answer：Okay, so I have this problem with four propositions about vectors, and I need to figure out how many of them are correct. Let me go through each one step by step.**Proposition 1:** If vectors a and b are both unit vectors and a is collinear with b, then a equals b.Hmm, unit vectors have a magnitude of 1. Collinear means they lie on the same line, so they can be in the same or opposite directions. If they are in the same direction, then yes, a would equal b. But if they are in opposite directions, then a would be equal to negative b. So, they don't necessarily have to be equal; they could also be negatives of each other. Therefore, this proposition isn't always true. So, I think Proposition 1 is **incorrect**.**Proposition 2:** The nonzero vector AB and BA are parallel vectors with equal magnitude.Vector AB goes from point A to point B, and vector BA goes from point B to point A. They are in exactly opposite directions, but their magnitudes are the same because the distance from A to B is the same as from B to A. So, they are parallel (since they lie on the same line) and have equal magnitude. This seems correct. So, Proposition 2 is **correct**.**Proposition 3:** If vector a is collinear with vector b, then the magnitude of (a + b) equals the sum of the magnitudes of a and b.Collinear vectors are either in the same direction or opposite directions. If they are in the same direction, then adding them would indeed result in a vector whose magnitude is the sum of their magnitudes. However, if they are in opposite directions, the magnitude of their sum would actually be the difference of their magnitudes, not the sum. So, this proposition isn't always true. Therefore, Proposition 3 is **incorrect**.**Proposition 4:** For vectors a, b, c, if a is parallel to b and b is parallel to c, then a is parallel to c.Parallel vectors mean they lie on the same line or parallel lines, so their directions are the same or opposite. If a is parallel to b, and b is parallel to c, then a should also be parallel to c. This seems to follow the transitive property of parallelism. So, Proposition 4 is **correct**.Wait, but hold on. What if one of the vectors is the zero vector? Because the zero vector is considered parallel to every vector. So, if b is the zero vector, then a being parallel to b doesn't necessarily mean a is parallel to c, unless c is also the zero vector or a is zero. Hmm, that complicates things. If b is zero, then a can be any vector, and c can be any vector, but a doesn't have to be parallel to c. So, in that case, the proposition doesn't hold. Therefore, Proposition 4 is **incorrect**.Wait, now I'm confused. Initially, I thought it was correct, but considering the zero vector, it might not always hold. Maybe the proposition assumes that the vectors are non-zero? The problem doesn't specify, so I have to consider the general case. Since the zero vector can be a case where the proposition fails, I think Proposition 4 is **incorrect**.So, summarizing:1. Incorrect2. Correct3. Incorrect4. IncorrectTherefore, only Proposition 2 is correct. So, the total number of correct propositions is 1.**Final Answer**boxed{A}

answer：Alright, so I've got these four math problems to solve. Let me take them one at a time and see if I can figure them out step by step. I'm going to write down my thoughts as I go along, just like I would if I were solving them on paper.**Problem 1: Calculate 6 + (-8) - (-5)**Okay, starting with the first problem: 6 plus negative 8 minus negative 5. Hmm, so that's like adding and subtracting positive and negative numbers. I remember that subtracting a negative is the same as adding a positive. So, let me rewrite this to make it clearer.6 + (-8) is the same as 6 minus 8. Then, minus (-5) is the same as plus 5. So, putting it all together: 6 - 8 + 5.Now, let's compute that step by step. 6 minus 8 is -2. Then, -2 plus 5 is 3. So, the result should be 3. That seems straightforward.**Problem 2: Calculate 5 3/5 + (-5 2/3) + 4 2/5 + (-1/3)**Alright, moving on to the second problem. This one has mixed numbers and fractions with different denominators. I need to add and subtract these. Let me write them down:5 3/5 + (-5 2/3) + 4 2/5 + (-1/3)First, I can think of this as adding all the positive numbers and then subtracting the negative ones. So, let's group them:(5 3/5 + 4 2/5) + (-5 2/3 - 1/3)Let me compute each group separately.Starting with the positive numbers: 5 3/5 + 4 2/5.Adding the whole numbers: 5 + 4 = 9.Adding the fractions: 3/5 + 2/5 = 5/5 = 1.So, 9 + 1 = 10.Now, the negative numbers: -5 2/3 - 1/3.Adding the fractions first: 2/3 + 1/3 = 3/3 = 1.So, -5 2/3 - 1/3 = -5 - 1 = -6.Now, combining both results: 10 + (-6) = 10 - 6 = 4.So, the result is 4.Wait, let me double-check that. Maybe I should convert the mixed numbers to improper fractions to ensure accuracy.5 3/5 is equal to (5*5 + 3)/5 = 28/5.4 2/5 is equal to (4*5 + 2)/5 = 22/5.So, 28/5 + 22/5 = 50/5 = 10.Similarly, -5 2/3 is equal to (-5*3 - 2)/3 = -17/3.-1/3 is just -1/3.Adding them: -17/3 + (-1/3) = -18/3 = -6.So, 10 + (-6) = 4. Yep, that checks out.**Problem 3: Calculate (-1/2 + 1/6 - 1/4) × 12**Okay, the third problem involves fractions inside parentheses multiplied by 12. Let's see:First, compute the expression inside the parentheses: -1/2 + 1/6 - 1/4.To add and subtract these fractions, I need a common denominator. The denominators are 2, 6, and 4. The least common multiple of 2, 6, and 4 is 12. So, let's convert each fraction to have 12 as the denominator.-1/2 = -6/121/6 = 2/12-1/4 = -3/12Now, adding them together: -6/12 + 2/12 - 3/12.Combine the numerators: (-6 + 2 - 3)/12 = (-7)/12.So, the expression inside the parentheses simplifies to -7/12.Now, multiply this by 12: (-7/12) × 12.The 12s cancel out, leaving -7.So, the result is -7.Alternatively, I could distribute the 12 to each term inside the parentheses:(-1/2) × 12 + (1/6) × 12 - (1/4) × 12Calculating each term:-1/2 × 12 = -61/6 × 12 = 2-1/4 × 12 = -3Now, adding them together: -6 + 2 - 3 = -7.Same result. Good.**Problem 4: Calculate -1^2022 + 27 × (-1/3)^2 - |-5|**Alright, the fourth problem looks a bit more complex with exponents and absolute value. Let's break it down step by step.First, let's rewrite the expression:-1^2022 + 27 × (-1/3)^2 - |-5|I need to remember the order of operations: parentheses, exponents, multiplication and division, addition and subtraction (PEMDAS).Let's handle each term separately.1. -1^2022: This is negative one raised to the power of 2022. Since 2022 is an even number, any negative number raised to an even power becomes positive. So, (-1)^2022 = 1. But wait, the negative sign is separate from the exponent. Is it -(1^2022) or (-1)^2022? In mathematics, exponentiation has higher precedence than the unary minus. So, it's interpreted as -(1^2022). Since 1 raised to any power is 1, this term is -1.Wait, let me confirm that. If it's written as -1^2022, it's ambiguous. But generally, without parentheses, it's interpreted as -(1^2022). So, yes, it's -1.2. 27 × (-1/3)^2: First, compute (-1/3)^2. Squaring a negative number gives a positive result. So, (-1/3)^2 = (1/3)^2 = 1/9. Now, multiply by 27: 27 × (1/9) = 3.3. -|-5|: The absolute value of -5 is 5. So, this term is -5.Now, putting it all together:-1 + 3 - 5Compute step by step:-1 + 3 = 22 - 5 = -3So, the result is -3.Wait, let me double-check the first term. If it's written as -1^2022, it's indeed -(1^2022) = -1. If it were (-1)^2022, it would be 1. But since there are no parentheses around -1, it's just -1.Alternatively, if someone interprets it as (-1)^2022, they might get 1, but that's not the standard interpretation. So, I think -1 is correct.So, combining all terms: -1 + 3 - 5 = -3.Alright, that seems solid.**Summary of Answers:**1. 32. 43. -74. -3I think I've got all four problems solved correctly. I went through each step carefully, especially making sure to handle the signs correctly and apply the order of operations properly. It's easy to make sign errors or misapply exponents, so double-checking each step was important.

