🧠:[<think>Okay, so I have this problem where I need to find the maximum value of the function ( f(x_1, x_2, ldots, x_n) = frac{x_1 sqrt{sum_{i=1}^{n} x_i}}{left( sum_{i=1}^{n-1} x_i right)^2 + x_n} ) given that all the ( x_i ) are non-negative and the sum ( sum_{i=1}^{n} x_i geq k ), where ( k geq 1 ). Hmm, let me think about how to approach this.First, I should note that the problem is asking for the maximum value of this function under the given constraints. Since all variables are non-negative and their total sum is at least ( k ), maybe the maximum occurs when the total sum is exactly ( k )? Because increasing the total sum beyond ( k ) might not necessarily increase the function value due to the denominator. Let me check that intuition.Suppose the total sum ( S = sum_{i=1}^n x_i ). Then, the function can be rewritten as ( f = frac{x_1 sqrt{S}}{(sum_{i=1}^{n-1} x_i)^2 + x_n} ). If we fix ( S = k ), maybe the maximum is achieved here? Let me assume that for now. So, we can set ( S = k ), and then try to maximize ( frac{x_1 sqrt{k}}{(k - x_n)^2 + x_n} ). Hmm, since ( sum_{i=1}^{n-1} x_i = k - x_n ).So now, the problem reduces to maximizing ( frac{x_1 sqrt{k}}{(k - x_n)^2 + x_n} ) under the constraints ( x_1 + sum_{i=2}^{n-1} x_i + x_n = k ), and all variables non-negative. Wait, but this is for fixed ( S = k ). Maybe it's better to consider variables ( x_1, x_2, ldots, x_{n-1}, x_n ), with ( x_1 + x_2 + ldots + x_{n-1} = T ), so ( T + x_n = S geq k ). But if we set ( S = k ), then ( T + x_n = k ), so ( x_n = k - T ). Then, the function becomes ( frac{x_1 sqrt{k}}{T^2 + (k - T)} ).But maybe I can consider this as a function of ( T ) and ( x_1 ). However, since ( x_1 leq T ), because ( x_1 ) is part of the sum ( T ). So, to maximize ( x_1 ), given ( T ), we would set ( x_1 = T ) and all other ( x_2, ldots, x_{n-1} = 0 ). That way, ( x_1 ) is as large as possible, which would help maximize the numerator. Let me test this idea.If we set ( x_2 = x_3 = ldots = x_{n-1} = 0 ), then ( T = x_1 ), and ( x_n = k - x_1 ). Then, the function becomes ( frac{x_1 sqrt{k}}{x_1^2 + (k - x_1)} ). So now, the problem is reduced to maximizing ( frac{x_1 sqrt{k}}{x_1^2 + k - x_1} ) with ( 0 leq x_1 leq k ).This seems like a single-variable optimization problem. Let me set ( g(x_1) = frac{x_1}{x_1^2 + k - x_1} ). Then, the maximum of ( g(x_1) sqrt{k} ) would be at the same point as the maximum of ( g(x_1) ).To find the maximum of ( g(x_1) ), take the derivative:( g'(x_1) = frac{(1)(x_1^2 + k - x_1) - x_1(2x_1 - 1)}{(x_1^2 + k - x_1)^2} )Simplify the numerator:( (x_1^2 + k - x_1) - x_1(2x_1 - 1) = x_1^2 + k - x_1 - 2x_1^2 + x_1 = -x_1^2 + k )So, ( g'(x_1) = frac{ -x_1^2 + k }{(x_1^2 + k - x_1)^2} )Set the derivative equal to zero:( -x_1^2 + k = 0 implies x_1^2 = k implies x_1 = sqrt{k} )But we need to check if ( x_1 = sqrt{k} ) is within the interval [0, k]. Since ( k geq 1 ), and ( sqrt(k) leq k ) when ( k geq 1 ), because for ( k geq 1 ), ( sqrt(k) leq k ). For example, if ( k=1 ), sqrt(1)=1; if ( k=2, sqrt(2)≈1.414 < 2, etc. So yes, ( sqrt(k) leq k ).Therefore, the critical point is at ( x_1 = sqrt(k) ). Then, the maximum value of ( g(x_1) ) is ( frac{sqrt(k)}{(sqrt(k))^2 + k - sqrt(k)} = frac{sqrt(k)}{k + k - sqrt(k)} = frac{sqrt(k)}{2k - sqrt(k)} ).So, the maximum value of ( f ) would be ( sqrt(k) * frac{sqrt(k)}{2k - sqrt(k)} = frac{k}{2k - sqrt(k)} ).Wait, but hold on. The function ( f ) is ( g(x_1) * sqrt(k) ), so actually:( f = frac{x_1 sqrt(k)}{x_1^2 + k - x_1} ). At ( x_1 = sqrt(k) ), it would be ( frac{sqrt(k) * sqrt(k)}{(sqrt(k))^2 + k - sqrt(k)} = frac{k}{k + k - sqrt(k)} = frac{k}{2k - sqrt(k)} ).Simplify that expression: divide numerator and denominator by k:( frac{1}{2 - 1/sqrt(k)} ).But is this really the maximum? Let me check endpoints.At ( x_1 = 0 ), the function is 0.At ( x_1 = k ), then ( x_n = k - x_1 = 0 ), so the denominator is ( k^2 + 0 = k^2 ). Then, the function is ( (k * sqrt(k)) / k^2 = sqrt(k)/k = 1/sqrt(k) ).Compare this with the critical point value ( frac{k}{2k - sqrt(k)} ). Let me compute both for a specific k, say k=1.If k=1: critical point value is ( 1/(2*1 - 1) = 1/1 = 1 ). Endpoint value is 1/sqrt(1)=1. So both are equal. Interesting.Wait, so at k=1, both the critical point and the endpoint give the same value. Maybe that's a case where the maximum occurs at both points? But when k=2, for example:Critical point value: ( 2/(4 - sqrt(2)) ≈ 2 / (4 - 1.414) ≈ 2 / 2.586 ≈ 0.774 ).Endpoint value: 1/sqrt(2) ≈ 0.707. So critical point is higher here. So for k>1, the critical point gives a higher value than the endpoint. But when k=1, they are equal. So perhaps the maximum is achieved at x1 = sqrt(k) when k >=1, but we have to check whether this is indeed the case.Wait, but when k=1, sqrt(k)=1, which is the same as the endpoint x1=k=1. So in that case, it's the same point. So maybe in general, the maximum is achieved at x1 = sqrt(k) if sqrt(k) <= k, which is always true for k>=1, since sqrt(k) <=k when k>=1.Therefore, the maximum value of the function is ( frac{k}{2k - sqrt(k)} ).But wait, let's check another example. Suppose k=4. Then, critical point x1=2. Then, denominator becomes (2^2) + 4 -2 =4 +4 -2=6. So f= (2*sqrt(4))/6 = (2*2)/6=4/6=2/3≈0.666. Endpoint at x1=4: denominator=16 +0=16, f=4*sqrt(4)/16=4*2/16=8/16=0.5. So critical point is higher. So yes, the critical point gives a higher value.But wait, wait, when k=1: x1=1, denominator=1 +1 -1=1, f=1*sqrt(1)/1=1. Which is equal to the endpoint. So in that case, both critical point and endpoint give the same value.But is this correct? Let me verify with k=1. If all variables except x1 and xn are zero, x1 + xn =1. Then, the function is (x1 *1)/(x1^2 + (1 -x1)). At x1=1, it's (1*1)/(1 +0)=1. At x1=sqrt(1)=1, same result. So yes, same value.So, in general, the maximum occurs at x1=sqrt(k), xn= k - sqrt(k), and the other variables zero. Therefore, the maximum value is k/(2k - sqrt(k)).But let me think again. Wait, when we set all variables except x1 and xn to zero, is that valid? The problem allows x_i >=0, so yes. By setting x2,...,x_{n-1} to zero, we can focus on x1 and xn. So that seems a valid approach.But perhaps there's another configuration where more variables are non-zero that could give a higher value? Let me check.Suppose n=3, k=1. So variables x1, x2, x3. Suppose I set x2 to some positive value. Let's say x1=a, x2=b, x3=1 -a -b. Then, the function is [a * sqrt(1)] / [(a + b)^2 + (1 -a -b)]. So denominator is (a + b)^2 +1 -a -b. Let's try a=1, b=0, which gives denominator=1 +0=1, so function=1/1=1. If I set a=sqrt(1)=1, but that's the same as above. If I set a=0.5, b=0.5, then denominator=(1)^2 + (1 -1)=1 +0=1, so function=0.5 /1=0.5, which is lower. If I set a=0.8, b=0.1, x3=0.1, denominator=(0.9)^2 +0.1=0.81 +0.1=0.91, so function=0.8 /0.91≈0.879, which is less than 1. So seems like setting other variables to zero gives the maximum. So maybe indeed the maximum is achieved when only x1 and xn are non-zero.Alternatively, suppose n=2. Then the function becomes [x1 sqrt(x1 +x2)] / [x1^2 +x2]. With x1 +x2 >=k. If n=2, then setting x2= k -x1. Then function is [x1 sqrt(k)]/[x1^2 +k -x1], same as before. So same result. So whether n=2 or higher, if we set the intermediate variables to zero, we can get the same expression.Therefore, regardless of n, as long as n >=2, the maximum is achieved when x1=sqrt(k), xn= k -sqrt(k), and the rest are zero. Therefore, the maximum value is k / (2k -sqrt(k)).But let me check the case when k=1. Then, maximum value is 1/(2 -1)=1. Which matches the example. For k=4, it's 4/(8 -2)=4/6=2/3≈0.666, which matched my earlier calculation.But wait, the problem states that k >=1. So this formula holds for k>=1. But let's check another value. Let's take k=2. Then, maximum value is 2/(4 -sqrt(2))≈2/(4 -1.414)=2/2.586≈0.774. Is this indeed the maximum?Let me plug into the function when x1=sqrt(2)≈1.414, xn=2 -1.414≈0.586, rest zero. Then, the numerator is x1*sqrt(k)=1.414*sqrt(2)=1.414*1.414≈2. The denominator is (sqrt(2))^2 +0.586≈2 +0.586≈2.586. Then, 2/2.586≈0.774. If I set x1=2, xn=0, then function value is 2*sqrt(2)/4≈2*1.414/4≈2.828/4≈0.707, which is lower. So yes, the critical point gives a higher value.Therefore, it seems that the maximum value is indeed k/(2k -sqrt(k)).But wait, let me check if this function can be further maximized by having S>k. Suppose instead of setting S=k, we set S>k. Let me see if that could lead to a higher value.Suppose we let S =k + t, where t>0. Then, the function becomes [x1 sqrt(k + t)] / [ (T)^2 + (k + t - T) ] where T =sum_{i=1}^{n-1}x_i. If we set x2,...,x_{n-1}=0, then T =x1, and x_n =k + t -x1. Then, the function is [x1 sqrt(k + t)] / [x1^2 + (k + t -x1)].To see if this can be higher than when t=0, let's take derivative with respect to x1.Let’s denote S =k + t. Then, function is x1 sqrt(S)/(x1^2 + S -x1). Let me compute its derivative with respect to x1.Numerator derivative: sqrt(S)[(x1^2 + S -x1) -x1(2x1 -1)]/(x1^2 + S -x1)^2Wait, similar to before. The derivative's numerator is sqrt(S)[(1)(x1^2 + S -x1) -x1(2x1 -1)] = sqrt(S)[x1^2 + S -x1 -2x1^2 +x1] = sqrt(S)[-x1^2 + S].Setting derivative zero: -x1^2 + S =0 => x1= sqrt(S). Then, maximum at x1= sqrt(S). Then, function value is [sqrt(S) * sqrt(S)] / [ S + S - sqrt(S) ]= S/(2S - sqrt(S))=1/(2 -1/sqrt(S)).So, as S increases, 1/sqrt(S) decreases, so denominator 2 -1/sqrt(S) increases, hence the whole expression decreases. Therefore, the maximum value of this function decreases as S increases. Therefore, the maximum occurs at the minimal S, which is S=k. Therefore, indeed, setting S=k gives the maximum value.Therefore, our initial assumption that the maximum occurs at S=k is correct. Therefore, the maximum value is k/(2k - sqrt(k)).But let me check another angle. Suppose we use the method of Lagrange multipliers for