💡Alright, so I have this problem about a hyperbola and some points on its asymptotes. Let me try to understand what it's asking. We have a hyperbola given by the equation (frac{x^2}{a^2} - frac{y^2}{b^2} = 1), where (a) and (b) are positive constants. Point (P) is any point on this hyperbola. There's a line passing through (P) that intersects the asymptotes (l_1: y = frac{b}{a}x) and (l_2: y = -frac{b}{a}x) at points (P_1) and (P_2) respectively. We're given a ratio (lambda = frac{P_1P}{PP_2}), and we need to prove that the area of triangle (OP_1P_2) is (frac{(1 + lambda)^2}{4|lambda|}ab).Okay, so first, I need to visualize this. The hyperbola has two asymptotes, which are straight lines that the hyperbola approaches but never touches. The asymptotes are symmetric with respect to both axes. Point (P) is somewhere on the hyperbola, and through this point, we draw a line that intersects both asymptotes at (P_1) and (P_2). The ratio (lambda) is defined as the ratio of the lengths from (P_1) to (P) and from (P) to (P_2).I think I should start by parametrizing point (P) on the hyperbola. Maybe using coordinates ((x, y)) that satisfy the hyperbola equation. Then, I can write the equation of the line passing through (P) and find its intersections with the asymptotes (l_1) and (l_2), which will give me points (P_1) and (P_2).Let me denote the coordinates of (P_1) as ((x_1, y_1)) and (P_2) as ((x_2, y_2)). Since (P_1) lies on (l_1), (y_1 = frac{b}{a}x_1). Similarly, since (P_2) lies on (l_2), (y_2 = -frac{b}{a}x_2).Now, the line passing through (P) and intersecting (P_1) and (P_2) can be written in parametric form. Let me assume that the line has a slope (m), so its equation is (y - y_P = m(x - x_P)), where ((x_P, y_P)) are the coordinates of (P). But since this line passes through both (P_1) and (P_2), I can express (x_1) and (x_2) in terms of (x_P) and (y_P).Alternatively, maybe using vectors or parametric equations would be better. Let me think about using parametric coordinates for the line. Let me set up a parameter (t) such that when (t = 0), we are at (P_1), and when (t = 1), we are at (P_2). Then, point (P) would correspond to some value of (t) between 0 and 1. But since (lambda = frac{P_1P}{PP_2}), this suggests that (P) divides the segment (P_1P_2) in the ratio (lambda:1). So, using the section formula, the coordinates of (P) can be expressed in terms of (P_1) and (P_2).Yes, that seems promising. The section formula in coordinate geometry states that if a point divides a line segment joining ((x_1, y_1)) and ((x_2, y_2)) in the ratio (m:n), then the coordinates of the point are (left(frac{mx_2 + nx_1}{m + n}, frac{my_2 + ny_1}{m + n}right)). In our case, since (lambda = frac{P_1P}{PP_2}), the ratio is (lambda:1). So, the coordinates of (P) would be:[x_P = frac{lambda x_2 + x_1}{1 + lambda}][y_P = frac{lambda y_2 + y_1}{1 + lambda}]But we know that (y_1 = frac{b}{a}x_1) and (y_2 = -frac{b}{a}x_2). Substituting these into the equation for (y_P):[y_P = frac{lambda left(-frac{b}{a}x_2right) + frac{b}{a}x_1}{1 + lambda} = frac{frac{b}{a}(x_1 - lambda x_2)}{1 + lambda}]So, we have expressions for (x_P) and (y_P) in terms of (x_1) and (x_2). Since (P) lies on the hyperbola, its coordinates must satisfy the hyperbola equation:[frac{x_P^2}{a^2} - frac{y_P^2}{b^2} = 1]Substituting the expressions for (x_P) and (y_P):[frac{left(frac{lambda x_2 + x_1}{1 + lambda}right)^2}{a^2} - frac{left(frac{frac{b}{a}(x_1 - lambda x_2)}{1 + lambda}right)^2}{b^2} = 1]Simplifying this equation might help us find a relationship between (x_1) and (x_2). Let's compute each