answer：Okay, so I have this equation: 2^x + 3^x + 6^x = 7^x. I need to find all values of x that satisfy this. Hmm, let me think about how to approach this.First, I know that exponential functions can sometimes be tricky because they grow or decay at different rates depending on the base. Here, the bases are 2, 3, 6, and 7. Since 7 is the largest base, 7^x will grow faster than the others as x increases. On the other hand, for negative x, 7^x will decay faster than the others because it's a larger base.Maybe I can try plugging in some integer values for x to see if they work. Let's start with x = 1.When x = 1:2^1 + 3^1 + 6^1 = 2 + 3 + 6 = 117^1 = 711 ≠ 7, so x = 1 is not a solution.Next, x = 2:2^2 + 3^2 + 6^2 = 4 + 9 + 36 = 497^2 = 4949 = 49, so x = 2 is a solution. Nice, that works.Let me try x = 3:2^3 + 3^3 + 6^3 = 8 + 27 + 216 = 2517^3 = 343251 ≠ 343, so x = 3 doesn't work.How about x = 0? Let's check:2^0 + 3^0 + 6^0 = 1 + 1 + 1 = 37^0 = 13 ≠ 1, so x = 0 isn't a solution.What about negative integers? Let's try x = -1:2^-1 + 3^-1 + 6^-1 = 1/2 + 1/3 + 1/6 = (3/6 + 2/6 + 1/6) = 6/6 = 17^-1 = 1/7 ≈ 0.1428571 ≠ 0.142857, so x = -1 doesn't work.Hmm, so x = 2 works. Maybe it's the only solution. But how can I be sure there aren't other solutions?Perhaps I can analyze the function f(x) = 2^x + 3^x + 6^x - 7^x and see where it crosses zero. If I can show that f(x) = 0 only at x = 2, then that's the only solution.Let me consider the behavior of f(x) as x approaches infinity and negative infinity.As x → ∞:- 7^x grows much faster than 2^x, 3^x, or 6^x, so f(x) will tend to negative infinity.As x → -∞:- 2^x, 3^x, and 6^x will tend to infinity because any positive number raised to a negative power becomes a large positive number. However, 7^x will also tend to infinity, but since 7 is larger than 6, 3, and 2, 7^x will dominate. Wait, no, actually, for negative x, 7^x = 1/(7^|x|), which tends to zero as |x| increases. Similarly, 2^x, 3^x, and 6^x tend to infinity as x → -∞. So f(x) will tend to positive infinity.So, f(x) goes from positive infinity at x = -∞ to negative infinity as x approaches ∞. Since f(x) is continuous, by the Intermediate Value Theorem, there must be at least one solution. We already found x = 2.But could there be more than one solution? Let's check the derivative to see if f(x) is strictly decreasing or if it has any turning points.f'(x) = ln(2)*2^x + ln(3)*3^x + ln(6)*6^x - ln(7)*7^xI need to determine if f'(x) is always negative or if it changes sign.At x = 2:f'(2) = ln(2)*4 + ln(3)*9 + ln(6)*36 - ln(7)*49Let me compute each term:ln(2) ≈ 0.6931, so 0.6931*4 ≈ 2.7724ln(3) ≈ 1.0986, so 1.0986*9 ≈ 9.8874ln(6) ≈ 1.7918, so 1.7918*36 ≈ 64.5048ln(7) ≈ 1.9459, so 1.9459*49 ≈ 95.3501Adding up the first three terms: 2.7724 + 9.8874 + 64.5048 ≈ 77.1646Subtracting the last term: 77.1646 - 95.3501 ≈ -18.1855So f'(2) ≈ -18.1855, which is negative.What about at x = 1:f'(1) = ln(2)*2 + ln(3)*3 + ln(6)*6 - ln(7)*7Compute each term:ln(2)*2 ≈ 1.3862ln(3)*3 ≈ 3.2958ln(6)*6 ≈ 10.7508ln(7)*7 ≈ 13.6213Adding up the first three: 1.3862 + 3.2958 + 10.7508 ≈ 15.4328Subtracting the last term: 15.4328 - 13.6213 ≈ 1.8115So f'(1) ≈ 1.8115, which is positive.So the derivative changes from positive at x = 1 to negative at x = 2. That means f(x) has a maximum somewhere between x = 1 and x = 2. Therefore, f(x) is increasing before that maximum and decreasing after it. Since f(x) crosses zero at x = 2 and tends to negative infinity as x increases, there can't be another solution for x > 2. What about for x < 2?Since f(x) tends to positive infinity as x approaches negative infinity and f(2) = 0, and f(x) is increasing up to some point and then decreasing, it's possible that f(x) could cross zero again for x < 2. Let me check f(0) again: f(0) = 3 - 1 = 2, which is positive. f(1) = 11 - 7 = 4, still positive. f(2) = 0. So from x = 0 to x = 2, f(x) decreases from 2 to 0. Since f(x) is positive at x = 0 and x = 1, and zero at x = 2, and the derivative is positive at x = 1, it suggests that f(x) is increasing from x = 1 to the maximum point, then decreasing after that. So between x = -∞ and x = 2, f(x) goes from positive infinity, reaches a maximum, then decreases to zero at x = 2. Therefore, it might cross zero only once at x = 2.Wait, but what about for x < 0? Since f(x) tends to positive infinity as x approaches negative infinity, and f(0) = 2, which is still positive, does f(x) ever cross zero for x < 0? Let me check f(-1):f(-1) = 2^-1 + 3^-1 + 6^-1 - 7^-1 = 1/2 + 1/3 + 1/6 - 1/7 ≈ 0.5 + 0.3333 + 0.1667 - 0.1429 ≈ 0.5 + 0.3333 = 0.8333 + 0.1667 = 1 - 0.1429 ≈ 0.8571, which is positive. So f(-1) ≈ 0.8571 > 0.What about x = -2:f(-2) = 2^-2 + 3^-2 + 6^-2 - 7^-2 = 1/4 + 1/9 + 1/36 - 1/49 ≈ 0.25 + 0.1111 + 0.0278 - 0.0204 ≈ 0.25 + 0.1111 = 0.3611 + 0.0278 = 0.3889 - 0.0204 ≈ 0.3685 > 0.Still positive. Maybe x = -3:f(-3) = 2^-3 + 3^-3 + 6^-3 - 7^-3 = 1/8 + 1/27 + 1/216 - 1/343 ≈ 0.125 + 0.0370 + 0.0046 - 0.0029 ≈ 0.125 + 0.0370 = 0.162 + 0.0046 = 0.1666 - 0.0029 ≈ 0.1637 > 0.Still positive. It seems like f(x) is always positive for x < 2 and only crosses zero at x = 2. Therefore, x = 2 is the only solution.But just to be thorough, let me check if f(x) could ever be zero for x < 2. Suppose x is very large negative, say x = -1000. Then f(x) would be approximately 2^(-1000) + 3^(-1000) + 6^(-1000) - 7^(-1000). All these terms are extremely small, but 2^(-1000) is larger than 7^(-1000) because 2 < 7. So f(x) would be a very small positive number minus an even smaller positive number, still positive. Therefore, f(x) is always positive for x < 2 and only touches zero at x = 2.Thus, the only solution is x = 2.

