💡Okay, so I have this problem about a parabola and a point on it, and I need to find the distance from that point to the focus of the parabola. Let me try to figure this out step by step.First, the parabola is given by the equation ( y^2 = 4x ). I remember that this is a standard form of a parabola that opens to the right. In general, the standard form is ( y^2 = 4ax ), where 'a' is the distance from the vertex to the focus. So, comparing this to the given equation, ( 4a = 4 ), which means ( a = 1 ). Therefore, the focus of this parabola should be at ( (a, 0) ), which is ( (1, 0) ). Got that.Now, the point M is given as ( M(x_0, 2sqrt{3}) ) and it lies on the parabola. So, I need to find the coordinates of M first, specifically ( x_0 ). Since M is on the parabola, its coordinates must satisfy the equation ( y^2 = 4x ). Let me plug in the y-coordinate of M into the equation.So, substituting ( y = 2sqrt{3} ) into ( y^2 = 4x ), we get:( (2sqrt{3})^2 = 4x_0 )Calculating the left side:( (2sqrt{3})^2 = 4 * 3 = 12 )So, ( 12 = 4x_0 )Dividing both sides by 4:( x_0 = 3 )Alright, so the coordinates of point M are ( (3, 2sqrt{3}) ).Now, I need to find the distance from M to the focus of the parabola. The focus is at ( (1, 0) ), as I found earlier. To find the distance between two points in a plane, I can use the distance formula:( text{Distance} = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} )Let me apply this formula to points M(3, 2√3) and the focus (1, 0).Substituting the coordinates into the formula:( text{Distance} = sqrt{(1 - 3)^2 + (0 - 2sqrt{3})^2} )Calculating each part:First, ( (1 - 3) = -2 ), so ( (-2)^2 = 4 )Second, ( (0 - 2sqrt{3}) = -2sqrt{3} ), so ( (-2sqrt{3})^2 = 4 * 3 = 12 )Adding these together:( 4 + 12 = 16 )Taking the square root of 16:( sqrt{16} = 4 )So, the distance from point M to the focus is 4 units.Wait, let me double-check if I did everything correctly. I found ( x_0 = 3 ) by plugging into the parabola equation, which seems right. Then, using the distance formula between (3, 2√3) and (1, 0), I got 4. That makes sense because the focus is at (1,0), and M is at (3, 2√3). The horizontal distance is 2 units, and the vertical distance is 2√3 units. Using Pythagoras, the distance should be √(4 + 12) = √16 = 4. Yep, that checks out.Alternatively, I remember that for a parabola ( y^2 = 4ax ), the distance from any point on the parabola to the focus is equal to the distance from that point to the directrix. The directrix of this parabola is the line x = -1. So, the distance from M to the directrix would be the horizontal distance from M to x = -1, which is ( x_0 - (-1) = x_0 + 1 ). Since ( x_0 = 3 ), this distance is 4. Therefore, the distance from M to the focus is also 4, which matches what I found earlier.So, both methods confirm that the distance is 4. I think that's solid.**Final Answer**The distance from point ( M ) to the focus of the parabola is boxed{4}.