term step by step.First, compute (x_P^2 / a^2):[frac{left(frac{lambda x_2 + x_1}{1 + lambda}right)^2}{a^2} = frac{(lambda x_2 + x_1)^2}{a^2(1 + lambda)^2}]Next, compute (y_P^2 / b^2):[frac{left(frac{frac{b}{a}(x_1 - lambda x_2)}{1 + lambda}right)^2}{b^2} = frac{left(frac{b}{a}(x_1 - lambda x_2)right)^2}{b^2(1 + lambda)^2} = frac{(x_1 - lambda x_2)^2}{a^2(1 + lambda)^2}]So, substituting back into the hyperbola equation:[frac{(lambda x_2 + x_1)^2}{a^2(1 + lambda)^2} - frac{(x_1 - lambda x_2)^2}{a^2(1 + lambda)^2} = 1]Factor out the common denominator:[frac{1}{a^2(1 + lambda)^2}left[(lambda x_2 + x_1)^2 - (x_1 - lambda x_2)^2right] = 1]Now, let's compute the numerator:[(lambda x_2 + x_1)^2 - (x_1 - lambda x_2)^2]Expanding both squares:First term: ((lambda x_2 + x_1)^2 = lambda^2 x_2^2 + 2lambda x_1 x_2 + x_1^2)Second term: ((x_1 - lambda x_2)^2 = x_1^2 - 2lambda x_1 x_2 + lambda^2 x_2^2)Subtracting the second term from the first:[(lambda^2 x_2^2 + 2lambda x_1 x_2 + x_1^2) - (x_1^2 - 2lambda x_1 x_2 + lambda^2 x_2^2) = 4lambda x_1 x_2]So, the numerator simplifies to (4lambda x_1 x_2). Therefore, the equation becomes:[frac{4lambda x_1 x_2}{a^2(1 + lambda)^2} = 1]Solving for (x_1 x_2):[x_1 x_2 = frac{a^2(1 + lambda)^2}{4lambda}]Okay, so now we have a relationship between (x_1) and (x_2). That's useful.Now, we need to find the area of triangle (OP_1P_2). The area of a triangle given three points can be found using the determinant formula:[text{Area} = frac{1}{2} |x_1 y_2 - x_2 y_1|]Since (O) is the origin, the coordinates are ((0,0)), (P_1) is ((x_1, y_1)), and (P_2) is ((x_2, y_2)). Plugging in the coordinates:[text{Area} = frac{1}{2} |x_1 y_2 - x_2 y_1|]We know that (y_1 = frac{b}{a}x_1) and (y_2 = -frac{b}{a}x_2), so substituting these:[text{Area} = frac{1}{2} |x_1 left(-frac{b}{a}x_2right) - x_2 left(frac{b}{a}x_1right)| = frac{1}{2} left| -frac{b}{a}x_1 x_2 - frac{b}{a}x_1 x_2 right| = frac{1}{2} left| -frac{2b}{a}x_1 x_2 right| = frac{1}{2} cdot frac{2b}{a} |x_1 x_2| = frac{b}{a} |x_1 x_2|]So, the area is (frac{b}{a} |x_1 x_2|). Earlier, we found that (x_1 x_2 = frac{a^2(1 + lambda)^2}{4lambda}). Plugging this into the area formula:[text{Area} = frac{b}{a} cdot left| frac{a^2(1 + lambda)^2}{4lambda} right| = frac{b}{a} cdot frac{a^2(1 + lambda)^2}{4|lambda|} = frac{a b (1 + lambda)^2}{4|lambda|}]Which simplifies to:[text{Area} = frac{(1 + lambda)^2}{4|lambda|} ab]So, that's the result we needed to prove. Wait, let me double-check my steps to make sure I didn't make a mistake. 1. I used the section formula correctly to express (x_P) and (y_P) in terms of (x_1), (x_2), and (lambda).2. Substituted these into the hyperbola equation and expanded correctly.3. Simplified the numerator correctly to (4lambda x_1 x_2).4. Solved for (x_1 x_2) correctly.5. Calculated the area using the determinant formula, substituted (y_1) and (y_2) correctly.6. Plugged in (x_1 x_2) into the area formula and simplified correctly.Everything seems to check out. So, the area of triangle (OP_1P_2) is indeed (frac{(1 + lambda)^2}{4|lambda|} ab).**Final Answer**The area of the triangle (triangle O P_{1} P_{2}) is (boxed{frac{(1 + lambda)^{2}}{4 |lambda|} a b}).

💡First, identify the number of apples Anna ate on each day.On Tuesday, Anna ate 4 apples.On Wednesday, she ate double the amount from Tuesday, which is 4 multiplied by 2, resulting in 8 apples.On Thursday, she ate half the amount from Tuesday, which is 4 divided by 2, resulting in 2 apples.Finally, add the number of apples eaten each day to find the total: 4 (Tuesday) + 8 (Wednesday) + 2 (Thursday) equals 14 apples.

