answer：Okay, so I have this problem to multiply two polynomials: (3x² - 4y³) and (9x⁴ + 12x²y³ + 16y⁶). Hmm, at first glance, it looks a bit complicated because of all the exponents and variables, but maybe there's a pattern or a formula I can use to simplify this.Let me write down the problem again to make sure I have it right:(3x² - 4y³)(9x⁴ + 12x²y³ + 16y⁶)Alright, so I need to multiply these two binomials. I remember that when multiplying two polynomials, I can use the distributive property, also known as the FOIL method for binomials. But wait, the second polynomial here has three terms, so it's actually a trinomial. That means I can't use the FOIL method directly; instead, I'll have to distribute each term in the first polynomial across each term in the second polynomial.Let me recall the distributive property: (a + b)(c + d + e) = a*c + a*d + a*e + b*c + b*d + b*e. So, applying that here, I'll need to multiply each term in (3x² - 4y³) by each term in (9x⁴ + 12x²y³ + 16y⁶).Let me break it down step by step.First, I'll distribute 3x² across each term in the second polynomial:3x² * 9x⁴ = ?3x² * 12x²y³ = ?3x² * 16y⁶ = ?Then, I'll distribute -4y³ across each term in the second polynomial:-4y³ * 9x⁴ = ?-4y³ * 12x²y³ = ?-4y³ * 16y⁶ = ?After calculating each of these products, I'll combine like terms if there are any.Let me calculate each multiplication one by one.Starting with 3x² * 9x⁴:Multiplying the coefficients: 3 * 9 = 27Multiplying the variables: x² * x⁴ = x^(2+4) = x⁶So, 3x² * 9x⁴ = 27x⁶Next, 3x² * 12x²y³:Coefficients: 3 * 12 = 36Variables: x² * x² = x^(2+2) = x⁴; and y³ remains as is since there's no y term in the first polynomial.So, 3x² * 12x²y³ = 36x⁴y³Now, 3x² * 16y⁶:Coefficients: 3 * 16 = 48Variables: x² remains as is, and y⁶ remains as is.So, 3x² * 16y⁶ = 48x²y⁶Alright, that's all the terms from distributing 3x². Now, moving on to distributing -4y³.First term: -4y³ * 9x⁴Coefficients: -4 * 9 = -36Variables: y³ * x⁴. Since multiplication is commutative, I can write this as x⁴y³.So, -4y³ * 9x⁴ = -36x⁴y³Second term: -4y³ * 12x²y³Coefficients: -4 * 12 = -48Variables: x² remains, and y³ * y³ = y^(3+3) = y⁶So, -4y³ * 12x²y³ = -48x²y⁶Third term: -4y³ * 16y⁶Coefficients: -4 * 16 = -64Variables: y³ * y⁶ = y^(3+6) = y⁹So, -4y³ * 16y⁶ = -64y⁹Now, let me list all the products I've calculated:From 3x²:1. 27x⁶2. 36x⁴y³3. 48x²y⁶From -4y³:4. -36x⁴y³5. -48x²y⁶6. -64y⁹Now, I need to combine these terms. Let's see if there are any like terms that can be combined.Looking at the terms:1. 27x⁶ – this is unique, no like terms.2. 36x⁴y³ – let's see if there's another x⁴y³ term.3. 48x²y⁶ – looking for another x²y⁶ term.4. -36x⁴y³ – this is the same as term 2.5. -48x²y⁶ – this is the same as term 3.6. -64y⁹ – unique.So, combining like terms:For x⁴y³: 36x⁴y³ + (-36x⁴y³) = 0For x²y⁶: 48x²y⁶ + (-48x²y⁶) = 0So, those middle terms cancel each other out. That's interesting.Therefore, the remaining terms are:27x⁶ and -64y⁹So, putting it all together, the product is:27x⁶ - 64y⁹Wait a minute, that seems too simple. Did I do that right? Let me double-check my calculations.First, when I multiplied 3x² by 9x⁴, I got 27x⁶. That seems correct.Then, 3x² * 12x²y³ = 36x⁴y³. Correct.3x² * 16y⁶ = 48x²y⁶. Correct.Now, -4y³ * 9x⁴ = -36x⁴y³. Correct.-4y³ * 12x²y³ = -48x²y⁶. Correct.-4y³ * 16y⁶ = -64y⁹. Correct.Then, combining like terms:36x⁴y³ - 36x⁴y³ = 048x²y⁶ - 48x²y⁶ = 0So, only 27x⁶ and -64y⁹ remain.Hmm, that seems right. But I recall that (a - b)(a² + ab + b²) = a³ - b³. Maybe this is a case of that formula.Let me check: Let a = 3x² and b = 4y³.Then, a³ = (3x²)³ = 27x⁶b³ = (4y³)³ = 64y⁹So, a³ - b³ = 27x⁶ - 64y⁹Yes, that's exactly what I got after expanding. So, it seems that the original expression is indeed a difference of cubes, which is why the middle terms canceled out.So, the final answer is 27x⁶ - 64y⁹.I think that's correct. I don't see any mistakes in my calculations, and the application of the difference of cubes formula confirms the result.