💡Okay, so Matt has five cousins coming to visit, and there are four identical rooms where they can stay. The question is asking how many different ways there are to put the cousins in the rooms, with the condition that any number of cousins can stay in one room. First, I need to understand the problem clearly. We have five cousins, let's call them A, B, C, D, and E for simplicity. There are four rooms, and these rooms are identical, meaning that it doesn't matter which specific room is which; what matters is how the cousins are grouped together. So, the key here is that the rooms are indistinct, so arrangements that can be transformed into each other by swapping rooms are considered the same.This problem seems related to partitioning a set into subsets, where each subset represents the cousins in a room. Since the rooms are identical, the order of the subsets doesn't matter. So, essentially, we're looking for the number of ways to partition five distinct cousins into up to four identical rooms, where each room can have any number of cousins, including zero. However, since all five cousins need to be accommodated, we can't have any room left empty if we're using all four rooms. Wait, actually, no—since the rooms are identical, having some rooms empty is allowed, but in this case, since we have five cousins and four rooms, it's possible that one room will have two cousins and the rest will have one each, or some other distribution.Wait, actually, the problem says "any number of cousins can stay in one room," so it's possible to have rooms with zero cousins as well. But since we have five cousins and four rooms, the maximum number of rooms that can be occupied is four, and the minimum is one. So, we need to consider all possible ways to distribute five cousins into one, two, three, or four rooms, considering that rooms are identical.This seems like a problem that can be approached using the concept of partitions in combinatorics, specifically the number of set partitions of a five-element set into at most four non-empty subsets. Since the rooms are identical, the order of the subsets doesn't matter, so we're dealing with unordered partitions.The number of ways to partition a set of n elements into k non-empty subsets is given by the Stirling numbers of the second kind, denoted as S(n, k). So, in this case, we need to calculate S(5, 1) + S(5, 2) + S(5, 3) + S(5, 4), because we can have the cousins distributed into 1, 2, 3, or 4 rooms.Let me recall the values of the Stirling numbers of the second kind for n=5:- S(5,1) = 1 (all cousins in one room)- S(5,2) = 15- S(5,3) = 25- S(5,4) = 10So, adding these up: 1 + 15 + 25 + 10 = 51.Wait, but let me make sure I'm not making a mistake here. The problem says there are four identical rooms, but it doesn't specify that all rooms must be used. So, actually, we can have partitions into 1, 2, 3, or 4 rooms, but since the rooms are identical, partitions into fewer rooms are still valid. So, the total number of ways is indeed the sum of Stirling numbers from k=1 to k=4 for n=5.Alternatively, another way to think about it is to consider the different possible distributions of cousins into rooms, considering that rooms are identical. So, we need to find all the integer partitions of 5 (the number of cousins) into at most 4 parts (since there are four rooms), and for each partition, calculate the number of distinct ways to assign the cousins accordingly.Let's list all the possible partitions of 5 into at most 4 parts:1. 5 (all cousins in one room)2. 4 + 1 (four cousins in one room, one cousin alone)3. 3 + 2 (three cousins in one room, two cousins in another)4. 3 + 1 + 1 (three cousins in one room, one cousin each in two other rooms)5. 2 + 2 + 1 (two cousins in two rooms, one cousin alone)6. 2 + 1 + 1 + 1 (two cousins in one room, one cousin each in three other rooms)These are all the possible partitions of 5 into at most 4 parts. Now, for each partition, we need to calculate the number of distinct ways to assign the cousins.1. For the partition 5: There's only 1 way, since all cousins are in one room.2. For the partition 4 + 1: We need to choose which one cousin is alone. Since the cousins are distinct, the number of ways is C(5,1) = 5.3. For the partition 3 + 2: We need to choose which three cousins are together. The number of ways is C(5,3) = 10.4. For the partition 3 + 1 + 1: We need to choose which three cousins are together, and the remaining two are each alone. Since the rooms are identical, the order of the single cousins doesn't matter. So, the number of ways is C(5,3) = 10.5. For the partition 2 + 2 + 1: We need to choose two pairs of cousins. However, since the rooms are identical, choosing pair A and then pair B is the same as choosing pair B and then pair A. So, we need to divide by 2 to avoid overcounting. The number of ways is [C(5,2) * C(3,2)] / 2 = (10 * 3) / 2 = 15.6. For the partition 2 + 1 + 1 + 1: We need to choose which two cousins are together, and the remaining three are each alone. The number of ways is C(5,2) = 10.Now, adding up all these possibilities: 1 + 5 + 10 + 10 + 15 + 10 = 51.So, the total number of different ways to put the five cousins into four identical rooms is 51.I think this matches with the earlier calculation using Stirling numbers of the second kind. The sum of S(5,1) + S(5,2) + S(5,3) + S(5,4) is indeed 1 + 15 + 25 + 10 = 51.Therefore, the final answer should be 51.