💡Alright, so I've got this equation to solve: (3cos^3 x - 7 cos^2 x + 3 cos x = 0). I need to find how many solutions there are between (0) and (pi). Hmm, okay. Let me think about how to approach this.First, I notice that all the terms have a (cos x) in them. Maybe I can factor that out. Let me try that. So, factoring out (cos x), the equation becomes:[cos x (3cos^2 x - 7cos x + 3) = 0]Okay, so now I have two factors multiplied together equal to zero. That means either (cos x = 0) or the quadratic part (3cos^2 x - 7cos x + 3 = 0) must be zero. Let me handle each case separately.**Case 1: (cos x = 0)**When does cosine equal zero? Well, in the range (0) to (pi), cosine is zero at (frac{pi}{2}). So that's one solution right there: (x = frac{pi}{2}).**Case 2: (3cos^2 x - 7cos x + 3 = 0)**This is a quadratic equation in terms of (cos x). Let me let (y = cos x) to make it clearer:[3y^2 - 7y + 3 = 0]Now, I can solve this quadratic for (y). Using the quadratic formula:[y = frac{7 pm sqrt{(-7)^2 - 4 cdot 3 cdot 3}}{2 cdot 3}]Calculating the discriminant:[(-7)^2 = 49][4 cdot 3 cdot 3 = 36][sqrt{49 - 36} = sqrt{13}]So, the solutions are:[y = frac{7 + sqrt{13}}{6} quad text{and} quad y = frac{7 - sqrt{13}}{6}]Let me approximate these values to see if they make sense for cosine.First, (sqrt{13}) is approximately 3.6055.So,[y_1 = frac{7 + 3.6055}{6} = frac{10.6055}{6} approx 1.7676][y_2 = frac{7 - 3.6055}{6} = frac{3.3945}{6} approx 0.5658]Wait a minute, cosine values can only be between -1 and 1. So, (y_1 approx 1.7676) is outside the valid range for cosine. That means this solution is not possible. So, we can discard (y_1).So, the only valid solution from the quadratic is (y = frac{7 - sqrt{13}}{6} approx 0.5658). Let's denote this as (y = cos x = frac{7 - sqrt{13}}{6}).Now, I need to find all (x) in (0 leq x leq pi) such that (cos x = frac{7 - sqrt{13}}{6}).Since (frac{7 - sqrt{13}}{6}) is positive (approximately 0.5658), and cosine is positive in the first and fourth quadrants. However, since our domain is (0) to (pi), which covers the first and second quadrants, and cosine is positive only in the first quadrant in this interval.Therefore, there is exactly one solution in (0 leq x leq pi) for (cos x = frac{7 - sqrt{13}}{6}). Let's denote this solution as (x = arccosleft(frac{7 - sqrt{13}}{6}right)).So, putting it all together:- From Case 1: (x = frac{pi}{2})- From Case 2: (x = arccosleft(frac{7 - sqrt{13}}{6}right))Therefore, there are two solutions in total.Wait, let me double-check if there are any other solutions I might have missed. Since we're dealing with a cubic equation in terms of (cos x), it's possible to have up to three solutions for (cos x), but one of them was invalid because it was greater than 1. The other two were (cos x = 0) and (cos x = frac{7 - sqrt{13}}{6}). So, in the interval (0) to (pi), each of these valid (cos x) values gives exactly one solution for (x). So, that's two solutions in total.I think that's it. I don't see any other possibilities here. So, the number of solutions is 2.**Final Answer**The number of solutions is boxed{2}.

💡First, I observe that both terms in the expression (2mn - 6m) share a common factor of (2m).Next, I factor out (2m) from each term:[2mn - 6m = 2m(n) - 2m(3)]Finally, I write the expression as a product of the common factor and the remaining terms:[2m(n - 3)]

💡Okay, so I'm trying to figure out how many days it took Frank to finish his favorite book. The book has 193 pages and is divided equally into 15 chapters. Frank reads 44 chapters per day. Hmm, wait a minute, that doesn't seem right. If there are only 15 chapters in the book, how can Frank read 44 chapters in a day? That must be a mistake. Maybe it's supposed to be 44 pages per day instead of chapters? That makes more sense because 44 chapters per day would mean he finishes the book in less than a day, which isn't practical.Alright, assuming Frank reads 44 pages per day, I need to find out how many days it took him to read all 193 pages. To do this, I'll divide the total number of pages by the number of pages he reads each day. So, 193 pages divided by 44 pages per day. Let me calculate that: 193 ÷ 44 equals approximately 4.386 days. But you can't have a fraction of a day when talking about reading, so I'll need to round up to the next whole number. That means it took Frank 5 days to finish the book.Wait, let me double-check my assumption. If the original problem said 44 chapters per day, and there are only 15 chapters, then Frank would finish the book in less than a day, which doesn't make sense. So, it's definitely more plausible that it's 44 pages per day. Yeah, I think that's the right approach.

