answer：Alright, so I've got this problem here with two parts. Let me try to tackle them one by one. First, the problem states that ( x, y in mathbb{R} ), ( m + n = 7 ), and ( f(x) = |x - 1| - |x + 1| ). **Part (1): Solve the inequality ( f(x) geq (m + n)x ).**Okay, so I need to solve ( |x - 1| - |x + 1| geq 7x ) because ( m + n = 7 ). Hmm, absolute value functions can be tricky because they change their behavior based on the value of ( x ). So, I think I should break this down into cases based on the critical points of the absolute value expressions. The expressions ( |x - 1| ) and ( |x + 1| ) change their behavior at ( x = 1 ) and ( x = -1 ), respectively. So, I can split the real line into three intervals:1. ( x leq -1 )2. ( -1 < x < 1 )3. ( x geq 1 )Let me handle each case separately.**Case 1: ( x leq -1 )**In this interval, ( x - 1 ) is negative because ( x leq -1 < 1 ), so ( |x - 1| = -(x - 1) = -x + 1 ). Similarly, ( x + 1 ) is also negative because ( x leq -1 ), so ( |x + 1| = -(x + 1) = -x - 1 ).Substituting these into ( f(x) ):[f(x) = (-x + 1) - (-x - 1) = -x + 1 + x + 1 = 2]So, the inequality becomes:[2 geq 7x]Dividing both sides by 7:[x leq frac{2}{7}]But wait, in this case, ( x leq -1 ), which is already more restrictive than ( x leq frac{2}{7} ). So, in this interval, all ( x leq -1 ) satisfy the inequality.**Case 2: ( -1 < x < 1 )**Here, ( x + 1 ) is positive because ( x > -1 ), so ( |x + 1| = x + 1 ). However, ( x - 1 ) is still negative because ( x < 1 ), so ( |x - 1| = -x + 1 ).Substituting into ( f(x) ):[f(x) = (-x + 1) - (x + 1) = -x + 1 - x - 1 = -2x]So, the inequality becomes:[-2x geq 7x]Adding ( 2x ) to both sides:[0 geq 9x]Dividing both sides by 9:[x leq 0]But in this case, ( -1 < x < 1 ). So, combining these, we get ( -1 < x leq 0 ).**Case 3: ( x geq 1 )**In this interval, both ( x - 1 ) and ( x + 1 ) are positive. So, ( |x - 1| = x - 1 ) and ( |x + 1| = x + 1 ).Substituting into ( f(x) ):[f(x) = (x - 1) - (x + 1) = x - 1 - x - 1 = -2]So, the inequality becomes:[-2 geq 7x]Dividing both sides by 7:[x leq -frac{2}{7}]But in this case, ( x geq 1 ), which doesn't overlap with ( x leq -frac{2}{7} ). So, there are no solutions in this interval.**Combining all cases:**From Case 1: ( x leq -1 )From Case 2: ( -1 < x leq 0 )From Case 3: No solutionsSo, combining these, the solution set is ( x leq 0 ).Wait, let me double-check that. In Case 2, we had ( -1 < x leq 0 ), and in Case 1, ( x leq -1 ). So, together, it's all ( x leq 0 ). That seems right.**Part (2): Find the minimum value of ( F = max{|x^2 - 4y + m|, |y^2 - 2x + n|} ).**Hmm, okay. So, ( F ) is the maximum of two absolute value expressions. We need to find the minimum possible value of this maximum. That is, we want to minimize the larger of the two values ( |x^2 - 4y + m| ) and ( |y^2 - 2x + n| ).Given that ( m + n = 7 ), perhaps we can express one variable in terms of the other. Let me write ( n = 7 - m ). So, the expressions become:1. ( |x^2 - 4y + m| )2. ( |y^2 - 2x + (7 - m)| )So, ( F = max{|x^2 - 4y + m|, |y^2 - 2x + 7 - m|} )I need to find the minimum value of ( F ) over all real numbers ( x, y ) and some ( m ) (since ( n ) is determined by ( m )).Wait, actually, is ( m ) a given constant or is it a variable? The problem says ( m + n = 7 ), but it doesn't specify whether ( m ) and ( n ) are given or if we can choose them. Hmm.Looking back at the problem statement: It says ( x, y in mathbb{R} ), ( m + n = 7 ), and ( f(x) = |x - 1| - |x + 1| ). Then, part (2) defines ( F ) as the maximum of those two absolute values.So, I think ( m ) and ( n ) are constants with ( m + n = 7 ), and we need to find the minimum of ( F ) over all ( x, y ). Or maybe ( m ) is a variable as well? Hmm, the problem isn't entirely clear.Wait, in part (1), ( m + n = 7 ) is given, and ( f(x) ) is defined. So, perhaps ( m ) and ( n ) are fixed constants with ( m + n = 7 ), and we need to find the minimum of ( F ) over all ( x, y ).Alternatively, maybe ( m ) is a parameter, and we need to find the minimum over ( x, y, m ). Hmm, the problem isn't entirely clear. Let me read it again."Let ( max{a, b} = begin{cases} a & (a geq b) b & (a < b) end{cases} ), find the minimum value of ( F = max{|x^2 - 4y + m|, |y^2 - 2x + n|} )."So, it says "find the minimum value of ( F )", without specifying over what variables. Since ( x ) and ( y ) are given as real numbers, and ( m ) and ( n ) are related by ( m + n = 7 ), perhaps ( m ) is a variable as well, so we can choose ( m ) and ( n ) such that ( m + n = 7 ) to minimize ( F ).Alternatively, maybe ( m ) and ( n ) are fixed, and we need to minimize over ( x ) and ( y ). Hmm, the problem is a bit ambiguous.Wait, in part (1), ( m + n = 7 ) is given, so perhaps ( m ) and ( n ) are fixed constants, and we need to find the minimum of ( F ) over ( x ) and ( y ).Alternatively, maybe ( m ) is a variable, and we can choose ( m ) and ( n ) such that ( m + n = 7 ) to minimize ( F ). Hmm.Given that in part (1), ( m + n = 7 ) is used, and in part (2), ( m ) and ( n ) are part of the expressions, I think it's more likely that ( m ) and ( n ) are fixed constants with ( m + n = 7 ), and we need to find the minimum of ( F ) over ( x ) and ( y ).But actually, the problem doesn't specify whether ( m ) is given or not. Hmm.Wait, let me think differently. Maybe ( m ) and ( n ) are constants, and we need to find the minimum of ( F ) over all ( x ) and ( y ). So, perhaps ( m ) is fixed, but since ( m + n = 7 ), ( n = 7 - m ). So, ( F ) is expressed in terms of ( m ), ( x ), and ( y ). So, maybe we can treat ( m ) as a variable as well, and find the minimum over all ( x, y, m ).Alternatively, perhaps ( m ) is fixed, and we just need to minimize over ( x ) and ( y ). Hmm.Wait, in the problem statement, it's written as:"Given ( x, y in mathbb{R} ), ( m + n = 7 ), ( f(x) = |x - 1| - |x + 1| ).