answer：Alright, so I've got this problem here about a parabola and a line intersecting it. Let me try to parse it step by step. First, the parabola is given by the equation ( y^2 = 2px ) where ( p > 0 ). I remember that for a parabola in this form, the focus is at ( (frac{p}{2}, 0) ) and the directrix is the line ( x = -frac{p}{2} ). The line given is ( y = -sqrt{3}(x - 1) ). It says this line passes through the focus of the parabola. So, if the focus is ( (frac{p}{2}, 0) ), then plugging this point into the line equation should satisfy it. Let me write that out:( 0 = -sqrt{3}(frac{p}{2} - 1) )Solving for ( p ):( 0 = -sqrt{3}(frac{p}{2} - 1) )Divide both sides by ( -sqrt{3} ):( 0 = frac{p}{2} - 1 )So,( frac{p}{2} = 1 )Multiply both sides by 2:( p = 2 )Okay, so that gives me ( p = 2 ). So statement A is true. That seems straightforward.Now, moving on. The line intersects the parabola at points M and N. I need to find these points of intersection. Since the parabola is ( y^2 = 4x ) (because ( p = 2 )), and the line is ( y = -sqrt{3}(x - 1) ). Let me substitute ( y ) from the line equation into the parabola equation:( (-sqrt{3}(x - 1))^2 = 4x )Simplify the left side:( 3(x - 1)^2 = 4x )Expand ( (x - 1)^2 ):( 3(x^2 - 2x + 1) = 4x )Multiply through:( 3x^2 - 6x + 3 = 4x )Bring all terms to one side:( 3x^2 - 6x + 3 - 4x = 0 )Combine like terms:( 3x^2 - 10x + 3 = 0 )So, the quadratic equation is ( 3x^2 - 10x + 3 = 0 ). Let me try to solve for ( x ) using the quadratic formula:( x = frac{10 pm sqrt{(-10)^2 - 4 cdot 3 cdot 3}}{2 cdot 3} )Calculate discriminant:( 100 - 36 = 64 )So,( x = frac{10 pm 8}{6} )Thus, two solutions:1. ( x = frac{10 + 8}{6} = frac{18}{6} = 3 )2. ( x = frac{10 - 8}{6} = frac{2}{6} = frac{1}{3} )So, the x-coordinates of M and N are 3 and 1/3. Now, let's find the corresponding y-coordinates using the line equation ( y = -sqrt{3}(x - 1) ).For ( x = 3 ):( y = -sqrt{3}(3 - 1) = -sqrt{3}(2) = -2sqrt{3} )So, point M is (3, -2√3).For ( x = 1/3 ):( y = -sqrt{3}(frac{1}{3} - 1) = -sqrt{3}(-frac{2}{3}) = frac{2sqrt{3}}{3} )So, point N is (1/3, 2√3/3).Alright, so now I have points M(3, -2√3) and N(1/3, 2√3/3).Next, statement B says |MN| = 8/3. Let me compute the distance between M and N.Distance formula:( |MN| = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} )Plugging in the coordinates:( |MN| = sqrt{(frac{1}{3} - 3)^2 + (frac{2sqrt{3}}{3} - (-2sqrt{3}))^2} )Simplify each part:First, ( frac{1}{3} - 3 = frac{1}{3} - frac{9}{3} = -frac{8}{3} )Second, ( frac{2sqrt{3}}{3} - (-2sqrt{3}) = frac{2sqrt{3}}{3} + 2sqrt{3} = frac{2sqrt{3}}{3} + frac{6sqrt{3}}{3} = frac{8sqrt{3}}{3} )So, now:( |MN| = sqrt{(-frac{8}{3})^2 + (frac{8sqrt{3}}{3})^2} )Calculate each square:( (-frac{8}{3})^2 = frac{64}{9} )( (frac{8sqrt{3}}{3})^2 = frac{64 cdot 3}{9} = frac{192}{9} )Add them:( frac{64}{9} + frac{192}{9} = frac{256}{9} )Take square root:( sqrt{frac{256}{9}} = frac{16}{3} )Hmm, so |MN| is 16/3, not 8/3. So statement B is false.Moving on to statement C: The circle with MN as its diameter is tangent to l, where l is the directrix.First, let's recall that the directrix of the parabola ( y^2 = 4x ) is ( x = -1 ).Now, the circle with diameter MN. To find if it's tangent to the directrix, I need to find the center and radius of the circle and see if the distance from the center to the directrix equals the radius.First, find the midpoint of MN, which is the center of the circle.Midpoint formula:( left( frac{x_1 + x_2}{2}, frac{y_1 + y_2}{2} right) )Plugging in:( left( frac{3 + frac{1}{3}}{2}, frac{-2sqrt{3} + frac{2sqrt{3}}{3}}{2} right) )Simplify:First coordinate:( frac{3 + frac{1}{3}}{2} = frac{frac{9}{3} + frac{1}{3}}{2} = frac{frac{10}{3}}{2} = frac{10}{6} = frac{5}{3} )Second coordinate:( frac{-2sqrt{3} + frac{2sqrt{3}}{3}}{2} = frac{-frac{6sqrt{3}}{3} + frac{2sqrt{3}}{3}}{2} = frac{-frac{4sqrt{3}}{3}}{2} = -frac{2sqrt{3}}{3} )So, the center is ( (frac{5}{3}, -frac{2sqrt{3}}{3}) ).The radius is half of |MN|, which we found to be 16/3, so radius is 8/3.Now, the directrix is ( x = -1 ). The distance from the center to the directrix is the horizontal distance between ( x = frac{5}{3} ) and ( x = -1 ).Distance = ( |frac{5}{3} - (-1)| = |frac{5}{3} + 1| = |frac{5}{3} + frac{3}{3}| = |frac{8}{3}| = frac{8}{3} )So, the distance from the center to the directrix is equal to the radius, which is 8/3. Therefore, the circle is tangent to the directrix. So statement C is true.Finally, statement D: Triangle OMN is an isosceles triangle.To check this, I need to see if any two sides are equal in length. The triangle is formed by points O(0,0), M(3, -2√3), and N(1/3, 2√3/3).Let's compute the lengths of OM, ON, and MN.We already know |MN| = 16/3.Compute |OM|:Distance from O(0,0) to M(3, -2√3):( |OM| = sqrt{(3 - 0)^2 + (-2sqrt{3} - 0)^2} = sqrt{9 + 12} = sqrt{21} )Compute |ON|:Distance from O(0,0) to N(1/3, 2√3/3):( |ON| = sqrt{(frac{1}{3} - 0)^2 + (frac{2sqrt{3}}{3} - 0)^2} = sqrt{frac{1}{9} + frac{12}{9}} = sqrt{frac{13}{9}} = frac{sqrt{13}}{3} )So, |OM| = √21 ≈ 4.583, |ON| ≈ 1.201, |MN| ≈ 5.333.None of these are equal. So, triangle OMN is not isosceles. Therefore, statement D is false.So, summarizing:A: TrueB: FalseC: TrueD: FalseTherefore, the correct statements are A and C.**Final Answer**boxed{AC}