constrained optimization.We need to maximize f(x1,...,xn)=x1 sqrt(S)/( (sum_{i=1}^{n-1}x_i)^2 +x_n ), where S=sum_{i=1}^n x_i >=k, and x_i >=0.But since we established that increasing S beyond k decreases the function, the maximum should be on the boundary where S=k.So we can set the constraint S=k, and use Lagrange multipliers.Let me set up the Lagrangian. Let me denote T =sum_{i=1}^{n-1}x_i. Then, the function is f= x1 sqrt(k)/(T^2 +x_n), with T +x_n =k.So substituting x_n =k - T, we have f= x1 sqrt(k)/(T^2 +k -T). We need to maximize x1/(T^2 +k -T) with x1 <=T (since x1 is part of T).So, to maximize x1/(T^2 +k -T), with x1 <=T and T <=k (since x_n=k -T >=0).But since x1 is part of T, the maximum of x1 for a given T is x1=T (setting all other x_i=0 for i=2,...,n-1). Therefore, maximum f is T sqrt(k)/(T^2 +k -T). Then, this reduces to the same single-variable problem as before, with maximum at T=sqrt(k). Therefore, confirming our previous result.Therefore, using Lagrange multipliers also leads us to the same conclusion.Therefore, the maximum value of the function is k/(2k -sqrt(k)).But let me check if this expression can be simplified or written in another form. For example, factor numerator and denominator:k/(2k -sqrt(k)) = [sqrt(k)]^2 / [2k - sqrt(k)] = sqrt(k) * sqrt(k) / [2k - sqrt(k)].Alternatively, multiply numerator and denominator by 1/sqrt(k):= sqrt(k)/ [2 sqrt(k) -1].So, another way to write it is sqrt(k)/(2 sqrt(k) -1).Which might be a simpler expression. Let me check:sqrt(k)/(2 sqrt(k) -1) = [sqrt(k)] / [2 sqrt(k) -1].Compare with previous: k/(2k -sqrt(k)).Multiply numerator and denominator of the second expression by 1/k:=1/(2 -1/sqrt(k)).Which is also a valid expression. So depending on how the answer is expected, either form is acceptable. But perhaps the problem expects it in terms of k, so k/(2k -sqrt(k)) or sqrt(k)/(2 sqrt(k)-1). Let me check for k=1, both expressions give 1/(2*1 -1)=1 and sqrt(1)/(2*sqrt(1)-1)=1/(2-1)=1. For k=4: 4/(8-2)=4/6=2/3 vs sqrt(4)/(4 -1)=2/3, same result. So both expressions are equivalent.Therefore, both forms are correct. Maybe the answer is presented as sqrt(k)/(2 sqrt(k)-1). Let me verify algebraically:sqrt(k)/(2 sqrt(k) -1) = [sqrt(k)] / [2 sqrt(k) -1]Multiply numerator and denominator by sqrt(k):= k / [2k - sqrt(k)]Which is the original expression. So yes, equivalent.Therefore, the maximum value is k/(2k - sqrt(k)) or equivalently sqrt(k)/(2 sqrt(k) -1). Either form is acceptable, but maybe the first one is better as it's in terms of k without square roots. However, the problem might prefer one over the other. Let me see the problem statement example. The example in the problem is Example 1, but there's no solution provided. Since the problem is stated in terms of k, I think both forms are acceptable, but perhaps we can rationalize the denominator for the second form.Alternatively, note that:sqrt(k)/(2 sqrt(k) -1) = [sqrt(k)] / [2 sqrt(k) -1] = [sqrt(k)(2 sqrt(k) +1)] / [ (2 sqrt(k) -1)(2 sqrt(k)+1) ] = [2k + sqrt(k)] / [4k -1]But this introduces a more complex expression. So probably better to leave it as k/(2k -sqrt(k)).Alternatively, check if the problem expects the answer in terms of a fraction with denominator rationalized. But given the problem statement, probably either form is acceptable.But let me check for k=1, the maximum value is 1/(2*1 -1)=1, which is correct as we saw. For k=2, it's 2/(4 -sqrt(2))≈0.774, which seems right.Alternatively, maybe the maximum is 1/2. Wait, for k=1, it's 1; for larger k, the value decreases. For example, as k approaches infinity, the expression k/(2k -sqrt(k)) ≈ k/(2k) =1/2. So the supremum as k approaches infinity is 1/2, but for finite k >=1, the maximum is k/(2k -sqrt(k)) which is greater than 1/2.But the problem asks for the maximum value given that k is a fixed constant >=1. So the answer should be in terms of k. Therefore, the maximum value is k/(2k -sqrt(k)) or sqrt(k)/(2 sqrt(k) -1).But perhaps to present it in a simplified fractional form, but I don't see a more simplified form. So I think either expression is correct, but maybe the problem expects the answer as sqrt(k)/(2 sqrt(k) -1). Let me check with an example.If k=4, sqrt(k)=2, so 2/(4 -1)=2/3, which matches 4/(8 -2)=4/6=2/3.Therefore, both forms are correct. Maybe the answer is written as boxed{dfrac{sqrt{k}}{2sqrt{k} - 1}}.Alternatively, boxed{dfrac{k}{2k - sqrt{k}}}.Both are correct. Let me check standard optimization problems. Often, rationalized forms are preferred, but since both are equivalent, it's up to the problem's preference. However, considering that the problem includes a square root in the numerator, writing the answer with sqrt(k) in the numerator might be more elegant. Let me confirm with k=1: sqrt(1)/(2 sqrt(1)-1)=1/(2 -1)=1, which is correct. So, yes, the answer can be written as sqrt(k)/(2 sqrt(k)-1). So I think that's the better form.Therefore, the maximum value is sqrt(k)/(2 sqrt(k) -1), so boxed answer is boxed{dfrac{sqrt{k}}{2sqrt{k} - 1}}.Final AnswerThe maximum value is boxed{dfrac{sqrt{k}}{2sqrt{k} - 1}}.</think>Given the problem of finding the maximum value of the function ( f(x_1, x_2, ldots, x_n) = frac{x_1 sqrt{sum_{i=1}^{n} x_i}}{left( sum_{i=1}^{n-1} x_i right)^2 + x_n} ) under the constraints ( x_1, x_2, ldots, x_n geq 0 ) and ( sum_{i=1}^{n} x_i geq k ) (where ( k geq 1 )), we proceed as follows:1. Assumption and Simplification: - We assume the total sum ( S = sum_{i=1}^n x_i = k ) since increasing ( S ) beyond ( k ) might not increase the function value. - Let ( T = sum_{i=1}^{n-1} x_i ). Then, ( x_n = k - T ). - The function simplifies to ( frac{x_1 sqrt{k}}{T^2 + (k - T)} ).2. Maximizing the Function: - To maximize ( x_1 ), we set ( x_2, x_3, ldots, x_{n-1} = 0 ), reducing ( T ) to ( x_1 ). - The function becomes ( frac{x_1 sqrt{k}}{x_1^2 + (k - x_1)} ).3. Single-Variable Optimization: - We find the critical points by differentiating ( g(x_1) = frac{x_1}{x_1^2 + k - x_1} ). - The critical point is found at ( x_1 = sqrt{k} ).4. Verification: - Substituting ( x_1 = sqrt{k} ) into the function, we get the value ( frac{sqrt{k} sqrt{k}}{(sqrt{k})^2 + k - sqrt{k}} = frac{k}{2k - sqrt{k}} ). - This simplifies to ( frac{sqrt{k}}{2sqrt{k} - 1} ).5. Conclusion: - The maximum value of the function is achieved when ( x_1 = sqrt{k} ), ( x_n = k - sqrt{k} ), and all other variables are zero.Thus, the maximum value is boxed{dfrac{sqrt{k}}{2sqrt{k} - 1}}.,<think>Okay, so I need to find the maximum value of this function ( f(x_1, x_2, ldots, x_n) = frac{x_1 sqrt{sum_{i=1}^{n} x_i}}{left(sum_{i=1}^{n-1} x_iright)^2 + x_n} ) given that all the ( x_i ) are non-negative and the sum of all ( x_i ) is at least ( k ), where ( k geq 1 ). Hmm, let's break this down step by step.First, the function looks a bit complicated. Let me rewrite it to see if I can simplify or find some structure. The numerator is ( x_1 ) times the square root of the total sum ( S = sum_{i=1}^{n} x_i ). The denominator is the square of the sum of the first ( n-1 ) variables plus the last variable ( x_n ). So, if I denote ( T = sum_{i=1}^{n-1} x_i ), then the function becomes ( f = frac{x_1 sqrt{S}}{T^2 + x_n} ). But ( S = T + x_n ), so substituting that in, the function becomes ( f = frac{x_1 sqrt{T + x_n}}{T^2 + x_n} ).Now, the constraints are ( x_i geq 0 ) for all ( i ), and ( S = T + x_n geq k ). We need to maximize ( f ).Let me consider if there's a way to use some optimization techniques here, like Lagrange multipliers, but since all variables are non-negative and the problem might have some symmetry or substitution possibilities, maybe I can reduce the number of variables.First, let's note that ( x_1 ) is in the numerator, so to maximize ( f ), we probably want ( x_1 ) to be as large as possible. However, increasing ( x_1 ) increases ( T ), which is in the denominator squared. So there's a trade-off here. Similarly, increasing ( x_n ) increases ( S ) (which is under a square root in the numerator) but also increases the denominator linearly. So perhaps there's a balance between these variables.Since all variables are non-negative, maybe the maximum occurs when some variables are zero? For example, if we set ( x_2, x_3, ldots, x_{n-1} ) to zero, then ( T = x_1 ), and ( S = x_1 + x_n ). Then the function simplifies to ( f = frac{x_1 sqrt{x_1 + x_n}}{x_1^2 + x_n} ). Maybe this is easier to handle. Let me check if this assumption is valid.If we set some variables to zero, does that restrict the maximum? Suppose we have ( x_2, ldots, x_{n-1} ) positive. Then, redistributing some of their values to ( x_1 ) or ( x_n ) might increase ( f ), since ( x_1 ) is in the numerator and others are only contributing to ( T ), which is squared in the denominator. Therefore, maybe setting them to zero does give a higher value. So perhaps the maximum occurs when ( x_2 = x_3 = ldots = x_{n-1} = 0 ).Assuming that, then the problem reduces to two variables: ( x_1 ) and ( x_n ), with ( x_1 + x_n geq k ), and ( x_1, x_n geq 0 ). The function becomes ( f(x_1, x_n) = frac{x_1 sqrt{x_1 + x_n}}{x_1^2 + x_n} ).Now, we can consider substituting ( S = x_1 + x_n geq k ). Let me denote ( S = x_1 + x_n ). Then ( x_n = S - x_1 ), so substituting into the function:( f = frac{x_1 sqrt{S}}{x_1^2 + S - x_1} ).So, ( f = frac{x_1 sqrt{S}}{x_1^2 - x_1 + S} ).But we have ( S geq k ). However, since we are trying to maximize ( f ), perhaps the maximum occurs at the minimal possible ( S ), which is ( S = k ). Because if we can achieve a higher ( f ) with a smaller ( S ), then since ( S geq k ), increasing ( S ) beyond ( k ) might not be beneficial. Let's check this intuition.Suppose ( S = k ), then ( x_n = k - x_1 ), and the function becomes ( f = frac{x_1 sqrt{k}}{x_1^2 + (k - x_1)} ). Let's denote ( g(x_1) = frac{x_1 sqrt{k}}{x_1^2 + k - x_1} ). To maximize ( g(x_1) ), we can take the derivative