sum those bounds.But perhaps a better idea is to compare with an integral or use induction.Alternatively, perhaps I can write the sum as:( S_n = sum_{k=1}^n frac{1}{2k(2k + 1)} ).I can write this as:( S_n = frac{1}{2} sum_{k=1}^n left( frac{1}{k} - frac{1}{2k + 1} right ) ).Wait, that might not be helpful. Let me try another partial fraction decomposition.Wait, I already did that. Alternatively, let me consider the sum:( S_n = sum_{k=1}^n left( frac{1}{2k} - frac{1}{2k + 1} right ) ).This can be written as:( left( frac{1}{2} + frac{1}{4} + ldots + frac{1}{2n} right ) - left( frac{1}{3} + frac{1}{5} + ldots + frac{1}{2n + 1} right ) ).Let me denote the first sum as ( S_1 ) and the second as ( S_2 ):( S_1 = sum_{k=1}^n frac{1}{2k} = frac{1}{2} sum_{k=1}^n frac{1}{k} ).( S_2 = sum_{k=1}^n frac{1}{2k + 1} ).So, ( S_n = S_1 - S_2 ).But ( S_2 = sum_{k=1}^n frac{1}{2k + 1} = sum_{m=3}^{2n + 1} frac{1}{m} ) where ( m ) is odd.Wait, perhaps I can relate this to the harmonic series.Recall that the harmonic series ( H_n = sum_{k=1}^n frac{1}{k} ).Also, the sum of reciprocals of odd numbers up to ( 2n + 1 ) is ( sum_{k=1}^{n + 1} frac{1}{2k - 1} ).But I'm not sure if this helps directly.Alternatively, let me think about the difference ( S_1 - S_2 ):( S_1 = frac{1}{2} H_n ).( S_2 = sum_{k=1}^n frac{1}{2k + 1} = sum_{m=3}^{2n + 1} frac{1}{m} ) where ( m ) is odd.Wait, actually, ( S_2 = sum_{k=1}^n frac{1}{2k + 1} = sum_{m=3}^{2n + 1} frac{1}{m} ) where ( m ) is odd.But the sum of reciprocals of odd numbers up to ( 2n + 1 ) can be written as:( sum_{k=1}^{2n + 1} frac{1}{k} - sum_{k=1}^n frac{1}{2k} ).Which is ( H_{2n + 1} - frac{1}{2} H_n ).But then ( S_2 = H_{2n + 1} - frac{1}{2} H_n ).But I'm not sure if this is helpful. Let me think differently.Alternatively, since each term ( frac{1}{2k(2k + 1)} ) is positive and decreasing, the sum ( S_n ) is bounded above by an integral.Consider the function ( f(x) = frac{1}{2x(2x + 1)} ). The sum ( S_n ) is less than the integral from 1 to ( n + 1 ) of ( f(x) ) dx.But integrating ( f(x) ) might be a bit involved. Let me compute the integral:( int frac{1}{2x(2x + 1)} dx ).Let me use substitution. Let ( u = 2x + 1 ), then ( du = 2 dx ), so ( dx = frac{du}{2} ).But the integral becomes:( int frac{1}{2x(u)} cdot frac{du}{2} ).Wait, maybe another substitution. Let me try partial fractions again for the integral.Express ( frac{1}{2x(2x + 1)} ) as ( frac{A}{2x} + frac{B}{2x + 1} ).Multiply both sides by ( 2x(2x + 1) ):( 1 = A(2x + 1) + B(2x) ).Expanding:( 1 = (2A + 2B)x + A ).So, ( 2A + 2B = 0 ) and ( A = 1 ).Thus, ( A = 1 ), ( B = -1 ).So, ( frac{1}{2x(2x + 1)} = frac{1}{2x} - frac{1}{2x + 1} ).Therefore, the integral becomes:( int left( frac{1}{2x} - frac{1}{2x + 1} right ) dx = frac{1}{2} ln |x| - frac{1}{2} ln |2x + 1| + C ).Simplify:( frac{1}{2} ln left( frac{x}{2x + 1} right ) + C ).Thus, the integral from 1 to ( n + 1 ) is:( frac{1}{2} ln left( frac{n + 1}{2(n + 1) + 1} right ) - frac{1}{2} ln left( frac{1}{3} right ) ).Simplify:( frac{1}{2} ln left( frac{n + 1}{2n + 3} right ) + frac{1}{2} ln 3 ).As ( n ) increases, ( frac{n + 1}{2n + 3} ) approaches ( frac{1}{2} ), so the integral approaches ( frac{1}{2} ln left( frac{1}{2} right ) + frac{1}{2} ln 3 = frac{1}{2} (ln 3 - ln 2) approx frac{1}{2} (1.0986 - 0.6931) = frac{1}{2} (0.4055) approx 0.20275 ).But wait, the sum ( S_n ) is approximately equal to this integral plus some error term. However, since each term of the sum is less than the integral over the interval, I'm not sure if this directly helps me bound ( S_n ) by ( frac{13}{42} ).Alternatively, let me compute the first few terms and see if I can find a pattern or a bound.Compute ( S_1 = frac{1}{2 times 1 (2 times 1 + 1)} = frac{1}{6} approx 0.1667 ).