answer：Okay, so I have this equation to solve:[frac{1}{(x - 3)(x - 4)} + frac{1}{(x - 4)(x - 5)} + frac{1}{(x - 5)(x - 6)} = frac{1}{8}.]Hmm, it's a rational equation with three terms on the left side. Each term is a fraction with a product of two linear factors in the denominator. I remember that sometimes these kinds of fractions can be simplified using partial fractions. Let me try that.Partial fraction decomposition is a method where we can express a complicated rational expression as a sum of simpler fractions. For each term like (frac{1}{(x - a)(x - b)}), we can write it as (frac{A}{x - a} + frac{B}{x - b}). Let me see if I can apply that here.Starting with the first term:[frac{1}{(x - 3)(x - 4)}.]I want to express this as (frac{A}{x - 3} + frac{B}{x - 4}). To find A and B, I can use the cover-up method or solve for them by combining the fractions.Let's set:[frac{1}{(x - 3)(x - 4)} = frac{A}{x - 3} + frac{B}{x - 4}.]Multiplying both sides by ((x - 3)(x - 4)) gives:[1 = A(x - 4) + B(x - 3).]To find A, I can cover up (x - 3) by setting (x = 3):[1 = A(3 - 4) + B(0) implies 1 = A(-1) implies A = -1.]Similarly, to find B, I cover up (x - 4) by setting (x = 4):[1 = A(0) + B(4 - 3) implies 1 = B(1) implies B = 1.]So, the first term becomes:[frac{-1}{x - 3} + frac{1}{x - 4}.]Wait, that's interesting. It's (frac{1}{x - 4} - frac{1}{x - 3}). Let me check the second term similarly.Second term:[frac{1}{(x - 4)(x - 5)}.]Expressing this as (frac{C}{x - 4} + frac{D}{x - 5}):Multiplying both sides by ((x - 4)(x - 5)):[1 = C(x - 5) + D(x - 4).]Covering up (x - 4) by setting (x = 4):[1 = C(4 - 5) + D(0) implies 1 = C(-1) implies C = -1.]Covering up (x - 5) by setting (x = 5):[1 = C(0) + D(5 - 4) implies 1 = D(1) implies D = 1.]So, the second term becomes:[frac{-1}{x - 4} + frac{1}{x - 5} = frac{1}{x - 5} - frac{1}{x - 4}.]Similarly, for the third term:[frac{1}{(x - 5)(x - 6)}.]Expressing as (frac{E}{x - 5} + frac{F}{x - 6}):Multiplying both sides by ((x - 5)(x - 6)):[1 = E(x - 6) + F(x - 5).]Covering up (x - 5) by setting (x = 5):[1 = E(5 - 6) + F(0) implies 1 = E(-1) implies E = -1.]Covering up (x - 6) by setting (x = 6):[1 = E(0) + F(6 - 5) implies 1 = F(1) implies F = 1.]So, the third term becomes:[frac{-1}{x - 5} + frac{1}{x - 6} = frac{1}{x - 6} - frac{1}{x - 5}.]Now, let me substitute all these back into the original equation:[left( frac{1}{x - 4} - frac{1}{x - 3} right) + left( frac{1}{x - 5} - frac{1}{x - 4} right) + left( frac{1}{x - 6} - frac{1}{x - 5} right) = frac{1}{8}.]Let me simplify this by combining like terms. Let's list out all the terms:- (frac{1}{x - 4})- (- frac{1}{x - 3})- (frac{1}{x - 5})- (- frac{1}{x - 4})- (frac{1}{x - 6})- (- frac{1}{x - 5})Looking at these, I can see that some terms will cancel out. Specifically:- The (frac{1}{x - 4}) and (- frac{1}{x - 4}) will cancel each other.