answer：Okay, so I have this inequality: |ax + 2| < 4, and the solution set is given as (-1, 3). I need to find the real number a. Hmm, let's see. I remember that absolute value inequalities can be rewritten without the absolute value by considering two cases: one where the expression inside is positive and one where it's negative. So, for |ax + 2| < 4, this should translate to -4 < ax + 2 < 4. Alright, so if I subtract 2 from all parts of the inequality, it becomes -6 < ax < 2. Now, to solve for x, I need to divide all parts by a. But wait, I don't know if a is positive or negative yet. If a is positive, dividing won't change the inequality signs, but if a is negative, I have to flip the inequality signs. Hmm, tricky.Given that the solution set is (-1, 3), which means x is between -1 and 3. So, if I write the inequality as -1 < x < 3, I can compare it to the inequality I have: -6/a < x < 2/a. Wait, but I don't know the sign of a yet. Maybe I can figure it out by looking at the direction of the inequalities.If a is positive, then -6/a would be negative and 2/a would be positive. So, the interval would be from a negative number to a positive number. But the given interval is from -1 to 3, which is also from a negative to a positive. So, a could be positive. But let's check if that works.If a is positive, then -6/a < x < 2/a. So, setting -6/a = -1 and 2/a = 3. Let's solve for a in both cases. First, -6/a = -1. Multiplying both sides by a gives -6 = -a, so a = 6. Second, 2/a = 3. Multiplying both sides by a gives 2 = 3a, so a = 2/3. Wait, but a can't be both 6 and 2/3 at the same time. That doesn't make sense. So, maybe my assumption that a is positive is wrong. Let me try assuming a is negative.If a is negative, then when I divide by a, the inequalities will flip. So, starting from -6 < ax < 2, dividing by a (negative) would give me -6/a > x > 2/a. Which can be rewritten as 2/a < x < -6/a. Now, comparing this to the given interval (-1, 3), I can set 2/a = -1 and -6/a = 3. Let's solve for a in both cases.First, 2/a = -1. Multiplying both sides by a gives 2 = -a, so a = -2.Second, -6/a = 3. Multiplying both sides by a gives -6 = 3a, so a = -2.Okay, this time both equations give me a = -2. That seems consistent. So, a must be -2. Let me double-check to make sure.If a = -2, then the inequality becomes |-2x + 2| < 4. Let's solve this:-4 < -2x + 2 < 4Subtract 2 from all parts:-6 < -2x < 2Divide by -2, remembering to flip the inequalities:3 > x > -1Which is the same as (-1, 3). Perfect, that matches the given solution set. So, a is indeed -2.I think that makes sense. I first assumed a was positive, but that led to inconsistent values for a. Then, by assuming a was negative, both boundary conditions gave me the same value for a, which worked when I checked it. So, I'm confident that a = -2 is the correct answer.

answer：First, I recognize that when ( n ) is divided by 7, the remainder is 2. This means that ( n ) can be expressed as ( n = 7k + 2 ) for some integer ( k ).Next, I need to find the remainder when ( n + 2023 ) is divided by 7. To do this, I'll add 2023 to the expression for ( n ):[n + 2023 = 7k + 2 + 2023][n + 2023 = 7k + 2025]Now, I'll determine how 2025 relates to multiples of 7. Dividing 2025 by 7 gives:[2025 div 7 = 289 text{ with a remainder of } 2]This means that ( 2025 = 7 times 289 + 2 ).Substituting this back into the expression for ( n + 2023 ):[n + 2023 = 7k + 7 times 289 + 2][n + 2023 = 7(k + 289) + 2]Since ( 7(k + 289) ) is a multiple of 7, the remainder when ( n + 2023 ) is divided by 7 is determined by the additional 2.Therefore, the remainder when ( n + 2023 ) is divided by 7 is 2.