(1) Solve the inequality ( f(x) geq (m + n)x );(2) Let ( max{a, b} = begin{cases} a & (a geq b) b & (a < b) end{cases} ), find the minimum value of ( F = max{|x^2 - 4y + m|, |y^2 - 2x + n|} )."So, in part (2), it's just given ( F ) as a function of ( x, y, m, n ), with ( m + n = 7 ). So, perhaps ( m ) and ( n ) are fixed, and we need to find the minimum of ( F ) over ( x ) and ( y ).Alternatively, perhaps ( m ) is a variable, and we can choose ( m ) and ( n ) such that ( m + n = 7 ) to minimize ( F ).Given that part (1) is about solving an inequality involving ( m + n ), and part (2) is about minimizing ( F ), which also involves ( m ) and ( n ), I think it's more likely that ( m ) and ( n ) are fixed constants with ( m + n = 7 ), and we need to find the minimum of ( F ) over ( x ) and ( y ).But actually, the problem doesn't specify whether ( m ) is given or not. Hmm, this is a bit confusing.Wait, maybe I should assume that ( m ) and ( n ) are fixed constants with ( m + n = 7 ), and we need to find the minimum of ( F ) over ( x ) and ( y ).Alternatively, perhaps ( m ) is a parameter, and we can choose ( m ) and ( n ) such that ( m + n = 7 ) to minimize ( F ). Hmm.Wait, let me think about the problem again. Since ( m + n = 7 ), and in part (2), ( F ) is expressed in terms of ( m ) and ( n ), perhaps we can treat ( m ) as a variable and choose it to minimize ( F ).Alternatively, maybe ( m ) is fixed, and we just need to minimize over ( x ) and ( y ).Given that the problem says "find the minimum value of ( F )", without specifying, I think it's safer to assume that ( m ) and ( n ) are fixed constants with ( m + n = 7 ), and we need to find the minimum of ( F ) over ( x ) and ( y ).But actually, in part (1), ( m + n = 7 ) is given, but ( m ) and ( n ) are not specified further. So, perhaps ( m ) and ( n ) are fixed, but their exact values are not given, so we might need to express the minimum in terms of ( m ) or something.Wait, no, the problem says "find the minimum value of ( F )", so it must be a numerical answer, not in terms of ( m ). Therefore, perhaps ( m ) and ( n ) are fixed, but since ( m + n = 7 ), maybe we can find the minimum over ( x ) and ( y ) regardless of ( m ) and ( n ).Wait, no, that doesn't make sense because ( F ) depends on ( m ) and ( n ). So, perhaps ( m ) and ( n ) are given, but their exact values are not specified, so maybe the minimum is achieved regardless of ( m ) and ( n ).Wait, this is getting confusing. Maybe I should proceed differently.Let me consider that ( m ) and ( n ) are fixed constants with ( m + n = 7 ), and we need to find the minimum of ( F ) over ( x ) and ( y ).So, ( F = max{|x^2 - 4y + m|, |y^2 - 2x + n|} ).We need to minimize this maximum.One approach is to set both expressions inside the max equal to each other, because the maximum will be minimized when both are equal. So, set:[|x^2 - 4y + m| = |y^2 - 2x + n|]But this might not always be possible, so perhaps we can consider the case where both expressions are equal and non-negative, or both are equal and non-positive.Alternatively, perhaps we can consider the case where both expressions are equal, and then find the minimum value.Alternatively, maybe we can use the triangle inequality or some other inequality to bound ( F ).Wait, another approach is to consider that ( F geq |x^2 - 4y + m| ) and ( F geq |y^2 - 2x + n| ). So, adding these two inequalities:[2F geq |x^2 - 4y + m| + |y^2 - 2x + n|]But I'm not sure if that helps directly.Alternatively, perhaps we can consider the sum inside the absolute values:[|x^2 - 4y + m| + |y^2 - 2x + n| geq |(x^2 - 4y + m) + (y^2 - 2x + n)|]By the triangle inequality.So,[|x^2 - 4y + m| + |y^2 - 2x + n| geq |x^2 + y^2 - 2x - 4y + (m + n)|]Since ( m + n = 7 ), this becomes:[|x^2 - 4y + m| + |y^2 - 2x + n| geq |x^2 + y^2 - 2x - 4y + 7|]Now, let's complete the squares for ( x ) and ( y ):For ( x ):( x^2 - 2x = (x - 1)^2 - 1 )For ( y ):( y^2 - 4y = (y - 2)^2 - 4 )So, substituting back:[x^2 + y^2 - 2x - 4y + 7 = (x - 1)^2 - 1 + (y - 2)^2 - 4 + 7 = (x - 1)^2 + (y - 2)^2 + 2]Therefore,[|x^2 - 4y + m| + |y^2 - 2x + n| geq |(x - 1)^2 + (y - 2)^2 + 2|]Since squares are always non-negative, ( (x - 1)^2 + (y - 2)^2 geq 0 ), so:[|(x - 1)^2 + (y - 2)^2 + 2| = (x - 1)^2 + (y - 2)^2 + 2 geq 2]Therefore,[|x^2 - 4y + m| + |y^2 - 2x + n| geq 2]But earlier, we had:[2F geq |x^2 - 4y + m| + |y^2 - 2x + n| geq 2]So,[2F geq 2 implies F geq 1]So, the minimum possible value of ( F ) is at least 1.Now, can we achieve ( F = 1 )?To check this, we need to see if there exist ( x ) and ( y ) such that both ( |x^2 - 4y + m| leq 1 ) and ( |y^2 - 2x + n| leq 1 ), and at least one of them equals 1.Wait, actually, since ( F = max{A, B} ), to have ( F = 1 ), both ( A ) and ( B ) must be less than or equal to 1, and at least one of them must be equal to 1.But in our case, we have:[|x^2 - 4y + m| + |y^2 - 2x + n| geq 2]And we have ( 2F geq 2 implies F geq 1 ). So, to achieve equality, we need:1. ( |x^2 - 4y + m| + |y^2 - 2x + n| = 2 )2. ( |x^2 - 4y + m| = |y^2 - 2x + n| = 1 )Wait, no, because if ( F = 1 ), then both ( |x^2 - 4y + m| leq 1 ) and ( |y^2 - 2x + n| leq 1 ), and their sum would be ( leq 2 ). But we have ( |x^2 - 4y + m| + |y^2 - 2x + n| geq 2 ), so the only way this can happen is if both are equal to 1.So, set:[|x^2 - 4y + m| = 1][|y^2 - 2x + n| = 1]And also,[(x - 1)^2 + (y - 2)^2 + 2 = 2 implies (x - 1)^2 + (y - 2)^2 = 0]Which implies ( x = 1 ) and ( y = 2 ).So, plugging ( x = 1 ) and ( y = 2 ) into the expressions:First expression:[|1^2 - 4*2 + m| = |1 - 8 + m| = |-7 + m| = |m - 7|]Second expression:[|2^2 - 2*1 + n| = |4 - 2 + n| = |2 + n| = |n + 2|]We need both of these to be equal to 1.So,1. ( |m - 7| = 1 )2. ( |n + 2| = 1 )Given that ( m + n = 7 ), let's solve these equations.From ( |m - 7| = 1 ), we have two cases:Case 1: ( m - 7 = 1 implies m = 8 )Then, ( n = 7 - m = 7 - 8 = -1 )Check ( |n + 2| = |-1 + 2| = |1| = 1 ). Okay, this works.Case 2: ( m - 7 = -1 implies m = 6 )Then, ( n = 7 - 6 = 1 )Check ( |n + 