- The (frac{1}{x - 5}) and (- frac{1}{x - 5}) will also cancel each other.So, after cancellation, we are left with:[- frac{1}{x - 3} + frac{1}{x - 6} = frac{1}{8}.]So, the equation simplifies to:[frac{1}{x - 6} - frac{1}{x - 3} = frac{1}{8}.]Now, I need to solve this equation. Let me write it again:[frac{1}{x - 6} - frac{1}{x - 3} = frac{1}{8}.]To solve for x, I can combine the two fractions on the left side. To do that, I need a common denominator, which would be ((x - 6)(x - 3)). Let's rewrite each fraction with this common denominator.First term:[frac{1}{x - 6} = frac{(x - 3)}{(x - 6)(x - 3)}.]Second term:[frac{1}{x - 3} = frac{(x - 6)}{(x - 6)(x - 3)}.]So, substituting back into the equation:[frac{(x - 3)}{(x - 6)(x - 3)} - frac{(x - 6)}{(x - 6)(x - 3)} = frac{1}{8}.]Combine the numerators:[frac{(x - 3) - (x - 6)}{(x - 6)(x - 3)} = frac{1}{8}.]Simplify the numerator:[(x - 3) - (x - 6) = x - 3 - x + 6 = (x - x) + (-3 + 6) = 0 + 3 = 3.]So, the equation becomes:[frac{3}{(x - 6)(x - 3)} = frac{1}{8}.]Now, cross-multiplying to solve for x:[3 times 8 = (x - 6)(x - 3).]Calculate the left side:[24 = (x - 6)(x - 3).]Let me expand the right side:[(x - 6)(x - 3) = x^2 - 3x - 6x + 18 = x^2 - 9x + 18.]So, the equation is now:[24 = x^2 - 9x + 18.]Let me bring all terms to one side to form a quadratic equation:[x^2 - 9x + 18 - 24 = 0 implies x^2 - 9x - 6 = 0.]Wait, hold on, that would be:[x^2 - 9x + 18 - 24 = x^2 - 9x - 6 = 0.]Wait, that seems correct. So, the quadratic equation is:[x^2 - 9x - 6 = 0.]Hmm, let me double-check my steps because sometimes I might make a mistake in signs.Starting from:[frac{3}{(x - 6)(x - 3)} = frac{1}{8}.]Cross-multiplying:[3 times 8 = (x - 6)(x - 3) implies 24 = x^2 - 9x + 18.]Yes, that's correct. Then moving 24 to the right:[x^2 - 9x + 18 - 24 = x^2 - 9x - 6 = 0.]Yes, that's correct.Now, to solve the quadratic equation (x^2 - 9x - 6 = 0), I can use the quadratic formula:[x = frac{-b pm sqrt{b^2 - 4ac}}{2a},]where (a = 1), (b = -9), and (c = -6).Plugging in the values:[x = frac{-(-9) pm sqrt{(-9)^2 - 4 times 1 times (-6)}}{2 times 1} = frac{9 pm sqrt{81 + 24}}{2} = frac{9 pm sqrt{105}}{2}.]Wait, hold on, because (b^2 - 4ac) is:[(-9)^2 - 4(1)(-6) = 81 + 24 = 105.]So, the solutions are:[x = frac{9 pm sqrt{105}}{2}.]Wait, but in the initial problem, the denominators are (x - 3), (x - 4), (x - 5), (x - 6). So, x cannot be 3, 4, 5, or 6 because that would make the denominators zero, which is undefined.So, I need to check if the solutions (frac{9 pm sqrt{105}}{2}) are equal to 3, 4, 5, or 6.Let me compute the approximate values of the solutions to check.First, compute (sqrt{105}):[sqrt{100} = 10, sqrt{121} = 11, so sqrt{105} approx 10.24695.]So,[x = frac{9 + 10.24695}{2} approx frac{19.24695}{2} approx 9.623475,]and[x = frac{9 - 10.24695}{2} approx frac{-1.24695}{2} approx -0.623475.]So, approximately, the solutions are 9.6235 and -0.6235.Now, checking if these are equal to 3, 4, 5, or 6.Clearly, 9.6235 is not equal to 3, 4, 5, or 6.Similarly, -0.6235 is not equal to 3, 4, 5, or 6.Therefore, both solutions are valid as they do not make any denominators zero.Wait, but let me check if I made a mistake in the quadratic equation.Wait, earlier, when