with respect to ( x_1 ).First, let's write ( g(x_1) = frac{x_1 sqrt{k}}{x_1^2 - x_1 + k} ). Let's compute the derivative ( g'(x_1) ):Let ( N = x_1 sqrt{k} ), ( D = x_1^2 - x_1 + k ).Then, ( g'(x_1) = frac{N' D - N D'}{D^2} ).Compute ( N' = sqrt{k} ).Compute ( D' = 2x_1 - 1 ).So,( g'(x_1) = frac{sqrt{k}(x_1^2 - x_1 + k) - x_1 sqrt{k}(2x_1 - 1)}{(x_1^2 - x_1 + k)^2} )Factor out ( sqrt{k} ):( g'(x_1) = frac{sqrt{k} [ (x_1^2 - x_1 + k) - x_1(2x_1 - 1) ] }{(x_1^2 - x_1 + k)^2} )Simplify the numerator inside the brackets:( x_1^2 - x_1 + k - 2x_1^2 + x_1 = -x_1^2 + k )So,( g'(x_1) = frac{sqrt{k}( -x_1^2 + k )}{(x_1^2 - x_1 + k)^2} )Set derivative to zero:( -x_1^2 + k = 0 Rightarrow x_1^2 = k Rightarrow x_1 = sqrt{k} ). Since ( x_1 geq 0 ).Now, we need to check if this critical point is within the domain. Since ( S = k ), ( x_1 leq k ). And since ( x_1 geq 0 ), and ( sqrt{k} leq k ) only if ( k geq 1 ), which is given (since ( k geq 1 )). So, ( x_1 = sqrt{k} ) is feasible.Therefore, the maximum when ( S = k ) is at ( x_1 = sqrt{k} ), then ( x_n = k - sqrt{k} ).Compute ( g(sqrt{k}) = frac{sqrt{k} cdot sqrt{k}}{ (sqrt{k})^2 - sqrt{k} + k } = frac{k}{k - sqrt{k} + k} = frac{k}{2k - sqrt{k}} ).Simplify denominator: ( 2k - sqrt{k} = sqrt{k}(2sqrt{k} - 1) ).Thus, ( g(sqrt{k}) = frac{k}{sqrt{k}(2sqrt{k} - 1)} = frac{sqrt{k}}{2sqrt{k} - 1} ).So that's the value when ( S = k ).But we need to check whether taking ( S > k ) can lead to a higher value. Let's suppose that ( S = m ), where ( m > k ). Then, we can write:( f = frac{x_1 sqrt{m}}{x_1^2 + m - x_1} ).Similarly, taking derivative with respect to ( x_1 ), same steps as before, set to zero:( -x_1^2 + m = 0 Rightarrow x_1 = sqrt{m} ). But then ( x_n = m - sqrt{m} ), and ( m geq k ).Then, substituting back, the maximum for each ( m ) is ( frac{sqrt{m}}{2sqrt{m} - 1} ).So the question becomes, does ( frac{sqrt{m}}{2sqrt{m} - 1} ) increase or decrease as ( m ) increases? Let's check by considering the function ( h(t) = frac{t}{2t - 1} ), where ( t = sqrt{m} geq sqrt{k} geq 1 ).Compute derivative ( h'(t) = frac{(2t - 1) cdot 1 - t cdot 2}{(2t - 1)^2} = frac{2t - 1 - 2t}{(2t - 1)^2} = frac{-1}{(2t - 1)^2} < 0 ). So ( h(t) ) is decreasing in ( t ). Therefore, as ( m ) increases, ( t = sqrt{m} ) increases, so ( h(t) ) decreases. Therefore, the maximum value of ( h(t) ) occurs at the minimal possible ( t ), which is ( t = sqrt{k} ). Therefore, the maximum of ( f ) occurs when ( m = k ), and ( x_1 = sqrt{k} ), ( x_n = k - sqrt{k} ), with the other variables set to zero.Therefore, the maximum value is ( frac{sqrt{k}}{2sqrt{k} - 1} ).Wait, but let's verify this. Suppose ( k = 1 ), then the maximum value would be ( frac{1}{2 cdot 1 - 1} = 1 ). Let's check if that makes sense.If ( k = 1 ), then ( x_1 = 1 ), ( x_n = 0 ), so ( S = 1 ), then ( f = frac{1 cdot sqrt{1}}{1^2 + 0} = 1 ). That seems correct. If we set ( x_1 = 1 ), ( x_n = 0 ), then indeed ( f = 1 ). If we tried to set ( S > 1 ), say ( S = 2 ), then the maximum for ( S = 2 ) is ( frac{sqrt{2}}{2sqrt{2} - 1} approx frac{1.414}{2.828 - 1} approx 1.414 / 1.828 ≈ 0.773 ), which is less than 1. So indeed, the maximum occurs at ( S = k ).Therefore, in the general case, the maximum value is ( frac{sqrt{k}}{2sqrt{k} - 1} ).But wait, let's check the denominator when ( x_1 = sqrt{k} ):Denominator is ( T^2 + x_n = x_1^2 + x_n = (sqrt{k})^2 + (k - sqrt{k}) = k + k - sqrt{k} = 2k - sqrt{k} ).Numerator is ( x_1 sqrt{S} = sqrt{k} cdot sqrt{k} = k ).Therefore, ( f = k / (2k - sqrt{k}) = sqrt{k}/(2sqrt{k} - 1) ). That's correct.But let's make sure that when we set other variables to zero, like ( x_2, ..., x_{n-1} ), that this is indeed permissible. For example, if ( n = 2 ), then the original function is ( f(x_1, x_2) = frac{x_1 sqrt{x_1 + x_2}}{x_1^2 + x_2} ), which matches our reduced case. So if ( n = 2 ), the maximum is as calculated. For ( n > 2 ), by setting the middle variables to zero, we get the same function, so the maximum would still be the same. Therefore, regardless of ( n ), the maximum occurs when all variables except ( x_1 ) and ( x_n ) are zero, and then ( x_1 = sqrt{k} ), ( x_n = k - sqrt{k} ).Therefore, the maximum value of the function is ( frac{sqrt{k}}{2sqrt{k} - 1} ).But let's check if this is valid when ( k = 1 ). As we saw earlier, yes, the maximum is 1, which is correct. For ( k = 4 ), the maximum would be ( 2/(4 - 1) = 2/3 ≈ 0.666 ). Wait, let's compute it properly. If ( k = 4 ), ( sqrt{k} = 2 ), so the maximum value is ( 2 / (2*2 - 1) = 2/3 ≈ 0.666 ). Let me test with actual numbers. If ( x_1 = 2 ), ( x_n = 4 - 2 = 2 ), then the function is ( 2 * sqrt(4) / (2^2 + 2) = 2*2 / (4 + 2) = 4/6 = 2/3 ≈ 0.666 ). Alternatively, if we set ( x_1 = 1 ), ( x_n = 3 ), then the function is ( 1 * sqrt(4) / (1 + 3) = 2 / 4 = 0.5 ), which is less. If we set ( x_1 = 3 ), ( x_n = 1 ), then denominator is ( 9 + 1 = 10 ), numerator is 3*2 = 6, so 6/10 = 0.6, still less than 2/3. So the maximum does occur at x1 = sqrt(k).Therefore, the conclusion seems consistent.However, there's one more check: when ( x_n = k - sqrt{k} ), we need to ensure that ( x_n geq 0 ). Since ( k geq 1 ), ( sqrt{k} leq k ), because for ( k geq 1 ), ( sqrt{k} leq k ). For example, k=1: sqrt(1)=1, so x_n=0. k=2: sqrt(2)≈1.414, x_n≈0.586 ≥0. So yes, x_n is non-negative.Therefore, all constraints are satisfied, and the maximum value is indeed ( frac{sqrt{k}}{2sqrt{k} - 1} ).Final AnswerThe maximum value is boxed{dfrac{sqrt{k}}{2sqrt{k} - 1}}.</think>Given ( x_{1}, x_{2}, ldots, x_{n} geq 0 ) and ( sum_{i=1}^{n} x_{i} geq k ) where ( k geq 1 ), we need to find the maximum value of the function:[fleft(x_{1}, x_{2}, ldots, x_{n}right) = frac{x_{1} sqrt{sum_{i=1}^{n} x_{i}}}{left(sum_{i=1}^{n-1} x_{i}right)^{2} + x_{n}}]First, we simplify the function by denoting ( T = sum_{i=1}^{n-1} x_{i} ) and ( S = T + x_{n} ). The function becomes:[f = frac{x_{1} sqrt{S}}{T^2 + x_{n}}]Assuming ( x_2, x_3, ldots, x_{n-1} = 0 ), we reduce the problem to two variables ( x_1 ) and ( x_n ). The function simplifies to:[f = frac{x_1 sqrt{x_1 + x_n}}{x_1^2 + x_n}]Given ( S = x_1 + x_n geq k ), we set ( S = k ) to minimize ( S ) and maximize the function. Substituting ( x_n = k - x_1 ), the function becomes:[f = frac{x_1 sqrt{k}}{x_1^2 + k - x_1}]Taking the derivative and finding the critical point, we determine ( x_1 = sqrt{k} ). Substituting ( x_1 = sqrt{k} ) and ( x_n = k - sqrt{k} ) into the function, we get:[f = frac{sqrt{k} sqrt{k}}{(sqrt{k})^2 + (k - sqrt{k})} = frac{k}{2k - sqrt{k}} = frac{sqrt{k}}{2sqrt{k} - 1}]Verification shows this value is indeed the maximum, as increasing ( S ) beyond ( k ) decreases the function value. Therefore, the maximum value is:[boxed{dfrac{sqrt{k}}{2sqrt{k} - 1}}]]

🔑:Let a=x_{1}, b=sum_{i=2}^{n-1} x_{i}, c=x_{n}, then a, b, c geqslant 0, a+b+c geqslant k. Up to this point, the original function is reduced from n variables to three variables, i.e.,begin{array}{l}fleft(x_{1}, x_{2}, cdots, x_{n}right) =frac{a sqrt{a+b+c}}{(a+b)^{2}+c}=g(a, b, c) .end{array}Let s=a+b+c, then s geqslant k. Thus,begin{array}{l}g(a, b, c)=frac{a sqrt{s}}{(s-c)^{2}+c} leqslant frac{(s-c) sqrt{s}}{(s-c)^{2}+c}=frac{sqrt{s}}{(s-c)+frac{c}{s-c}} .end{array}Notice thatbegin{array}{l}(s-c)+frac{c}{s-c} geqslant 2 sqrt{s}-1 Leftrightarrow(c-s+sqrt{s})^{2} geqslant 0,end{array}Therefore, g(a, b, c) leqslant frac{sqrt{s}}{2 sqrt{s}-1}=frac{1}{2-frac{1}{sqrt{s}}}leqslant frac{1}{2-frac{1}{sqrt{k}}}=frac{sqrt{k}}{2 sqrt{k}-1} text {. }When b=0, c=k-sqrt{k}, a=sqrt{k}, i.e.,x_{1}=sqrt{k}, x_{i}=0(i=2,3, cdots, n-1), x_{n}=k-sqrt{k}the equality holds.Therefore, the maximum value of fleft(x_{1}, x_{2}, cdots, x_{n}right) isfrac{sqrt{k}}{2 sqrt{k}-1} text {. }

🧠:[<think>Okay, so I need to find the eccentricity of the ellipse given by the equation x²/25 + y²/16 = 1. Hmm, let me recall what I know about ellipses. The standard form of an ellipse is (x²/a²) + (y²/b²) = 1, where 'a' and 'b' are the lengths of the semi-major and semi-minor axes, respectively. But wait, I need to remember which one is which. If a > b, then the major axis is along the x-axis, and if b > a, then it's along the y-axis. Looking at the given equation, the denominator under x² is 25 and under y² is 16. Since 25 > 16, that means a² = 25 and b² = 16. So, a = 5 and b = 4. Since a > b, the major axis is along the x-axis. Now, the formula for the eccentricity (e) of an ellipse is e = c/a, where c is the distance from the center to a focus. But how do I find c? I remember there's a relationship between a, b, and c for an ellipse: c² = a² - b². Let me check that. Yes, that's right. So, c = sqrt(a² - b²). Plugging in the values, c = sqrt(25 - 16) = sqrt(9) = 3. Therefore, the eccentricity e = c/a = 3/5. Let me verify that. Wait, the options given are A: 4/5, B: 3/5, C: 3/4, D: 4/3. So, 3/5 is option B. That seems correct. Let me go through the steps again to be sure. The equation is x²/25 + y²/16 = 1. Comparing with standard form (x²/a² + y²/b² = 1), so a² = 25, hence a = 5, and b² = 16, so b = 4. Since a > b, major axis is along the x-axis. Then c² = a² - b² = 25 - 16 = 9, so c = 3. Eccentricity e = c/a = 3/5. Yep, that matches option B. I think that's right. Maybe the answer is B: 3/5. Let me just check if I mixed up a and b. If by mistake I took b as the larger one, then maybe I would have messed up. But no, since a² is under x² and since 25 is bigger, a is indeed larger. So, no problem there. Alternatively, if the major axis were along the y-axis, then the formula would have been y²/a² + x²/b² = 1, but in this case, it's x² over the larger denominator, so major axis is x. Hence, correct formula applies. So, yes, e = 3/5. I think that's solid. The answer should be B.Final Answerboxed{B}</think>To find the eccentricity of the ellipse given by the equation (dfrac{x^2}{25} + dfrac{y^2}{16} = 1), we start by identifying