2| = |1 + 2| = |3| = 3 ). This does not equal 1. So, this case doesn't work.Therefore, the only solution is ( m = 8 ) and ( n = -1 ).So, when ( x = 1 ) and ( y = 2 ), we have ( F = max{1, 1} = 1 ).Therefore, the minimum value of ( F ) is 1.Wait, but does this mean that for ( m = 8 ) and ( n = -1 ), the minimum ( F ) is 1? Or is this the minimum regardless of ( m ) and ( n )?Wait, actually, in this case, we found that when ( x = 1 ) and ( y = 2 ), and ( m = 8 ), ( n = -1 ), then ( F = 1 ). But if ( m ) and ( n ) are fixed, then this might not be achievable. Hmm.Wait, but in the problem, ( m + n = 7 ) is given, but ( m ) and ( n ) are not specified further. So, perhaps ( m ) and ( n ) can be chosen such that ( m = 8 ) and ( n = -1 ), which satisfies ( m + n = 7 ). Therefore, the minimum value of ( F ) is 1.Alternatively, if ( m ) and ( n ) are fixed, then this might not be possible. But since the problem doesn't specify ( m ) and ( n ) further, I think it's safe to assume that ( m ) and ( n ) can be chosen to achieve this minimum.Therefore, the minimum value of ( F ) is 1.**Final Answer**(1) boxed{x leq 0}(2) boxed{1}

answer：Alright, so I have this problem where a sequence of real numbers ( x_0, x_1, x_2, ldots ) is defined with ( x_0 = 1989 ) and for each ( n geq 1 ), [x_n = -frac{1989}{n} sum_{k=0}^{n-1} x_k.]I need to calculate the value of ( sum_{n=0}^{1989} 2^n x_n ). Hmm, okay. Let me try to unpack this step by step.First, I know that ( x_0 ) is given as 1989. For ( n = 1 ), the formula becomes:[x_1 = -frac{1989}{1} sum_{k=0}^{0} x_k = -1989 cdot x_0 = -1989 times 1989.]So, ( x_1 = -1989^2 ). Got that.Now, for ( n = 2 ):[x_2 = -frac{1989}{2} sum_{k=0}^{1} x_k = -frac{1989}{2} (x_0 + x_1).]Substituting the known values:[x_2 = -frac{1989}{2} (1989 - 1989^2) = -frac{1989}{2} times 1989 (1 - 1989) = -frac{1989^2}{2} times (-1988).]Wait, that seems a bit messy. Maybe I should look for a pattern or a general formula instead of computing each term individually, especially since I need to go up to ( n = 1989 ).Let me try to express ( x_n ) in terms of ( x_{n-1} ). From the given recurrence:[x_n = -frac{1989}{n} sum_{k=0}^{n-1} x_k.]But the sum ( sum_{k=0}^{n-1} x_k ) is related to ( x_{n-1} ) as well. Let me denote ( S_n = sum_{k=0}^{n} x_k ). Then, ( S_n = S_{n-1} + x_n ).From the recurrence, we have:[x_n = -frac{1989}{n} S_{n-1}.]But also, ( S_n = S_{n-1} + x_n = S_{n-1} - frac{1989}{n} S_{n-1} = S_{n-1} left(1 - frac{1989}{n}right) ).So,[S_n = S_{n-1} left(1 - frac{1989}{n}right).]This looks like a recursive relation for ( S_n ). Maybe I can express ( S_n ) in terms of ( S_0 ).Starting from ( S_0 = x_0 = 1989 ).For ( n = 1 ):[S_1 = S_0 left(1 - frac{1989}{1}right) = 1989 (1 - 1989) = 1989 times (-1988).]For ( n = 2 ):[S_2 = S_1 left(1 - frac{1989}{2}right) = 1989 times (-1988) times left(1 - frac{1989}{2}right).]Hmm, this seems to be getting complicated. Maybe there's a better way to approach this.Wait, perhaps I can write ( S_n ) in terms of factorials or binomial coefficients? Let me see.Looking back at the recurrence:[S_n = S_{n-1} left(1 - frac{1989}{n}right).]This can be rewritten as:[S_n = S_{n-1} left(frac{n - 1989}{n}right).]So, recursively,[S_n = S_0 times prod_{k=1}^{n} frac{k - 1989}{k}.]Since ( S_0 = 1989 ), we have:[S_n = 1989 times prod_{k=1}^{n} frac{k - 1989}{k}.]Simplifying the product:[prod_{k=1}^{n} frac{k - 1989}{k} = frac{(-1988)(-1987)cdots(n - 1989)}{n!}.]Notice that the numerator is ( (-1)^{1989} times frac{(1988 + n - 1)!}{(n - 1)!} ) but I'm not sure. Wait, actually, the numerator is:[prod_{k=1}^{n} (k - 1989) = (-1988)(-1987)cdots(n - 1989).]This is similar to a factorial but shifted. Specifically, it's ( (-1)^{1989} times frac{(1988 + n)!}{(n - 1)!} ) but I need to check.Wait, actually, if I let ( m = k - 1989 ), then when ( k = 1 ), ( m = -1988 ), and when ( k = n ), ( m = n - 1989 ). So, the product becomes:[prod_{m=-1988}^{n - 1989} m.]But this is a bit messy. Maybe another approach.Alternatively, recognizing that the product ( prod_{k=1}^{n} frac{k - 1989}{k} ) can be written as:[frac{(-1988)(-1987)cdots(n - 1989)}{n!} = frac{(-1)^{1989} times 1988 times 1987 times cdots times (1989 - n)}{n!}.]Wait, that might not be correct. Let me think.Actually, ( prod_{k=1}^{n} (k - 1989) = (-1)^{1989} times frac{(1988 + n)!}{(n - 1)!} ) but I'm not entirely sure. Maybe it's better to think in terms of binomial coefficients.Wait, let's consider the expression ( prod_{k=1}^{n} frac{k - 1989}{k} ). This is similar to ( frac{( -1988)( -1987)cdots(n - 1989)}{n!} ).Notice that ( (-1988)(-1987)cdots(n - 1989) ) is equal to ( (-1)^{1989} times (1988)(1987)cdots(1989 - n) ).But this is getting too convoluted. Maybe I should think about the general term ( x_n ).From the recurrence, we have:[x_n = -frac{1989}{n} S_{n-1}.]But ( S_{n-1} = sum_{k=0}^{n-1} x_k ). If I can express ( S_{n-1} ) in terms of ( x_{n-1} ), maybe I can find a pattern.Wait, earlier I had:[S_n = S_{n-1} left(1 - frac{1989}{n}right).]So, ( S_n = S_{n-1} - frac{1989}{n} S_{n-1} ).But ( x_n = -frac{1989}{n} S_{n-1} ), so ( S_n = S_{n-1} + x_n = S_{n-1} - frac{1989}{n} S_{n-1} = S_{n-1} left(1 - frac{1989}{n}right) ).This seems consistent.Maybe I can write ( S_n ) as:[S_n = S_0 times prod_{k=1}^{n} left(1 - frac{1989}{k}right).]Since ( S_0 = 1989 ), we have:[S_n = 1989 times prod_{k=1}^{n} left(1 - frac{1989}{k}right).]This product can be simplified. Let's write it as:[prod_{k=1}^{n} left(frac{k - 1989}{k}right) = frac{(-1988)(-1987)cdots(n - 1989)}{n!}.]This is similar to a binomial coefficient. Specifically, it resembles the expression for ( binom{-1989}{n} ) or something similar, but I'm not sure.Wait, actually, the product ( prod_{k=1}^{n} frac{k - 1989}{k} ) can be written as:[frac{(-1)^{1989} times 1988 times 1987 times cdots times (1989 - n)}{n!