( S_2 = frac{1}{6} + frac{1}{2 times 2 (2 times 2 + 1)} = frac{1}{6} + frac{1}{20} = frac{10}{60} + frac{3}{60} = frac{13}{60} approx 0.2167 ).( S_3 = frac{13}{60} + frac{1}{2 times 3 (2 times 3 + 1)} = frac{13}{60} + frac{1}{42} approx 0.2167 + 0.0238 = 0.2405 ).( S_4 = frac{13}{60} + frac{1}{42} + frac{1}{2 times 4 (2 times 4 + 1)} = 0.2405 + frac{1}{70} approx 0.2405 + 0.0143 = 0.2548 ).( S_5 = 0.2548 + frac{1}{2 times 5 (2 times 5 + 1)} = 0.2548 + frac{1}{110} approx 0.2548 + 0.0091 = 0.2639 ).Hmm, it seems like the sum is increasing, but how does it behave as ( n ) approaches infinity?The infinite sum ( S = sum_{k=1}^infty frac{1}{2k(2k + 1)} ).As we've decomposed earlier, this is ( sum_{k=1}^infty left( frac{1}{2k} - frac{1}{2k + 1} right ) ).This is an alternating series, but it's actually a telescoping series in disguise. Let me write the partial sums:( S_n = sum_{k=1}^n left( frac{1}{2k} - frac{1}{2k + 1} right ) ).Let me write out the terms:( left( frac{1}{2} - frac{1}{3} right ) + left( frac{1}{4} - frac{1}{5} right ) + ldots + left( frac{1}{2n} - frac{1}{2n + 1} right ) ).This doesn't telescope in the usual sense, but if I consider the infinite sum, it converges to ( ln 2 - 1 ) or something similar? Wait, let me think.Actually, recall that ( sum_{k=1}^infty left( frac{1}{2k} - frac{1}{2k + 1} right ) = sum_{k=1}^infty frac{1}{2k(2k + 1)} ).This can be related to the digamma function or known series, but perhaps a better approach is to relate it to the natural logarithm.Consider the Taylor series for ( ln(1 + x) ) around ( x = 0 ):( ln(1 + x) = sum_{k=1}^infty (-1)^{k + 1} frac{x^k}{k} ).But I'm not sure if this directly helps.Alternatively, consider the integral representation:( frac{1}{2k(2k + 1)} = int_0^1 x^{2k - 1} dx - int_0^1 x^{2k} dx ).Wait, let me check:( int_0^1 x^{2k - 1} dx = frac{1}{2k} ).( int_0^1 x^{2k} dx = frac{1}{2k + 1} ).Thus, ( frac{1}{2k} - frac{1}{2k + 1} = int_0^1 (x^{2k - 1} - x^{2k}) dx = int_0^1 x^{2k - 1}(1 - x) dx ).Therefore, the sum ( S_n = sum_{k=1}^n int_0^1 x^{2k - 1}(1 - x) dx ).Interchange sum and integral (if allowed):( S_n = int_0^1 (1 - x) sum_{k=1}^n x^{2k - 1} dx ).The sum inside is a geometric series:( sum_{k=1}^n x^{2k - 1} = x sum_{k=1}^n x^{2(k - 1)} = x sum_{m=0}^{n - 1} x^{2m} = x cdot frac{1 - x^{2n}}{1 - x^2} ).Thus,( S_n = int_0^1 (1 - x) cdot x cdot frac{1 - x^{2n}}{1 - x^2} dx ).Simplify:( S_n = int_0^1 frac{x(1 - x)(1 - x^{2n})}{1 - x^2} dx ).Note that ( 1 - x^2 = (1 - x)(1 + x) ), so we can cancel out ( (1 - x) ):( S_n = int_0^1 frac{x(1 - x^{2n})}{1 + x} dx ).This simplifies to:( S_n = int_0^1 frac{x - x^{2n + 1}}{1 + x} dx ).Now, let me split the fraction:( frac{x}{1 + x} = 1 - frac{1}{1 + x} ).Similarly, ( frac{x^{2n + 1}}{1 + x} = frac{x^{2n + 1}}{1 + x} ).Thus,( S_n = int_0^1 left( 1 - frac{1}{1 + x} - frac{x^{2n + 1}}{1 + x} right ) dx ).Integrate term by term:1. ( int_0^1 1 dx = 1 ).2. ( int_0^1 frac{1}{1 + x} dx = ln(1 + x) bigg|_0^1 = ln 2 ).3. ( int_0^1 frac{x^{2n + 1}}{1 + x} dx ).This integral is a bit tricky. Let me denote it as ( I_n ):( I_n = int_0^1 frac{x^{2n + 1}}{1 + x} dx ).Perhaps I can use substitution or recognize it as a known integral.Let me make substitution ( u = x ), then ( du = dx ). Not helpful.Alternatively, note that ( frac{x^{2n + 1}}{1 + x} = x^{2n} - x^{2n - 1} + x^{2n - 2} - ldots - x + frac{1}{1 + x} ) for even powers? Wait, no, that's for ( frac{1}{1 + x} ) expansion.Wait, perhaps I can write ( frac{x^{2n + 1}}{1 + x} = x^{2n} - x^{2n - 1} + x^{2n - 2} - ldots + x^2 - x + frac{1}{1 + x} ). Let me check:Multiply both sides by ( 1 + x ):Left side: ( x^{2n + 1} ).Right side: ( (x^{2n} - x^{2n - 1} + ldots - x + frac{1}{1 + x})(1 + x) ).Wait, actually, that might not work. Alternatively, consider that:( frac{x^{2n + 1}}{1 + x} = x^{2n} - x^{2n - 1} + x^{2n - 2} - ldots + x^2 - x + frac{1}{1 + x} ).But when I multiply by ( 1 + x ), I get:( x^{2n + 1} = (x^{2n} - x^{2n - 1} + ldots - x + frac{1}{1 + x})(1 + x) ).But this seems complicated. Maybe another approach.Alternatively, note that ( frac{x^{2n + 1}}{1 + x} = frac{x^{2n + 1} + x^{2n + 2} - x^{2n + 2}}{1 + x} = x^{2n + 1}(1 + x) - x^{2n + 2} ) divided by ( 1 + x ).Wait, that seems messy. Alternatively, consider that ( frac{x^{2n + 1}}{1 + x} = x^{2n} cdot frac{x}{1 + x} ).We already know that ( frac{x}{1 + x} = 1 - frac{1}{1 + x} ), so:( I_n = int_0^1 x^{2n} left( 1 - frac{1}{1 + x} right ) dx = int_0^1 x^{2n} dx - int_0^1 frac{x^{2n}}{1 + x} dx ).Compute the first integral:( int_0^1 x^{2n} dx = frac{1}{2n + 1} ).The second integral is similar to ( I_n ) but with exponent ( 2n ):Let me denote ( J_n = int_0^1 frac{x^{2n}}{1 + x} dx ).So, ( I_n = frac{1}{2n + 1} - J_n ).But ( J_n = int_0^1 frac{x^{2n}}{1 + x} dx ).This is similar to ( I_n ) but with exponent ( 2n ). I can relate ( J_n ) to ( I_n ) perhaps.Wait, perhaps we can find a recursive relation.Alternatively, note that ( J_n = int_0^1 frac{x^{2n}}{1 + x} dx = int_0^1 frac{x^{2n} + x^{2n + 1} - x^{2n + 1}}{1 + x} dx ).But this might not help. Alternatively, observe that ( frac{x^{2n}}{1 + x} = x^{2n - 1} - x^{2n - 2} + ldots + x^2 - x + frac{1}{1 + x} ), similar to earlier.But this seems too involved. Maybe instead of trying to compute ( I_n ), I can bound it.Since ( 0 leq x leq 1 ), ( frac{x^{2n + 1}}{1 + x} leq x^{2n + 1} ).Thus, ( I_n leq int_0^1 x^{2n + 1} dx = frac{1}{2n + 2} ).Therefore, ( S_n = 1 - ln 2 - I_n geq 1 - ln 2 - frac{1}{2n + 2} ).But we need an upper bound, not a lower bound. Alternatively, since ( frac{x^{2n + 1}}{1 + x} geq 0 ), ( I_n geq 0 ), so:( S_n = 1 - ln 2 - I_n leq 1 - ln 2 ).But ( 1 - ln 2 approx 1 - 0.6931 = 0.3069 ), which is greater than ( frac{13}{42} approx 0.3095 ). Wait, actually, ( 1 - ln 2 approx 0.3069 ) is less than ( frac{13}{42} approx 0.3095 ).Wait, so ( S_n leq 1 - ln 2 approx 0.3069 < frac{13}{42} approx 0.3095 ).But wait, the problem states that the sum is less than ( frac{13}{42} ), but according to this, the infinite sum is approximately 0.3069, which is indeed less than ( frac{13}{42} approx 0.3095 ).Therefore, for all ( n ), ( S_n < 1 - ln 2 < frac{13}{42} ).But let me verify ( 1 - ln 2 ) vs ( frac{13}{42} ):Compute ( 1 - ln 2 approx 1 - 0.6931 = 0.3069 ).( frac{13}{42} approx 0.3095 ).So, ( 0.3069 < 0.3095 ), which is true.Thus, ( S_n < 1 - ln 2 < frac{13}{42} ).But wait, the problem asks to prove ( S_n < frac{13}{42} ). So, since ( 1 - ln 2 < frac{13}{42} ), it's even stronger.Alternatively, perhaps I can compute the exact value of the infinite sum and show it's less than ( frac{13}{42} ).Wait, let me compute ( S_n ) as ( n to infty ):( S = sum_{k=1}^infty frac{1}{2k(2k + 1)} ).We can write this as:( S = sum_{k=1}^infty left( frac{1}{2k} - frac{1}{2k + 1} right ) ).This is actually a known series. It converges to ( 1 - ln 2 ).Yes, because:( sum_{k=1}^infty left( frac{1}{2k} - frac{1}{2k + 1} right ) = sum_{k=1}^infty frac{1}{2k(2k + 1)} ).And this is equal to ( 1 - ln 2 ).Thus, the infinite sum is ( 1 - ln 2 approx 0.3069 ), which is less than ( frac{13}{42} approx 0.3095 ).Therefore, for all finite ( n ), the sum ( S_n ) is less than ( 1 - ln 2 ), which is less than ( frac{13}{42} ).Hence, the inequality holds.</think>