💡Okay, so I have this problem with two functions: f(x) = log base a of x, and g(x) = x squared minus 6x plus 9. The condition is that there exists a unique real number x such that both f(g(x)) and g(f(x)) equal zero. I need to find the value of a.First, let me make sure I understand the functions. f(x) is a logarithmic function with base a, which means it's only defined for positive x. g(x) is a quadratic function, which I can probably factor or complete the square to make it easier to work with.Let me write down the given equations:1. f(g(x)) = 02. g(f(x)) = 0Starting with the first equation, f(g(x)) = 0. Since f is a logarithm, log base a of something equals zero means that something must be 1 because a^0 = 1. So, g(x) must equal 1.So, g(x) = x² - 6x + 9 = 1. Let me solve this equation for x.x² - 6x + 9 = 1 Subtract 1 from both sides: x² - 6x + 8 = 0Now, factor this quadratic: Looking for two numbers that multiply to 8 and add to -6. Hmm, -2 and -4. So, (x - 2)(x - 4) = 0 Thus, x = 2 or x = 4.So, from f(g(x)) = 0, we get two solutions: x = 2 and x = 4.Now, moving on to the second equation, g(f(x)) = 0. Let's substitute f(x) into g.g(f(x)) = (f(x))² - 6f(x) + 9 = 0 Let me set y = f(x) to simplify: y² - 6y + 9 = 0 This is another quadratic equation. Let me factor it: (y - 3)² = 0 So, y = 3.Since y = f(x) = log base a of x, this means: log_a x = 3 Which implies x = a³.So, from g(f(x)) = 0, we get x = a³.Now, the problem states that there exists a unique real number x that satisfies both equations. That means the solutions from f(g(x)) = 0 and g(f(x)) = 0 must intersect at exactly one point.From f(g(x)) = 0, we have x = 2 and x = 4. From g(f(x)) = 0, we have x = a³. For there to be a unique solution, a³ must be equal to either 2 or 4, but not both. Because if a³ were equal to both 2 and 4, that would imply 2 = 4, which is impossible.Wait, actually, hold on. If a³ equals 2, then x = 2 is the solution, and if a³ equals 4, then x = 4 is the solution. So, depending on the value of a, either x = 2 or x = 4 will satisfy both equations. But the problem says there exists a unique real number x. So, does that mean that only one of these can be a solution? Or is there a way for both to be solutions, but somehow only one is valid?Wait, let me think again. If a³ = 2, then x = 2 is the solution. But does x = 4 also satisfy g(f(x)) = 0? Let's check.If a³ = 2, then a = cube root of 2. Then, f(4) = log base a of 4. Since a = 2^(1/3), log base a of 4 is log_{2^(1/3)} 4. Using change of base formula, that's (ln 4)/(ln 2^(1/3)) = (2 ln 2)/( (1/3) ln 2 ) = 6. So, f(4) = 6. Then, g(f(4)) = g(6) = 6² - 6*6 + 9 = 36 - 36 + 9 = 9, which is not zero. So, x = 4 doesn't satisfy g(f(x)) = 0 when a³ = 2.Similarly, if a³ = 4, then a = cube root of 4. Then, f(2) = log base a of 2. Since a = 4^(1/3) = (2²)^(1/3) = 2^(2/3). So, log base 2^(2/3) of 2 is (ln 2)/( (2/3) ln 2 ) = 3/2. So, f(2) = 3/2. Then, g(f(2)) = g(3/2) = (3/2)² - 6*(3/2) + 9 = 9/4 - 9 + 9 = 9/4, which is not zero. So, x = 2 doesn't satisfy g(f(x)) = 0 when a³ = 4.Therefore, if a³ = 2, then only x = 2 satisfies both equations. If a³ = 4, then only x = 4 satisfies both equations. So, in both cases, there is a unique solution. But the problem says "there exists a unique real number x". So, does that mean that both a³ = 2 and a³ = 4 are possible solutions? Or is there something else I'm missing?Wait, the problem says "there exists a unique real number x". So, for each a, there is either x = 2 or x = 4 as a solution, but not both. So, both a = cube root of 2 and a = cube root of 4 would satisfy the condition that there's a unique x. But the problem is asking for the value of a. So, is there only one possible a? Or are there two possible a's?Wait, let me check the problem statement again: "if there exists a unique real number x such that f(g(x)) = g(f(x)) = 0, find the value of the real number a."So, it's saying that there is exactly one x that satisfies both equations. So, if a is such that a³ = 2, then x = 2 is the unique