}.]But I'm not sure if that's helpful. Maybe another approach.Let me consider the generating function for the sequence ( x_n ). Let ( G(x) = sum_{n=0}^{infty} x_n t^n ). Maybe I can find a closed-form expression for ( G(t) ) and then evaluate it at ( t = 2 ).Given the recurrence:[x_n = -frac{1989}{n} sum_{k=0}^{n-1} x_k quad text{for } n geq 1.]Multiply both sides by ( t^n ) and sum over ( n geq 1 ):[sum_{n=1}^{infty} x_n t^n = -1989 sum_{n=1}^{infty} left( sum_{k=0}^{n-1} x_k right) frac{t^n}{n}.]The left-hand side is ( G(t) - x_0 = G(t) - 1989 ).The right-hand side is more complicated. Let me denote ( S_n = sum_{k=0}^{n} x_k ), so ( sum_{k=0}^{n-1} x_k = S_{n-1} ).Thus, the right-hand side becomes:[-1989 sum_{n=1}^{infty} frac{S_{n-1}}{n} t^n.]Let me change the index of summation by letting ( m = n - 1 ):[-1989 sum_{m=0}^{infty} frac{S_m}{m+1} t^{m+1} = -1989 t sum_{m=0}^{infty} frac{S_m}{m+1} t^m.]Now, ( S_m = sum_{k=0}^{m} x_k ), so ( frac{S_m}{m+1} ) is the average of the first ( m+1 ) terms. Hmm, not sure if that helps.Wait, perhaps integrating the generating function. Let me recall that:[sum_{m=0}^{infty} S_m t^m = frac{G(t)}{1 - t}.]But I have ( sum_{m=0}^{infty} frac{S_m}{m+1} t^m ). Let me denote this as ( sum_{m=0}^{infty} frac{S_m}{m+1} t^m ).Notice that ( frac{1}{m+1} = int_{0}^{1} t^m dt ). So,[sum_{m=0}^{infty} frac{S_m}{m+1} t^m = sum_{m=0}^{infty} S_m int_{0}^{1} t^m dt cdot t^m = int_{0}^{1} sum_{m=0}^{infty} S_m (t tau)^m dtau,]where ( tau ) is a dummy variable. Then,[int_{0}^{1} frac{G(t tau)}{1 - t tau} dtau.]So, putting it all together, the right-hand side becomes:[-1989 t int_{0}^{1} frac{G(t tau)}{1 - t tau} dtau.]This seems quite involved. Maybe there's a simpler approach.Wait, going back to the original recurrence:[x_n = -frac{1989}{n} S_{n-1}.]But ( S_{n-1} = sum_{k=0}^{n-1} x_k ). So, if I can express ( S_{n-1} ) in terms of ( x_{n-1} ), maybe I can find a relation between ( x_n ) and ( x_{n-1} ).From the recurrence, we have:[x_n = -frac{1989}{n} S_{n-1}.]But ( S_{n-1} = S_{n-2} + x_{n-1} ). And from the recurrence for ( x_{n-1} ):[x_{n-1} = -frac{1989}{n-1} S_{n-2}.]So,[S_{n-1} = S_{n-2} + x_{n-1} = S_{n-2} - frac{1989}{n-1} S_{n-2} = S_{n-2} left(1 - frac{1989}{n-1}right).]Therefore,[x_n = -frac{1989}{n} S_{n-1} = -frac{1989}{n} S_{n-2} left(1 - frac{1989}{n-1}right).]But ( x_{n-1} = -frac{1989}{n-1} S_{n-2} ), so ( S_{n-2} = -frac{(n-1)}{1989} x_{n-1} ).Substituting back,[x_n = -frac{1989}{n} left(-frac{(n-1)}{1989} x_{n-1}right) left(1 - frac{1989}{n-1}right).]Simplify:[x_n = frac{(n-1)}{n} x_{n-1} left(1 - frac{1989}{n-1}right) = frac{(n-1)}{n} x_{n-1} left(frac{n - 1 - 1989}{n - 1}right).]Simplify further:[x_n = frac{(n-1)}{n} x_{n-1} times frac{n - 1990}{n - 1} = frac{n - 1990}{n} x_{n-1}.]So, we have:[x_n = frac{n - 1990}{n} x_{n-1}.]This is a much simpler recurrence relation! So, each term is a multiple of the previous term. Let me write this as:[x_n = left(1 - frac{1990}{n}right) x_{n-1}.]Wait, that seems a bit off. Let me double-check:From earlier steps:[x_n = frac{n - 1990}{n} x_{n-1}.]Yes, that's correct. So, starting from ( x_0 = 1989 ), we can express ( x_n ) as:[x_n = x_0 times prod_{k=1}^{n} frac{k - 1990}{k}.]Simplifying the product:[x_n = 1989 times prod_{k=1}^{n} frac{k - 1990}{k} = 1989 times frac{(-1989)(-1988)cdots(n - 1990)}{n!}.]This product in the numerator is ( (-1)^{1990} times frac{(1989 + n - 1)!}{(n - 1)!} ) but I'm not sure. Wait, actually, it's similar to a binomial coefficient.Wait, ( prod_{k=1}^{n} (k - 1990) = (-1989)(-1988)cdots(n - 1990) ).This is equal to ( (-1)^{1990} times frac{(1989 + n - 1)!}{(n - 1)!} ) but I'm not sure. Alternatively, recognizing that this is ( (-1)^{1990} times binom{1989 + n - 1}{n} times n! ) but I'm getting confused.Wait, perhaps it's better to write it as:[x_n = 1989 times frac{(-1989)(-1988)cdots(n - 1990)}{n!} = 1989 times frac{(-1)^{1990} times 1989 times 1988 times cdots times (1990 - n)}{n!}.]Wait, but ( 1990 - n ) is positive only if ( n < 1990 ). Since we're summing up to ( n = 1989 ), which is less than 1990, this makes sense.So, simplifying,[x_n = 1989 times (-1)^{1990} times frac{1989!}{(1989 - n)! n!} = 1989 times (-1)^{1990} times binom{1989}{n}.]But ( (-1)^{1990} = 1 ) since 1990 is even. So,[x_n = 1989 times binom{1989}{n}.]Wait, but earlier terms were negative. For example, ( x_1 = -1989^2 ). Let me check with ( n = 1 ):[x_1 = 1989 times binom{1989}{1} = 1989 times 1989 = 1989^2.]But earlier, I found ( x_1 = -1989^2 ). There's a discrepancy here. So, I must have made a mistake in the sign.Going back, when I expressed ( x_n = frac{n - 1990}{n} x_{n-1} ), let's check the sign for ( n = 1 ):For ( n = 1 ):[x_1 = frac{1 - 1990}{1} x_0 = frac{-1989}{1} times 1989 = -1989^2.]Which matches the earlier result. So, the general formula should include a negative sign. Let me correct that.From the recurrence:[x_n = frac{n - 1990}{n} x_{n-1} = left(1 - frac{1990}{n}right) x_{n-1}.]But when ( n = 1 ), it's ( x_1 = frac{1 - 1990}{1} x_0 = -1989 x_0 ).So, the general term should be:[x_n = (-1)^n 1989 binom{1989}{n}.]Let me verify this for ( n = 1 ):[x_1 = (-1)^1 1989 binom{1989}{1} = -1989 times 1989 = -1989^2.]Correct.For ( n = 2 ):[x_2 = (-1)^2 1989 binom{1989}{2} = 1989 times frac{1989 times 1988}{2} = frac{1989^2 times 1988}{2}.]But earlier, I had:[x_2 = -frac{1989}{2} (1989 - 1989^2) = -frac{1989}{2} times 1989 (1 - 1989) = -frac{1989^2}{2} times (-1988) = frac{1989^2 times 1988}{2}.]Which matches. So, the general formula is:[x_n = (-1)^n 1989 binom{1989}{n}.]Great, so now I can express ( x_n ) as ( (-1)^n 1989 binom{1989}{n} ).Now, the sum I need to compute is:[sum_{n=0}^{1989} 2^n x_n = sum_{n=0}^{1989} 2^n (-1)^n 1989 binom{1989}{n}.]Factor out the constant 1989:[1989 sum_{n=0}^{1989} (-2)^n binom{1989}{n}.]Now, this sum resembles the binomial expansion of ( (1 + (-2))^{1989} ). Recall that:[sum_{n=0}^{m} binom{m}{n} a^n = (1 + a)^m.]In this case, ( a = -2 ) and ( m = 1989 ). So,[sum_{n=0}^{1989} binom{1989}{n} (-2)^n = (1 - 2)^{1989} = (-1)^{1989} = -1.]Therefore, the sum becomes:[1989 times (-1) = -1989.]So, the value of ( sum_{n=0}^{1989} 2^n x_n ) is ( -1989 ).**Final Answer**boxed{-1989}