I simplified:[frac{3}{(x - 6)(x - 3)} = frac{1}{8},]I cross-multiplied to get:[24 = (x - 6)(x - 3).]But expanding ((x - 6)(x - 3)):[x^2 - 3x - 6x + 18 = x^2 - 9x + 18.]So, 24 = x^2 - 9x + 18.Then, moving 24 to the right:[x^2 - 9x + 18 - 24 = x^2 - 9x - 6 = 0.]Yes, that's correct.So, the quadratic equation is correct, and the solutions are (x = frac{9 pm sqrt{105}}{2}).Wait, but in the initial problem, the user had a different quadratic equation, which was (x^2 - 9x + 6 = 0), leading to solutions (frac{9 pm sqrt{57}}{2}). But in my calculation, I got (x^2 - 9x - 6 = 0), leading to (frac{9 pm sqrt{105}}{2}).Hmm, that suggests that perhaps I made a mistake in the earlier steps.Let me go back and check.Starting from:[frac{1}{(x - 3)(x - 4)} + frac{1}{(x - 4)(x - 5)} + frac{1}{(x - 5)(x - 6)} = frac{1}{8}.]I decomposed each term into partial fractions:First term: (frac{1}{x - 4} - frac{1}{x - 3}).Second term: (frac{1}{x - 5} - frac{1}{x - 4}).Third term: (frac{1}{x - 6} - frac{1}{x - 5}).Adding them up:[left( frac{1}{x - 4} - frac{1}{x - 3} right) + left( frac{1}{x - 5} - frac{1}{x - 4} right) + left( frac{1}{x - 6} - frac{1}{x - 5} right).]Combining like terms:- The (frac{1}{x - 4}) and (- frac{1}{x - 4}) cancel.- The (frac{1}{x - 5}) and (- frac{1}{x - 5}) cancel.Left with:[- frac{1}{x - 3} + frac{1}{x - 6} = frac{1}{8}.]So, that's correct.Then, combining the two fractions:[frac{1}{x - 6} - frac{1}{x - 3} = frac{1}{8}.]Expressed as:[frac{(x - 3) - (x - 6)}{(x - 6)(x - 3)} = frac{1}{8}.]Simplify numerator:[(x - 3) - (x - 6) = x - 3 - x + 6 = 3.]So,[frac{3}{(x - 6)(x - 3)} = frac{1}{8}.]Cross-multiplying:[3 times 8 = (x - 6)(x - 3) implies 24 = x^2 - 9x + 18.]So,[x^2 - 9x + 18 - 24 = x^2 - 9x - 6 = 0.]Yes, that's correct. So, the quadratic is (x^2 - 9x - 6 = 0), leading to solutions (frac{9 pm sqrt{105}}{2}).But in the initial problem, the user had a different result. Wait, perhaps I misread the problem.Wait, looking back, the user's problem was:[frac{1}{(x - 3)(x - 4)} + frac{1}{(x - 4)(x - 5)} + frac{1}{(x - 5)(x - 6)} = frac{1}{8}.]And in their solution, they had:After partial fractions, they got:[frac{1}{x - 6} - frac{1}{x - 3} = frac{1}{8}.]Then, multiplying both sides by (8(x - 6)(x - 3)):[8(x - 3) - 8(x - 6) = (x - 6)(x - 3).]Simplifying:[8x - 24 - 8x + 48 = x^2 - 9x + 18.]Which simplifies to:[24 = x^2 - 9x + 18.]Then,[x^2 - 9x - 6 = 0.]Wait, but in the user's solution, they wrote:"Multiplying both sides by 8(x - 6)(x - 3), we get:8(x - 3) - 8(x - 6) = (x - 6)(x - 3).Simplifying,8 × 3 = x² - 9x + 18,x² - 9x + 6 = 0."Wait, hold on, that seems incorrect.Because when they multiplied both sides by 8(x - 6)(x - 3), the left side becomes:8(x - 3) - 8(x - 6).But 8(x - 3) - 8(x - 6) is not 8 × 3.Wait, let's compute:8(x - 3) - 8(x - 6) = 8x - 24 - 8x + 48 = (8x - 8x) + (-24 + 48) = 0 + 24 = 24.So, 24 = (x - 6)(x - 3).Which is the same as I had earlier, leading to x² - 9x + 18 = 24, so x² - 9x - 6 = 0.But in the user's solution, they wrote:"8 × 3 = x² - 9x + 18,x² - 9x + 