💡Okay, so I have this sequence problem to solve. It's about a sequence {a_n} where a_1 is 1, and each subsequent term is defined by a_{n+1} = 2a_n + 2. There are two parts to the problem. First, I need to prove that the sequence {a_n + 2} is a geometric sequence and then find the general term formula for {a_n}. Second, I have to define another sequence {b_n} where each term is b_n = n(a_n + 2), and find the sum of the first n terms of {b_n}, denoted as T_n.Starting with the first part. I remember that a geometric sequence is one where each term is a constant multiple of the previous term. So, if I can show that {a_n + 2} has a common ratio, then it's geometric.Given a_{n+1} = 2a_n + 2, maybe I can manipulate this equation to involve a_n + 2. Let me try adding 2 to both sides:a_{n+1} + 2 = 2a_n + 2 + 2 = 2a_n + 4.Hmm, that simplifies to a_{n+1} + 2 = 2(a_n + 2). Oh, that's interesting! So, each term a_{n+1} + 2 is exactly twice the previous term a_n + 2. That means the ratio between consecutive terms is 2, which is constant. Therefore, {a_n + 2} is indeed a geometric sequence with common ratio 2.Now, to find the general term formula for {a_n}, I need the first term of the geometric sequence {a_n + 2}. Since a_1 = 1, then a_1 + 2 = 3. So, the first term of the geometric sequence is 3, and the common ratio is 2.The general term of a geometric sequence is given by:a_n + 2 = 3 * 2^{n-1}.Therefore, solving for a_n:a_n = 3 * 2^{n-1} - 2.Alright, that takes care of the first part. Now, moving on to the second part. We have b_n = n(a_n + 2). From the first part, we know that a_n + 2 = 3 * 2^{n-1}, so substituting that in:b_n = n * 3 * 2^{n-1} = 3n * 2^{n-1}.Now, we need to find the sum T_n of the first n terms of {b_n}. So,T_n = b_1 + b_2 + ... + b_n = 3*1*2^{0} + 3*2*2^{1} + 3*3*2^{2} + ... + 3n*2^{n-1}.Factor out the 3:T_n = 3*(1*2^{0} + 2*2^{1} + 3*2^{2} + ... + n*2^{n-1}).Let me denote the sum inside the parentheses as S:S = 1*2^{0} + 2*2^{1} + 3*2^{2} + ... + n*2^{n-1}.I need to find a way to compute S. I recall that there's a formula for sums of the form sum_{k=1}^{n} k*r^{k-1}, which is similar to S. Let me see if I can derive it.Consider S = sum_{k=1}^{n} k*r^{k-1}.Multiply both sides by r:r*S = sum_{k=1}^{n} k*r^{k}.Subtract the original S from this equation:r*S - S = sum_{k=1}^{n} k*r^{k} - sum_{k=1}^{n} k*r^{k-1}.Simplify the right-hand side:= sum_{k=1}^{n} k*r^{k} - sum_{k=1}^{n} k*r^{k-1}= sum_{k=1}^{n} k*r^{k} - sum_{k=0}^{n-1} (k+1)*r^{k}.Wait, let me adjust the indices:Let me write the second sum as sum_{k=0}^{n-1} (k+1)*r^{k}.So, subtracting:sum_{k=1}^{n} k*r^{k} - sum_{k=0}^{n-1} (k+1)*r^{k} = sum_{k=1}^{n} k*r^{k} - sum_{k=0}^{n-1} (k+1)*r^{k}.Let me write out the terms:For the first sum: 1*r^1 + 2*r^2 + ... + n*r^n.For the second sum: 1*r^0 + 2*r^1 + ... + n*r^{n-1}.Subtracting