answer：Okay, so I have this problem here: If ( x, y in mathbb{R} ), and ( 2y ) is the geometric mean of ( 1 + x ) and ( 1 - x ), then the trajectory of the moving point ( (x, y) ), excluding points on the x-axis, is which of the following? The options are a straight line, a circle, one branch of a hyperbola, or an ellipse.Alright, let me break this down. First, I remember that the geometric mean of two numbers ( a ) and ( b ) is ( sqrt{ab} ). So, if ( 2y ) is the geometric mean of ( 1 + x ) and ( 1 - x ), that means:[2y = sqrt{(1 + x)(1 - x)}]Hmm, okay. So, to get rid of the square root, I can square both sides of the equation. Let me do that:[(2y)^2 = (1 + x)(1 - x)]Calculating the left side, ( (2y)^2 ) is ( 4y^2 ). On the right side, ( (1 + x)(1 - x) ) is a difference of squares, so that simplifies to ( 1 - x^2 ). So now the equation becomes:[4y^2 = 1 - x^2]Let me rearrange this to get all terms on one side. I'll add ( x^2 ) to both sides:[x^2 + 4y^2 = 1]Hmm, this looks familiar. It's a quadratic equation in both ( x ) and ( y ). Let me recall the standard forms of conic sections. The general form of an ellipse is ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ). If I divide both sides of my equation by 1 to get it into that form, it becomes:[frac{x^2}{1} + frac{y^2}{frac{1}{4}} = 1]So, that simplifies to:[x^2 + frac{y^2}{frac{1}{4}} = 1]Which is indeed an ellipse with semi-major axis 1 along the x-axis and semi-minor axis ( frac{1}{2} ) along the y-axis.Wait, but the question mentions excluding points on the x-axis. So, points where ( y = 0 ). Let me see, if ( y = 0 ), then from the original equation ( 2y = sqrt{(1 + x)(1 - x)} ), we have ( 0 = sqrt{(1 + x)(1 - x)} ). Squaring both sides, ( 0 = (1 + x)(1 - x) ), so ( 1 - x^2 = 0 ), which gives ( x = pm 1 ). So, the points ( (1, 0) ) and ( (-1, 0) ) are excluded.But in the equation ( x^2 + 4y^2 = 1 ), when ( y = 0 ), ( x = pm 1 ), which are exactly the points we need to exclude. So, the trajectory is the ellipse ( x^2 + 4y^2 = 1 ) without the points ( (1, 0) ) and ( (-1, 0) ).Looking back at the options, A is a straight line, which doesn't fit because our equation is quadratic. B is a circle, but our equation isn't a circle because the coefficients of ( x^2 ) and ( y^2 ) are different. C is one branch of a hyperbola, but our equation is an ellipse, not a hyperbola. D is an ellipse, which matches our equation.Wait, but just to make sure, let me think again. The equation ( x^2 + 4y^2 = 1 ) is definitely an ellipse. The standard form is ( frac{x^2}{1^2} + frac{y^2}{(1/2)^2} = 1 ), so it's an ellipse centered at the origin with major axis length 2 and minor axis length 1.So, unless there's a trick here, the trajectory should be an ellipse. The only thing is that we're excluding points on the x-axis, but those are just two points on the ellipse, so the rest of the ellipse remains.I don't think it's a hyperbola because hyperbola equations have a minus sign between the terms, like ( x^2 - y^2 = 1 ), but here we have a plus sign. So, it's definitely an ellipse.Therefore, the answer should be D: An ellipse.**Final Answer**boxed{D}

answer：Okay, so I have this problem to prove that the difference of the squares of adjacent sides of a parallelogram is less than the product of its diagonals. Hmm, let me try to visualize a parallelogram first. A parallelogram has opposite sides equal and parallel, right? So, if I consider a parallelogram ABCD, then AB is equal to CD, and AD is equal to BC. The diagonals are AC and BD.I need to find the difference of the squares of adjacent sides. Let's say the sides are AB and AD. So, if AB is of length 'a' and AD is of length 'b', then the difference of their squares would be (a^2 - b^2). Wait, but the problem says "the difference of the squares of adjacent sides," so it could be either (a^2 - b^2) or (b^2 - a^2). But since it's a difference, I guess it's the absolute value, so (|a^2 - b^2|).Now, the product of the diagonals. The diagonals are AC and BD. Let me denote their lengths as 'd1' and 'd2'. So, the product would be (d1 times d2). The problem states that (|a^2 - b^2| < d1 times d2).Hmm, how do I relate the sides and the diagonals in a parallelogram? I remember there's a formula that relates the sides and diagonals in a parallelogram. It's something like the sum of the squares of the diagonals equals twice the sum of the squares of the sides. Let me write that down:(d1^2 + d2^2 = 2(a^2 + b^2))But wait, how does that help me with the product of the diagonals? Maybe I can use some inequality here. I know that for any two positive numbers, the arithmetic mean is greater than or equal to the geometric mean. So, (frac{d1^2 + d2^2}{2} geq sqrt{d1^2 times d2^2}), which simplifies to (frac{d1^2 + d2^2}{2} geq d1 times d2).But from the formula above, (d1^2 + d2^2 = 2(a^2 + b^2)), so substituting that in, we get:(frac{2(a^2 + b^2)}{2} geq d1 times d2)Which simplifies to:(a^2 + b^2 geq d1 times d2)Hmm, but I need to relate (|a^2 - b^2|) to (d1 times d2). Maybe I can find another relationship or use another inequality.Let me think about the vectors involved. In a parallelogram, the diagonals can be expressed as the sum and difference of the adjacent sides. If I consider vectors, let’s say vector AB is (vec{u}) and vector AD is (vec{v}). Then, the diagonals AC and BD can be written as (vec{u} + vec{v}) and (vec{v} - vec{u}), respectively.The product of the diagonals would then be (|vec{u} + vec{v}| times |vec{v} - vec{u}|). I recall that the product of the magnitudes of two vectors can be related to their dot product. The dot product of (vec{u} + vec{v}) and (vec{v} - vec{u}) is:((vec{u} + vec{v}) cdot (vec{v} - vec{u}) = vec{u} cdot vec{v} - vec{u} cdot vec{u} + vec{v} cdot vec{v} - vec{v} cdot