6 = 0."Wait, that seems incorrect because 8(x - 3) - 8(x - 6) is 24, not 8 × 3.So, the user made a mistake in their calculation, leading to an incorrect quadratic equation.Therefore, the correct quadratic equation is x² - 9x - 6 = 0, leading to solutions (frac{9 pm sqrt{105}}{2}).Wait, but in my earlier steps, I had:After cross-multiplying, I got 24 = (x - 6)(x - 3), which is x² - 9x + 18 = 24, so x² - 9x - 6 = 0.Yes, that's correct.So, the correct solutions are (frac{9 pm sqrt{105}}{2}).But the user's solution had x² - 9x + 6 = 0, leading to (frac{9 pm sqrt{57}}{2}).So, the user made a mistake in their calculation when they wrote:"8 × 3 = x² - 9x + 18,x² - 9x + 6 = 0."That step is incorrect because 8(x - 3) - 8(x - 6) is 24, not 8 × 3.Therefore, the correct quadratic equation is x² - 9x - 6 = 0, leading to solutions (frac{9 pm sqrt{105}}{2}).So, to summarize, the correct solutions are:[x = frac{9 + sqrt{105}}{2} quad text{and} quad x = frac{9 - sqrt{105}}{2}.]These are approximately 9.6235 and -0.6235, which are valid as they do not make any denominators zero.Therefore, the real solutions are (frac{9 + sqrt{105}}{2}) and (frac{9 - sqrt{105}}{2}).

answer：Alright, so I have this problem about an ellipse. Let me try to figure it out step by step. First, I remember that an ellipse has two axes: the major axis and the minor axis. The problem mentions the short axis, which I think is the minor axis. It says the short axis is √5. So, the length of the minor axis is √5. I also recall that the minor axis is related to the semi-minor axis, which is usually denoted as 'b'. So, if the minor axis is √5, then the semi-minor axis 'b' would be half of that, right? So, b = √5 / 2. Let me write that down: b = √5 / 2.Next, the problem gives the eccentricity of the ellipse as 2/3. Eccentricity, denoted as 'e', is a measure of how "stretched" the ellipse is. For an ellipse, the formula for eccentricity is e = c/a, where 'c' is the distance from the center to each focus, and 'a' is the semi-major axis. So, e = 2/3 = c/a.I also remember that in an ellipse, the relationship between a, b, and c is given by the equation a² = b² + c². So, if I can find 'a' and 'c', I can figure out more about the ellipse.Given that e = c/a = 2/3, I can express c as (2/3)a. So, c = (2/3)a.Now, plugging this into the equation a² = b² + c², we get:a² = (√5 / 2)² + ( (2/3)a )²Let me compute each term:(√5 / 2)² = (5) / 4 = 1.25( (2/3)a )² = (4/9)a²So, substituting back into the equation:a² = 1.25 + (4/9)a²Hmm, let's solve for a². I'll subtract (4/9)a² from both sides:a² - (4/9)a² = 1.25This simplifies to:(5/9)a² = 1.25To solve for a², I can multiply both sides by 9/5:a² = 1.25 * (9/5) = (5/4) * (9/5) = 9/4So, a² = 9/4, which means a = 3/2. So, the semi-major axis is 3/2.Now, since a = 3/2, I can find c:c = (2/3)a = (2/3)*(3/2) = 1So, c = 1. Therefore, the distance from the center to each focus is 1.Now, the problem mentions two foci, F₁ and F₂. A line passes through F₁ and intersects the ellipse at points A and B. We