answer：First, I need to understand what it means for two graphs to be symmetric with respect to a particular shape. Symmetry in this context means that if I reflect one graph over the specified shape, it will coincide with the other graph.I'll start by examining the two functions given: ( y = 3^x ) and ( y = 3^{-x} ). To determine the symmetry, I'll consider a general point ( (a, 3^a) ) on the graph of ( y = 3^x ). If I reflect this point over the y-axis, the x-coordinate changes sign, resulting in the point ( (-a, 3^a) ).Now, I'll check if this reflected point lies on the graph of ( y = 3^{-x} ). Substituting ( x = -a ) into ( y = 3^{-x} ), I get:[y = 3^{-(-a)} = 3^a]This confirms that the reflected point ( (-a, 3^a) ) is indeed on the graph of ( y = 3^{-x} ).Since this holds true for any arbitrary point ( (a, 3^a) ) on ( y = 3^x ), it means that every point on ( y = 3^x ) has a corresponding point on ( y = 3^{-x} ) that is symmetric with respect to the y-axis.Therefore, the graphs of ( y = 3^x ) and ( y = 3^{-x} ) are symmetric with respect to the y-axis.

answer：Okay, so I have this problem about an arithmetic sequence and another sequence related to it. Let me try to break it down step by step.First, part (1) says that the arithmetic sequence {a_n} satisfies the equation: (a₁ + a₂) + (a₂ + a₃) + … + (a_n + a_{n+1}) = 2n(n + 1) for any positive integer n. I need to find the general term formula for {a_n}.Hmm, arithmetic sequence. So, I remember that in an arithmetic sequence, each term is the previous term plus a common difference, d. So, a_n = a₁ + (n - 1)d.Now, the given equation is a sum of consecutive pairs of terms from a₁ to a_{n+1}. Let me write that out for a few terms to see the pattern.For n = 1: (a₁ + a₂) = 2*1*(1 + 1) = 4For n = 2: (a₁ + a₂) + (a₂ + a₃) = 2*2*(2 + 1) = 12For n = 3: (a₁ + a₂) + (a₂ + a₃) + (a₃ + a₄) = 2*3*(3 + 1) = 24Wait, so for each n, the sum is 2n(n + 1). Let me think about how this sum relates to the arithmetic sequence.Each term in the sum is a pair of consecutive terms: (a₁ + a₂), (a₂ + a₃), ..., (a_n + a_{n+1}). So, there are n such pairs. Each pair is the sum of two consecutive terms of the arithmetic sequence.Let me express each pair in terms of a₁ and d. Since a_k = a₁ + (k - 1)d, then a_{k+1} = a₁ + kd. So, a_k + a_{k+1} = 2a₁ + (2k - 1)d.So, the sum from k=1 to n of (a_k + a_{k+1}) is the sum from k=1 to n of [2a₁ + (2k - 1)d]. Let me write that as:Sum = Σ_{k=1}^n [2a₁ + (2k - 1)d] = 2n a₁ + d Σ_{k=1}^n (2k - 1)I can compute the sum Σ_{k=1}^n (2k - 1). That's the sum of the first n odd numbers. I remember that the sum of the first n odd numbers is n². Let me verify:For n=1: 1 = 1²For n=2: 1 + 3 = 4 = 2²For n=3: 1 + 3 + 5 = 9 = 3²Yes, that seems right.So, Σ_{k=1}^n (2k - 1) = n².Therefore, the sum becomes:Sum = 2n a₁ + d * n²But we know this sum is equal to 2n(n + 1). So,2n a₁ + d n² = 2n(n + 1)Let me simplify this equation:Divide both sides by n (since n ≠ 0):2 a₁ + d n = 2(n + 1)So,2 a₁ + d n = 2n + 2Hmm, this equation must hold for all positive integers n. That suggests that the coefficients of n and the constants must match on both sides.So, let's equate the coefficients:Coefficient of n: d = 2Constant term: 2 a₁ = 2So, from the constant term: 2 a₁ = 2 ⇒ a₁ = 1From the coefficient of n: d = 2Therefore, the arithmetic sequence has first term a₁ = 1 and common difference d = 2.So, the general term is:a_n = a₁ + (n - 1)d = 1 + (n - 1)*2 = 1 + 2n - 2 = 2n - 1Let me check this with n=1,2,3:a₁ = 2*1 - 1 = 1a₂ = 2*2 - 1 = 3a₃ = 2*3 - 1 = 5a₄ = 2*4 - 1 = 7Now, let's