💡First, I recognize that the function ( f ) is defined such that ( f(cos x) = cos 2x ). To find ( f(sin 15^circ) ), I need to express ( sin 15^circ ) in terms of a cosine function.Using the co-function identity, I know that ( sin theta = cos(90^circ - theta) ). Applying this identity to ( sin 15^circ ), I get:[sin 15^circ = cos(90^circ - 15^circ) = cos 75^circ]Now, I can substitute ( cos 75^circ ) into the function ( f ):[f(sin 15^circ) = f(cos 75^circ)]According to the given function definition, ( f(cos x) = cos 2x ). Therefore:[f(cos 75^circ) = cos(2 times 75^circ) = cos 150^circ]Next, I need to evaluate ( cos 150^circ ). Since ( 150^circ ) is in the second quadrant where cosine values are negative, and knowing that ( cos 150^circ = -cos 30^circ ), I can use the known value of ( cos 30^circ ):[cos 150^circ = -cos 30^circ = -frac{sqrt{3}}{2}]Therefore, the value of ( f(sin 15^circ) ) is ( -frac{sqrt{3}}{2} ).

💡Alright, so I have this problem about a factory with 140 workers. Each worker can make either 25 bolts or 20 nuts per day. The goal is to figure out how many workers should make bolts and how many should make nuts to produce the maximum number of sets, where one set consists of one bolt and two nuts.Okay, first, I need to understand the problem clearly. We have two products: bolts and nuts. Each bolt needs two nuts to make a set. So, the number of sets we can make is limited by whichever product runs out first—the bolts or the nuts. To maximize the number of sets, we need to balance the production of bolts and nuts so that we don't have an excess of one without the other.Let me define some variables to make this clearer. Let’s say:- Let ( x ) be the number of workers assigned to produce bolts.- Let ( y ) be the number of workers assigned to produce nuts.Since the total number of workers is 140, we have the equation:[x + y = 140]Now, each worker producing bolts makes 25 bolts per day, so the total number of bolts produced per day is ( 25x ).Similarly, each worker producing nuts makes 20 nuts per day, so the total number of nuts produced per day is ( 20y ).Since each set requires one bolt and two nuts, the number of sets we can make is determined by the limiting factor. That is, the number of sets can't exceed the number of bolts, and it also can't exceed half the number of nuts (because each set needs two nuts). So, mathematically, the number of sets ( S ) is:[S = min(25x, frac{20y}{2}) = min(25x, 10y)]To maximize ( S ), we need to ensure that ( 25x = 10y ). This way, the number of bolts and the number of nuts are perfectly balanced, and neither is in excess. So, setting ( 25x = 10y ), we can solve for one variable in terms of the other.Let’s solve for ( y ) in terms of ( x ):[25x = 10y Rightarrow y = frac{25}{10}x Rightarrow y = 2.5x]But since the number of workers has to be an integer, ( y ) must be a whole number. This suggests that ( x ) should be chosen such that ( 2.5x ) is also an integer. To avoid dealing with fractions, perhaps I can express this ratio differently.Alternatively, I can express the ratio of workers producing bolts to nuts. From ( y = 2.5x ), we can write:[frac{x}{y} = frac{1}{2.5} = frac{2}{5}]So, the ratio of workers producing bolts to nuts is 2:5. This means for every 2 workers making bolts, there are 5 workers making nuts.Given that the total number of workers is 140, we can set up the ratio as:[x : y = 2 : 5]Let’s let the number of parts be ( k ). Then:[x = 2k y = 5k]And since ( x + y = 140 ):[2k + 5k = 140 7k = 140 k = 20]Therefore:[x = 2k = 40 y = 5k = 100]So, 40 workers should be assigned to produce bolts, and 100 workers should be assigned to produce nuts.Let me double-check this to make sure it makes sense. If 40 workers make bolts, they produce ( 40 times 25 = 1000 ) bolts per day. If 100 workers make nuts, they produce ( 100 times 20 = 2000 ) nuts per day. Since each set requires one bolt and two nuts, the number of sets we can make is:[min(1000, frac{2000}{2}) = min(1000, 1000) = 1000]So, we can produce 1000 sets per day, which seems to be the maximum possible given the constraints.I think this makes sense. By balancing the production of bolts and nuts according to the ratio needed for the sets, we ensure that neither resource is wasted, and we maximize the number of complete sets produced each day.Just to explore another angle, suppose we didn't balance them. For example, if all workers made bolts, we would have ( 140 times 25 = 3500 ) bolts, but only ( 0 ) nuts, so we couldn't make any sets. Similarly, if all workers made nuts, we would have ( 140 times 20 = 2800 ) nuts, which could make ( frac{2800}{2} = 1400 ) sets, but we would have no bolts to match them with, so again, no sets. Therefore, balancing is essential to maximize the number of sets.Another thought: what if we tried a different ratio? Suppose we tried 30 workers on bolts and 110 on nuts. Then, bolts produced would be ( 30 times 25 = 750 ), and nuts would be ( 110 times 20 = 2200 ). The number of sets would be ( min(750, 1100) = 750 ), which is less than 1000. So, clearly, the balanced approach gives a better result.Similarly, if we tried 50 workers on bolts and 90 on nuts, bolts would be ( 50 times 25 = 1250 ), nuts would be ( 90 times 20 = 1800 ), and sets would be ( min(1250, 900) = 900 ), still less than 1000.This further confirms that the balanced ratio of 2:5 for bolts to nuts is the optimal solution.In summary, by setting up the problem with variables, defining the constraints, and ensuring that the production of bolts and nuts is balanced according to the set requirements, we arrive at the optimal number of workers assigned to each task to maximize the number of sets produced daily.