answer：Okay, so I have this problem about a quadrilateral with sides 60, 100, 120, and 80. It's both cyclic (inscribed in a circle) and tangential (has an inscribed circle). The question is about finding the absolute difference between the two segments, x and y, created by the point where the incircle touches the side of length 120. First, let me recall some properties of tangential quadrilaterals. I remember that in a tangential quadrilateral, the sums of the lengths of the opposite sides are equal. So, that means AB + CD should equal BC + AD. Let me check that with the given sides: AB is 60, BC is 100, CD is 120, and DA is 80. So, AB + CD is 60 + 120 = 180, and BC + DA is 100 + 80 = 180. Okay, that checks out, so it's indeed a tangential quadrilateral.Since it's also cyclic, it's a bicentric quadrilateral. I remember that for bicentric quadrilaterals, there are some special relationships between the radii of the incircle and circumcircle, but I don't know if that's relevant here. Maybe not, since the problem is about the incircle's point of tangency.In a tangential quadrilateral, the lengths from the vertices to the points of tangency can be expressed in terms of the semiperimeter. The semiperimeter, s, is (AB + BC + CD + DA)/2. Let me calculate that: (60 + 100 + 120 + 80)/2 = (360)/2 = 180. So, s is 180.In a tangential quadrilateral, the lengths from each vertex to the points of tangency can be found using the formula: if we denote the tangent lengths as p, q, r, t, then we have:AB = p + q,BC = q + r,CD = r + t,DA = t + p.And since the semiperimeter is s, each of these tangent lengths can be found as:p = (AB + DA - BC)/2,q = (AB + BC - DA)/2,r = (BC + CD - AB)/2,t = (CD + DA - BC)/2.Wait, let me make sure I have that right. Alternatively, since the sum of the two opposite sides are equal, each tangent length can be expressed as s - opposite side.Wait, no, that's for the lengths of the tangents. Let me think again.In a tangential quadrilateral, the lengths from each vertex to the point of tangency are equal for adjacent sides. So, if I denote the tangent lengths as follows: starting from vertex A, the tangent lengths to the two adjacent sides AB and AD are both equal to some value, say, w. Similarly, from vertex B, the tangent lengths to sides AB and BC are equal to some value, say, x. From vertex C, the tangent lengths to sides BC and CD are equal to y, and from vertex D, the tangent lengths to sides CD and DA are equal to z.So, we have:AB = w + x = 60,BC = x + y = 100,CD = y + z = 120,DA = z + w = 80.And since the semiperimeter is s = 180, we have:w + x + y + z = s = 180.But actually, since AB + BC + CD + DA = 2s, which is 360, so that's consistent.So, we can set up the equations:1. w + x = 60,2. x + y = 100,3. y + z = 120,4. z + w = 80.We have four equations with four variables: w, x, y, z.Let me try to solve these equations step by step.From equation 1: w = 60 - x.From equation 2: y = 100 - x.From equation 3: z = 120 - y = 120 - (100 - x) = 20 + x.From equation 4: z + w = 80.Substituting z and w from above:(20 + x) + (60 - x) = 80.Simplify:20 + x + 60 - x = 80,80 = 80.Hmm, that's an identity, which means the system is consistent, but we need another way to find the variables.Wait, maybe I need to use the fact that the quadrilateral is cyclic as well. For a bicentric quadrilateral, there is a relationship between the inradius, the circumradius, and the distance between the centers, but I'm not sure if that's helpful here.Alternatively, since it's cyclic, we can use Brahmagupta's formula to find the area, and since it's tangential, the area can also be expressed as r*s, where r is the inradius and s is the semiperimeter.So, let me compute the area using Brahmagupta's formula first.Brahmagupta's formula states that the area K of a cyclic quadrilateral with sides a, b, c, d is:K = sqrt[(s - a)(s - b)(s - c)(s - d)].Given a = 60, b = 100, c = 120, d = 80, and s = 180.So,K = sqrt[(180 - 60)(180 - 100)(180 - 120)(180 - 80)]= sqrt[120 * 80 * 60 * 100].Let me compute that:First, multiply 120 and 80: 120*80 = 9600.Then, multiply 60 and 100: 60*100 = 6000.Now, multiply 9600 and 6000: 9600*6000 = 57,600,000.So, K = sqrt[57,600,000].Let me compute sqrt(57,600,000).Note that 57,600,000 = 576 * 100,000.sqrt(576) = 24, and sqrt(100,000) = 316.227766... but wait, 100,000 is 10^5, so sqrt(10^5) = 10^(2.5) = 10^2 * sqrt(10) = 100 * 3.16227766 ≈ 316.227766.But actually, 57,600,000 = 576 * 100,000 = (24^2) * (10^5).So, sqrt(57,600,000) = 24 * sqrt(10^5) = 24 * (10^(2.5)) = 24 * 10^2 * sqrt(10) = 2400 * sqrt(10).Wait, that doesn't seem right because 2400^2 = 5,760,000, which is much less than 57,600,000.Wait, maybe I made a mistake in breaking it down.Wait, 57,600,000 = 576 * 100,000.But 576 is 24^2, and 100,000 is 10^5, which is (10^2.5)^2.Wait, no, 10^5 is (10^(2.5))^2? Wait, 10^5 is (10^(2.5))^2? Let me check:(10^(2.5))^2 = 10^(5), yes, correct.So, sqrt(576 * 10^5) = sqrt(576) * sqrt(10^5) = 24 * (10^(2.5)).But 10^(2.5) is 10^2 * 10^0.5 = 100 * sqrt(10) ≈ 100 * 3.16227766 ≈ 316.227766.So, 24 * 316.227766 ≈ 24 * 316.227766 ≈ 7589.466384.Wait, but let me compute 24 * 316.227766:24 * 300 = 7200,24 * 16.227766 ≈ 24 * 16 = 384, and 24 * 0.227766 ≈ 5.466.So, total ≈ 7200 + 384 + 5.466 ≈ 7589.466.But wait, that seems too large because the sides are 60, 100, 120, 80, so the area shouldn't be that big.Wait, maybe I made a mistake in the calculation.Wait, 120 * 80 = 9600,60 * 100 = 6000,9600 * 6000 = 57,600,000,sqrt(57,600,000) = sqrt(576 * 100,000) = 24 * sqrt(100,000).But sqrt(100,000) is 316.227766,So, 24 * 316.227766 ≈ 7589.466.Wait, but let me check with another approach.Alternatively, 57,600,000 = 576 * 100,000 = 24^2 * (10^5).So, sqrt(24^2 * 10^5) = 24 * sqrt(10^5).But sqrt(10^5) = 10^(5/2) = 10^2 * 10^(1/2) = 100 * sqrt(10) ≈ 100 * 3.16227766 ≈ 316.227766.So, 24 * 316.227766 ≈ 7589.466.But let me check if that makes sense. The sides are 60, 100, 120, 80, so the maximum possible area for a cyclic quadrilateral with these sides is indeed around 7589.466.Wait, but let me check with another formula. Maybe using Heron's formula for a triangle, but since it's a quadrilateral, Brahmagupta's formula is the way to go.Alternatively, maybe I can compute it as 2400 * sqrt(10), because 24 * 100 * sqrt(10) = 2400 * sqrt(10), and sqrt(10) ≈ 3.16227766, so 2400 * 3.16227766 ≈ 7589.466.Yes, that seems correct.So, the area K is 2400 * sqrt(10).Now, since the quadrilateral is tangential, the area can also be expressed as K = r * s, where r is the inradius and s is the semiperimeter.We have K = 2400 * sqrt(10), and s = 180.So, 2400 * sqrt(10) = r * 180,Therefore, r = (2400 * sqrt(10)) / 180.Simplify:2400 / 180 = 240 / 18 = 40 / 3 ≈ 13.3333.So, r = (40/3) * sqrt(10).Wait, that seems correct.But wait, in the initial problem, we are asked about the segments x and y on the side of length 120. Let me recall that in a tangential quadrilateral, the lengths from the vertices to the points of tangency can be found using the semiperimeter.Wait, earlier, I set up the equations:w + x = 60,x + y = 100,y + z = 120,z + w = 80.And from these, we found that:w = 60 - x,y = 100 - x,z = 20 + x,and substituting into equation 4: (20 + x) + (60 - x) = 80, which simplifies to 80 = 80, which is always true, so we need another approach.Wait, maybe we can use the fact that the inradius is related to the area and the semiperimeter, which we have already done.But how does that help us find x and y?Wait, perhaps I can use the formula for the area in terms of the inradius and the sides.Wait, in a tangential quadrilateral, the area is also equal to r * s, which we have as 2400 * sqrt(10) = r * 180, so r = (2400 * sqrt(10))/180 = (40/3) * sqrt(10).But how does that help us find x and y?Wait, maybe I can use the fact that the inradius is related to the distances from the center to the sides, but I'm not sure.Alternatively, perhaps I can use the formula for the lengths of the tangent segments in terms of the sides and the semiperimeter.Wait, in a tangential quadrilateral, the lengths of the tangent segments from each vertex are given by:w = (AB + AD - BC)/2,x = (AB + BC - AD)/2,y = (BC + CD - AB)/2,z = (CD + DA - BC)/2.Wait, let me check that.Yes, in a tangential quadrilateral, the lengths of the tangent segments can be calculated as:w = (AB + AD - BC)/2,x = (AB + BC - AD)/2,y = (BC + CD - AB)/2,z = (CD + DA - BC)/2.So, let's compute these:w = (60 + 80 - 100)/2 = (140 - 100)/2 = 40/2 = 20,x = (60 + 100 - 80)/2 = (160 - 80)/2 = 80/2 = 40,y = (100 + 120 - 60)/2 = (220 - 60)/2 = 160/2 = 80,z = (120 + 80 - 100)/2 = (200 - 100)/2 = 100/2 = 50.Wait, let me check these calculations:w = (60 + 80 - 100)/2 = (140 - 100)/2 = 40/2 = 20,x = (60 + 100 - 80)/2 = (160 - 80)/2 = 80/2 = 40,y = (100 + 120 - 60)/2 = (220 - 60)/2 = 160/2 = 80,z = (120 + 80 - 100)/2 = (200 - 100)/2 = 100/2 = 50.So, w = 20, x = 40, y = 80, z = 50.Wait, but let me check if these satisfy the original equations:w + x = 20 + 40 = 60 ✔️,x + y = 40 + 80 = 120 ✔️,y + z = 80 + 50 = 130 ≠ 120 ❌,Wait, that's a problem. Because CD is 120, so y + z should be 120, but according to this, y + z = 130, which is incorrect.Wait, so I must have made a mistake in the formula.Wait, let me double-check the formula for the tangent lengths.I think the correct formula is:w = (AB + AD - BC)/2,x = (AB + BC - AD)/2,y = (BC + CD - AB)/2,z = (CD + DA - BC)/2.Wait, let me compute y again:y = (BC + CD - AB)/2 = (100 + 120 - 60)/2 = (220 - 60)/2 = 160/2 = 80.z = (CD + DA - BC)/2 = (120 + 80 - 100)/2 = (200 - 100)/2 = 100/2 = 50.So, y + z = 80 + 50 = 130, but CD is 120, which is a contradiction.Therefore, my formula must be incorrect.Wait, maybe I got the formula wrong. Let me look it up in my mind.In a tangential quadrilateral, the lengths of the tangent segments from each vertex are equal to (s - a), where a is the opposite side.Wait, no, that's not correct. Wait, in a tangential quadrilateral, the lengths of the tangent segments from each vertex are equal to (s - a), where a is the adjacent side.Wait, no, let me think again.Actually, in a tangential quadrilateral, the lengths of the tangent segments from each vertex are equal to (s - a), where a is the opposite side.Wait, let me check that.Wait, if we denote the tangent lengths as follows:From vertex A: w,From vertex B: x,From vertex C: y,From vertex D: z.Then, we have:AB = w + x,BC = x + y,CD = y + z,DA = z + w.So, solving for w, x, y, z in terms of s and the sides.We can write:w = (AB + DA - BC)/2,x = (AB + BC - DA)/2,y = (BC + CD - AB)/2,z = (CD + DA - BC)/2.Wait, let me compute y again:y = (BC + CD - AB)/2 = (100 + 120 - 60)/2 = (220 - 60)/2 = 160/2 = 80,z = (CD + DA - BC)/2 = (120 + 80 - 100)/2 = (200 - 100)/2 = 100/2 = 50.So, y + z = 80 + 50 = 130, but CD is 120, which is a problem.Wait, so perhaps the formula is different.Wait, maybe I should use the formula where each tangent length is (s - a), where a is the side opposite.Wait, let me think. In a tangential quadrilateral, the tangent lengths can be expressed as:w = s - AB,x = s - BC,y = s - CD,z = s - DA.Wait, let me test that.s = 180,w = 180 - 60 = 120,x = 180 - 100 = 80,y = 180 - 120 = 60,z = 180 - 80 = 100.But then, AB = w + x = 120 + 80 = 200 ≠ 60 ❌,So that's not correct.Wait, maybe it's the other way around: each tangent length is (s - opposite side).Wait, let me try:w = s - CD = 180 - 120 = 60,x = s - DA = 180 - 80 = 100,y = s - AB = 180 - 60 = 120,z = s - BC = 180 - 100 = 80.Then, AB = w + x = 60 + 100 = 160 ≠ 60 ❌,No, that's not correct either.Wait, perhaps I need to think differently.Wait, in a tangential quadrilateral, the sum of the two opposite sides are equal, which we have already confirmed: AB + CD = BC + DA = 180.So, the semiperimeter s = 180.In a tangential quadrilateral, the lengths of the tangent segments can be found as:w = (AB + AD - BC)/2,x = (AB + BC - AD)/2,y = (BC + CD - AB)/2,z = (CD + DA - BC)/2.Wait, let me compute these again:w = (60 + 80 - 100)/2 = (140 - 100)/2 = 40/2 = 20,x = (60 + 100 - 80)/2 = (160 - 80)/2 = 80/2 = 40,y = (100 + 120 - 60)/2 = (220 - 60)/2 = 160/2 = 80,z = (120 + 80 - 100)/2 = (200 - 100)/2 = 100/2 = 50.So, w = 20, x = 40, y = 80, z = 50.Now, let's check the sides:AB = w + x = 20 + 40 = 60 ✔️,BC = x + y = 40 + 80 = 120 ✔️,CD = y + z = 80 + 50 = 130 ≠ 120 ❌,DA = z + w = 50 + 20 = 70 ≠ 80 ❌.Wait, that's not correct either. There's a mistake