🧠:<think>Okay, the user is asking if I believe in the existence of other forms of life in the universe. I should start by explaining that as an AI, I don't have beliefs, but I can discuss the possibilities based on scientific evidence and reasoning.First, I'll mention the vastness of the universe. There are billions of galaxies, each with countless stars and planets. The sheer scale suggests that there might be other planets capable of supporting life.Next, I'll talk about the discovery of exoplanets, especially those in the habitable zone where liquid water could exist. This increases the likelihood of life elsewhere.Then, I'll consider the conditions necessary for life as we know it, like liquid water, stable environment, and organic molecules. Since these conditions might exist on other planets, it's plausible that life could form there.I should also mention extremophiles on Earth, which thrive in extreme conditions. This shows that life can adapt to various environments, making it more likely to exist in different forms elsewhere.Additionally, the presence of organic molecules in space, like in comets and interstellar clouds, supports the idea that the building blocks for life are widespread.I'll touch on the Fermi Paradox, which questions why we haven't encountered intelligent life despite the high probability. This could be due to the vast distances, short duration of human civilization, or other factors.Finally, I'll conclude that while there's no concrete evidence yet, the scientific community remains optimistic, and ongoing missions and research are helping us explore this possibility.</think>

🧠:<think>Okay, so I'm trying to figure out whether the premise "A dog runs through the grass" allows us to conclude that the hypothesis "A dog taking a nap outside" is true. Let me break this down step by step.First, I'll look at the premise: a dog is running through the grass. That tells me that the dog is active, moving around, maybe playing or exercising. The action here is running, which is a dynamic activity.Now, the hypothesis is about the dog taking a nap outside. Taking a nap is the opposite of running; it's a resting activity. If a dog is running, it's unlikely to be napping at the same time. These two actions are mutually exclusive in the moment they're happening.But wait, could the dog have taken a nap before or after running? The premise doesn't specify the timeline. It just states that the dog is running through the grass at a particular moment. So, without more information about what happened before or after, we can't assume the dog was napping.Also, the premise doesn't mention anything about the dog's state before or after running. It's possible the dog was napping before, but it's also possible it wasn't. Since the premise only gives us information about the dog running, we can't make any definitive conclusions about napping.Therefore, based solely on the premise, we don't have enough information to confirm that the dog was taking a nap. The two activities are different and don't necessarily relate to each other in the context provided.So, the answer should be that it's not possible to tell because the premise doesn't provide information about the dog napping, only about it running.</think>