solution. Similarly, if a³ = 4, then x = 4 is the unique solution. So, both a = cube root of 2 and a = cube root of 4 would satisfy the condition.But wait, let me think again. If a is cube root of 2, then x = 2 is the unique solution. If a is cube root of 4, then x = 4 is the unique solution. So, both are valid. But the problem is asking for "the value of the real number a". So, does that mean there are two possible values for a? Or is there something else?Wait, maybe I need to consider whether both x = 2 and x = 4 can be solutions for the same a. But from earlier, if a³ = 2, then x = 4 doesn't satisfy g(f(x)) = 0, and if a³ = 4, then x = 2 doesn't satisfy g(f(x)) = 0. So, for each a, only one x works. Therefore, both a = cube root of 2 and a = cube root of 4 are valid solutions.But the problem says "there exists a unique real number x". So, if a is such that both x = 2 and x = 4 satisfy the equations, then there wouldn't be a unique x. But in our case, for each a, only one x works. So, both a = cube root of 2 and a = cube root of 4 are valid.Wait, but let me think again. If a is such that a³ = 2 and a³ = 4 simultaneously, which is impossible, then x would be both 2 and 4, but since a can't be both, each a gives only one x. So, both a = cube root of 2 and a = cube root of 4 are valid.But the problem is asking for "the value of the real number a". So, maybe both are acceptable. But let me check the problem statement again: "find the value of the real number a". It says "the value", implying maybe only one. Hmm.Wait, perhaps I made a mistake earlier. Let me re-examine the equations.From f(g(x)) = 0, we have x = 2 or x = 4.From g(f(x)) = 0, we have x = a³.So, for x to satisfy both equations, x must be equal to a³ and also be either 2 or 4. So, a³ must be equal to 2 or 4. Therefore, a can be cube root of 2 or cube root of 4.But the problem says "there exists a unique real number x". So, if a³ is 2, then x = 2 is the unique solution. If a³ is 4, then x = 4 is the unique solution. So, both are valid. Therefore, a can be either cube root of 2 or cube root of 4.But the problem is asking for "the value of the real number a". So, maybe both are acceptable. But perhaps I need to check if both are valid or if only one is.Wait, let me think about the uniqueness. If a³ is 2, then x = 2 is the unique solution. Similarly, if a³ is 4, x = 4 is the unique solution. So, both are valid. Therefore, the possible values of a are cube root of 2 and cube root of 4.But the problem says "find the value of the real number a". So, maybe both are acceptable, but perhaps I need to write both.Alternatively, maybe I need to consider that if a³ = 2, then x = 2 is the solution, and if a³ = 4, x = 4 is the solution. So, both are valid, but the problem is asking for a, so both are possible.Wait, but in the problem statement, it's "there exists a unique real number x". So, for each a, there is exactly one x. So, both a = cube root of 2 and a = cube root of 4 are valid.But let me check if there's any other constraints. For example, when a = cube root of 2, does x = 2 satisfy both equations? Yes, because f(g(2)) = f(1) = 0, and g(f(2)) = g(3) = 0. Similarly, when a = cube root of 4, x = 4 satisfies both equations.Wait, hold on. When a = cube root of 2, f(2) = log_{cube root of 2} 2 = 3, because (cube root of 2)^3 = 2. So, f(2) = 3. Then, g(f(2)) = g(3) = 3² - 6*3 + 9 = 9 - 18 + 9 = 0. So, yes, x = 2 works.Similarly, when a = cube root of 4, f(4) = log_{cube root of 4} 4 = 3, because (cube root of 4)^3 = 4. So, f(4) = 3. Then, g(f(4)) = g(3) = 0. So, x = 4 works.Therefore, both a = cube root of 2 and a = cube root of 4 are valid solutions. So, the answer is both.But the problem is asking for "the value of the real number a". So, maybe both are acceptable. But perhaps the problem expects both answers.Alternatively, maybe I need to consider that a is positive and not equal to 1, which both cube roots satisfy.So, in conclusion, the possible values of a are cube root of 2 and cube root of 4.