here.Wait, perhaps the formula is different. Maybe it's:w = (AB + BC - CD)/2,x = (BC + CD - DA)/2,y = (CD + DA - AB)/2,z = (DA + AB - BC)/2.Wait, let me try that.w = (60 + 100 - 120)/2 = (160 - 120)/2 = 40/2 = 20,x = (100 + 120 - 80)/2 = (220 - 80)/2 = 140/2 = 70,y = (120 + 80 - 60)/2 = (200 - 60)/2 = 140/2 = 70,z = (80 + 60 - 100)/2 = (140 - 100)/2 = 40/2 = 20.Now, let's check the sides:AB = w + x = 20 + 70 = 90 ≠ 60 ❌,So that's not correct either.Wait, I'm getting confused. Maybe I should approach this differently.Since the quadrilateral is both cyclic and tangential, it's bicentric. There's a formula that relates the inradius, the circumradius, and the distance between the centers, but I don't think that's helpful here.Alternatively, perhaps I can use the fact that in a bicentric quadrilateral, the sum of the squares of the sides is equal to 8r^2 + 4R^2, but I'm not sure.Wait, maybe I can use the formula for the area in two ways: Brahmagupta's formula and r*s.We have K = 2400*sqrt(10) = r*180,So, r = (2400*sqrt(10))/180 = (40/3)*sqrt(10).Now, how does this relate to the segments x and y on the side of length 120.Wait, in a tangential quadrilateral, the lengths of the tangent segments from each vertex are equal. So, on side CD of length 120, the point of tangency divides it into two segments, let's say, from C to the point is y, and from D to the point is z.Wait, earlier, I denoted y and z as the tangent lengths from C and D, respectively.Wait, but in the initial setup, I had:AB = w + x = 60,BC = x + y = 100,CD = y + z = 120,DA = z + w = 80.So, in this case, the side CD is divided into y and z, where y is adjacent to BC and z is adjacent to DA.So, the segments are y and z, with y + z = 120.We need to find |y - z|.Wait, but earlier, when I tried to compute y and z, I got y = 80 and z = 50, which would make y + z = 130, which is incorrect because CD is 120.So, that approach must be wrong.Wait, perhaps I need to use the inradius and the formula for the area in terms of the inradius and the sides.Wait, the area can also be expressed as the sum of the areas of the four triangles formed by the inradius and the sides.Each triangle has a height of r, and the base as the side length.So, the area K = r*(AB + BC + CD + DA)/2 = r*s.Which is consistent with K = r*s.But how does that help me find y and z?Wait, perhaps I can use the formula for the area in terms of the tangent segments.Wait, in a tangential quadrilateral, the area can also be expressed as:K = sqrt[(w*x*y*z)].Wait, no, that's not correct. Wait, in a tangential quadrilateral, the area is K = r*s, and in a bicentric quadrilateral, there's a relation between r, R, and the distance between centers, but I don't think that's helpful here.Wait, perhaps I can use the formula for the tangent segments in terms of the inradius and the angles.Wait, in a tangential polygon, the length of the tangent from a vertex is equal to (r)/(tan(theta/2)), where theta is the angle at that vertex.But since the quadrilateral is cyclic, the sum of opposite angles is 180 degrees.Wait, maybe I can use that.Let me denote the angles at vertices A, B, C, D as α, β, γ, δ respectively.Since it's cyclic, α + γ = 180°, and β + δ = 180°.In a tangential quadrilateral, the sum of the two opposite angles is 180°, which is already satisfied because it's cyclic.Wait, but in a tangential quadrilateral, the sum of the two opposite angles is not necessarily 180°, unless it's also cyclic.Wait, actually, in a tangential quadrilateral, the sum of the two opposite angles is not necessarily 180°, but in a cyclic quadrilateral, it is. So, in a bicentric quadrilateral, both properties hold.So, perhaps I can use the formula for the tangent lengths in terms of the angles.In a tangential quadrilateral, the tangent lengths can be expressed as:w = r / tan(α/2),x = r / tan(β/2),y = r / tan(γ/2),z = r / tan(δ/2).Since the quadrilateral is cyclic, we have α + γ = 180°, and β + δ = 180°.So, tan(γ/2) = tan((180° - α)/2) = tan(90° - α/2) = cot(α/2).Similarly, tan(δ/2) = tan((180° - β)/2) = tan(90° - β/2) = cot(β/2).So, y = r / tan(γ/2) = r / cot(α/2) = r * tan(α/2),Similarly, z = r / tan(δ/2) = r * tan(β/2).Wait, but we also have:From the tangent lengths:w = r / tan(α/2),x = r / tan(β/2),y = r * tan(α/2),z = r * tan(β/2).So, we can write:w = r / tan(α/2),y = r * tan(α/2),Similarly,x = r / tan(β/2),z = r * tan(β/2).So, from these, we can see that:w * y = r^2,x * z = r^2.So, w * y = x * z.Given that, let's compute w, x, y, z in terms of r.But we also have the equations:w + x = 60,x + y = 100,y + z = 120,z + w = 80.And we have w * y = x * z = r^2.Given that, let's denote:From w + x = 60,and z + w = 80,Subtracting these two equations:(z + w) - (w + x) = 80 - 60,z - x = 20,So, z = x + 20.Similarly, from x + y = 100,and y + z = 120,Subtracting these two equations:(y + z) - (x + y) = 120 - 100,z - x = 20,Which is consistent with z = x + 20.So, z = x + 20.Now, from w + x = 60,w = 60 - x.From z + w = 80,z + w = (x + 20) + (60 - x) = 80,Which simplifies to 80 = 80, which is always true.Now, from x + y = 100,y = 100 - x.From y + z = 120,(100 - x) + (x + 20) = 120,Which simplifies to 120 = 120, which is always true.So, we have:w = 60 - x,y = 100 - x,z = x + 20.Now, we also have w * y = x * z.So,(60 - x) * (100 - x) = x * (x + 20).Let me expand the left side:(60 - x)(100 - x) = 60*100 - 60x - 100x + x^2 = 6000 - 160x + x^2.The right side is:x(x + 20) = x^2 + 20x.So, setting them equal:6000 - 160x + x^2 = x^2 + 20x.Subtract x^2 from both sides:6000 - 160x = 20x.Add 160x to both sides:6000 = 180x.So, x = 6000 / 180 = 100/3 ≈ 33.3333.So, x = 100/3.Now, let's find the other variables:w = 60 - x = 60 - 100/3 = (180 - 100)/3 = 80/3 ≈ 26.6667,y = 100 - x = 100 - 100/3 = (300 - 100)/3 = 200/3 ≈ 66.6667,z = x + 20 = 100/3 + 20 = 100/3 + 60/3 = 160/3 ≈ 53.3333.Now, let's check if w * y = x * z:w * y = (80/3) * (200/3) = 