need to find the perimeter of triangle ABF₂.Hmm, okay. So, the perimeter of triangle ABF₂ is the sum of the lengths of sides AB, BF₂, and AF₂.Wait, but how do I find these lengths? Maybe there's a property of ellipses that can help here.I remember that one of the defining properties of an ellipse is that the sum of the distances from any point on the ellipse to the two foci is constant and equal to 2a. So, for any point P on the ellipse, PF₁ + PF₂ = 2a.In this case, points A and B are on the ellipse, so for both points, the sum of distances to F₁ and F₂ is 2a.So, for point A: AF₁ + AF₂ = 2aSimilarly, for point B: BF₁ + BF₂ = 2aGiven that the line passes through F₁ and intersects the ellipse at A and B, so F₁ is one of the points on that line, and A and B are the other two intersection points.Wait, actually, if the line passes through F₁, then F₁ is one of the points where the line intersects the ellipse. But the problem says the line intersects the ellipse at points A and B. So, does that mean F₁ is either A or B? Or is F₁ a separate point?Wait, the problem says "a line passing through F₁ intersects the ellipse at points A and B." So, the line passes through F₁ and intersects the ellipse at A and B. So, F₁ is one of the points on the line, and the line intersects the ellipse at A and B. So, is F₁ one of the intersection points? Or is it just that the line passes through F₁ and intersects the ellipse at A and B, meaning F₁ is not necessarily one of the intersection points?Wait, that's confusing. Let me think.If a line passes through F₁ and intersects the ellipse at A and B, then the line passes through F₁ and cuts the ellipse at A and B. So, F₁ is a point on the line, but it's not necessarily one of the intersection points with the ellipse. So, A and B are two distinct points where the line intersects the ellipse, and the line also passes through F₁.But wait, if the line passes through F₁ and intersects the ellipse at A and B, then F₁ must lie on the line, but it's not necessarily on the ellipse. Wait, but F₁ is a focus, which is inside the ellipse, not on it. So, the line passes through F₁ and intersects the ellipse at A and B. So, A and B are two points on the ellipse, and the line connecting them passes through F₁.So, in this case, the line AB passes through F₁, which is inside the ellipse.So, we have points A and B on the ellipse, and the line AB passes through F₁.So, we need to find the perimeter of triangle ABF₂.So, the perimeter is AB + AF₂ + BF₂.Hmm.But how do we find AB, AF₂, and BF₂?Wait, maybe we can use the property of the ellipse that the sum of distances from any point on the ellipse to the two foci is 2a.So, for point A: AF₁ + AF₂ = 2aSimilarly, for point B: BF₁ + BF₂ = 2aSo, AF₂ = 2a - AF₁Similarly, BF₂ = 2a - BF₁So, AF₂ + BF₂ = 2*(2a) - (AF₁ + BF₁)But AF₁ + BF₁ is the sum of distances from A and B to F₁.But since A and B are on the line passing through F₁, maybe AF₁ + BF₁ is equal to AB?Wait, no. If A and B are on a line passing through F₁, then F₁ is somewhere