vec{u})Simplifying this, the cross terms cancel out:(vec{u} cdot vec{v} - vec{v} cdot vec{u} = 0)So, we're left with:(vec{v} cdot vec{v} - vec{u} cdot vec{u} = |vec{v}|^2 - |vec{u}|^2 = b^2 - a^2)So, the dot product of the diagonals is (b^2 - a^2). But the product of the magnitudes of the diagonals is (|vec{u} + vec{v}| times |vec{v} - vec{u}|). I need to relate this to (|a^2 - b^2|).I remember that for any two vectors, the absolute value of their dot product is less than or equal to the product of their magnitudes. That is:(|(vec{u} + vec{v}) cdot (vec{v} - vec{u})| leq |vec{u} + vec{v}| times |vec{v} - vec{u}|)Substituting the dot product we found earlier:(|b^2 - a^2| leq |vec{u} + vec{v}| times |vec{v} - vec{u}|)Which is exactly what we wanted to prove:(|a^2 - b^2| < d1 times d2)Wait, but the inequality I used was less than or equal to. The problem says "less than." When does equality hold? It holds when the vectors are parallel or one is a scalar multiple of the other. In the case of a parallelogram, when would the diagonals be parallel? Only if the parallelogram is a rectangle or a rhombus? Hmm, maybe in a rectangle, the diagonals are equal, but they are not necessarily parallel. Wait, in a rectangle, the diagonals are equal in length but still intersect at some angle. So, maybe equality holds only when the parallelogram is a rectangle or a rhombus? Or perhaps when it's a square.But in general, for a non-rectangular and non-rhombus parallelogram, the inequality should be strict. So, we can say that (|a^2 - b^2| < d1 times d2) for a general parallelogram.Let me check with an example. Suppose I have a parallelogram with sides 3 and 4. Let's compute the diagonals. Using the formula (d1^2 + d2^2 = 2(a^2 + b^2)), so (d1^2 + d2^2 = 2(9 + 16) = 50). Let's say d1 = 5 and d2 = 5, then their product is 25. The difference of squares is (16 - 9 = 7), which is less than 25. If d1 = 1 and d2 = 7, their product is 7, which is still greater than 7. Wait, but can the diagonals be 1 and 7? Let me check if that satisfies (d1^2 + d2^2 = 50). 1 + 49 = 50, yes. So, even in this case, 7 < 7? No, that's equal. Hmm, so in this case, the product equals the difference of squares. But in reality, can the diagonals be 1 and 7? Because in a parallelogram, the length of the diagonals must satisfy the triangle inequality. For sides 3 and 4, the diagonals must satisfy |d1 - d2| < 2a < d1 + d2 or something like that. Wait, no, the triangle inequality for the diagonals would be that each diagonal must be less than the sum of the sides and greater than the difference. So, for a parallelogram, each diagonal must be greater than |a - b| and less than a + b. So, for sides 3 and 4, the diagonals must be greater than 1 and less than 7. So, d1 = 1 is not possible because it's equal to |4 - 3|, but in reality, the diagonals must be greater than that. So, maybe my example is invalid because d1 cannot be 1.Let me choose another example where d1 and d2 are both greater than 1. Suppose d1 = 5 and d2 = 5, as before. Then, the product is 25, and the difference of squares is 7, which is less than 25. Another example: sides 5 and 12. Then, (d1^2 + d2^2 = 2(25 + 144) = 338). Let's say d1 = 13 and d2 = 13, then their product is 169. The difference of squares is (144 - 25 = 119), which is less than 169. If d1 = 10 and d2 = sqrt(338 - 100) = sqrt(238) ≈ 15.43, then the product is 10 * 15.43 ≈ 154.3, which is still greater than 119.So, in these examples, the inequality holds. It seems that the product of the diagonals is always greater than the difference of the squares of the sides. Therefore, the proof using the Cauchy-Schwarz inequality (which is what I used with the dot product) shows that (|a^2 - b^2| leq d1 times d2), and in most cases, it's strictly less than, except when the parallelogram is a rectangle or a rhombus, where equality might hold. But in those cases, the difference of squares would be zero or something else, so maybe equality doesn't actually occur in the context of the problem.I think I have a good grasp now. The key was to express the diagonals as vectors, compute their dot product, and then apply the Cauchy-Schwarz inequality to relate the difference of squares to the product of the diagonals.

answer：Okay, so I need to solve this problem about triangles ABC where A is the origin, and points B and C are distinct points with nonnegative integer coordinates lying on the line 37x + y = 1853. I have to find how many such triangles have an area that's a positive integer.First, let me understand the problem step by step.1. **Points B and C**: Both points lie on the line 37x + y = 1853. So, for each point, if I know x, I can find y as y = 1853 - 37x. Since x and y must be nonnegative integers, x has to be such that y is also a nonnegative integer. So, x can range from 0 up to the maximum value where y is still nonnegative.2. **Coordinates of B and C**: Let me denote B as (x1, y1) and C as (x2, y2). So, y1 = 1853 - 37x1 and y2 = 1853 - 37x2.3. **Area of Triangle ABC**: The area can be calculated using the determinant formula for the area of a triangle given by three points. Since A is the origin (0,0), the area is (1/2)|x1*y2 - x2*y1|. Let me write that down:Area = (1/2)|x1*y2 - x2*y1|Substituting y1 and y2:Area = (1/2)|x1*(1853 - 37x2) - x2*(1853 - 37x1)|Let me expand this expression:= (1/2)|1853x1 - 37x1x2 - 1853x2 + 37x1x2|Wait, the -37x1x2 and +37x1x2 cancel each other out. So, that simplifies to:= (1/2)|1853x1 - 1853x2|Factor out 1853:= (1/2)*1853|x1 - x2|So, Area = (1853/2)|x1 - x2|Hmm, so for the area to be an integer, (1853/2)|x1 - x2| must be an integer. Since 1853 is an odd number (because 37*50 = 1850, so 1853 is 1850 + 3, which is odd), dividing it by 2 gives a non-integer. Therefore, to make the entire expression an integer, |x1 - x2| must be even. Because 1853 is odd, multiplying it by an even number will give an even number, which when divided by 2 gives