💡Alright, so I'm trying to solve this geometry problem about an isosceles triangle with some angle bisectors and a circle passing through certain points. Let me try to unpack this step by step.First, the problem states: In an isosceles triangle ( ABC ), the bisectors ( AD, BE, CF ) are drawn. We need to find ( BC ), given that ( AB = AC = 1 ), and the vertex ( A ) lies on the circle passing through the points ( D, E, ) and ( F ).Okay, so let's start by visualizing the triangle. Since ( ABC ) is isosceles with ( AB = AC = 1 ), that means ( ABC ) has two equal sides, ( AB ) and ( AC ), so the base is ( BC ). Let me denote ( BC ) as ( x ). So, ( BC = x ), and ( AB = AC = 1 ).Now, angle bisectors ( AD, BE, CF ) are drawn. So, ( AD ) is the bisector of angle ( A ), ( BE ) is the bisector of angle ( B ), and ( CF ) is the bisector of angle ( C ). These bisectors meet the opposite sides, so ( D ) is on ( BC ), ( E ) is on ( AC ), and ( F ) is on ( AB ).The key piece of information here is that vertex ( A ) lies on the circle passing through points ( D, E, ) and ( F ). So, points ( D, E, F ) lie on a circle, and ( A ) is also on this circle. That means ( A ) is concyclic with ( D, E, F ).I think this implies that quadrilateral ( ADEF ) is cyclic. Wait, but ( D, E, F ) are points on the sides of the triangle, so maybe it's triangle ( DEF ) with ( A ) lying on its circumcircle.I need to recall some properties of angle bisectors and cyclic quadrilaterals. Maybe I can use the Angle Bisector Theorem or some properties related to cyclic quadrilaterals.Let me try to sketch the triangle. Let me denote:- ( AB = AC = 1 )- ( BC = x )- ( AD ) is the angle bisector of angle ( A ), so it splits ( BC ) into segments ( BD ) and ( DC ). Since ( ABC ) is isosceles, ( BD = DC = x/2 ).Wait, is that correct? Actually, in an isosceles triangle, the angle bisector from the apex (which is ( A ) here) also serves as the median and the altitude. So, yes, ( AD ) will bisect ( BC ) into two equal parts, each of length ( x/2 ).Similarly, ( BE ) and ( CF ) are angle bisectors. Since ( ABC ) is isosceles, angles at ( B ) and ( C ) are equal. Therefore, the angle bisectors ( BE ) and ( CF ) will have some symmetrical properties.Let me denote the points:- ( D ) is on ( BC ), so ( BD = DC = x/2 ).- ( E ) is on ( AC ), so let's denote ( AE = y ) and ( EC = 1 - y ).- ( F ) is on ( AB ), so let's denote ( AF = z ) and ( FB = 1 - z ).Since ( BE ) and ( CF ) are angle bisectors, by the Angle Bisector Theorem, we can relate the lengths of the sides.For angle bisector ( BE ) in triangle ( ABC ), it divides side ( AC ) into segments proportional to the adjacent sides. So, ( frac{AE}{EC} = frac{AB}{BC} ).Similarly, for angle bisector ( CF ), ( frac{AF}{FB} = frac{AC}{BC} ).Given that ( AB = AC = 1 ) and ( BC = x ), we can write:For ( BE ):( frac{AE}{EC} = frac{AB}{BC} = frac{1}{x} )So, ( frac{y}{1 - y} = frac{1}{x} )Which implies ( y = frac{1}{x + 1} )Similarly, for ( CF ):( frac{AF}{FB} = frac{AC}{BC} = frac{1}{x} )So, ( frac{z}{1 - z} = frac{1}{x} )Which implies ( z = frac{1}{x + 1} )So, both ( E ) and ( F ) divide ( AC ) and ( AB ) respectively in the same ratio ( frac{1}{x} ).Now, since points ( D, E, F ) lie on a circle, and ( A ) also lies on this circle, we can use the property that four points are concyclic if the power of a point with respect to the circle is zero.Alternatively, since ( A ) lies on the circumcircle of triangle ( DEF ), we can use some cyclic quadrilateral properties.Let me recall that if four points are concyclic, then the cross ratio is real, or the power of point ( A ) with respect to the circle through ( D, E, F ) is zero.Alternatively, maybe using the Power of a Point theorem.Wait, the Power of a Point theorem states that for a point ( A ) outside a circle, the product of the lengths from ( A ) to the points of intersection with the circle is equal for any two lines through ( A ). But in this case, ( A ) is on the circle, so the power of ( A ) with respect to the circle is zero.Alternatively, since ( A ) is on the circumcircle of ( DEF ), we can use cyclic quadrilateral properties, such as opposite angles summing to 180 degrees.Alternatively, maybe using coordinates to model the