on the line segment AB or extended beyond.Wait, since F₁ is a focus inside the ellipse, and the line passes through F₁ and intersects the ellipse at A and B, then F₁ must lie between A and B on the line.So, the points are arranged as A---F₁---B on the line.Therefore, AF₁ + BF₁ = AB.Wait, no. If F₁ is between A and B, then AF₁ + BF₁ = AB.Wait, let me think.If F₁ is between A and B, then the distance from A to B is AF₁ + F₁B, which is equal to AF₁ + BF₁, since F₁B is the same as BF₁.So, AB = AF₁ + BF₁.Therefore, AF₂ + BF₂ = 2*(2a) - (AF₁ + BF₁) = 4a - ABSo, AF₂ + BF₂ = 4a - ABTherefore, the perimeter of triangle ABF₂ is AB + AF₂ + BF₂ = AB + (4a - AB) = 4aSo, the perimeter is 4a.Wait, that's interesting. So, regardless of where the line passes through F₁, the perimeter of triangle ABF₂ is always 4a.Is that correct? Let me verify.Given that AF₂ + BF₂ = 4a - AB, so when we add AB, we get 4a.Yes, that seems to make sense.Therefore, the perimeter is 4a.We already found that a = 3/2, so 4a = 4*(3/2) = 6.Therefore, the perimeter is 6.So, the answer should be 6, which is option C.Wait, let me double-check.We have:Perimeter = AB + AF₂ + BF₂But AF₂ = 2a - AF₁BF₂ = 2a - BF₁So, AF₂ + BF₂ = 4a - (AF₁ + BF₁)But AF₁ + BF₁ = AB, since F₁ is between A and B.Therefore, AF₂ + BF₂ = 4a - ABTherefore, perimeter = AB + (4a - AB) = 4aSo, yes, perimeter is 4a.Given a = 3/2, 4a = 6.Therefore, the perimeter is 6.So, the answer is C: 6.**Final Answer**boxed{C}

answer：First, I know that Mitchell has 30 pencils.Since Mitchell has 6 more pencils than Antonio, I can find out how many pencils Antonio has by subtracting 6 from Mitchell's total.So, Antonio has 30 - 6 = 24 pencils.To find out how many pencils they have together, I add Mitchell's pencils to Antonio's pencils: 30 + 24 = 54.Therefore, Mitchell and Antonio have a total of 54 pencils together.

answer：First, I need to recall the formula for calculating the volume of a cuboid. The volume is found by multiplying the area of the base by the height of the cuboid.Next, I'll identify the given values from the problem. The base area is 14 square centimeters, and the height is 13 centimeters.Now, I'll substitute these values into the formula:Volume = Base Area × HeightVolume = 14 cm² × 13 cmFinally, I'll perform the multiplication to find the volume:14 × 13 = 182So, the volume of the cuboid is 182 cubic centimeters.

answer：Okay, so I have this problem about a sequence {a_n} where the product of the first n terms is T_n, and it's given that T_n = 2 - 2a_n for any natural number n. The problem has two parts: first, I need to prove that the sequence {1/T_n} is an arithmetic sequence, and second, I need to find the sum of the first n terms of another sequence {b_n}, where b_n is defined as (1 - a_n)(1 - a_{n+1}).Starting with part (I). I know that T_n is the product of the first n terms of {a_n}, so T_n = a_1 * a_2 * ... * a_n. It's given that T_n = 2 - 2a_n. Hmm, that seems interesting because T_n is expressed in terms of a_n, which is the nth term of