an integer.So, the key condition is that |x1 - x2| must be even. That is, x1 and x2 must be both even or both odd. Because if two numbers are both even or both odd, their difference is even.So, now I need to find how many pairs of points (B, C) there are such that x1 and x2 are both even or both odd.First, let me figure out how many possible x-values there are for points on the line 37x + y = 1853 with nonnegative integer coordinates.We have y = 1853 - 37x ≥ 0, so 37x ≤ 1853. Therefore, x ≤ 1853/37.Let me compute 1853 divided by 37.37*50 = 1850, so 1853 - 1850 = 3. So, 1853/37 = 50 + 3/37 ≈ 50.081. So, x can be from 0 to 50, inclusive.Therefore, x can take integer values from 0, 1, 2, ..., 50. So, there are 51 possible x-values.Now, among these 51 x-values, how many are even and how many are odd?From 0 to 50 inclusive:Number of even x's: Let's see, starting from 0, which is even, then 2, 4, ..., 50. So, how many terms? The nth term is 0 + (n-1)*2 = 50. So, 50 = 2(n-1) => n-1 = 25 => n=26. So, 26 even numbers.Similarly, number of odd x's: From 1 to 49. So, 1, 3, ..., 49. The nth term is 1 + (n-1)*2 = 49. So, 49 = 1 + 2(n-1) => 48 = 2(n-1) => n-1=24 => n=25. So, 25 odd numbers.So, total points: 51, which is 26 even x's and 25 odd x's.Now, since we need to choose two distinct points B and C such that their x-coordinates are both even or both odd.The number of ways to choose two points with even x's: C(26, 2) = (26*25)/2 = 325.Similarly, the number of ways to choose two points with odd x's: C(25, 2) = (25*24)/2 = 300.Therefore, total number of such pairs is 325 + 300 = 625.But wait, hold on. The problem says "distinct triangles". So, does this count each triangle once? Because choosing B and C is the same as choosing C and B for the triangle. So, in combinations, C(n, 2) counts unordered pairs, which is correct because triangle ABC is the same as triangle ACB.So, 625 is the number of unordered pairs of points B and C with both x's even or both odd, which would result in the area being an integer.But let me double-check my reasoning.First, the area is (1853/2)|x1 - x2|. So, for this to be integer, |x1 - x2| must be even, as 1853 is odd. So, yes, x1 and x2 must be both even or both odd.Number of even x's: 26, number of odd x's:25.Number of pairs with both even: C(26,2)=325.Number of pairs with both odd: C(25,2)=300.Total: 325+300=625.Therefore, the number of such triangles is 625.Wait, but just to make sure, is there any possibility that two different pairs (B,C) could result in the same triangle? But since B and C are distinct points, each pair defines a unique triangle with A. So, each pair corresponds to exactly one triangle, and since we are counting unordered pairs, each triangle is counted once.Therefore, 625 is the correct number.But let me think again: the line 37x + y = 1853 has 51 points with integer coordinates (x from 0 to 50). So, total number of triangles is C(51,2)=1275. But we are only counting those where the area is integer, which is 625.So, 625 is less than 1275, which makes sense because only half of the pairs would have x's both even or both odd.Wait, 26 even and 25 odd, so the number of pairs with both even is C(26,2)=325, both odd is C(25,2)=300, so total 625. The rest of the pairs would be one even and one odd, which would result in |x1 -x2| being odd, so the area would be (1853/2)*odd, which is not integer because 1853 is odd, so 1853*odd is odd, and dividing by 2 would give a non-integer.Therefore, yes, only 625 triangles have integer area.So, I think that is the answer.**Final Answer**boxed{625}

answer：Okay, so I need to show that there are infinitely many positive integers ( n ) such that ( n ) divides ( 2^n + 1 ). Hmm, that sounds a bit tricky, but let me try to break it down.First, I should understand what it means for ( n ) to divide ( 2^n + 1 ). That means when you divide ( 2^n + 1 ) by ( n ), there's no remainder. In other words, ( 2^n + 1 ) is a multiple of ( n ). So, mathematically, this can be written as:[2^n equiv -1 pmod{n}]Which implies:[2^{2n} equiv 1 pmod{n}]Because if ( 2^n equiv -1 pmod{n} ), then squaring both sides gives ( 2^{2n} equiv 1 pmod{n} ).This tells me that the order of 2 modulo ( n ) divides ( 2n ). The order of an element ( a ) modulo ( n ) is the smallest positive integer ( k ) such that ( a^k equiv 1 pmod{n} ). So, in this case, the order of 2 modulo ( n ) must divide ( 2n ).But I'm not sure how to use this directly. Maybe I should look for specific forms of ( n ) that satisfy this condition. Let me think about small values of ( n ) and see if I can spot a pattern.Let's try ( n = 1 ):[2^1 + 1 = 3]Does 1 divide 3? Yes, because any number divides 1. So, ( n = 1 ) works.Next, ( n = 2 ):[2^2 + 1 = 5]Does 2 divide 5? No, because 5 divided by 2 is 2.5, which is not an integer. So, ( n = 2 ) doesn't work.How about ( n = 3 ):[2^3 + 1 = 9]Does 3 divide 9? Yes, because 9 divided by 3 is 3. So, ( n = 3 ) works.Moving on to ( n = 4 ):[2^4 + 1 = 17]Does 4 divide 17? No, because 17 divided by 4 is 4.25. So, ( n = 4 ) doesn't work.Next, ( n = 5 ):[2^5 + 1 = 33]Does 5 divide 33? No, because 33 divided by 5 is 6.6. So, ( n = 5 ) doesn't work.How about ( n = 6 ):[2^6 + 1 = 65]Does 6 divide 65? No, because 65 divided by 6 is approximately 10.833. So, ( n = 6 ) doesn't work.Let's try ( n = 7 ):[2^7 + 1 = 129]Does 7 divide 129? Let's see, 7 times 18 is 126, and 129 minus 126 is 3. So, no, 7 doesn't divide 129.Hmm, so far, only ( n = 1 ) and ( n = 3 ) work. Maybe I need to look for a pattern or a specific type of number.Wait, ( n = 1 ) is trivial because 1 divides everything. So, the interesting case is ( n = 3 ). Let me see if I can find more numbers like 3.What if I consider powers of 3? Let's try ( n = 9 ):[2^9 + 1 = 513]Does 9 divide 513? Let's check: 9 times 57 is 513. Yes, so ( n = 9 ) works.Interesting! So, ( n = 3 ) and ( n = 9 ) both work. Let me try ( n = 27 ):[2^{27} + 1]Calculating ( 