💡Okay, so I have this problem where I'm given two positive real numbers, x and y, and they add up to 4. I need to find the range of the real number m such that the inequality 1/x + 4/y is always greater than or equal to m. The options are given as A, B, C, D, but I think I should figure it out myself first.Alright, let's start by understanding what's given. We have x and y are positive real numbers, and x + y = 4. So, y is equal to 4 - x. That might be useful because I can substitute y in terms of x or vice versa.The inequality we're dealing with is 1/x + 4/y ≥ m. Since y = 4 - x, maybe I can rewrite the inequality in terms of x only. Let me try that.So, substituting y, we get 1/x + 4/(4 - x) ≥ m. Now, I need to find the minimum value of the left-hand side expression because m has to be less than or equal to that minimum for the inequality to always hold. So, essentially, I need to find the minimum of f(x) = 1/x + 4/(4 - x) for x in (0,4).Hmm, how do I find the minimum of this function? I remember that for functions, one way to find minima or maxima is by taking derivatives. Maybe I can use calculus here.Let's compute the derivative of f(x). So, f(x) = 1/x + 4/(4 - x). The derivative f’(x) would be the derivative of 1/x plus the derivative of 4/(4 - x).The derivative of 1/x is -1/x², and the derivative of 4/(4 - x) is 4/(4 - x)² because when you have 1/(a - x), the derivative is 1/(a - x)². Wait, actually, let me double-check that.Using the chain rule, the derivative of 4/(4 - x) is 4 * derivative of 1/(4 - x). The derivative of 1/(4 - x) is (0 - (-1))/(4 - x)², which is 1/(4 - x)². So, the derivative is 4/(4 - x)².So, putting it together, f’(x) = -1/x² + 4/(4 - x)².To find the critical points, set f’(x) = 0.So, -1/x² + 4/(4 - x)² = 0.Let me rearrange this equation:4/(4 - x)² = 1/x²Multiply both sides by x²(4 - x)² to eliminate denominators:4x² = (4 - x)²Let me expand (4 - x)²:(4 - x)² = 16 - 8x + x²So, 4x² = 16 - 8x + x²Bring all terms to one side:4x² - x² + 8x - 16 = 0Simplify:3x² + 8x - 16 = 0Now, solve this quadratic equation. Let's use the quadratic formula:x = [-b ± sqrt(b² - 4ac)] / (2a)Here, a = 3, b = 8, c = -16.Discriminant D = b² - 4ac = 64 - 4*3*(-16) = 64 + 192 = 256sqrt(D) = 16So, x = [-8 ± 16]/(2*3)First solution: (-8 + 16)/6 = 8/6 = 4/3 ≈ 1.333Second solution: (-8 - 16)/6 = -24/6 = -4But x must be positive and less than 4 because y = 4 - x must also be positive. So, x = 4/3 is the critical point.Now, let's check if this is a minimum. We can use the second derivative test or analyze the behavior around this point.Let me compute the second derivative f''(x).First, f’(x) = -1/x² + 4/(4 - x)²So, f''(x) is the derivative of f’(x):Derivative of -1/x² is 2/x³Derivative of 4/(4 - x)² is 4 * 2/(4 - x)³ = 8/(4 - x)³So, f''(x) = 2/x³ + 8/(4 - x)³At x = 4/3, let's compute f''(4/3):First, x = 4/3, so 4 - x = 4 - 4/3 = 8/3So, f''(4/3) = 2/( (4/3)³ ) + 8/( (8/3)³ )Compute each term:(4/3)³ = 64/27, so 2/(64/27) = 2 * 