the semi-major axis (a) and the semi-minor axis (b). The standard form of an ellipse is (dfrac{x^2}{a^2} + dfrac{y^2}{b^2} = 1) where (a > b) indicates the major axis is along the x-axis. Here, (a^2 = 25) and (b^2 = 16), so (a = 5) and (b = 4).The eccentricity (e) of an ellipse is given by (e = dfrac{c}{a}), where (c) is the distance from the center to a focus. Using the relationship (c^2 = a^2 - b^2), we calculate:[c = sqrt{a^2 - b^2} = sqrt{25 - 16} = sqrt{9} = 3]Thus, the eccentricity is:[e = dfrac{c}{a} = dfrac{3}{5}]The correct answer is (boxed{B}).,<think>Okay, so I need to find the eccentricity of the ellipse given by the equation x²/25 + y²/16 = 1. Hmm, let me recall what eccentricity means for an ellipse. From what I remember, the eccentricity (e) measures how "stretched" the ellipse is. It's a value between 0 and 1, where 0 is a circle and closer to 1 means it's more elongated.First, I should probably figure out which is the major axis and minor axis of the ellipse. The general form of an ellipse is x²/a² + y²/b² = 1, where a and b are the semi-major and semi-minor axes. But wait, I need to be careful here. If a > b, then the major axis is along the x-axis, and if b > a, then it's along the y-axis. In this equation, under x² we have 25 and under y² we have 16. Since 25 is larger than 16, that means a² = 25 and b² = 16, so a = 5 and b = 4. Therefore, the major axis is along the x-axis because the larger denominator is under x². Now, the formula for eccentricity of an ellipse is e = c/a, where c is the distance from the center to each focus, and a is the semi-major axis length. But I need to remember how to find c. There's a relationship between a, b, and c in an ellipse: a² = b² + c². Wait, is that right? Let me double-check. If a is the semi-major axis and b is the semi-minor axis, then yes, the formula should be a² = b² + c². So c would be sqrt(a² - b²). Then, the eccentricity e = c/a = sqrt(a² - b²)/a.Let me plug in the values here. a² is 25, so a is 5. b² is 16, so b is 4. Then c = sqrt(a² - b²) = sqrt(25 - 16) = sqrt(9) = 3. Therefore, the eccentricity e is c/a = 3/5. Looking at the answer choices: A is 4/5, B is 3/5, C is 3/4, D is 4/3. So 3/5 is option B. Wait a second, let me make sure I didn't mix up a and b. If the major axis is along the x-axis, then a is indeed 5 (the square root of 25) and b is 4. So yes, c = sqrt(a² - b²) = sqrt(25 -16) = 3. Therefore, e = 3/5. So the answer should be B. But just to be thorough, let me check the formula once more. Sometimes I get confused between the major and minor axes. So, for an ellipse, when it's written as x²/a² + y²/b² = 1, if a > b, the major axis is along the x-axis. Then, the standard formula for eccentricity is e = sqrt(1 - (b²/a²)). Let me compute that. So, 1 - (b²/a²) = 1 - (16/25) = (25 -16)/25 = 9/25. Then sqrt(9/25) is 3/5. So same result. Yep, that confirms it. The eccentricity is 3/5, which is answer B. Just to make sure I didn't make any miscalculations, let me check each step again:1. The given equation is x²/25 + y²/16 =1. 2. Comparing with standard form x²/a² + y²/b² =1, since 25 >16, so a²=25, hence a=5, major axis along x-axis.3. b²=16, so b=4.4. Eccentricity e= sqrt(1 - (b²/a²)) = sqrt(1 -16/25)=sqrt(9/25)=3/5.Yes, that all lines up. So the answer is definitely B.Final Answerboxed{B}</think>To find the eccentricity of the ellipse given by the equation (dfrac{x^{2}}{25} + dfrac{y^{2}}{16} = 1), we start by identifying the semi-major axis (a) and the semi-minor axis (b). Given the equation (dfrac{x^{2}}{25} + dfrac{y^{2}}{16} = 1), we see that (a^2 = 25) and (b^2 = 16). Since (25 > 16), the major axis is along the x-axis, so (a = 5) and (b = 4).The formula for the eccentricity (e) of an ellipse is (e = dfrac{c}{a}), where (c) is the distance from the center to each focus. The relationship between (a), (b), and (c) is given by (c = sqrt{a^2 - b^2}).Calculating (c):[c = sqrt{a^2 - b^2} = sqrt{25 - 16} = sqrt{9} = 3]Thus, the eccentricity (e) is:[e = dfrac{c}{a} = dfrac{3}{5}]Therefore, the eccentricity of the ellipse is (boxed{B}).]

🔑:From the given information, we know that the foci of the ellipse dfrac{x^{2}}{25} + dfrac{y^{2}}{16} = 1 are on the x-axis, with a=5, b=4, and c= sqrt{a^{2}-b^{2}}=3.Therefore, the eccentricity of the ellipse e= dfrac{c}{a}= dfrac{3}{5}.The eccentricity of the ellipse dfrac{x^{2}}{25} + dfrac{y^{2}}{16} = 1 is dfrac{3}{5}.Hence, the correct option is boxed{B}.The foci of the ellipse dfrac{x^{2}}{25} + dfrac{y^{2}}{16} = 1 are on the x-axis, with a=5, b=4, and c= sqrt{a^{2}-b^{2}}=3. The eccentricity of the ellipse e= dfrac{c}{a}= dfrac{3}{5}, which allows us to find the answer.This question examines the standard equation of an ellipse and its simple geometric properties, focusing on the application of the formula for the eccentricity of an ellipse. It is considered a basic question.

🧠:[<think>Okay, so I need to solve this inequality involving the function f(x) = 1/(2^x + 2). The problem states that the inequality f(x) + f(a - 2x) ≤ 1/2 should hold for any x in the interval [2, 3]. Then I have to find the range of real number a from the given options. Let me think step by step.First, let me write down what's given. The function is f(x) = 1/(2^x + 2). The inequality is f(x) + f(a - 2x) ≤ 1/2, and this must be true for all x in [2, 3]. The question is about finding the possible values of a.So, maybe I should start by substituting f(x) into the inequality. Let's do that. The inequality becomes:1/(2^x + 2) + 1/(2^{a - 2x} + 2) ≤ 1/2.Hmm, this looks a bit complex. Let me see if I can manipulate it. Let me denote y = 2^x. Since x is in [2, 3], then y = 2^x will be in [2^2, 2^3] = [4, 8]. So y ∈ [4, 8].Then, 2^{a - 2x} can be rewritten as 2^{a}/2^{2x} = 2^a / (2^x)^2 = 2^a / y^2. So, substituting y into the inequality, the second term becomes 1/(2^a / y^2 + 2) = 1/(2 + 2^a / y^2) = y^2 / (2^a + 2 y^2).Wait, maybe that's complicating things. Let me check again. Let's see:2^{a - 2x} = 2^a * 2^{-2x} = 2^a / 2^{2x}. Since 2^{2x} is (2^x)^2, which is y^2. So 2^{a - 2x} = 2^a / y^2.Therefore, the second term in the inequality is 1/(2^{a - 2x} + 2) = 1/(2^a / y^2 + 2). Let me write that as 1/( (2^a + 2 y^2)/y^2 ) = y^2 / (2^a + 2 y^2). Yes, that seems right.So the inequality becomes:1/(y + 2) + y^2 / (2^a + 2 y^2) ≤ 1/2.Hmm, that might not be the easiest way to handle it. Maybe instead, I can set t = 2^x, so when x is in [2, 3], t is in [4, 8]. Then, let's express everything in terms of t.So, f(x) = 1/(t + 2). Then, f(a - 2x) = 1/(2^{a - 2x} + 2). Let's compute 2^{a - 2x}:2^{a - 2x} = 2^a / 2^{2x} = 2^a / t^2. Therefore, f(a - 2x) = 1/(2^a / t^2 + 2) = t^2 / (2^a + 2 t^2).So, substituting into the inequality:1/(t + 2) + t^2 / (2^a + 2 t^2) ≤ 1/2.So, for each t in [4, 8], we need this inequality to hold. So the problem reduces to finding a such that for all t in [4, 8], 1/(t + 2) + t^2/(2^a + 2 t^2) ≤ 1/2.Hmm. Let's see if we can rearrange this inequality. Let me denote the left-hand side as LHS.LHS = 1/(t + 2) + t^2/(2^a + 2 t^2) ≤ 1/2.Let me subtract 1/(t + 2) from both sides:t^2/(2^a + 2 t^2) ≤ 1/2 - 1/(t + 2).Compute the right-hand side (RHS):1/2 - 1/(t + 2) = ( (t + 2) - 2 ) / [2(t + 2)] ) = t / [2(t + 2)].So, the inequality becomes:t^2/(2^a + 2 t^2) ≤ t / [2(t + 2)].Multiply both sides by (2^a + 2 t^2) and 2(t + 2), assuming these are positive, which they are since t is positive and 2^a is always positive.So, cross-multiplying:t^2 * 2(t + 2) ≤ t * (2^a + 2 t^2).Simplify left side: 2 t^2 (t + 2) = 2 t^3 + 4 t^2.Right side: t * (2^a + 2 t^2) = 2^a t + 2 t^3.So, the inequality becomes:2 t^3 + 4 t^2 ≤ 2^a t + 2 t^3.Subtract 2 t^3 from both sides:4 t^2 ≤ 2^a t.Divide both sides by t (since t ≥ 4 > 0, we can divide):4 t ≤ 2^a.So, 2^a ≥ 4 t.Since this must hold for all t in [4, 8], then 2^a must be greater than or equal to 4 t for all t in [4, 8]. Therefore, 2^a ≥ max_{t ∈ [4,8]} 4 t.But the maximum of 4 t over t ∈ [4,8] is 4*8=32. So 2^a ≥ 32.But 32 is 2^5, so 2^a ≥ 2^5 implies a ≥5.Therefore, the range of a is [5, ∞). Which is option A.Wait, but let me verify this because sometimes when you cross-multiply inequalities, you can make mistakes if the denominators are negative, but in this case, denominators are positive because 2^a + 2 t^2 is always positive, and t + 2 is positive. So the cross-multiplication is safe.So the steps are:1. Start with f(x) + f(a - 2x) ≤ 1/2.2. Substitute f(x) = 1/(2^x + 2) and express 2^{a - 2x} as 2^a / (2^x)^2.3. Let t = 2^x, so t ∈ [4,8].4. Rewrite the inequality in terms of t and manipulate algebraically.5. After simplifying, arrive at the condition 2^a ≥ 4 t for all t ∈ [4,8].6. Therefore, 2^a must be at least 32 (since the maximum t is 8), so a ≥5.Therefore, the answer is option A: [5, ∞). But let me check again.Wait, maybe I made a miscalculation when simplifying.Let me recheck the steps:Original inequality:1/(t + 2) + t^2/(2^a + 2 t^2) ≤ 1/2.Subtract 1/(t + 2):t^2/(2^a + 2 t^2) ≤ 1/2 - 1/(t + 2).Compute RHS:1/2 - 1/(t + 2) = [ (t + 2) - 2 ] / [2(t + 2)] ) = t / [2(t + 2)].So, inequality is:t^2/(2^a + 2 t^2) ≤ t / [2(t + 2)].Multiply both sides by denominators:Left side: t^2 * 2(t + 2)Right side: t * (2^a + 2 t^2)Therefore, 2 t^2 (t + 2) ≤ t (2^a + 2 t^2)Divide both sides by t (since t >0):2 t (t + 2) ≤ 2^a + 2 t^2Expand left side:2 t^2 + 4 t ≤ 2^a + 2 t^2Subtract 2 t^2 from both sides:4 t ≤ 2^aThus, 2^a ≥ 4 tTherefore, since t ∈ [4,8], then 4 t ∈ [16,32]. Hence, 2^a must be greater than or equal to 32. So 2^a ≥32 => a ≥5.Thus, a ≥5. So the answer is A: [5, ∞). So that's correct.But let me test with a specific value. Suppose a =5. Then, 2^a =32. So for t=8, 4t=32, so 32=32, which is equal. So equality holds when t=8. Let's check the original inequality.For a=5 and x=3 (since t=8 when x=3), then:f(3) + f(5 - 2*3) = f(3) + f(-1).Compute f(3)=1/(2^3 +2)=1/(8 +2)=1/10.Compute f(-1)=1/(2^{-1} +2)=1/(0.5 +2)=1/2.5=0.4=2/5.So total is 1/10 + 2/5 = 1/10 +4/10=5/10=1/2. So equality holds. For a=5, when x=3, the sum is exactly 1/2. For x in [2,3), t is less than 8, so 4t <32, so 2^a=32>4t, so the inequality would be strict. Let's check x=2, t=4. Then a=5.f(2) + f(5 -4)=f(2)+f(1). f(2)=1/(4 +2)=1/6≈0.1667. f(1)=1/(2 +2)=1/4=0.25. Sum≈0.4167, which is less than 0.5. So yes, the inequality holds.If a is less than 5, say a=4, then 2^a=16. Then 4t when t=8 is 32, but 16<32. So 2^a=16 <32, which violates 2^a ≥4t. Therefore, the inequality would not hold when a=4 and t=8. Let's check:For a=4 and x=3:f(3) + f(4 -6)=f(3)+f(-2). f(3)=1/10, f(-2)=1/(2^{-2}+2)=1/(0.25 +2)=1/2.25≈0.4444. Sum≈0.1 +0.444≈0.5444>0.5, which violates the inequality. So a=4 is invalid. Hence, a must be at least 5.Similarly, if a=6, then 2^a=64, which is greater than 4*8=32, so it's safe. So all a≥5 are valid.Therefore, the answer is A: [5, ∞).But wait, the options given are:A: [5, +∞)B: [4, +∞)C: (-∞,6]D: (-∞,4]So according to my reasoning, the answer is A. But let me check the answer again because sometimes test questions can be tricky.Wait, let me check when x is in [2,3], a-2x varies. When x=2, a-2x = a-4. When x=3, a-2x = a-6. So, for the argument of the second function f(a-2x), the input is a-2x. Since f is defined as 1/(2^x + 2), but in the second term, it's 2^{a - 2x} +2. Wait, but 2^{a - 2x} is defined for any real a and x, so there's no restriction on a-2x? Wait, no, but the function f(y)=1/(2^y +2) is defined for all real numbers y, because 2^y +2 is always positive. So even if a-2x is negative, it's okay. So even if a-2x is negative, like when a=5 and x=3, a-2x=5-6=-1, which is okay. So no domain issues here.But in the problem statement, the inequality must hold for x in [2,3]. So the critical point is that the maximum of 4t occurs at t=8, which is x=3. Therefore, the minimum value of a that satisfies 2^a ≥32 is a=5, and any a greater than 5 would make 2^a even larger, which is still satisfies 2^a ≥4t for t up to 8. So yes, the answer is A.But wait, the options given include B: [4, +∞). Why is that an option? Let me check again my algebra.Let me verify when a=5. Then, as before, for x=3, f(3)+f(-1)=1/10 +2/5=1/2. For x=2, f(2)+f(5-4)=f(2)+f(1)=1/6 +1/4= (2 +3)/12=5/12≈0.4167<0.5. So it's okay.If a=6, then 2^a=64. For x=3, 4t=32, 64>32, so the inequality holds. For any x in between, the term 4t is between 16 and 32, so 64 is always larger. So the sum will be even smaller. So yes, a≥5 is correct.But let me see why the answer might be B. Maybe if there was an error in the algebra steps. Let me retrace.Original inequality:1/(2^x +2) +1/(2^{a -2x}+2) ≤1/2.We set t=2^x, t∈[4,8]. Then, 2^{a -2x}=2^a/(2^{2x})=2^a/t^2. Therefore, the second term is 1/(2^a/t^2 +2)=t^2/(2^a +2t^2).So the inequality becomes:1/(t +2) + t^2/(2^a +2t^2) ≤1/2.Then, as before, subtract 1/(t +2):t^2/(2^a +2t^2) ≤1/2 -1/(t +2)= t/[2(t +2)].Multiply both sides by denominators:t^2 * 2(t +2) ≤ t*(2^a +2t^2).Divide by t:2t(t +2) ≤2^a +2t^2.Expand left side:2t^2 +4t ≤2^a +2t^2.Cancel 2t^2:4t ≤2^a.Thus, 2^a ≥4t for all t∈[4,8]. Therefore, 2^a must be ≥32, so a≥5.Therefore, answer is A.Therefore, the correct answer is A. So I think that's correct. Maybe the options are given correctly. So the answer should be A.But in the problem statement, the options are A: [5, ∞), B: [4, ∞), etc. So unless there is a miscalculation here.Wait, perhaps I made a mistake in substituting variables. Let me check again.Wait, when we set t=2^x, then x ∈ [2,3] gives t ∈ [4,8]. Then, 2^{a -2x} is 2^a / 2^{2x}=2^a / (2^x)^2=2^a / t^2. So yes, that substitution is correct.Then the second term in the inequality is 1/(2^{a -2x}+2)=1/(2^a/t^2 +2)= t^2/(2^a +2t^2). Yes.Thus, the rest of the steps follow. So unless I messed up the cross-multiplication step.Wait, after getting:t^2/(2^a +2t^2) ≤ t/(2(t +2)).Multiply both sides by (2^a +2t^2) and 2(t +2), which are positive:Left side: t^2 *2(t +2).Right side: t*(2^a +2t^2).Thus: 2t^2(t +2) ≤t(2^a +2t^2).Divide both sides by t (t>0):2t(t +2) ≤2^a +2t^2.Then expand left side: 2t^2 +4t ≤2^a +2t^2.Subtract 2t^2: 4t ≤2^a.Hence, 2^a ≥4t.Yes, that's correct.So, if the answer is A, then the options must have A as [5, ∞). But let me check if the problem might have a typo or if I misread.Wait, the original problem states: the inequality holds for any x ∈ [2,3]. So the maximum t is 8, hence 4t=32, so a≥5. Therefore, the answer is A.Therefore, I think the answer is option A. So maybe the initial options are correct.Final Answerboxed{A}</think>Given the function ( f(x) = frac{1}{2^x + 2} ), we need to determine the range of real number ( a ) such that the inequality ( f(x) + f(a - 2x) leq frac{1}{2} ) holds for any ( x in [2, 3] ).First, we substitute ( f(x) ) and ( f(a - 2x) ) into the inequality:[frac{1}{2^x + 2} + frac{1}{2^{a - 2x} + 2} leq frac{1}{2}]Let ( t = 2^x ). Since ( x in [2, 3] ), ( t ) ranges from ( 4 ) to ( 8 ). Rewriting the inequality in terms of ( t ):[frac{1}{t + 2} + frac{1}{2^a / t^2 + 2} leq frac{1}{2}]Simplify the second term:[frac{1}{2^a / t^2 + 2} = frac{t^2}{2^a + 2t^2}]Substituting back, the inequality becomes:[frac{1}{t + 2} + frac{t^2}{2^a + 2t^2} leq frac{1}{2}]Subtract (frac{1}{t + 2}) from both sides:[frac{t^2}{2^a + 2t^2} leq frac{1}{2} - frac{1}{t + 2}]Simplify the right-hand side:[frac{1}{2} - frac{1}{t + 2} = frac{t}{2(t + 2)}]Thus, the inequality becomes:[frac{t^2}{2^a + 2t^2} leq frac{t}{2(t + 2)}]Cross-multiplying and simplifying:[2t^2(t + 2) leq t(2^a + 2t^2)]Divide both sides by ( t ):[2t(t + 2) leq 2^a + 2t^2]Expand and simplify:[2t^2 + 4t leq 2^a + 2t^2]Subtract ( 2t^2 ) from both sides:[4t leq 2^a]Since ( t in [4, 8] ), the maximum value of ( 4t ) is ( 32 ). Therefore:[2^a geq 32 implies a geq 5]Thus, the range of ( a ) is ([5, +infty)), which corresponds to option (boxed{A}).,<think>Okay, let's see. I have this problem where I need to find the range of real number ( a ) such that the inequality ( f(x) + f(a - 2x) leq frac{1}{2} ) holds for any ( x in [2, 3] ). The function given is ( f(x) = frac{1}{2^x + 2} ). The answer choices are A through D, with different intervals for ( a ). Hmm, I need to figure out which one is correct.First, let me understand the problem. The function ( f(x) ) is defined as ( frac{1}{2^x + 2} ). So, for each ( x ), we take 2 to the power ( x ), add 2, then take the reciprocal. Then, we have to consider ( f(x) + f(a - 2x) ), which is the sum of this function evaluated at ( x ) and at ( a - 2x ). The inequality needs to hold for any ( x ) between 2 and 3. Therefore, the maximum value of ( f(x) + f(a - 2x) ) over the interval ( [2, 3] ) must be less than or equal to ( frac{1}{2} ). So, my job is to find all ( a ) such that this condition is satisfied.Let me start by writing out the inequality:[frac{1}{2^x + 2} + frac{1}{2^{a - 2x} + 2} leq frac{1}{2}]This needs to be true for all ( x in [2, 3] ). So, I need to ensure that for each ( x ) in that interval, the sum of those two fractions is at most ( frac{1}{2} ).To tackle this, maybe I can analyze the function ( g(x) = f(x) + f(a - 2x) ) and find its maximum over ( [2, 3] ), then set that maximum less than or equal to ( frac{1}{2} ). The challenge is to find the maximum of ( g(x) ) in terms of ( a ), then solve for ( a ).First, let's note that ( f(t) = frac{1}{2^t + 2} ). This is a decreasing function because as ( t ) increases, ( 2^t ) increases, so the denominator increases, making the whole fraction decrease. So, ( f(t) ) is decreasing in ( t ).Therefore, ( f(x) ) is decreasing in ( x ), and ( f(a - 2x) ) is decreasing in ( (a - 2x) ). However, since ( a - 2x ) is a linear function of ( x ), with coefficient -2, that means ( a - 2x ) is decreasing in ( x ). Therefore, ( f(a - 2x) ) is increasing in ( x ), because the argument ( a - 2x ) is decreasing, and since ( f ) is decreasing, applying a decreasing function to a decreasing argument gives an increasing function. So, ( f(a - 2x) ) is increasing in ( x ).So, ( f(x) ) is decreasing, and ( f(a - 2x) ) is increasing. Therefore, the sum ( g(x) = f(x) + f(a - 2x) ) is a combination of a decreasing and an increasing function. The sum might have a maximum somewhere in the interval or at the endpoints. Since it's continuous, the maximum will occur either at a critical point inside the interval or at the endpoints.Therefore, to find the maximum of ( g(x) ), I can check the values at the endpoints ( x = 2 ) and ( x = 3 ), and also check if there is any critical point in between where the derivative is zero.Alternatively, since the problem states that the inequality must hold for all ( x in [2, 3] ), perhaps it's sufficient to ensure that the inequality holds at both endpoints. But I need to verify this because sometimes the maximum could occur inside the interval. However, given that ( f(x) ) is decreasing and ( f(a - 2x) ) is increasing, their sum could have a single maximum or minimum. Let me check.Alternatively, maybe I can substitute ( y = 2x ), but not sure. Let's proceed step by step.First, let's consider the endpoints.When ( x = 2 ):( f(2) + f(a - 4) leq frac{1}{2} )When ( x = 3 ):( f(3) + f(a - 6) leq frac{1}{2} )So, if the inequality holds at both endpoints, does it hold in between? Maybe not necessarily. Because the sum could peak somewhere inside. So, perhaps I need to analyze the derivative.Let me compute the derivative of ( g(x) = f(x) + f(a - 2x) ).First, ( f(t) = frac{1}{2^t + 2} ), so the derivative of ( f(t) ) with respect to ( t ) is:( f'(t) = frac{ -2^t ln 2 }{(2^t + 2)^2} )Therefore, the derivative of ( g(x) ) with respect to ( x ) is:( g'(x) = f'(x) + f'(a - 2x) cdot (-2) )So:( g'(x) = frac{ -2^x ln 2 }{(2^x + 2)^2 } + (-2) cdot frac{ -2^{a - 2x} ln 2 }{(2^{a - 2x} + 2)^2 } )Simplify:( g'(x) = - frac{2^x ln 2}{(2^x + 2)^2} + frac{2 cdot 2^{a - 2x} ln 2}{(2^{a - 2x} + 2)^2} )Factor out ( ln 2 ):( g'(x) = ln 2 left[ - frac{2^x}{(2^x + 2)^2} + frac{2 cdot 2^{a - 2x}}{(2^{a - 2x} + 2)^2} right] )To find critical points, set ( g'(x) = 0 ):( - frac{2^x}{(2^x + 2)^2} + frac{2 cdot 2^{a - 2x}}{(2^{a - 2x} + 2)^2} = 0 )Which implies:( frac{2 cdot 2^{a - 2x}}{(2^{a - 2x} + 2)^2} = frac{2^x}{(2^x + 2)^2} )Let me denote ( u = 2^x ), then ( 2^{a - 2x} = 2^{a} cdot 2^{-2x} = 2^a cdot (2^x)^{-2} = 2^a / u^2 ).So substituting:Left-hand side: ( frac{2 cdot (2^a / u^2)}{(2^a / u^2 + 2)^2} )Right-hand side: ( frac{u}{(u + 2)^2} )Therefore:( frac{2 cdot (2^a / u^2)}{(2^a / u^2 + 2)^2} = frac{u}{(u + 2)^2} )Multiply numerator and denominator on the left by ( u^4 ):Numerator becomes ( 2 cdot 2^a cdot u^2 )Denominator becomes ( (2^a + 2u^2)^2 )So:( frac{2 cdot 2^a cdot u^2}{(2^a + 2u^2)^2} = frac{u}{(u + 2)^2} )Cross-multiplying:( 2 cdot 2^a cdot u^2 cdot (u + 2)^2 = u cdot (2^a + 2u^2)^2 )Simplify both sides:Left: ( 2^{a + 1} cdot u^2 cdot (u + 2)^2 )Right: ( u cdot (2^a + 2u^2)^2 )Divide both sides by ( u ) (since ( u = 2^x > 0 )):( 2^{a + 1} cdot u cdot (u + 2)^2 = (2^a + 2u^2)^2 )This seems complicated. Maybe we can set ( v = u^2 ), so ( v = 2^{2x} ). Let me see.Alternatively, let's think if there's a specific value of ( a ) that could make this equation hold for some ( x in [2, 3] ). But perhaps this is too involved. Maybe instead of trying to find critical points, which might be difficult, I can analyze the behavior of the function ( g(x) ).Given that ( f(x) ) is decreasing and ( f(a - 2x) ) is increasing, their sum might have a single critical point where the derivative is zero. However, verifying this requires more work.Alternatively, perhaps the maximum occurs at one of the endpoints, which would make the problem easier. Let me check for specific values of ( a ).Suppose ( a ) is such that ( a - 2x ) is within a certain range. When ( x in [2, 3] ), then ( 2x in [4, 6] ), so ( a - 2x in [a - 6, a - 4] ). So the argument of the second function is between ( a - 6 ) and ( a - 4 ). Depending on ( a ), this interval can be in different parts of the real line.Since ( f(t) ) is defined for all real numbers, but its behavior depends on ( t ). Since ( 2^t ) is always positive, ( f(t) ) is always positive and decreasing.If ( a - 6 ) is very large (i.e., ( a ) is large), then ( a - 2x ) is large, so ( f(a - 2x) ) is approaching ( 0 ). Similarly, if ( a - 4 ) is very small (i.e., ( a ) is small), then ( a - 2x ) is small, so ( f(a - 2x) ) approaches ( frac{1}{2^{a - 2x} + 2} ), which would be larger.Therefore, when ( a ) is very large, ( f(a - 2x) ) is very small, so ( g(x) ) is approximately ( f(x) ), which is at most ( f(2) = frac{1}{2^2 + 2} = frac{1}{6} ) when ( x = 2 ). Since ( 1/6 + 0 = 1/6 leq 1/2 ), so for large ( a ), the inequality holds. But when ( a ) decreases, ( f(a - 2x) ) increases, so the sum might exceed ( 1/2 ).Therefore, there might be a minimal value of ( a ) such that the inequality holds for all ( x in [2, 3] ). The answer options suggest that ( a ) is either bounded below or above. Since the options A and B are lower bounds, and C and D are upper bounds. From the previous reasoning, since large ( a ) are acceptable, but small ( a ) might not be, the correct answer is likely a lower bound, so either A or B.But let's check.Suppose ( a ) is minimal such that ( f(x) + f(a - 2x) leq frac{1}{2} ) for all ( x in [2, 3] ). To find the minimal ( a ), we need to ensure that the maximum of ( g(x) = f(x) + f(a - 2x) ) is ( frac{1}{2} ). So, set the maximum of ( g(x) ) equal to ( frac{1}{2} ) and solve for ( a ).But how?Alternatively, perhaps the maximum occurs when ( x = 2 ) or ( x = 3 ), so let's check the values at the endpoints.At ( x = 2 ):( f(2) + f(a - 4) = frac{1}{2^2 + 2} + frac{1}{2^{a - 4} + 2} = frac{1}{6} + frac{1}{2^{a - 4} + 2} )At ( x = 3 ):( f(3) + f(a - 6) = frac{1}{2^3 + 2} + frac{1}{2^{a - 6} + 2} = frac{1}{10} + frac{1}{2^{a - 6} + 2} )We need both sums to be ( leq frac{1}{2} ).So let's first analyze ( x = 2 ):( frac{1}{6} + frac{1}{2^{a - 4} + 2} leq frac{1}{2} )Subtract ( frac{1}{6} ):( frac{1}{2^{a - 4} + 2} leq frac{1}{2} - frac{1}{6} = frac{1}{3} )Take reciprocals (inequality reverses because both sides are positive):( 2^{a - 4} + 2 geq 3 )So:( 2^{a - 4} geq 1 )Which implies:( a - 4 geq 0 Rightarrow a geq 4 )Similarly, check at ( x = 3 ):( frac{1}{10} + frac{1}{2^{a - 6} + 2} leq frac{1}{2} )Subtract ( frac{1}{10} ):( frac{1}{2^{a - 6} + 2} leq frac{1}{2} - frac{1}{10} = frac{2}{5} )Take reciprocals:( 2^{a - 6} + 2 geq frac{5}{2} )So:( 2^{a - 6} geq frac{5}{2} - 2 = frac{1}{2} )Which implies:( a - 6 geq -1 Rightarrow a geq 5 )Wait, this is conflicting. At ( x = 2 ), we get ( a geq 4 ), and at ( x = 3 ), we get ( a geq 5 ). So, to satisfy both, ( a geq 5 ). However, the answer options don't have 5 as the lower bound except for option A. But option B is [4, ∞). Wait, but according to the calculation at x=3, we need a≥5. However, perhaps the maximum occurs somewhere else?But perhaps my analysis is flawed. Let me check.Wait, the reciprocal step:Starting with ( frac{1}{2^{a - 6} + 2} leq frac{2}{5} ).Taking reciprocals (since both sides are positive, inequality reverses):( 2^{a - 6} + 2 geq frac{5}{2} )Therefore:( 2^{a - 6} geq frac{5}{2} - 2 = frac{5}{2} - frac{4}{2} = frac{1}{2} )So ( 2^{a - 6} geq frac{1}{2} )Taking log base 2:( a - 6 geq -1 Rightarrow a geq 5 )Yes, that's correct. So, for the inequality to hold at x=3, we need a≥5. But at x=2, we need a≥4. So, the stricter condition is a≥5. Therefore, if we set a=5, then both inequalities hold? Wait, but maybe for a=5, somewhere in between x=2 and x=3, the sum exceeds 1/2.Therefore, perhaps the critical maximum is not at the endpoints but somewhere in the middle. Therefore, even if we set a=5, ensuring that at x=2 and x=3, the sum is ≤1/2, there might be a point in between where it's higher. Therefore, my approach of checking only the endpoints is insufficient.Therefore, I need to analyze further.Alternatively, maybe substitute a specific value of a to test. Let's try a=5.So, let's take a=5. Then, for x ∈ [2,3], the argument of the second function is a - 2x = 5 - 2x. When x=2, 5 - 4=1; when x=3, 5 -6= -1. So, the argument of the second function is in [-1,1]. Therefore, 2^{a - 2x} would be 2^{1}=2 when x=2, decreasing to 2^{-1}=0.5 when x=3.Therefore, f(a - 2x) ranges from 1/(2 + 2) = 1/4 to 1/(0.5 + 2)=1/2.5=0.4.Meanwhile, f(x) when x ∈ [2,3] is between 1/(2^3 +2)=1/10 and 1/(2^2 +2)=1/6 ≈0.1667.Therefore, when a=5, the sum f(x) + f(5 - 2x) ranges between 1/10 + 1/4 = 0.1 + 0.25 = 0.35 and 1/6 + 0.4 ≈0.1667 +0.4=0.5667. Wait, 0.5667 is more than 0.5. So, when a=5, the sum at x=3 is 1/10 + 1/(2^{-1} +2)=1/10 +1/(0.5 +2)=1/10 + 1/2.5=0.1 +0.4=0.5. So exactly 0.5. But in between, say at x=2.5, let's compute.At x=2.5, a -2x=5 -5=0. So, f(0)=1/(1 +2)=1/3≈0.333. Then f(2.5)=1/(2^{2.5} +2). 2^{2.5}=sqrt(2^5)=sqrt(32)≈5.656. So 5.656 +2≈7.656, so 1/7.656≈0.1306. So the sum is≈0.1306 +0.333≈0.4636 <0.5.At x=2.25: a-2x=5 -4.5=0.5. f(0.5)=1/(sqrt(2) +2)≈1/(1.414 +2)=1/3.414≈0.292. f(2.25)=1/(2^{2.25} +2). 2^2=4, 2^0.25≈1.189, so 2^{2.25}=4*1.189≈4.756. So denominator≈4.756 +2=6.756. So f(2.25)≈1/6.756≈0.148. Sum≈0.292 +0.148≈0.44 <0.5.At x=2.75: a-2x=5 -5.5= -0.5. f(-0.5)=1/(2^{-0.5} +2)=1/(1/sqrt(2) +2)≈1/(0.707 +2)=1/2.707≈0.369. f(2.75)=1/(2^{2.75}+2). 2^{2.75}=2^{11/4}= (2^{2})*(2^{3/4})=4*(cube root of 2^{3/4})? Wait, 2^{3/4}=e^{(3/4)ln2}≈e^{0.519}≈1.681. So 4*1.681≈6.724. So denominator≈6.724 +2=8.724. So f(2.75)≈1/8.724≈0.1146. Sum≈0.369 +0.1146≈0.4836 <0.5.Wait, so at x=3, the sum is exactly 0.5, and in between, it's less. So, maybe when a=5, the maximum is exactly 0.5 at x=3, and elsewhere it's less. Therefore, a=5 is acceptable.But according to the previous calculation, at x=2, when a=5, the sum is 1/6 +1/(2^{1} +2)=1/6 +1/4≈0.1667 +0.25=0.4167 <0.5. So, the maximum occurs at x=3 when a=5. So, if we take a=5, the inequality holds.But according to the answer choices, option A is [5, ∞), and option B is [4, ∞). But earlier, when a=4, let's check.Take a=4. Then, at x=2: f(2) + f(4 -4)=f(2) + f(0)=1/6 +1/3=0.1667 +0.333≈0.5, which is exactly 0.5. At x=3: f(3) + f(4 -6)=f(3) +f(-2)=1/10 +1/(2^{-2} +2)=1/10 +1/(0.25 +2)=1/10 +1/2.25≈0.1 +0.444≈0.544>0.5. So, the inequality fails at x=3 when a=4. Therefore, a=4 is not acceptable.But if we check in between, maybe when a=4, at some x in [2,3], the sum is greater than 0.5. Indeed, as shown at x=3, it's 0.544. Therefore, a=4 is too small.Therefore, the minimal value of a is 5, and the answer is [5, ∞), which is option A.Wait, but according to the previous analysis, when a=5, the sum at x=3 is exactly 0.5, and at other points it's less. But when I took a=4, the sum at x=2 is exactly 0.5, but at x=3 it's higher. So, the problem is that when a is decreased below 5, the sum at x=3 increases beyond 0.5. Therefore, the critical point is x=3, requiring a≥5.However, this contradicts my initial thought that when a is increased, the second term becomes smaller. Wait, but when a increases, the argument a -2x increases, so 2^{a -2x} increases, making f(a -2x) decrease. Therefore, increasing a causes f(a -2x) to decrease, hence the sum f(x) +f(a -2x) decreases. Therefore, the larger a is, the smaller the sum. Thus, the minimal a for which the maximum of the sum is 0.5 is a=5. So, for all a≥5, the inequality holds. For a<5, the sum exceeds 0.5 at x=3.But in that case, the answer should be [5, ∞), which is option A. However, the answer choices include B: [4, ∞). But according to the calculation above, when a=4, the sum at x=3 is over 0.5, which violates the inequality. Therefore, option A is correct.Wait, but let's check again. When a=5, x=3: f(3) + f(5 -6)=f(3) +f(-1)=1/10 +1/(2^{-1} +2)=1/10 +1/(0.5 +2)=1/10 +1/2.5=0.1 +0.4=0.5.But when a=5, at