16000/9 ≈ 1777.7778,x * z = (100/3) * (160/3) = 16000/9 ≈ 1777.7778,So, that's correct.Now, we have the tangent lengths:w = 80/3,x = 100/3,y = 200/3,z = 160/3.Now, the side CD is divided into y and z, which are 200/3 and 160/3.So, the segments are y = 200/3 and z = 160/3.Therefore, the absolute difference |y - z| = |200/3 - 160/3| = |40/3| = 40/3 ≈ 13.3333.Wait, but the problem asks for |x - y|, where x and y are the segments on the side of length 120.Wait, in my notation, y and z are the segments on CD, which is 120. So, y = 200/3 ≈ 66.6667 and z = 160/3 ≈ 53.3333.So, |y - z| = |200/3 - 160/3| = 40/3 ≈ 13.3333.But the problem asks for |x - y|, but in my notation, x is a tangent length on side AB, not on CD.Wait, perhaps I misread the problem. Let me check.The problem says: "The point of tangency of the inscribed circle to the side of length 120 divides that side into segments of length x and y. Find |x - y|."So, in this case, the side of length 120 is CD, which is divided into segments y and z, which are 200/3 and 160/3.So, the segments are y = 200/3 and z = 160/3.Thus, |y - z| = |200/3 - 160/3| = 40/3.But 40/3 is approximately 13.3333, which is 13 and 1/3.Wait, but in the initial problem, the user wrote:"Find |x-y|."But in my solution, I found |y - z| = 40/3.Wait, perhaps in the problem, x and y are the two segments on the side of length 120, so in my notation, x and y would be y and z.So, the problem is asking for |x - y|, where x and y are the two segments on the side of length 120, which are y = 200/3 and z = 160/3.So, |x - y| = |200/3 - 160/3| = 40/3.But 40/3 is approximately 13.3333, which is 13 and 1/3.Wait, but in the initial problem, the user wrote:"Find |x-y|."But in my solution, I found |y - z| = 40/3.Wait, perhaps the problem is asking for |x - y| where x and y are the two segments on the side of length 120, which are y and z in my notation.So, the answer is 40/3, which is approximately 13.3333.But the initial problem's solution that the user provided had an answer of 400/3, which is approximately 133.3333, which seems too large.Wait, perhaps I made a mistake in the calculation.Wait, let me go back.We have:From the equations:w + x = 60,x + y = 100,y + z = 120,z + w = 80.We found:w = 80/3,x = 100/3,y = 200/3,z = 160/3.So, on side CD, which is 120, the segments are y = 200/3 and z = 160/3.So, |y - z| = |200/3 - 160/3| = 40/3.So, the answer should be 40/3, which is approximately 13.3333.But the initial problem's solution had 400/3, which is much larger.Wait, perhaps I made a mistake in the earlier steps.Wait, let me check the calculation where I solved for x.We had:(60 - x)(100 - x) = x(x + 20),Which expanded to:6000 - 160x + x^2 = x^2 + 20x,Subtracting x^2 from both sides:6000 - 160x = 20x,Adding 160x to both sides:6000 = 180x,So, x = 6000 / 180 = 100/3 ≈ 33.3333.That seems correct.Then, w = 60 - x = 60 - 100/3 = 80/3 ≈ 26.6667,y = 100 - x = 100 - 100/3 = 200/3 ≈ 66.6667,z = x + 20 = 100/3 + 20 = 160/3 ≈ 53.3333.So, on side CD, the segments are y = 200/3 and z = 160/3,Thus, |y - z| = 40/3.So, the answer should be 40/3, which is approximately 13.3333.But in the initial problem's solution, the user wrote:"Let CR = x and RD = y. Since Q and R are tangent points, CQ = CR = x and DR = DS = y. Also, x + y = 120. The inradius r forms right angles with the sides, hence x = y + 400/3. Solving x + y = 120 and x = y + 400/3 simultaneously, substitute y = x - 400/3:x + x - 400/3 = 120,2x = 120 + 400/3,2x = 760/3,x = 380/3,y = 380/3 - 400/3 = -20/3.The absolute difference:|x - y| = |380/3 - (-20/3)| = 400/3 ≈ 133.3333."Wait, that seems incorrect because it leads to a negative y, which is impossible.Moreover, the approach seems flawed because the inradius doesn't directly relate to the segments in that way.So, perhaps the initial solution was incorrect, and the correct answer is 40/3.But let me double-check.Wait, in the initial problem, the user wrote:"Let CR = x and RD = y. Since Q and R are tangent points, CQ = CR = x and DR = DS = y. Also, x + y = 120. The inradius r forms right angles with the sides, hence x = y + 400/3."Wait, that seems incorrect because the inradius is perpendicular to the sides, but it doesn't directly translate to x = y + r or something like that.Wait, perhaps the user confused the inradius with some other length.In reality, the inradius is related to the area and the semiperimeter, but not directly to the segments x and y in that way.So, I think the correct answer is 40/3, which is approximately 13.3333.But let me check with another approach.Since we have the tangent lengths w, x, y, z as 80/3, 100/3, 200/3, 160/3,Then, on side CD, which is 120, the segments are y = 200/3 and z = 160/3,So, |y - z| = 40/3.Therefore, the answer is 40/3.But the initial problem's solution had 400/3, which is 133.3333, which is much larger than the side length of 120, so that can't be correct.Therefore, I think the correct answer is 40/3, which is approximately 13.3333.So, the absolute difference is 40/3.But let me check once more.We have:w = 80/3 ≈ 26.6667,x = 100/3 ≈ 33.3333,y = 200/3 ≈ 66.6667,z = 160/3 ≈ 53.3333.So, on side CD, which is 120, the segments are y = 200/3 ≈ 66.6667 and z = 160/3 ≈ 53.3333,Thus, |y - z| = 40/3 ≈ 13.3333.Yes, that seems correct.Therefore, the answer is 40/3, which is approximately 13.3333.But the problem asks for |x - y|, where x and y are the segments on the side of length 120.So, if x = 200/3 and y = 160/3, then |x - y| = 40/3.Alternatively, if x and y are the two segments, regardless of order, the absolute difference is 40/3.So, the answer is 40/3.But let me check if 40/3 is correct.Wait, 40/3 is approximately 13.3333, which is less than the side length of 120, which makes sense.Therefore, I think the correct answer is 40/3.But the initial problem's solution had 400/3, which is incorrect.So, in conclusion, the absolute difference |x - y| is 40/3.