the sequence. I wonder if I can find a relationship between T_n and T_{n-1}.Since T_n is the product up to the nth term, T_n = T_{n-1} * a_n. So, substituting the given expression for T_n, we have:T_n = 2 - 2a_n But also, T_n = T_{n-1} * a_nSo, setting these equal: T_{n-1} * a_n = 2 - 2a_nI can solve for a_n here. Let's see:T_{n-1} * a_n + 2a_n = 2 a_n (T_{n-1} + 2) = 2 a_n = 2 / (T_{n-1} + 2)But from the given, T_{n-1} = 2 - 2a_{n-1}. Wait, is that correct? No, actually, T_{n-1} = 2 - 2a_{n-1} only for n-1, but in our case, we have T_{n-1} in terms of a_n. Maybe I need a different approach.Alternatively, let's consider the reciprocal of T_n, which is 1/T_n. The problem wants to show that {1/T_n} is an arithmetic sequence. An arithmetic sequence has a constant difference between consecutive terms. So, if I can show that 1/T_n - 1/T_{n-1} is a constant, then it's arithmetic.Let me try to express 1/T_n in terms of 1/T_{n-1}.From T_n = 2 - 2a_n, and since T_n = T_{n-1} * a_n, we have:T_n = T_{n-1} * a_n => a_n = T_n / T_{n-1}Substituting into T_n = 2 - 2a_n:T_n = 2 - 2*(T_n / T_{n-1})Let me rearrange this equation:T_n = 2 - (2 T_n) / T_{n-1} Bring the term with T_n to the left:T_n + (2 T_n) / T_{n-1} = 2 Factor T_n:T_n (1 + 2 / T_{n-1}) = 2 Divide both sides by (1 + 2 / T_{n-1}):T_n = 2 / (1 + 2 / T_{n-1}) Simplify the denominator:1 + 2 / T_{n-1} = (T_{n-1} + 2) / T_{n-1} So, T_n = 2 / ((T_{n-1} + 2) / T_{n-1}) ) = 2 * (T_{n-1} / (T_{n-1} + 2)) Thus, T_n = (2 T_{n-1}) / (T_{n-1} + 2)Hmm, that's a recursive formula for T_n in terms of T_{n-1}. Maybe taking reciprocals will help.Let me take reciprocals of both sides:1 / T_n = (T_{n-1} + 2) / (2 T_{n-1}) Simplify:1 / T_n = (T_{n-1} + 2) / (2 T_{n-1}) = (1 + 2 / T_{n-1}) / 2 Which is:1 / T_n = (1 / 2) + (1 / T_{n-1})So, rearranged:1 / T_n - 1 / T_{n-1} = 1/2That's exactly the condition for an arithmetic sequence! The difference between consecutive terms is constant and equal to 1/2. So, {1/T_n} is an arithmetic sequence with common difference 1/2.But wait, I should also check the first term to make sure. Let's compute T_1.For n=1, T_1 = a_1. From the given, T_1 = 2 - 2a_1. So:a_1 = 2 - 2a_1 Bring terms together:a_1 + 2a_1 = 2 3a_1 = 2 a_1 = 2/3Therefore, T_1 = a_1 = 2/3, so 1/T_1 = 3/2.So, the first term of the sequence {1/T_n} is 3/2, and each subsequent term increases by 1/2. Therefore, the sequence is indeed arithmetic.Alright, part (I) seems done.Moving on to part (II). We need to find the sum S_n of the first n terms of the sequence {b_n}, where b_n = (1 - a_n)(1 - a_{n+1}).First, let's try to express a_n in terms of n. Since we know that {1/T_n} is an arithmetic sequence, we can find an expression for T_n.From part (I), we have:1/T_n = 3/2 + (n - 1)*(1/2) Simplify:1/T_n = 3/2 + n/2 - 1/2 = (3/2 - 1/2) + n/2 = 1 + n/2 = (n + 2)/2Therefore, T_n = 2 / (n + 2)So, T_n = 2 / (n + 2). But T_n is the product of the first n terms, so T_n = a_1 * a_2 * ... * a_n = 2 / (n + 2)We can also express a_n in terms of T_n and T_{n-1}:a_n = T_n / T_{n-1}Since T_n = 2 / (n + 2) and T_{n-1} = 2 / (n + 1), then:a_n = (2 / (n + 2)) / (2 / (n + 1)) = (n + 1) / (n + 2)So, a_n = (n + 1)/(n + 2)Therefore, 1 - a_n = 1 - (n + 1)/(n + 2) = (n + 2 - n - 1)/(n + 2) = 1/(n + 2)Similarly, 1 - a_{n+1} = 1 - (n + 2)/(n + 3) = 1/(n + 3)Therefore, b_n = (1 - a_n)(1 - a_{n+1}) = [1/(n + 2)] * [1/(n + 3)] = 1 / [(n + 2)(n + 3)]So, b_n = 1 / [(n + 2)(n + 3)]Now, we need to find S_n, the sum of the first n terms of {b_n}.So, S_n = b_1 + b_2 + ... + b_nGiven that b_k = 1 / [(k + 2)(k + 3)], so:S_n = sum_{k=1}^{n} [1 / ((k + 2)(k + 3))]This looks like a telescoping series. Let me try to decompose it using partial fractions.We can write 1 / [(k + 2)(k + 3)] as A / (k + 2) + B / (k + 3)Solving for A and B:1 = A(k + 3) + B(k + 2)Let me set k = -3: 1 = A(0) + B(-1) => 1 = -B => B = -1Set k = -2: 1 = A(1) + B(0) => 1 = A => A = 1So, 1 / [(k + 2)(k + 3)] = 1/(k + 2) - 1/(k + 3)Therefore, S_n = sum_{k=1}^{n} [1/(k + 2) - 1/(k + 3)]Let's write out the terms:For k=1: 1/3 - 1/4 For k=2: 1/4 - 1/5 For k=3: 1/5 - 1/6 ... For k=n: 1/(n + 2) - 1/(n + 3)Adding these up, most terms cancel out:S_n = (1/3 - 1/4) + (1/4 - 1/5) + (1/5 - 1/6) + ... + (1/(n + 2) - 1/(n + 3))All the intermediate terms cancel, leaving:S_n = 1/3 - 1/(n + 3)Simplify:S_n = (n + 3 - 3) / [3(n + 3)] = n / [3(n + 3)]So, S_n = n / (3n + 9)Therefore, the sum of the first n terms of {b_n} is n / (3n + 9).Let me double-check my steps to make sure I didn't make any mistakes.1. Found that {1/T_n} is arithmetic with common difference 1/2 and first term 3/2. That seems correct.2. Expressed T_n as 2 / (n + 2). Then a_n = T_n / T_{n-1} = (n + 1)/(n + 2). That also seems correct.3. Calculated 1 - a_n = 1/(n + 2) and 1 - a_{n+1} = 1/(n + 3). So, b_n = 1 / [(n + 2)(n + 3)]. Correct.4. Decomposed b_n into partial fractions: 1/(k + 2) - 1/(k + 3). That's standard for telescoping series.5. Summed the series, saw cancellation, ended up with S_n = 1/3 - 1/(n + 3) = n / (3n + 9). That seems correct.I think all steps are valid. So, the final answer for part (II) is S_n = n / (3n + 9).**Final Answer**The sum of the first ( n ) terms of the sequence ({b_n}) is (boxed{dfrac{n}{3n + 9}}).

answer：First, I need to determine the daily wages of a, b, and c based on the given ratio and the known wage of c.Given that the daily wages are in the ratio 3:4:5 and c's daily wage is 105, I can set up the equation 5k = 105 to find the value of k.Solving for k, I get k = 21.Using this value of k, I can find the daily wages of a and b:- a's daily wage = 3k = 63- b's daily wage = 4k = 84Next, I calculate the total earnings contributed by b and c:- b's total earning = 84 * 9 = 756- c's total earning = 105 * 4 = 420Adding these together, the combined earning of b and c is 756 + 420 = 1176.Subtracting this from the total earning of 1554, I find the earning contributed by a:1554 - 1176 = 378.Finally, to find the number of days a worked, I divide a's total earning by a's daily wage:378 / 63 = 6.Therefore, a worked for 6 days.