27/64 = 54/64 = 27/32Similarly, (8/3)³ = 512/27, so 8/(512/27) = 8 * 27/512 = 216/512 = 27/64So, f''(4/3) = 27/32 + 27/64 = (54/64 + 27/64) = 81/64 > 0Since the second derivative is positive, this critical point is a local minimum. Therefore, the function f(x) has a minimum at x = 4/3.Now, let's compute f(4/3):f(4/3) = 1/(4/3) + 4/(4 - 4/3) = 3/4 + 4/(8/3) = 3/4 + (4 * 3)/8 = 3/4 + 12/8 = 3/4 + 3/2Convert to common denominator:3/4 + 6/4 = 9/4 = 2.25So, the minimum value of f(x) is 9/4. Therefore, the inequality 1/x + 4/y ≥ m will always hold if m is less than or equal to 9/4.Looking back at the options:A: (2, +∞) – This is incorrect because m needs to be less than or equal to 9/4, not greater.B: (-∞, 7/4] – 7/4 is 1.75, which is less than 9/4, so this is too restrictive.C: (3, +∞) – Again, m needs to be less than or equal to 9/4, so this is incorrect.D: (-∞, 9/4] – This is correct because m must be less than or equal to 9/4.So, the correct answer is D.**Final Answer**boxed{D}

💡Okay, so I have this equation: ( x^2 + y^2 + 6x - 4y - 11 = 0 ). The question is asking for the area of the region defined by this equation. Hmm, I remember that equations involving ( x^2 ) and ( y^2 ) often represent circles, ellipses, or other conic sections. Since both ( x^2 ) and ( y^2 ) are present and have the same coefficient (which is 1 in this case), it's likely a circle.Alright, so if it's a circle, I need to rewrite the equation in the standard form of a circle's equation, which is ( (x - h)^2 + (y - k)^2 = r^2 ), where ( (h, k) ) is the center and ( r ) is the radius. To do that, I think I need to complete the square for both the ( x ) terms and the ( y ) terms.Let me start by grouping the ( x ) terms and the ( y ) terms together:( x^2 + 6x + y^2 - 4y = 11 ).Now, I'll complete the square for the ( x ) terms. The coefficient of ( x ) is 6. To complete the square, I take half of 6, which is 3, and then square it to get 9. So, I add and subtract 9 to complete the square:( (x^2 + 6x + 9) - 9 + y^2 - 4y = 11 ).Similarly, for the ( y ) terms, the coefficient of ( y ) is -4. Half of -4 is -2, and squaring that gives 4. So, I add and subtract 4 for the ( y ) terms:( (x^2 + 6x + 9) - 9 + (y^2 - 4y + 4) - 4 = 11 ).Now, simplifying this, the equation becomes:( (x + 3)^2 - 9 + (y - 2)^2 - 4 = 11 ).Combining the constants on the left side:( (x + 3)^2 + (y - 2)^2 - 13 = 11 ).Adding 13 to both sides to isolate the squared terms:( (x + 3)^2 + (y - 2)^2 = 24 ).Okay, so now the equation is in the standard form of a circle. The center of the circle is at ( (-3, 2) ) and the radius ( r ) is the square root of 24. Let me simplify that:( r = sqrt{24} = sqrt{4 times 6} = 2sqrt{6} ).Now, to find the area of the circle, I use the formula ( A = pi r^2 ). Plugging in the radius:( A = pi (2sqrt{6})^2 ).Calculating the square:( (2sqrt{6})^2 = 4 times 6 = 24 ).So, the area is:( A = 24pi ).Therefore, the area of the region defined by the given equation is ( 24pi ).