answer：Alright, so I've got this problem about sequences and differences. It's divided into three parts, and I need to tackle each one step by step. Let me start with the first part.**Problem 1:** If ( a_n = n^2 - n ), determine whether ( {triangle a_n} ) is an arithmetic sequence, and explain the reason.Okay, so I remember that the first-order difference sequence ( triangle a_n ) is defined as ( a_{n+1} - a_n ). So, I need to compute this difference and see if it forms an arithmetic sequence.Let me compute ( triangle a_n ):( triangle a_n = a_{n+1} - a_n = ( (n+1)^2 - (n+1) ) - (n^2 - n) )Let me expand ( (n+1)^2 ):( (n+1)^2 = n^2 + 2n + 1 )So, substituting back:( triangle a_n = (n^2 + 2n + 1 - n - 1) - (n^2 - n) )Simplify inside the first parentheses:( n^2 + 2n + 1 - n - 1 = n^2 + n )So now:( triangle a_n = (n^2 + n) - (n^2 - n) = n^2 + n - n^2 + n = 2n )So, ( triangle a_n = 2n ). Now, is this an arithmetic sequence?An arithmetic sequence has a constant difference between consecutive terms. Let me check the difference between ( triangle a_{n+1} ) and ( triangle a_n ):( triangle a_{n+1} - triangle a_n = 2(n+1) - 2n = 2n + 2 - 2n = 2 )Since the difference is constant (2), ( {triangle a_n} ) is indeed an arithmetic sequence with the first term ( triangle a_1 = 2*1 = 2 ) and common difference 2.Wait, hold on, let me check ( triangle a_1 ):( triangle a_1 = a_2 - a_1 = (2^2 - 2) - (1^2 - 1) = (4 - 2) - (1 - 1) = 2 - 0 = 2 )Yes, so the first term is 2, and the common difference is 2. So, it's an arithmetic sequence.Okay, that seems straightforward.**Problem 2:** If ( a_1 = 1 ) and ( triangle a_n - a_n = 2^n ), find the general formula for the sequence ( {a_n} ).Alright, so ( triangle a_n = a_{n+1} - a_n ), and we have ( triangle a_n - a_n = 2^n ).So, substituting ( triangle a_n ):( a_{n+1} - a_n - a_n = 2^n )Simplify:( a_{n+1} - 2a_n = 2^n )So, we have a recurrence relation:( a_{n+1} = 2a_n + 2^n )This is a linear nonhomogeneous recurrence relation. I think I can solve this using the method of solving linear recursions.First, let's write the homogeneous part:( a_{n+1} - 2a_n = 0 )The characteristic equation is ( r - 2 = 0 ), so ( r = 2 ). Therefore, the general solution to the homogeneous equation is ( a_n^{(h)} = C cdot 2^n ).Now, we need a particular solution ( a_n^{(p)} ) to the nonhomogeneous equation.The nonhomogeneous term is ( 2^n ). Since the homogeneous solution already includes ( 2^n ), we need to multiply by n to find a particular solution.So, let's assume ( a_n^{(p)} = D cdot n cdot 2^n ).Let's plug this into the recurrence:( a_{n+1}^{(p)} = 2a_n^{(p)} + 2^n )Compute ( a_{n+1}^{(p)} = D cdot (n+1) cdot 2^{n+1} = D cdot (n+1) cdot 2 cdot 2^n = 2D(n+1)2^n )Compute ( 2a_n^{(p)} = 2 cdot D cdot n cdot 2^n = 2Dn cdot 2^n )Now, plug into the equation:( 2D(n+1)2^n = 2Dn2^n + 2^n )Simplify:Left side: ( 2D(n+1)2^n )Right side: ( 2Dn2^n + 2^n )Subtract right side from left side:( 2D(n+1)2^n - 2Dn2^n - 2^n = 0 )Factor out ( 2^n ):( 2^n [2D(n+1) - 2Dn - 1] = 0 )Simplify inside the brackets:( 2Dn + 2D - 2Dn - 1 = 2D - 1 )Set equal to zero:( 2D - 1 = 0 ) => ( D = 1/2 )So, the particular solution is ( a_n^{(p)} = frac{1}{2} n 2^n = n 2^{n-1} )Therefore, the general solution is:( a_n = a_n^{(h)} + a_n^{(p)} = C cdot 2^n + n 2^{n-1} )Now, apply the initial condition ( a_1 = 1 ).Compute ( a_1 ):( a_1 = C cdot 2^1 + 1 cdot 2^{0} = 2C + 1 )Set equal to 1:( 2C + 1 = 1 ) => ( 2C = 0 ) => ( C = 0 )So, the general formula is:( a_n = n 2^{n-1} )Let me verify this with n=1:( a_1 = 1 * 2^{0} = 1 ) which matches.n=2:( a_2 = 2 * 2^{1} = 4 )Compute using the recurrence:( a_2 = 2a_1 + 2^1 = 2*1 + 2 = 4 ) which matches.n=3:( a_3 = 3 * 2^{2} = 12 )Using recurrence:( a_3 = 2a_2 + 2^2 = 2*4 + 4 = 12 ) which matches.Good, seems correct.**Problem 3:** For the sequence ( {a_n} ) in question (2), determine if there exists an arithmetic sequence ( {b_n} ) such that ( b_1 C_n^1 + b_2 C_n^2 + ldots + b_n C_n^n = a_n ) for all ( n in mathbb{N}^* ), and if so, find the general formula for the sequence ( {b_n} ). If not, please explain the reason.Hmm, okay. So, we have ( a_n = n 2^{n-1} ), and we need to find an arithmetic sequence ( {b_n} ) such that the sum ( sum_{k=1}^n b_k C_n^k = a_n ) for all n.First, let's recall that ( C_n^k ) is the binomial coefficient, which is ( binom{n}{k} ).So, the equation is:( sum_{k=1}^n b_k binom{n}{k} = n 2^{n-1} )We need to find an arithmetic sequence ( {b_n} ) such that this holds for all n.An arithmetic sequence has the form ( b_k = b_1 + (k-1)d ), where d is the common difference.But maybe there's a simpler form. Let me think.Alternatively, perhaps ( b_k ) is linear in k, since it's an arithmetic sequence. So, ( b_k = A k + B ). But since it's an arithmetic sequence, it's actually ( b_k = b_1 + (k-1)d ), which is linear in k.But let's see if we can find such coefficients.Let me consider the sum ( sum_{k=1}^n b_k binom{n}{k} ).Given that ( sum_{k=0}^n binom{n}{k} x^k = (1 + x)^n ). Maybe generating functions can help here.But we have ( sum_{k=1}^n b_k binom{n}{k} = n 2^{n-1} ).Let me think about generating functions.Suppose we consider the generating function for ( b_k ). Let me denote ( B(x) = sum_{k=0}^infty b_k x^k ).But since ( b_k ) is an arithmetic sequence, it's linear in k, so ( b_k = A k + B ). Wait, but in our case, ( b_k ) is an arithmetic sequence, so ( b_k = b_1 + (k-1)d ).But let's see if we can express the sum in terms of generating functions.We have:( sum_{k=1}^n b_k binom{n}{k} = n 2^{n-1} )Let me denote ( S(n) = sum_{k=1}^n b_k binom{n}{k} )We need ( S(n) = n 2^{n-1} )I recall that ( sum_{k=0}^n binom{n}{k} x^k = (1 + x)^n )Also, ( sum_{k=0}^n k binom{n}{k} x^{k-1} = n(1 + x)^{n-1} )Wait, that's a standard identity.Let me see:Differentiating both sides of ( (1 + x)^n = sum_{k=0}^n binom{n}{k} x^k ) with respect to x:( n(1 + x)^{n-1} = sum_{k=1}^n k binom{n}{k} x^{k-1} )Multiply both sides by x:( n x (1 + x)^{n-1} = sum_{k=1}^n k binom{n}{k} x^{k} )So, ( sum_{k=1}^n k binom{n}{k} x^{k} = n x (1 + x)^{n-1} )Interesting. Now, in our case, we have ( sum_{k=1}^n b_k binom{n}{k} ). If ( b_k ) is linear in k, say ( b_k = A k + B ), then:( sum_{k=1}^n (A k + B) binom{n}{k} = A sum_{k=1}^n k binom{n}{k} + B sum_{k=1}^n binom{n}{k} )We know that:( sum_{k=1}^n binom{n}{k} = 2^n - 1 )And:( sum_{k=1}^n k binom{n}{k} = n 2^{n-1} )So, substituting:( A (n 2^{n-1}) + B (2^n - 1) = n 2^{n-1} )We need this to hold for all n. So, let's write:( A n 2^{n-1} + B (2^n - 1) = n 2^{n-1} )Let me factor out ( 2^{n-1} ):( 2^{n-1} (A n + 2 B) - B = n 2^{n-1} )So, equate coefficients:For the ( 2^{n-1} ) term:( A n + 2 B = n )For the constant term:( -B = 0 )From the constant term: ( -B = 0 ) => ( B = 0 )Substitute B=0 into the first equation:( A n = n ) => ( A = 1 )So, ( b_k = A k + B = 1 * k + 0 = k )Therefore, the arithmetic sequence ( {b_n} ) is ( b_k = k )Let me verify this.Given ( b_k = k ), then:( sum_{k=1}^n k binom{n}{k} = n 2^{n-1} )Which is exactly the identity we used earlier. So, yes, it holds.Therefore, such an arithmetic sequence exists, and it's ( b_k = k )So, summarizing:1. ( triangle a_n = 2n ), which is an arithmetic sequence with first term 2 and common difference 2.2. The general formula for ( a_n ) is ( n 2^{n-1} )3. The arithmetic sequence ( {b_n} ) exists and is given by ( b_n = n )