term by term:(1*r^1 - 1*r^0) + (2*r^2 - 2*r^1) + ... + (n*r^n - n*r^{n-1}) - n*r^n.Wait, actually, when subtracting, the last term of the second sum is n*r^{n-1}, so when subtracted from the first sum, which goes up to n*r^n, we have:= [1*r^1 - 1*r^0] + [2*r^2 - 2*r^1] + ... + [n*r^n - n*r^{n-1}] - n*r^n.Simplify each bracket:= (r - 1) + (2r^2 - 2r) + (3r^3 - 3r^2) + ... + (n r^n - n r^{n-1}) - n r^n.Notice that each term cancels with the next one:= (r - 1) + (2r^2 - 2r) + (3r^3 - 3r^2) + ... + (n r^n - n r^{n-1}) - n r^n.So, most terms cancel out:= -1 + (0) + (0) + ... + (0) - n r^n.Therefore, r*S - S = -1 - n r^n.Factor out S:S*(r - 1) = -1 - n r^n.Therefore,S = (-1 - n r^n)/(r - 1).But in our case, r = 2, so plug that in:S = (-1 - n*2^n)/(2 - 1) = (-1 - n*2^n)/1 = -1 - n*2^n.But wait, that seems negative. Let me check my steps.Wait, when I subtracted, I got:r*S - S = -1 - n r^n.Which is S*(r - 1) = -1 - n r^n.So, S = (-1 - n r^n)/(r - 1).But with r = 2, that's (-1 - n*2^n)/(2 - 1) = -1 - n*2^n.But S is a sum of positive terms, so getting a negative result doesn't make sense. I must have made a mistake in the subtraction.Let me go back.We have:r*S = sum_{k=1}^{n} k*r^{k}.S = sum_{k=1}^{n} k*r^{k-1}.Subtracting S from r*S:r*S - S = sum_{k=1}^{n} k*r^{k} - sum_{k=1}^{n} k*r^{k-1}.Let me write both sums:First sum: 1*r^1 + 2*r^2 + 3*r^3 + ... + n*r^n.Second sum: 1*r^0 + 2*r^1 + 3*r^2 + ... + n*r^{n-1}.Subtracting term by term:(1*r^1 - 1*r^0) + (2*r^2 - 2*r^1) + (3*r^3 - 3*r^2) + ... + (n*r^n - n*r^{n-1}).So, each term is k*r^{k} - k*r^{k-1} = k*r^{k-1}(r - 1).But when we subtract the entire sums, it's:sum_{k=1}^{n} [k*r^{k} - k*r^{k-1}] = sum_{k=1}^{n} k*r^{k-1}(r - 1).Factor out (r - 1):(r - 1)*sum_{k=1}^{n} k*r^{k-1} = (r - 1)*S.But we also have r*S - S = (r - 1)*S.Wait, that's just restating the same thing. Maybe I need another approach.Alternatively, let me consider writing S as:S = sum_{k=1}^{n} k*2^{k-1}.I can write this as (1/2)*sum_{k=1}^{n} k*2^{k}.Let me denote Q = sum_{k=1}^{n} k*2^{k}.Then, S = (1/2)*Q.So, let's compute Q.Q = sum_{k=1}^{n} k*2^{k}.Again, use the technique of multiplying by r and subtracting.Let me write Q = 2 + 2*2^2 + 3*2^3 + ... + n*2^n.Multiply both sides by 2:2Q = 2^2 + 2*2^3 + 3*2^4 + ... + n*2^{n+1}.Subtract the original Q from this:2Q - Q = Q = (2^2 + 2*2^3 + 3*2^4 + ... + n*2^{n+1}) - (2 + 2*2^2 + 3*2^3 + ... + n*2^n).Simplify term by term:= (2^2 - 2) + (2*2^3 - 2*2^2) + (3*2^4 - 3*2^3) + ... + (n*2^{n+1} - n*2^n).Factor out the common terms:= (4 - 2) + 2*(8 - 4) + 3*(16 - 8) + ... + n*(2^{n+1} - 2^n).Simplify each term:= 2 + 2*4 + 3*8 + ... + n*2^n.Wait, that's similar to the original Q but shifted.Wait, let me write it as:Q = 2 + 4 + 8 + ... + 2^n + n*2^{n+1} - sum_{k=1}^{n} 2^k.Wait, maybe another approach.Wait, let me think differently. The standard