2^{27} ) is a bit tedious, but let's see. ( 2^{10} = 1024 ), ( 2^{20} = 1,048,576 ), so ( 2^{27} = 134,217,728 ). Adding 1 gives 134,217,729.Does 27 divide 134,217,729? Let's check: 27 times 5,000,000 is 135,000,000, which is more than 134,217,729. So, let's try 27 times 4,970,000: 27 * 4,970,000 = 134,190,000. Subtracting that from 134,217,729 gives 27,729. Now, 27,729 divided by 27 is 1,027. So, 27 * 4,971,027 = 134,217,729. Therefore, yes, 27 divides ( 2^{27} + 1 ).Wow, so ( n = 3 ), ( n = 9 ), and ( n = 27 ) all work. It seems like powers of 3 might be a good candidate for ( n ). Let me test ( n = 81 ):[2^{81} + 1]This is a huge number, but I can use modular arithmetic to check divisibility by 81.I know that ( 2^4 = 16 equiv 16 pmod{81} )( 2^5 = 32 equiv 32 pmod{81} )( 2^6 = 64 equiv 64 pmod{81} )( 2^7 = 128 equiv 128 - 81 = 47 pmod{81} )( 2^8 = 94 equiv 94 - 81 = 13 pmod{81} )( 2^9 = 26 pmod{81} )( 2^{10} = 52 pmod{81} )( 2^{11} = 104 equiv 104 - 81 = 23 pmod{81} )( 2^{12} = 46 pmod{81} )( 2^{13} = 92 equiv 92 - 81 = 11 pmod{81} )( 2^{14} = 22 pmod{81} )( 2^{15} = 44 pmod{81} )( 2^{16} = 88 equiv 88 - 81 = 7 pmod{81} )( 2^{17} = 14 pmod{81} )( 2^{18} = 28 pmod{81} )( 2^{19} = 56 pmod{81} )( 2^{20} = 112 equiv 112 - 81 = 31 pmod{81} )( 2^{21} = 62 pmod{81} )( 2^{22} = 124 equiv 124 - 81 = 43 pmod{81} )( 2^{23} = 86 equiv 86 - 81 = 5 pmod{81} )( 2^{24} = 10 pmod{81} )( 2^{25} = 20 pmod{81} )( 2^{26} = 40 pmod{81} )( 2^{27} = 80 pmod{81} )( 2^{28} = 160 equiv 160 - 2*81 = 160 - 162 = -2 equiv 79 pmod{81} )( 2^{29} = 158 equiv 158 - 2*81 = 158 - 162 = -4 equiv 77 pmod{81} )( 2^{30} = 154 equiv 154 - 1*81 = 73 pmod{81} )( 2^{31} = 146 equiv 146 - 1*81 = 65 pmod{81} )( 2^{32} = 130 equiv 130 - 1*81 = 49 pmod{81} )( 2^{33} = 98 equiv 98 - 1*81 = 17 pmod{81} )( 2^{34} = 34 pmod{81} )( 2^{35} = 68 pmod{81} )( 2^{36} = 136 equiv 136 - 1*81 = 55 pmod{81} )( 2^{37} = 110 equiv 110 - 1*81 = 29 pmod{81} )( 2^{38} = 58 pmod{81} )( 2^{39} = 116 equiv 116 - 1*81 = 35 pmod{81} )( 2^{40} = 70 pmod{81} )( 2^{41} = 140 equiv 140 - 1*81 = 59 pmod{81} )( 2^{42} = 118 equiv 118 - 1*81 = 37 pmod{81} )( 2^{43} = 74 pmod{81} )( 2^{44} = 148 equiv 148 - 1*81 = 67 pmod{81} )( 2^{45} = 134 equiv 134 - 1*81 = 53 pmod{81} )( 2^{46} = 106 equiv 106 - 1*81 = 25 pmod{81} )( 2^{47} = 50 pmod{81} )( 2^{48} = 100 equiv 100 - 1*81 = 19 pmod{81} )( 2^{49} = 38 pmod{81} )( 2^{50} = 76 pmod{81} )( 2^{51} = 152 equiv 152 - 1*81 = 71 pmod{81} )( 2^{52} = 142 equiv 142 - 1*81 = 61 pmod{81} )( 2^{53} = 122 equiv 122 - 1*81 = 41 pmod{81} )( 2^{54} = 82 equiv 82 - 1*81 = 1 pmod{81} )Oh, interesting! So, ( 2^{54} equiv 1 pmod{81} ). That means the order of 2 modulo 81 is 54. But we're interested in ( 2^{81} + 1 ). Let's see:Since ( 2^{54} equiv 1 pmod{81} ), then ( 2^{54k} equiv 1 pmod{81} ) for any integer ( k ). So, ( 2^{81} = 2^{54 + 27} = 2^{54} cdot 2^{27} equiv 1 cdot 2^{27} pmod{81} ).From earlier, we saw that ( 2^{27} equiv 80 pmod{81} ). So, ( 2^{81} equiv 80 pmod{81} ). Therefore, ( 2^{81} + 1 equiv 80 + 1 = 81 equiv 0 pmod{81} ). So, yes, 81 divides ( 2^{81} + 1 ).Wow, so ( n = 3 ), ( n = 9 ), ( n = 27 ), and ( n = 81 ) all work. It seems like powers of 3 are a good candidate. Let me try to generalize this.Suppose ( n = 3^k ) for some positive integer ( k ). I want to show that ( 3^k ) divides ( 2^{3^k} + 1 ).Let's use mathematical induction to prove this.**Base Case:** ( k = 1 ). Then ( n = 3 ). We saw earlier that ( 2^3 + 1 = 9 ), and 3 divides 9. So, the base case holds.**Inductive Step:** Assume that for some integer ( k geq 1 ), ( 3^k ) divides ( 2^{3^k} + 1 ). We need to show that ( 3^{k+1} ) divides ( 2^{3^{k+1}} + 1 ).Let me write ( n = 3^{k+1} ). Then, ( 2^{n} + 1 = 2^{3^{k+1}} + 1 ).Notice that ( 3^{k+1} = 3 cdot 3^k ). So, ( 2^{3^{k+1}} = (2^{3^k})^3 ).From the inductive hypothesis, ( 2^{3^k} equiv -1 pmod{3^k} ). Let's see if we can lift this to modulo ( 3^{k+1} ).Using the lifting the exponent lemma (LTE), which states that if ( p ) is an odd prime, ( p ) divides ( a + b ), and ( p ) doesn't divide ( a ) or ( b ), then:[v_p(a^n + b^n) = v_p(a + b) + v_p(n)]where ( v_p ) is the exponent of the highest power of ( p ) dividing the number.In our case, ( p = 3 ), ( a = 2 ), ( b = 1 ), and ( n = 3^k ). Wait, but ( 2 + 1 = 3 ), which is divisible by 3. Also, 3 doesn't divide 2 or 1. So, LTE applies.Thus,[v_3(2^{3^k} + 1^{3^k}) = v_3(2 + 1) + v_3(3^k) = v_3(3) + v_3(3^k) = 1 + k]This means that ( 3^{k+1} ) divides ( 2^{3^k} + 1 ).But wait, we need to show that ( 3^{k+1} ) divides ( 2^{3^{k+1}} + 1 ), not ( 2^{3^k} + 1 ). Hmm, maybe I need to adjust my approach.Let me consider ( 2^{3^{k+1}} + 1 = (2^{3^k})^3 + 1 ). Let me denote ( x = 2^{3^k} ). Then, the expression becomes ( x^3 + 1 ).Factorizing ( x^3 + 1 ), we get:[x^3 + 1 = (x + 1)(x^2 - x + 1)]From the inductive hypothesis, ( x equiv -1 pmod{3^k} ). So, ( x + 1 equiv 0 pmod{3^k} ).We need to find the highest power of 3 dividing ( x^3 + 1 ). Since ( x equiv -1 pmod{3^k} ), let's write ( x = -1 + 3^k m ) for some integer ( m ).Then,[x^3 + 1 = (-1 + 3^k m)^3 + 1]Expanding this:[(-1)^3 + 3(-1)^2(3^k m) + 3(-1)(3^k m)^2 + (3^k m)^3 + 1]Simplifying:[-1 + 3 cdot 1 cdot 3^k m + 3 cdot (-1) cdot 3^{2k} m^2 + 3^{3k} m^3 + 1]The -1 and +1 cancel out:[3^{k+1} m - 3^{2k + 1} m^2 + 3^{3k} m^3]Now, factor out ( 3^{k+1} ):[3^{k+1} left( m - 3^{k} m^2 + 3^{2k - 1} m^3 right)]Since ( k geq 1 ), ( 3^{k} ) and ( 3^{2k - 1} ) are integers. Therefore, ( x^3 + 1 ) is divisible by ( 3^{k+1} ).Thus, ( 2^{3^{k+1}} + 1 ) is divisible by ( 3^{k+1} ), completing the inductive step.By the principle of mathematical induction, for all positive integers ( k ), ( n = 3^k ) satisfies ( n ) divides ( 2^n + 1 ). Therefore, there are infinitely many such ( n ).I think this makes sense. I started by checking small cases, noticed a pattern with powers of 3, and then used induction with the lifting the exponent lemma to generalize it. It seems solid!