x=2.5, we had approximately 0.4636, which is less. So, the maximum occurs at x=3 when a=5. Thus, a=5 is the minimal a such that the maximum is exactly 0.5, so for all a≥5, the maximum will be less than or equal to 0.5. Because as a increases beyond 5, the term f(a -2x) becomes even smaller, so the sum decreases.Therefore, the correct answer should be A: [5, ∞). But the options don't have that. Wait, wait, no. Wait the options are:A: [5, +∞)B: [4, +∞)C: (-∞,6]D: (-∞,4]So, yes, option A is [5, ∞). But maybe my analysis is incorrect. Let me verify with a=5 and a=6.Take a=6. Then, at x=3: f(3) +f(6 -6)=f(3) +f(0)=1/10 +1/3≈0.1 +0.333≈0.433 <0.5. At x=2: f(2) +f(6 -4)=f(2) +f(2)=1/6 +1/6≈0.1667 +0.1667≈0.333 <0.5. So, for a=6, it's even smaller.If a=5, as we saw, the maximum is exactly 0.5. So, the minimal a is 5, hence the answer is A.But why did the problem's options include B? Maybe I made a mistake. Let me check the calculation again.Wait, when a=5, x=3: f(3) + f(5 -6)=f(3) +f(-1)=1/10 +1/(2^{-1} +2)=1/10 +1/(0.5 +2)=0.1 +0.4=0.5. Correct.But when a=5, let's check x=2. Then, f(2) +f(5 -4)=f(2)+f(1)=1/6 +1/(2 +2)=1/6 +1/4≈0.1667 +0.25≈0.4167 <0.5. So, the maximum is indeed at x=3 when a=5.However, when a=4.5 (which is between 4 and 5), check at x=3: a -6= -1.5. f(-1.5)=1/(2^{-1.5} +2)=1/(1/(2^{1.5}) +2)=1/(1/(2.828) +2)=1/(0.353 +2)=1/2.353≈0.425. Then f(3)=1/10=0.1. Sum≈0.1 +0.425=0.525>0.5. Therefore, violates the inequality. Hence, a=4.5 is invalid.Similarly, a=5 is the minimal a where at x=3 the sum is exactly 0.5, and for a>5, the sum is less.But wait, let me check a=5 and x=3.5 (wait, x is in [2,3], so x cannot be 3.5). Wait, x is only up to 3. So, the maximal x is 3. Therefore, when a=5, the critical point is at x=3, where the sum is exactly 0.5. So, for all a≥5, the sum at x=3 will be ≤0.5, because f(a -6) =1/(2^{a -6} +2). As a increases beyond 5, 2^{a -6} increases, so f(a -6) decreases, making the sum smaller.Hence, the minimal a is 5, so the answer is A.But initially, when I checked at x=2, a≥4 gives equality at x=2. However, since x=3 requires a higher a (a≥5), that is the stricter condition.Therefore, the answer should be A: [5, ∞). However, in the problem's options, A is [5, +∞), B is [4, +∞). The answer provided in the options is likely A. But let me check once more.Wait, maybe my mistake is in substituting a -2x when x is in [2,3]. For example, when a=4, then at x=2, the second term is f(4 -4)=f(0)=1/3, as before, and at x=3, f(4 -6)=f(-2). Wait, f(-2)=1/(2^{-2} +2)=1/(1/4 +2)=1/(2.25)=4/9≈0.444. Then f(3)=1/10, so sum≈0.1 +0.444≈0.544>0.5. So, a=4 is invalid.If a=5, then at x=3, the sum is 0.5. So, to ensure that at x=3, the sum is ≤0.5, a must be ≥5.But here's another thought: maybe the maximum of the function g(x) occurs not at x=3, but somewhere else when a is less than 5. Let's check for a=5, the maximum is at x=3. For a slightly less than 5, say a=4.5, the maximum might shift to somewhere else.Take a=4.5. At x=3: sum is f(3) +f(4.5 -6)=f(3) +f(-1.5)=0.1 + ≈0.425≈0.525>0.5. At x=2.5: a -2x=4.5 -5= -0.5. So f(-0.5)=1/(2^{-0.5} +2)=1/(1/√2 +2)≈1/(0.707 +2)=1/2.707≈0.369. f(2.5)=1/(2^{2.5} +2)≈1/(5.656 +2)=1/7.656≈0.1306. Sum≈0.369 +0.1306≈0.499≈0.5. So, almost 0.5. Therefore, for a=4.5, at x=2.5, the sum is≈0.5, and at x=3, it's≈0.525. Therefore, the maximum is still at x=3.Similarly, for a=4.75, let's check.a=4.75. At x=3: f(3) +f(4.75 -6)=f(3) +f(-1.25)=0.1 +1/(2^{-1.25} +2). 2^{-1.25}=1/(2^{1.25})=1/(2*2^{0.25})≈1/(2*1.189)≈1/2.378≈0.420. So denominator≈0.420 +2≈2.420. Thus, f(-1.25)≈1/2.420≈0.413. So sum≈0.1 +0.413≈0.513>0.5.At x=2.75: a -2x=4.75 -5.5= -0.75. f(-0.75)=1/(2^{-0.75} +2)=1/(1/(2^{0.75}) +2). 2^{0.75}=sqrt(2^{1.5})=sqrt(2.828)≈1.682. So 1/1.682≈0.594. So denominator≈0.594 +2≈2.594. Therefore, f(-0.75)≈1/2.594≈0.385. f(2.75)=1/(2^{2.75}+2)≈1/(6.727 +2)=1/8.727≈0.1146. Sum≈0.385 +0.1146≈0.499≈0.5.So, for a=4.75, the maximum is at x=3,≈0.513. Thus, even when a is slightly less than 5, the maximum exceeds 0.5. Therefore, the minimal a is indeed 5.Therefore, the answer is A: [5, ∞). However, looking back at the answer options, this is option A. But the original problem in Chinese might have different options? Wait, no. The user provided the options as:A: [5,+∞)B: [4,+∞)C: (−∞,6]D: (−∞,4]Therefore, based on the calculations, the correct answer is A. However, according to some test problems I've seen before, sometimes the minimal a is determined by the other endpoint. But according to this analysis, no. So, perhaps the answer is A.But let me think again. Perhaps there is an alternative approach. Maybe by using inequalities or convexity.Note that f(x) = 1/(2^x +2). Let me consider substituting t = 2^x. Since x ∈ [2,3], then t ∈ [4,8]. Then, f(x) = 1/(t + 2). Similarly, 2^{a - 2x} = 2^a / 2^{2x} = 2^a / t^2. So, f(a - 2x) = 1/(2^{a - 2x} +2) = 1/(2^a / t^2 + 2).Therefore, the inequality becomes:1/(t + 2) + 1/(2^a / t^2 + 2) ≤ 1/2, for t ∈ [4,8].Let me denote s = t, so s ∈ [4,8].So the inequality is:1/(s + 2) + 1/(2^a / s^2 + 2) ≤ 1/2.Let me denote the second term as 1/( (2^a)/s^2 + 2 ). Let me combine the two fractions.But this might not be helpful. Alternatively, let's set y = s^2. Wait, perhaps not. Alternatively, cross-multiplying.But another idea: Let me define u = s (so u ∈ [4,8]), then the inequality is:1/(u + 2) + 1/(2^a / u^2 + 2) ≤ 1/2.Let me let v = 2^a. Then, the inequality becomes:1/(u + 2) + 1/(v / u^2 + 2) ≤ 1/2.We need to find v such that this holds for all u ∈ [4,8]. Then, since v = 2^a, solving for v will give us a condition on a.Let's rewrite the inequality:1/(u + 2) + 1/(v/u^2 + 2) ≤ 1/2.Let me denote the second denominator as (v + 2u^2)/u^2. Therefore, the second term is u^2/(v + 2u^2).So, the inequality is:1/(u + 2) + u^2/(v + 2u^2) ≤ 1/2.Multiply both sides by 2(u + 2)(v + 2u^2) to eliminate denominators:2(v + 2u^2) + 2u^2(u + 2) ≤ (u + 2)(v + 2u^2).Expand both sides:Left side: 2v + 4u^2 + 2u^3 + 4u^2 = 2v + 8u^2 + 2u^3Right side: (u + 2)(v + 2u^2) = u*v + 2u^3 + 2v + 4u^2Subtract left side from right side:(u*v + 2u^3 + 2v + 4u^2) - (2v + 8u^2 + 2u^3) = u*v + 0u^3 -4u^2 -0vTherefore:u*v -4u^2 ≥0So:u*(v -4u) ≥0Since u ∈ [4,8], u >0, so the inequality reduces to:v -4u ≥0 ⇒ v ≥4u.Therefore, for the original inequality to hold for all u ∈ [4,8], we must have v ≥4u for all u ∈ [4,8]. The minimal v that satisfies this for all u ∈ [4,8] is v ≥4*8=32. Because if v ≥32, then since u ≤8, 4u ≤32 ≤v. However, if v=32, then for u=8, 4u=32, equality holds. For u <8, 4u <32 ≤v, so v ≥32 ensures v ≥4u for all u ∈ [4,8].But wait, if we set v=32, then the original inequality becomes equality when u=8. But in our previous analysis, when a=5, the inequality holds with equality at x=3 (u=8 when x=3, since s=2^x. Wait, no, wait. Wait, s = t =2^x. When x=3, s=2^3=8. So, u=8. Then, v=2^a.So, according to this, we need v=2^a ≥4u for all u ∈ [4,8]. The maximum of 4u on u ∈ [4,8] is 32. Therefore, 2^a ≥32 ⇒ a ≥5. Since 32=2^5. Therefore, a ≥5.Therefore, this confirms our previous result: a ≥5, so the answer is A: [5, ∞).Hence, the correct answer is boxed{A}.Final Answerboxed{A}</think>Given the function ( f(x) = frac{1}{2^x + 2} ), we need to find the range of real number ( a ) such that the inequality ( f(x) + f(a - 2x) leq frac{1}{2} ) holds for any ( x in [2, 3] ).First, we analyze the inequality at the endpoints ( x = 2 ) and ( x = 3 ):1. At ( x = 2 ): [ f(2) + f(a - 4) leq frac{1}{2} ] Substituting ( f(2) = frac{1}{6} ): [ frac{1}{6} + frac{1}{2^{a - 4} + 2} leq frac{1}{2} ] Solving this, we find: [ frac{1}{2^{a - 4} + 2} leq frac{1}{3} implies 2^{a - 4} geq 1 implies a geq 4 ]2. At ( x = 3 ): [ f(3) + f(a - 6) leq frac{1}{2} ] Substituting ( f(3) = frac{1}{10} ): [ frac{1}{10} + frac{1}{2^{a - 6} + 2} leq frac{1}{2} ] Solving this, we find: [ frac{1}{2^{a - 6} + 2} leq frac{2}{5} implies 2^{a - 6} geq frac{1}{2} implies a geq 5 ]Next, we consider the function ( g(x) = f(x) + f(a - 2x) ) and analyze its behavior. We find that the maximum value of ( g(x) ) occurs at ( x = 3 ) when ( a ) is minimized. For ( a = 5 ), the inequality holds exactly at ( x = 3 ):[f(3) + f(5 - 6) = frac{1}{10} + frac{1}{2^{-1} + 2} = frac{1}{10} + frac{1}{2.5} = 0.1 + 0.4 = 0.5]For ( a > 5 ), the term ( f(a - 2x) ) decreases, ensuring the inequality holds. For ( a < 5 ), the inequality fails at ( x = 3 ).Thus, the range of ( a ) is ( [5, +infty) ).[boxed{A}]]

🔑:To solve the problem, we start by understanding the behavior of the function f(x)=frac{1}{{{2^x}+2}}. Since y=2^{x} is monotonically increasing on mathbb{R}, it follows that f(x), being the reciprocal of a monotonically increasing function plus a constant, is monotonically decreasing on mathbb{R}. Given the inequality f(x)+f(a-2x) leq frac{1}{2}, we need to find the range of a for which this holds true for any x in [2,3]. First, let's examine a specific case where f(x)+f(2-x)=frac{1}{2} to understand the behavior of f(x)+f(a-2x):begin{align*}f(x)+f(2-x) &= frac{1}{{{2^x}+2}} + frac{1}{{{2^{2-x}}+2}} &= frac{1}{{{2^x}+2}} + frac{{{2^x}}}{{2({{2^x}+2})}} &= frac{1}{2}.end{align*}Since f(x) is monotonically decreasing, for f(x)+f(a-2x) leq frac{1}{2} to hold, it must be that f(a-2x) leq f(2-x). This is because f(x)+f(a-2x) reaching frac{1}{2} is a critical point, and any deviation in the argument of f that increases the value (due to the decreasing nature of f) would violate the inequality. Thus, we have:begin{align*}f(a-2x) &leq f(2-x) Rightarrow a-2x &geq 2-x quad text{(since f is decreasing, reversing inequality)} Rightarrow a &geq 2+x.end{align*}Given x in [2,3], to find the minimum value of a, we consider the maximum value of 2+x over this interval, which is 2+3=5. Therefore, we have:begin{align*}a &geq 5.end{align*}This implies that the range of real number a is boxed{left[5,+infty right)}. Hence, the correct answer is boxed{text{A}}.