problem.Let me consider setting up coordinate axes to model triangle ( ABC ). Let me place point ( A ) at the origin ( (0, 0) ), and since ( AB = AC = 1 ), let me place ( B ) at ( (b, c) ) and ( C ) at ( (-b, c) ) to maintain the isosceles nature with ( AB = AC ).Wait, maybe it's simpler to place ( A ) at ( (0, h) ), and ( B ) at ( (-d, 0) ), ( C ) at ( (d, 0) ), so that ( BC ) is along the x-axis from ( (-d, 0) ) to ( (d, 0) ), and ( A ) is at ( (0, h) ).Given that ( AB = AC = 1 ), we can compute ( h ) and ( d ) in terms of ( x = BC = 2d ).So, ( AB = sqrt{(d)^2 + (h)^2} = 1 )So, ( d^2 + h^2 = 1 )Since ( BC = 2d = x ), so ( d = x/2 ), and ( h = sqrt{1 - (x/2)^2} )So, coordinates:- ( A = (0, sqrt{1 - (x/2)^2}) )- ( B = (-x/2, 0) )- ( C = (x/2, 0) )Now, we need to find the coordinates of points ( D, E, F ).Point ( D ) is the midpoint of ( BC ), since ( AD ) is the angle bisector and also the median in an isosceles triangle. So, ( D = (0, 0) ).Wait, that's interesting. So, point ( D ) is at the origin.Point ( E ) is on ( AC ), and point ( F ) is on ( AB ). We already found earlier that ( AE = frac{1}{x + 1} ) and ( AF = frac{1}{x + 1} ).So, let's compute the coordinates of ( E ) and ( F ).First, parametrize ( AC ). Point ( A ) is at ( (0, h) ), and point ( C ) is at ( (x/2, 0) ). So, the parametric equation of ( AC ) can be written as:( x(t) = t cdot (x/2) )( y(t) = h - t cdot h )where ( t ) ranges from 0 to 1.Similarly, for point ( E ), which divides ( AC ) in the ratio ( AE : EC = 1 : x ). So, ( t = frac{1}{x + 1} ).Therefore, coordinates of ( E ):( x_E = frac{1}{x + 1} cdot frac{x}{2} = frac{x}{2(x + 1)} )( y_E = h - frac{1}{x + 1} cdot h = h cdot left(1 - frac{1}{x + 1}right) = h cdot frac{x}{x + 1} )Similarly, for point ( F ) on ( AB ). Point ( A ) is at ( (0, h) ), and point ( B ) is at ( (-x/2, 0) ). The parametric equation of ( AB ) is:( x(s) = -s cdot frac{x}{2} )( y(s) = h - s cdot h )where ( s ) ranges from 0 to 1.Since ( AF : FB = 1 : x ), ( s = frac{1}{x + 1} ).Therefore, coordinates of ( F ):( x_F = -frac{1}{x + 1} cdot frac{x}{2} = -frac{x}{2(x + 1)} )( y_F = h - frac{1}{x + 1} cdot h = h cdot frac{x}{x + 1} )So, now we have coordinates for ( D, E, F ):- ( D = (0, 0) )- ( E = left( frac{x}{2(x + 1)}, frac{h x}{x + 1} right) )- ( F = left( -frac{x}{2(x + 1)}, frac{h x}{x + 1} right) )Now, we need to find the equation of the circle passing through ( D, E, F ), and then ensure that point ( A ) lies on this circle.Since ( D ) is at the origin, the general equation of a circle passing through ( D ) is ( x^2 + y^2 + L x + M y + N = 0 ). But since ( D = (0, 0) ) lies on the circle, substituting gives ( 0 + 0 + 0 + 0 + N = 0 ), so ( N = 0 ).Thus, the equation simplifies to ( x^2 + y^2 + L x + M y = 0 ).Now, we need to find ( L ) and ( M ) such that points ( E ) and ( F ) lie on this circle.Let's substitute point ( E ) into the equation:( left( frac{x}{2(x + 1)} right)^2 + left( frac{h x}{x + 1} right)^2 + L cdot frac{x}{2(x + 1)} + M cdot frac{h x}{x + 1} = 0 )Similarly, substituting point ( F ):( left( -frac{x}{2(x + 1)} right)^2 + left( frac{h x}{x + 1} right)^2 + L cdot left(-frac{x}{2(x + 1)}right) + M cdot frac{h x}{x + 1} = 0 )Let me compute these equations.First, for point ( E ):( left( frac{x^2}{4(x + 1)^2} right) + left( frac{h^2 x^2}{(x + 1)^2} right) + L cdot frac{x}{2(x + 1)} + M cdot frac{h x}{x + 1} = 0 )Similarly, for point ( F ):( left( frac{x^2}{4(x + 1)^2} right) + left( frac{h^2 x^2}{(x + 1)^2} right) - L cdot frac{x}{2(x + 1)} + M cdot frac{h x}{x + 1} = 0 )Now, subtracting the equation for ( F ) from the equation for ( E ):( left[ frac{x^2}{4(x + 1)^2} + frac{h^2 x^2}{(x + 1)^2} + L cdot frac{x}{2(x + 1)} + M cdot frac{h x}{x + 1} right] - left[ frac{x^2}{4(x + 1)^2} + frac{h^2 x^2}{(x + 1)^2} - L cdot frac{x}{2(x + 1)} + M cdot frac{h x}{x + 1} right] = 0 - 0 )Simplifying:( L cdot frac{x}{2(x + 1)} + M cdot frac{h x}{x + 1} - (- L cdot frac{x}{2(x + 1)} + M cdot frac{h x}{x + 1}) = 0 )Wait, actually, let me compute it step by step.Subtracting term by term:- The first two terms cancel out: ( frac{x^2}{4(x + 1)^2} - frac{x^2}{4(x + 1)^2} = 0 )- Similarly, ( frac{h^2 x^2}{(x + 1)^2} - frac{h^2 x^2}{(x + 1)^2} = 0 )- For the ( L ) terms: ( L cdot frac{x}{2(x + 1)} - (- L cdot frac{x}{2(x + 1)}) = L cdot frac{x}{2(x + 1)} + L cdot frac{x}{2(x + 1)} = L cdot frac{x}{(x + 1)} )- For the ( M ) terms: ( M cdot frac{h x}{x + 1} - M cdot frac{h x}{x + 1} = 0 )So, overall:( L cdot frac{x}{(x + 1)} = 0 )Since ( x ) is positive and ( x + 1 ) is positive, this implies ( L = 0 ).Now, knowing that ( L = 0 ), we can substitute back into one of the equations for ( E ) or ( F ) to find ( M ).Let's use the equation for ( E ):( frac{x^2}{4(x + 1)^2} + frac{h^2 x^2}{(x + 1)^2} + 0 + M cdot frac{h x}{x + 1} = 0 )Combine the terms:( left( frac{x^2}{4(x + 1)^2} + frac{h^2 x^2}{(x + 1)^2} right) + M cdot frac{h x}{x + 1} = 0 )Factor out ( frac{x^2}{(x + 1)^2} ):( frac{x^2}{(x + 1)^2} left( frac{1}{4} + h^2 right) + M cdot frac{h x}{x + 1} = 0 )Let me denote ( frac{1}{4} + h^2 = k ), but perhaps it's better to express ( h^2 ) in terms of ( x ).Recall that ( h^2 = 1 - left( frac{x}{2} right)^2 ), since ( AB = 1 ) and ( AB ) is the hypotenuse of the right triangle formed by splitting ( ABC ) at ( D ).So, ( h^2 = 1 - frac{x^2}{4} )Therefore, ( frac{1}{4} + h^2 = frac{1}{4} + 1 - frac{x^2}{4} = frac{5}{4} - frac{x^2}{4} = frac{5 - x^2}{4} )So, substituting back:( frac{x^2}{(x + 1)^2} cdot frac{5 - x^2}{4} + M cdot frac{h x}{x + 1} = 0 )Let me write this as:( frac{x^2 (5 - x^2)}{4(x + 1)^2} + M cdot frac{h x}{x + 1} = 0 )Multiply both sides by ( 4(x + 1)^2 ) to eliminate denominators:( x^2 (5 - x^2) + 4 M h x (x + 1) = 0 )Now, solve for ( M ):( 4 M h x (x + 1) = -x^2 (5 - x^2) )( M = frac{ -x^2 (5 - x^2) }{4 h x (x + 1)} )Simplify:( M = frac{ -x (5 - x^2) }{4 h (x + 1)} )Now, recall that ( h = sqrt{1 - frac{x^2}{4}} ), so:( M = frac{ -x (5 - x^2) }{4 sqrt{1 - frac{x^2}{4}} (x + 1)} )So, now we have ( L = 0 ) and ( M ) expressed in terms of ( x ). Thus, the equation of the circle is:( x^2 + y^2 + 0 cdot x + M y = 0 )Simplify:( x^2 + y^2 + M y = 0 )Now, we need to ensure that point ( A ) lies on this circle. Point ( A ) is at ( (0, h) ). Let's substitute ( x = 0 ) and ( y = h ) into the circle equation:( 0^2 + h^2 + M cdot h = 0 )Simplify:( h^2 + M h = 0 )Factor:( h (h + M) = 0 )Since ( h neq 0 ) (as ( ABC ) is a triangle with positive area), we have:( h + M = 0 )Thus:( M = -h )But earlier, we found:( M = frac{ -x (5 - x^2) }{4 h (x + 1)} )So, setting ( M = -h ):( -h = frac{ -x (5 - x^2) }{4 h (x + 1)} )Multiply both sides by ( 4 h (x + 1) ):( -h cdot 4 h (x + 1) = -x (5 - x^2) )Simplify:( -4 h^2 (x + 1) = -x (5 - x^2) )Multiply both sides by ( -1 ):( 4 h^2 (x + 1) = x (5 - x^2) )Now, substitute ( h^2 = 1 - frac{x^2}{4} ):( 4 left(1 - frac{x^2}{4}right) (x + 1) = x (5 - x^2) )Simplify the left side:( 4 left(1 - frac{x^2}{4}right) (x + 1) = (4 - x^2)(x + 1) )So, the equation becomes:( (4 - x^2)(x + 1) = x (5 - x^2) )Let me expand both sides:Left side:( (4 - x^2)(x + 1) = 4x + 4 - x^3 - x^2 )Right side:( x (5 - x^2) = 5x - x^3 )So, set them equal:( 4x + 4 - x^3 - x^2 = 5x - x^3 )Simplify:Subtract ( 5x - x^3 ) from both sides:( 4x + 4 - x^3 - x^2 - 5x + x^3 = 0 )Simplify terms:- ( 4x - 5x = -x )- ( -x^3 + x^3 = 0 )- ( -x^2 )- ( +4 )So, the equation becomes:( -x - x^2 + 4 = 0 )Multiply both sides by ( -1 ):( x + x^2 - 4 = 0 )Rearrange:( x^2 + x - 4 = 0 )Now, solve this quadratic equation for ( x ):Using the quadratic formula:( x = frac{ -b pm sqrt{b^2 - 4ac} }{2a} )Here, ( a = 1 ), ( b = 1 ), ( c = -4 ):( x = frac{ -1 pm sqrt{1 + 16} }{2} = frac{ -1 pm sqrt{17} }{2} )Since ( x ) represents a length, it must be positive. Therefore, we discard the negative root:( x = frac{ -1 + sqrt{17} }{2} )So, ( BC = frac{ sqrt{17} - 1 }{2} )Let me just verify the steps to ensure I didn't make any mistakes.1. Set up coordinate system: Seems correct.2. Found coordinates of ( D, E, F ): Correct, using angle bisector theorem.3. Derived equations for the circle: Correct, using general circle equation and substituting points.4. Subtracted equations for ( E ) and ( F ) to find ( L = 0 ): Correct.5. Substituted back to find ( M ): Correct.6. Ensured point ( A ) lies on the circle, leading to equation involving ( h ) and ( M ): Correct.7. Solved the resulting equation to find ( x ): Correct.Therefore, the length of ( BC ) is ( frac{ sqrt{17} - 1 }{2} ).