verify the given condition for n=1,2,3.For n=1: (a₁ + a₂) = 1 + 3 = 4, which equals 2*1*(1+1)=4. Good.For n=2: (a₁ + a₂) + (a₂ + a₃) = (1+3) + (3+5) = 4 + 8 = 12, which equals 2*2*(2+1)=12. Good.For n=3: (a₁ + a₂) + (a₂ + a₃) + (a₃ + a₄) = 4 + 8 + 12 = 24, which equals 2*3*(3+1)=24. Perfect.So, part (1) seems solved: a_n = 2n - 1.Moving on to part (2). It says: In sequence {b_n}, b₁=1, b₂=2. If we take the b_n-th term from sequence {a_n} and denote it as c_n, and {c_n} is a geometric sequence, find the sum of the first n terms of {b_n}, denoted as T_n.Alright, let's parse this.We have another sequence {b_n}, with b₁=1, b₂=2. So, the first two terms are given. We need to find the rest of the terms such that when we take the b_n-th term from {a_n}, we get a geometric sequence {c_n}.So, c_n = a_{b_n}.Given that {c_n} is a geometric sequence, which means each term is a multiple of the previous term by a common ratio r.Given that, we can write c_{n+1} / c_n = r, a constant.Given that, let's write down what we know.First, c₁ = a_{b₁} = a₁ = 1 (since b₁=1)c₂ = a_{b₂} = a₂ = 3 (since b₂=2)So, c₁ = 1, c₂ = 3.So, the common ratio r = c₂ / c₁ = 3 / 1 = 3.Therefore, {c_n} is a geometric sequence with first term 1 and common ratio 3. So, c_n = 1 * 3^{n-1} = 3^{n-1}.But also, c_n = a_{b_n} = 2 b_n - 1, since a_k = 2k - 1.So, we have:2 b_n - 1 = 3^{n - 1}Therefore, solving for b_n:2 b_n = 3^{n - 1} + 1b_n = (3^{n - 1} + 1) / 2So, the general term for {b_n} is b_n = (3^{n - 1} + 1)/2.Now, we need to find the sum of the first n terms of {b_n}, which is T_n = b₁ + b₂ + ... + b_n.Given that b_n = (3^{n - 1} + 1)/2, let's write T_n as:T_n = Σ_{k=1}^n b_k = Σ_{k=1}^n [(3^{k - 1} + 1)/2] = (1/2) Σ_{k=1}^n 3^{k - 1} + (1/2) Σ_{k=1}^n 1Let me compute each sum separately.First, Σ_{k=1}^n 3^{k - 1} is a geometric series with first term 1 (when k=1, 3^{0}=1) and common ratio 3. The sum of the first n terms of a geometric series is (r^n - 1)/(r - 1). So,Σ_{k=1}^n 3^{k - 1} = (3^n - 1)/(3 - 1) = (3^n - 1)/2Second, Σ_{k=1}^n 1 = n, since we're adding 1 n times.Therefore, putting it all together:T_n = (1/2) * [(3^n - 1)/2] + (1/2) * n = (3^n - 1)/4 + n/2To combine these terms, let's express n/2 as 2n/4:T_n = (3^n - 1 + 2n)/4So, T_n = (3^n + 2n - 1)/4Let me verify this with n=1 and n=2.For n=1: T₁ = b₁ = 1. Plugging into the formula: (3^1 + 2*1 - 1)/4 = (3 + 2 -1)/4 = 4/4 = 1. Correct.For n=2: T₂ = b₁ + b₂ = 1 + 2 = 3. Plugging into the formula: (3^2 + 2*2 -1)/4 = (9 + 4 -1)/4 = 12/4 = 3. Correct.For n=3: Let's compute b₃ first. b₃ = (3^{2} + 1)/2 = (9 + 1)/2 = 10/2 = 5. So, T₃ = 1 + 2 + 5 = 8. Plugging into the formula: (3^3 + 2*3 -1)/4 = (27 + 6 -1)/4 = 32/4 = 8. Correct.Looks good. So, the sum T_n is (3^n + 2n -1)/4.**Final Answer**(1) The general term is boxed{2n - 1}.(2) The sum of the first ( n ) terms of ( {b_n} ) is boxed{dfrac{3^n + 2n - 1}{4}}.

answer：First, I need to determine the number of water bottles in the box. The problem states there are 2 dozen water bottles. Since one dozen equals 12 bottles, 2 dozen would be 2 multiplied by 12, which equals 24 water bottles.Next, the problem mentions there are half a dozen more apple bottles than water bottles. Half a dozen is equal to 6 bottles. Therefore, the number of apple bottles is the number of water bottles plus 6, which is 24 plus 6, resulting in 30 apple bottles.Finally, to find the total number of bottles in the box, I add the number of water bottles and apple bottles together. This means adding 24 water bottles to 30 apple bottles, which gives a total of 54 bottles.