formula for sum_{k=1}^{n} k*r^k is r*(1 - (n+1)*r^n + n*r^{n+1}) / (1 - r)^2.But since r ≠ 1.In our case, r = 2, so plugging in:sum_{k=1}^{n} k*2^k = 2*(1 - (n+1)*2^n + n*2^{n+1}) / (1 - 2)^2.Compute denominator: (1 - 2)^2 = 1.So,sum_{k=1}^{n} k*2^k = 2*(1 - (n+1)*2^n + n*2^{n+1}) / 1 = 2 - 2*(n+1)*2^n + 2*n*2^{n+1}.Wait, that seems complicated. Maybe I made a mistake in recalling the formula.Alternatively, let me derive it.Let me consider Q = sum_{k=1}^{n} k*2^k.Let me write Q = 2 + 2*2^2 + 3*2^3 + ... + n*2^n.Multiply both sides by 2:2Q = 2^2 + 2*2^3 + 3*2^4 + ... + n*2^{n+1}.Subtract Q from 2Q:2Q - Q = Q = (2^2 + 2*2^3 + 3*2^4 + ... + n*2^{n+1}) - (2 + 2*2^2 + 3*2^3 + ... + n*2^n).Simplify term by term:= (4 - 2) + (8 - 4) + (16 - 8) + ... + (2^{n+1} - 2^n) - n*2^{n+1}.Wait, actually, let me see:First term: 4 - 2 = 2.Second term: 8 - 4 = 4.Third term: 16 - 8 = 8....Last term: 2^{n+1} - 2^n = 2^n.But also, there's a -n*2^{n+1} at the end.So, Q = 2 + 4 + 8 + ... + 2^n - n*2^{n+1}.The sum 2 + 4 + 8 + ... + 2^n is a geometric series with first term 2, ratio 2, and n terms.Sum = 2*(2^n - 1)/(2 - 1) = 2*(2^n - 1) = 2^{n+1} - 2.Therefore,Q = (2^{n+1} - 2) - n*2^{n+1} = 2^{n+1} - 2 - n*2^{n+1} = (1 - n)*2^{n+1} - 2.Thus,Q = (1 - n)*2^{n+1} - 2.But Q = sum_{k=1}^{n} k*2^k, so S = (1/2)*Q = (1/2)*[(1 - n)*2^{n+1} - 2] = (1 - n)*2^n - 1.Therefore, S = (1 - n)*2^n - 1.So, going back to T_n:T_n = 3*S = 3*[(1 - n)*2^n - 1] = 3*(1 - n)*2^n - 3.Simplify:= 3*2^n - 3n*2^n - 3.Factor out 3*2^n:= 3*2^n(1 - n) - 3.Alternatively, factor out -3:= -3n*2^n + 3*2^n - 3.But let me write it as:T_n = 3*(1 - n)*2^n - 3.Alternatively, factor out 3:= 3[(1 - n)*2^n - 1].But maybe it's better to write it as:T_n = 3*(2^n - n*2^n) - 3 = 3*2^n(1 - n) - 3.Alternatively, factor out 3:= 3[2^n(1 - n) - 1].But perhaps another form is better. Let me see.Alternatively, let's compute it step by step.We have S = (1 - n)*2^n - 1.So,T_n = 3*S = 3*(1 - n)*2^n - 3.= 3*2^n - 3n*2^n - 3.= 3*2^n(1 - n) - 3.Alternatively, factor out 3:= 3[2^n(1 - n) - 1].But maybe we can write it as:= 3*(2^n - n*2^n - 1).= 3*(2^n(1 - n) - 1).Alternatively, factor out 2^n:= 3*2^n(1 - n) - 3.I think that's as simplified as it gets. So, the sum T_n is 3*(2^n - n*2^n) - 3, which can be written as 3*(1 - n)*2^n - 3.Alternatively, factor out 3:= 3[ (1 - n)2^n - 1 ].But perhaps another way to write it is:= 3*(2^n - n*2^n - 1).= 3*(2^n(1 - n) - 1).Alternatively, factor out 2^n:= 3*2^n(1 - n) - 3.I think that's the most straightforward form.So, summarizing:1. {a_n + 2} is a geometric sequence with first term 3 and ratio 2, so a_n = 3*2^{n-1} - 2.2. T_n = 3*(1 - n)*2^n - 3.Alternatively, T_n = 3*(2^n - n*2^n) - 3.But to make it look cleaner, perhaps factor out 2^n:T_n = 3*2^n(1 - n) - 3.Alternatively, factor out 3:T_n = 3[2^n(1 - n) - 1].Either way, that's the sum.