💡Okay, so I have this system of linear equations:[begin{cases}x + 3y = 2 + a 3x + y = -4aend{cases}]And I need to find the range of ( a ) such that the solution ( (x, y) ) satisfies ( x + y > 2 ).Hmm, let me think. I remember that to solve a system of equations, I can use either substitution or elimination. Maybe elimination would be easier here because the coefficients are manageable.Looking at the equations:1. ( x + 3y = 2 + a )2. ( 3x + y = -4a )I notice that if I add these two equations together, the coefficients of ( x ) and ( y ) will add up in a way that might make things simpler.Let me try adding them:( (x + 3y) + (3x + y) = (2 + a) + (-4a) )Simplifying the left side:( x + 3x + 3y + y = 4x + 4y )And the right side:( 2 + a - 4a = 2 - 3a )So, after adding, I get:( 4x + 4y = 2 - 3a )Hmm, that's interesting. I can factor out a 4 from the left side:( 4(x + y) = 2 - 3a )Then, dividing both sides by 4:( x + y = frac{2 - 3a}{4} )Oh, wait! The problem asks for ( x + y > 2 ). So, if I set up the inequality:( frac{2 - 3a}{4} > 2 )Now, I can solve this inequality for ( a ).First, multiply both sides by 4 to eliminate the denominator:( 2 - 3a > 8 )Then, subtract 2 from both sides:( -3a > 6 )Now, divide both sides by -3. But wait, I remember that when you divide or multiply both sides of an inequality by a negative number, the inequality sign flips.So, dividing both sides by -3:( a < -2 )So, the range of ( a ) is all real numbers less than -2.Let me double-check my steps to make sure I didn't make a mistake.1. I added the two equations correctly: ( x + 3y + 3x + y = 2 + a - 4a ) simplifies to ( 4x + 4y = 2 - 3a ).2. Factoring out 4 gives ( 4(x + y) = 2 - 3a ), which seems right.3. Dividing both sides by 4: ( x + y = frac{2 - 3a}{4} ). That looks correct.4. Setting up the inequality ( frac{2 - 3a}{4} > 2 ). Makes sense because we need ( x + y > 2 ).5. Multiplying both sides by 4: ( 2 - 3a > 8 ). Correct.6. Subtracting 2: ( -3a > 6 ). Right.7. Dividing by -3 and flipping the inequality: ( a < -2 ). Yep, that's correct.So, I think my solution is correct. The range of ( a ) is all real numbers less than -2.**Final Answer**The range of ( a ) is boxed{a < -2}.

💡Okay, so I have this problem about a sequence defined by a recursive formula. It's divided into three parts: I need to find the first few terms, then prove that another sequence is geometric, and finally work with some series involving that geometric sequence. Let me try to break this down step by step.Starting with part (I): Find ( a_2 ), ( a_3 ), ( a_4 ).Given ( a_1 = 1 ), and the recursive formula:[a_{n+1} = begin{cases}frac{1}{2}a_n + n, & text{if } n text{ is odd} a_n - 2n, & text{if } n text{ is even}end{cases}]So, to find ( a_2 ), since ( n = 1 ) is odd, we use the first case:[a_2 = frac{1}{2}a_1 + 1 = frac{1}{2}(1) + 1 = frac{1}{2} + 1 = frac{3}{2}]Alright, ( a_2 = frac{3}{2} ).Next, ( a_3 ). Here, ( n = 2 ) is even, so we use the second case:[a_3 = a_2 - 2(2) = frac{3}{2} - 4 = frac{3}{2} - frac{8}{2} = -frac{5}{2}]So, ( a_3 = -frac{5}{2} ).Moving on to ( a_4 ). Now, ( n = 3 ) is odd, so again the first case applies:[a_4 = frac{1}{2}a_3 + 3 = frac{1}{2}left(-frac{5}{2}right) + 3 = -frac{5}{4} + 3 = -frac{5}{4} + frac{12}{4} = frac{7}{4}]Therefore, ( a_4 = frac{7}{4} ).So, part (I) is done. I think I did that correctly, but let me double-check:- ( a_1 = 1 )- ( a_2 = frac{3}{2} ) ✔️- ( a_3 = -frac{5}{2} ) ✔️- ( a_4 = frac{7}{4} ) ✔️Looks good.Moving on to part (II): Prove that the sequence ( {b_n} ) is a geometric sequence and find its general term formula. Given ( b_n = a_{2n} - 2 ).Hmm, so ( b_n ) is defined in terms of the even-indexed terms of ( a_n ). I need to show that ( b_n ) is geometric, meaning each term is a constant multiple of the previous one.First, let's compute ( b_1 ), ( b_2 ), ( b_3 ) to see if we can spot a pattern.From part (I):- ( a_2 = frac{3}{2} ), so ( b_1 = a_2 - 2 = frac{3}{2} - 2 = -frac{1}{2} )- ( a_4 = frac{7}{4} ), so ( b_2 = a_4 - 2 = frac{7}{4} - 2 = frac{7}{4} - frac{8}{4} = -frac{1}{4} )- Let me compute ( a_6 ) to get ( b_3 ). Wait, maybe I should find a general relation instead.But perhaps it's better to find a recursive formula for ( b_n ).Given ( b_n = a_{2n} - 2 ), so ( b_{n+1} = a_{2(n+1)} - 2 = a_{2n + 2} - 2 ).I need to express ( a_{2n + 2} ) in terms of ( a_{2n} ).Looking at the recursive formula for ( a_{n+1} ):Since ( 2n ) is even, when ( n ) is even, the next term is ( a_{n+1} = a_n - 2n ). Wait, but here ( n ) is the index, so when computing ( a_{2n + 1} ) and ( a_{2n + 2} ), we need to see whether ( 2n ) is even or odd.Wait, ( 2n ) is always even, so ( a_{2n + 1} ) is computed using the case for even ( n ):[a_{2n + 1} = a_{2n} - 2(2n) = a_{2n} - 4n]Then, ( a_{2n + 2} ) is computed from ( a_{2n + 1} ). Since ( 2n + 1 ) is odd, we use the first case:[a_{2n + 2} = frac{1}{2}a_{2n + 1} + (2n + 1)]Substituting ( a_{2n + 1} = a_{2n} - 4n ):[a_{2n + 2} = frac{1}{2}(a_{2n} - 4n) + 2n + 1 = frac{1}{2}a_{2n} - 2n + 2n + 1 = frac{1}{2}a_{2n} + 1]Therefore, ( a_{2n + 2} = frac{1}{2}a_{2n} + 1 ).Now, let's express ( b_{n+1} = a_{2n + 2} - 2 ):[b_{n+1} = left( frac{1}{2}a_{2n} + 1 right) - 2 = frac{1}{2}a_{2n} - 1]But ( b_n = a_{2n} - 2 ), so ( a_{2n} = b_n + 2 ). Substitute this into the expression for ( b_{n+1} ):[b_{n+1} = frac{1}{2}(b_n + 2) - 1 = frac{1}{2}b_n + 1 - 1 = frac{1}{2}b_n]So, ( b_{n+1} = frac{1}{2}b_n ). This shows that ( {b_n} ) is a geometric sequence with common ratio ( frac{1}{2} ).Now, the first term ( b_1 = a_2 - 2 = frac{3}{2} - 2 = -frac{1}{2} ).Therefore, the general term formula is:[b_n = b_1 times left( frac{1}{2} right)^{n - 1} = -frac{1}{2} times left( frac{1}{2} right)^{n - 1} = -left( frac{1}{2} right)^n]So, ( b_n = -left( frac{1}{2} right)^n ).Let me verify this with the earlier computed terms:- ( b_1 = -frac{1}{2} ) ✔️- ( b_2 = -frac{1}{4} ) ✔️- If I compute ( b_3 ), it should be ( -frac{1}{8} ). Let's check by computing ( a_6 ):Compute ( a_5 ): since ( n = 4 ) is even, ( a_5 = a_4 - 2(4) = frac{7}{4} - 8 = frac{7}{4} - frac{32}{4} = -frac{25}{4} )Then, ( a_6 ): ( n = 5 ) is odd, so ( a_6 = frac{1}{2}a_5 + 5 = frac{1}{2}left( -frac{25}{4} right) + 5 = -frac{25}{8} + 5 = -frac{25}{8} + frac{40}{8} = frac{15}{8} )Thus, ( b_3 = a_6 - 2 = frac{15}{8} - 2 = frac{15}{8} - frac{16}{8} = -frac{1}{8} ), which matches ( -left( frac{1}{2} right)^3 = -frac{1}{8} ). So, the formula holds.Alright, part (II) seems solid.Moving on to part (III): Given ( left( frac{3}{4} right)^n cdot C_n = -n b_n ), and ( S_n = C_1 + C_2 + dots + C_n ), prove that ( S_n < 6 ).First, let's express ( C_n ) in terms of ( b_n ):From ( left( frac{3}{4} right)^n cdot C_n = -n b_n ), we can solve for ( C_n ):[C_n = -n b_n cdot left( frac{4}{3} right)^n]But from part (II), we know ( b_n = -left( frac{1}{2} right)^n ). Substitute this into the expression:[C_n = -n cdot left( -left( frac{1}{2} right)^n right) cdot left( frac{4}{3} right)^n = n cdot left( frac{1}{2} right)^n cdot left( frac{4}{3} right)^n]Simplify the expression:[C_n = n cdot left( frac{1}{2} cdot frac{4}{3} right)^n = n cdot left( frac{2}{3} right)^n]So, ( C_n = n left( frac{2}{3} right)^n ).Now, ( S_n = C_1 + C_2 + dots + C_n = sum_{k=1}^n k left( frac{2}{3} right)^k ).We need to find ( S_n ) and show that it's less than 6.I remember that the sum ( sum_{k=1}^infty k x^k ) converges to ( frac{x}{(1 - x)^2} ) for ( |x| < 1 ). Since ( frac{2}{3} < 1 ), the infinite series converges, but we need the partial sum up to ( n ).However, the problem asks to prove ( S_n < 6 ). Since the infinite series converges to a value, let's compute that first to see if it's less than 6.Compute ( S = sum_{k=1}^infty k left( frac{2}{3} right)^k ):Using the formula ( sum_{k=1}^infty k x^k = frac{x}{(1 - x)^2} ), plug in ( x = frac{2}{3} ):[S = frac{frac{2}{3}}{left(1 - frac{2}{3}right)^2} = frac{frac{2}{3}}{left(frac{1}{3}right)^2} = frac{frac{2}{3}}{frac{1}{9}} = frac{2}{3} times 9 = 6]So, the infinite series converges to 6. Therefore, the partial sum ( S_n ) must be less than 6 for all ( n ).But to make this rigorous, I should show that ( S_n < 6 ) for all ( n ). Let me think about how to approach