💡Okay, so I have this problem here where I'm given a sequence defined by a function. Let me try to understand it step by step.First, the problem says that a₁ = 2, and the point (aₙ, aₙ₊₁) lies on the graph of the function f(x) = x² + 2x. So, that means for each n, aₙ₊₁ is equal to f(aₙ). So, substituting, that would mean aₙ₊₁ = (aₙ)² + 2aₙ. Got that.Part (I) asks me to prove that the sequence {lg(1 + aₙ)} is a geometric sequence. Hmm, okay. So, I need to show that the ratio between consecutive terms is constant. That is, I need to show that lg(1 + aₙ₊₁) divided by lg(1 + aₙ) is a constant for all n.Let me write down what I know:aₙ₊₁ = (aₙ)² + 2aₙSo, 1 + aₙ₊₁ = (aₙ)² + 2aₙ + 1. Wait, that looks like a perfect square. Let me check:(aₙ + 1)² = (aₙ)² + 2aₙ + 1. Yes, exactly! So, 1 + aₙ₊₁ = (1 + aₙ)².So, taking the logarithm on both sides, I get:lg(1 + aₙ₊₁) = lg[(1 + aₙ)²] = 2 lg(1 + aₙ)So, that means lg(1 + aₙ₊₁) = 2 lg(1 + aₙ). Therefore, the ratio of lg(1 + aₙ₊₁) to lg(1 + aₙ) is 2, which is a constant. Hence, the sequence {lg(1 + aₙ)} is a geometric sequence with common ratio 2.Cool, that wasn't too bad. Let me just recap:1. Start with aₙ₊₁ = (aₙ)² + 2aₙ.2. Add 1 to both sides: 1 + aₙ₊₁ = (1 + aₙ)².3. Take log base 10: lg(1 + aₙ₊₁) = 2 lg(1 + aₙ).4. Therefore, the ratio is 2, so it's a geometric sequence.Alright, part (I) seems done.Now, part (II) asks me to find Tₙ = (1 + a₁)(1 + a₂)...(1 + aₙ) and also find the general formula for the sequence {aₙ}.Let me tackle the general formula for {aₙ} first. From part (I), I know that {lg(1 + aₙ)} is a geometric sequence with the first term lg(1 + a₁) and common ratio 2.Given that a₁ = 2, so 1 + a₁ = 3. Therefore, lg(1 + a₁) = lg 3.Since it's a geometric sequence, the nth term is given by:lg(1 + aₙ) = lg 3 * 2^{n-1}So, exponentiating both sides to get rid of the logarithm:1 + aₙ = 10^{lg 3 * 2^{n-1}} = (10^{lg 3})^{2^{n-1}}} = 3^{2^{n-1}}Therefore, aₙ = 3^{2^{n-1}} - 1.So, that's the general formula for aₙ. Let me check for n=1:a₁ = 3^{2^{0}} - 1 = 3^1 - 1 = 3 - 1 = 2. Correct.For n=2:a₂ = 3^{2^{1}} - 1 = 3^2 - 1 = 9 - 1 = 8.Wait, let's verify using the original recurrence:a₂ = a₁² + 2a₁ = 2² + 2*2 = 4 + 4 = 8. Correct.Similarly, a₃ = a₂² + 2a₂ = 8² + 2*8 = 64 + 16 = 80.Using the formula: a₃ = 3^{2^{2}} - 1 = 3^4 - 1 = 81 - 1 = 80. Correct. So, the formula seems solid.Now, moving on to Tₙ = (1 + a₁)(1 + a₂)...(1 + aₙ).From the earlier step, we have 1 + aₖ = 3^{2^{k-1}} for each k.Therefore, Tₙ = product from k=1 to n of 3^{2^{k-1}}.Which is equal to 3^{sum from k=1 to n of 2^{k-1}}}.The sum from k=1 to n of 2^{k-1} is a geometric series with first term 1 and ratio 2, so sum is 2^{n} - 1.Therefore, Tₙ = 3^{2^{n} - 1}.Let me verify this for n=1:T₁ = 1 + a₁ = 3. According to the formula, 3^{2^1 - 1} = 3^{2 - 1} = 3^1 = 3. Correct.For n=2:T₂ = (1 + a₁)(1 + a₂) = 3 * 9 = 27.Formula: 3^{2^2 - 1} = 3^{4 - 1} = 3^3 = 27. Correct.For n=3:T₃ = 3 * 9 * 81 = 2187.Formula: 3^{2^3 - 1} = 3^{8 - 1} = 3^7 = 2187. Correct.So, the formula for Tₙ is 3^{2^{n} - 1}.Therefore, summarizing:(I) The sequence {lg(1 + aₙ)} is geometric with common ratio 2.(II) Tₙ = 3^{2^{n} - 1}, and aₙ = 3^{2^{n-1}} - 1.**Final Answer**(I) The sequence {lg(1+a_n)} is a geometric sequence with common ratio 2.(II) The expression for T_n is boxed{3^{2^n-1}} and the general formula for the sequence {a_n} is boxed{a_n=3^{2^{n-1}}-1}.