this.One way is to consider the difference between the infinite sum and the partial sum. Since all terms ( C_k ) are positive (because ( n ) is positive and ( left( frac{2}{3} right)^n ) is positive), the partial sum ( S_n ) is less than the infinite sum ( S ), which is 6.Therefore, ( S_n < S = 6 ).But let me write this out more formally.Consider the infinite series ( S = sum_{k=1}^infty C_k = 6 ). Since each ( C_k > 0 ), the partial sums ( S_n ) form an increasing sequence bounded above by 6. Therefore, ( S_n < 6 ) for all ( n ).Alternatively, I can use the formula for the partial sum of ( sum_{k=1}^n k x^k ). Let me recall that formula.The partial sum ( S_n = sum_{k=1}^n k x^k ) can be expressed as:[S_n = frac{x - (n + 1)x^{n + 1} + n x^{n + 2}}{(1 - x)^2}]Let me verify this formula.Yes, the formula for the finite sum is:[sum_{k=1}^n k x^k = frac{x - (n + 1)x^{n + 1} + n x^{n + 2}}{(1 - x)^2}]So, plugging ( x = frac{2}{3} ):[S_n = frac{frac{2}{3} - (n + 1)left( frac{2}{3} right)^{n + 1} + n left( frac{2}{3} right)^{n + 2}}{left(1 - frac{2}{3}right)^2}]Simplify the denominator:[left(1 - frac{2}{3}right)^2 = left( frac{1}{3} right)^2 = frac{1}{9}]So,[S_n = frac{frac{2}{3} - (n + 1)left( frac{2}{3} right)^{n + 1} + n left( frac{2}{3} right)^{n + 2}}{frac{1}{9}} = 9 left[ frac{2}{3} - (n + 1)left( frac{2}{3} right)^{n + 1} + n left( frac{2}{3} right)^{n + 2} right]]Simplify each term:First term: ( 9 times frac{2}{3} = 6 )Second term: ( -9(n + 1)left( frac{2}{3} right)^{n + 1} )Third term: ( +9n left( frac{2}{3} right)^{n + 2} = +9n left( frac{2}{3} right)^{n} left( frac{4}{9} right) = +4n left( frac{2}{3} right)^n )Wait, let me compute the third term correctly:( left( frac{2}{3} right)^{n + 2} = left( frac{2}{3} right)^n times left( frac{2}{3} right)^2 = left( frac{2}{3} right)^n times frac{4}{9} )Thus,Third term: ( 9n times frac{4}{9} left( frac{2}{3} right)^n = 4n left( frac{2}{3} right)^n )So, putting it all together:[S_n = 6 - 9(n + 1)left( frac{2}{3} right)^{n + 1} + 4n left( frac{2}{3} right)^n]Let me factor out ( left( frac{2}{3} right)^n ):First, note that ( left( frac{2}{3} right)^{n + 1} = frac{2}{3} left( frac{2}{3} right)^n ). So,[S_n = 6 - 9(n + 1)frac{2}{3} left( frac{2}{3} right)^n + 4n left( frac{2}{3} right)^n]Simplify the coefficients:- ( 9 times frac{2}{3} = 6 )So,[S_n = 6 - 6(n + 1)left( frac{2}{3} right)^n + 4n left( frac{2}{3} right)^n]Combine the terms with ( left( frac{2}{3} right)^n ):[S_n = 6 + left[ -6(n + 1) + 4n right] left( frac{2}{3} right)^n]Simplify inside the brackets:[-6(n + 1) + 4n = -6n - 6 + 4n = (-6n + 4n) - 6 = -2n - 6]Thus,[S_n = 6 + (-2n - 6)left( frac{2}{3} right)^n = 6 - (2n + 6)left( frac{2}{3} right)^n]Since ( left( frac{2}{3} right)^n > 0 ) and ( 2n + 6 > 0 ), the term ( -(2n + 6)left( frac{2}{3} right)^n ) is negative. Therefore,[S_n = 6 - text{(positive term)} < 6]Hence, ( S_n < 6 ) for all ( n ).Alternatively, another approach is to use generating functions or consider the series as a geometric series multiplied by ( k ), but I think the partial sum formula suffices here.Just to recap:1. Expressed ( C_n ) in terms of ( b_n ).2. Found that ( C_n = n left( frac{2}{3} right)^n ).3. Recognized that ( S_n ) is the partial sum of ( sum_{k=1}^n k left( frac{2}{3} right)^k ).4. Used the formula for the partial sum of such a series.5. Simplified the expression to show that ( S_n = 6 - text{something positive} ), hence ( S_n < 6 ).I think this is thorough enough. I don't see any mistakes in the steps, and the logic flows correctly. The key was recognizing the form of the series and applying the appropriate partial sum formula.**Final Answer**(I) ( a_2 = boxed{dfrac{3}{2}} ), ( a_3 = boxed{-dfrac{5}{2}} ), ( a_4 = boxed{dfrac{7}{4}} )(II) The sequence ( {b_n} ) is a geometric sequence with the general term formula ( b_n = boxed{-left( dfrac{1}{2} right)^n} )(III) ( S_n < boxed{6} )

💡Okay, so I have this problem here about two vectors, a and b, in the plane. They form an angle of 120 degrees, and their dot product is -1. I need to find the minimum value of the magnitude of their difference, which is |a - b|. Hmm, let me try to figure this out step by step.First, I remember that the dot product of two vectors can be expressed in terms of their magnitudes and the cosine of the angle between them. The formula is:a · b = |a| |b| cosθGiven that the angle θ is 120 degrees, and the dot product is -1, I can plug those values into the formula:-1 = |a| |b| cos(120°)I know that cos(120°) is equal to -1/2 because 120 degrees is in the second quadrant where cosine is negative, and cos(120°) = cos(180° - 60°) = -cos(60°) = -1/2. So substituting that in:-1 = |a| |b| (-1/2)Let me solve for |a| |b|. Multiplying both sides by -2:(-1) * (-2) = |a| |b| (-1/2) * (-2)Which simplifies to:2 = |a| |b|So, the product of the magnitudes of vectors a and b is 2. That's a useful piece of information.Now, I need to find the minimum value of |a - b|. I recall that the magnitude of the difference of two vectors can be found using the formula:|a - b| = sqrt(|a|² + |b|² - 2(a · b))Wait, let me verify that. Yes, because (a - b) · (a - b) = |a|² - 2a·b + |b|², so taking the square root gives the magnitude.But in this case, we already know that a · b is -1, so substituting that in:|a - b| = sqrt(|a|² + |b|² - 2*(-1)) = sqrt(|a|² + |b|² + 2)So, |a - b| = sqrt(|a|² + |b|² + 2)Now, I need to find the minimum value of this expression. I know that |a| |b| = 2 from earlier. So, perhaps I can express |a|² + |b|² in terms of |a| |b|.I remember that for any two real numbers, |a|² + |b|² is always greater than or equal to 2|a||b|, by the AM-GM inequality. Specifically, (|a|² + |b|²)/2 ≥ (|a||b|)². Wait, no, that's not quite right. The AM-GM inequality states that (|a|² + |b|²)/2 ≥ (|a||b|). Hmm, let me double-check.Wait, actually, the AM-GM inequality for two non-negative real numbers x and y states that (x + y)/2 ≥ sqrt(xy). So, if I let x = |a|² and y = |b|², then:(|a|² + |b|²)/2 ≥ sqrt(|a|² |b|²) = |a||b|So, multiplying both sides by 2:|a|² + |b|² ≥ 2|a||b|We know that |a||b| = 2, so:|a|² + |b|² ≥ 2*2 = 4Therefore, |a|² + |b|² is at least 4. So, substituting back into the expression for |a - b|:|a - b| = sqrt(|a|² + |b|² + 2) ≥ sqrt(4 + 2) = sqrt(6)So, the minimum value of |a - b| is sqrt(6). But wait, is this achievable? That is, does there exist vectors a and b such that |a|² + |b|² = 4 and |a||b| = 2?Let me check. If |a|² + |b|² = 4 and |a||b| = 2, then let me denote |a| = x and |b| = y. So, we have:x² + y² = 4andxy = 2I can try to solve these equations. From the second equation, y = 2/x. Substitute into the first equation:x² + (2/x)² = 4x² + 4/x² = 4Multiply both sides by x²:x⁴ + 4 = 4x²Bring all terms to one side:x⁴ - 4x² + 4 = 0This is a quadratic in terms of x². Let me set z = x²:z² - 4z + 4 = 0Solving this quadratic equation:z = [4 ± sqrt(16 - 16)] / 2 = [4 ± 0]/2 = 2So, z = 2, which means x² = 2, so x = sqrt(2). Therefore, y = 2/x = 2/sqrt(2) = sqrt(2). So, both |a| and |b| are sqrt(2).Therefore, when |a| = |b| = sqrt(2), we have |a|² + |b|² = 2 + 2 = 4, and |a||b| = 2, which satisfies both conditions. Therefore, the minimum value of |a - b| is indeed sqrt(6).Wait, let me just visualize this. If both vectors have the same magnitude, sqrt(2), and the angle between them is 120 degrees, then the triangle formed by vectors a, b, and a - b would be an equilateral triangle? Hmm, no, because the angle is 120 degrees, not 60. So, it's more like a triangle with two sides of length sqrt(2) and the included angle of 120 degrees. The third side would be |a - b|, which we found to be sqrt(6). That makes sense because in such a triangle, using the law of cosines:c² = a² + b² - 2ab cosθHere, c is |a - b|, a and b are the magnitudes sqrt(2), and θ is 120 degrees. So:c² = (sqrt(2))² + (sqrt(2))² - 2*(sqrt(2))*(sqrt(2))*cos(120°)Simplify:c² = 2 + 2 - 2*2*(-1/2) = 4 - (-2) = 4 + 2 = 6Therefore, c = sqrt(6). Yep, that checks out.So, putting it all together, the minimum value of |a - b| is sqrt(6), achieved when both vectors have magnitude sqrt(2).**Final Answer**The minimum value of (|overrightarrow{a} - overrightarrow{b}|) is (boxed{sqrt{6}}).