🧠:[<think>Okay, let's try to tackle this problem step by step. The problem states that for the sequence {a_n}, which consists of positive terms, the sum of the first n terms is S_n. For any natural number n, the terms a_n, S_n, and a_n squared form an arithmetic sequence. We need to find the value of a_{2009} from the given options.First, let me recall what an arithmetic sequence is. In an arithmetic sequence, the difference between consecutive terms is constant. So, if three terms x, y, z form an arithmetic sequence, then the middle term y is the average of x and z. That is, 2y = x + z.Applying this to the problem, for each n, the three terms a_n, S_n, a_n^2 form an arithmetic sequence. Therefore, we can write the equation:2S_n = a_n + a_n^2.Hmm, that seems straightforward. Let me write that down:2S_n = a_n + a_n². (1)Now, S_n is the sum of the first n terms of the sequence, so S_n = a_1 + a_2 + ... + a_n. Also, note that S_{n} = S_{n-1} + a_n for n ≥ 2, and S_1 = a_1.Since the problem states that the relation holds for any n ∈ ℕ^*, including n = 1, let's check that first. For n = 1:2S_1 = a_1 + a_1².But S_1 is just a_1, so:2a_1 = a_1 + a_1².Subtracting 2a_1 from both sides:0 = -a_1 + a_1².Which simplifies to:a_1² - a_1 = 0.Factoring:a_1(a_1 - 1) = 0.Since the terms are positive, a_1 ≠ 0, so a_1 = 1.Alright, so we found that a_1 = 1. Let's check this. If a_1 is 1, then S_1 = 1. Then, the arithmetic sequence would be a_1=1, S_1=1, a_1²=1. So, 1, 1, 1. That's an arithmetic sequence with common difference 0. Makes sense. Good.Now, let's move on to n = 2. For n = 2, the arithmetic sequence is a_2, S_2, a_2². So, according to the condition:2S_2 = a_2 + a_2². (2)But S_2 = S_1 + a_2 = 1 + a_2. Substitute into equation (2):2(1 + a_2) = a_2 + a_2².Expanding the left side:2 + 2a_2 = a_2 + a_2².Bringing all terms to one side:a_2² + a_2 - 2 - 2a_2 = 0.Simplify:a_2² - a_2 - 2 = 0.This is a quadratic equation in terms of a_2. Let's solve it:a_2 = [1 ± √(1 + 8)] / 2 = [1 ± 3]/2.Since a_2 is positive, we take the positive root:a_2 = (1 + 3)/2 = 2.So, a_2 = 2. Then S_2 = 1 + 2 = 3. Let's check if 2, 3, 4 (since a_2²=4) form an arithmetic sequence. The differences are 3 - 2 = 1 and 4 - 3 = 1. Yes, it's an arithmetic sequence. Good.Moving on to n = 3. For n = 3, the arithmetic sequence is a_3, S_3, a_3². So:2S_3 = a_3 + a_3². (3)But S_3 = S_2 + a_3 = 3 + a_3. Substitute into equation (3):2(3 + a_3) = a_3 + a_3².Left side: 6 + 2a_3.Right side: a_3 + a_3².Bring all terms to left:6 + 2a_3 - a_3 - a_3² = 0.Simplify:6 + a_3 - a_3² = 0.Rearranged:-a_3² + a_3 + 6 = 0.Multiply both sides by -1:a_3² - a_3 - 6 = 0.Solving quadratic equation:a_3 = [1 ± √(1 + 24)] / 2 = [1 ± 5]/2.Again, positive term, so:a_3 = (1 + 5)/2 = 3.Check: S_3 = 3 + 3 = 6. Then, the terms a_3 = 3, S_3 = 6, a_3² = 9. Check arithmetic sequence: 3, 6, 9. The differences are 3 and 3. Correct. So arithmetic sequence with common difference 3. Perfect.Continuing this pattern, seems like a_n = n? Let's test n = 4.For n = 4, arithmetic sequence a_4, S_4, a_4². So:2S_4 = a_4 + a_4². (4)S_4 = S_3 + a_4 = 6 + a_4. Substitute:2(6 + a_4) = a_4 + a_4².Left side: 12 + 2a_4.Right side: a_4 + a_4².Bring all terms to left:12 + 2a_4 - a_4 - a_4² = 0.Simplify:12 + a_4 - a_4² = 0.Rearranged:-a_4² + a_4 + 12 = 0.Multiply by -1:a_4² - a_4 - 12 = 0.Solutions:a_4 = [1 ± √(1 + 48)] / 2 = [1 ± 7]/2.Positive solution: (1 + 7)/2 = 4.Check: S_4 = 6 + 4 = 10. Then, terms a_4 = 4, S_4 = 10, a_4² = 16. Check arithmetic sequence: 4, 10, 16. The differences are 6 and 6. Correct. So yes, arithmetic sequence with common difference 6. So a_4 = 4.So, the pattern here is that a_n = n. Let's check if this is true in general.Assume that a_k = k for all k ≤ n. Then S_n = 1 + 2 + ... + n = n(n + 1)/2.Then for n + 1, according to the problem, we have:2S_{n+1} = a_{n+1} + a_{n+1}².But S_{n+1} = S_n + a_{n+1} = [n(n + 1)/2] + a_{n+1}.Substituting into the equation:2([n(n + 1)/2] + a_{n+1}) = a_{n+1} + a_{n+1}².Simplify left side:n(n + 1) + 2a_{n+1} = a_{n+1} + a_{n+1}².Bring all terms to left:n(n + 1) + 2a_{n+1} - a_{n+1} - a_{n+1}² = 0.Simplify:n(n + 1) + a_{n+1} - a_{n+1}² = 0.Rearranged:-a_{n+1}² + a_{n+1} + n(n + 1) = 0.Multiply by -1:a_{n+1}² - a_{n+1} - n(n + 1) = 0.This is a quadratic equation in a_{n+1}:a_{n+1} = [1 ± √(1 + 4n(n + 1))]/2.Let's compute the discriminant:√(1 + 4n(n + 1)) = √(4n² + 4n + 1) = √( (2n + 1)^2 ) = 2n + 1.So, the solutions are:a_{n+1} = [1 ± (2n + 1)] / 2.Since a_{n+1} is positive, take the positive root:a_{n+1} = [1 + 2n + 1]/2 = (2n + 2)/2 = n + 1.Therefore, by induction, if a_k = k for all k ≤ n, then a_{n+1} = n + 1. Since we have established the base case (a_1 = 1), the sequence is indeed a_n = n for all n ∈ ℕ^*.Therefore, a_{2009} = 2009. Checking the options, option D is 2009. So the answer should be D.But wait, let me verify once more. Maybe I made a mistake in the induction step.Wait, when we assumed a_k = k up to n, then computed a_{n+1} and found it to be n + 1. So the induction holds. So the sequence is a_n = n for all n. Therefore, a_{2009} = 2009. Option D.But let me check with n=5 just to be thorough.For n=5: Assuming a_1=1, a_2=2, a_3=3, a_4=4, so S_4=10. Then for a_5:2S_5 = a_5 + a_5².S_5 = S_4 + a_5 = 10 + a_5.Thus, 2(10 + a_5) = a_5 + a_5².Left: 20 + 2a_5.Right: a_5 + a_5².Bring over:20 + 2a_5 - a_5 - a_5² = 0.Simplify: 20 + a_5 - a_5² = 0.Quadratic: a_5² - a_5 -20=0.Solutions: [1 ±√(1 +80)]/2 = [1 ±9]/2. Positive solution: (1+9)/2=5. So a_5=5. Correct.Yes, seems solid. So the general term is a_n = n, hence a_{2009}=2009, which is option D. So the answer is D.Final Answerboxed{D}</think>Given the sequence ({a_n}) where (a_n), (S_n), and (a_n^2) form an arithmetic sequence for any (n in mathbb{N}^*), we need to find (a_{2009}).1. Base Case (n = 1): - For (n = 1), the terms are (a_1), (S_1), and (a_1^2). - Since (S_1 = a_1), the equation becomes (2a_1 = a_1 + a_1^2). - Solving this, we get (a_1(a_1 - 1) = 0). Since (a_1) is positive, (a_1 = 1).2. Inductive Step: - Assume (a_k = k) for all (k leq n). Then (S_n = frac{n(n+1)}{2}). - For (n + 1), the equation is (2S_{n+1} = a_{n+1} + a_{n+1}^2). - Substituting (S_{n+1} = S_n + a_{n+1}) into the equation, we get: [ 2left(frac{n(n+1)}{2} + a_{n+1}right) = a_{n+1} + a_{n+1}^2 ] - Simplifying, we get: [ n(n+1) + 2a_{n+1} = a_{n+1} + a_{n+1}^2 ] - Rearranging terms, we get the quadratic equation: [ a_{n+1}^2 - a_{n+1} - n(n+1) = 0 ] - Solving this quadratic equation, we find (a_{n+1} = n + 1).3. Conclusion: - By induction, (a_n = n) for all (n in mathbb{N}^*). - Therefore, (a_{2009} = 2009).The final answer is (boxed{D}).,<think>Alright, let's tackle this problem step by step. The problem says that for the sequence {a_n}, which has positive terms, the sum of the first n terms is S_n. For any natural number n, it's given that a_n, S_n, a_n² form an arithmetic sequence. We need to find a_{2009} and the options are given as A:1, B:4018, C:2010, D:2009.First, let me recall what an arithmetic sequence is. In an arithmetic sequence, the difference between consecutive terms is constant. So, if three terms x, y, z are in arithmetic sequence, then the middle term y is the average of x and z. That is, 2y = x + z. So, applying that to the given problem: since a_n, S_n, a_n² form an arithmetic sequence, we must have 2S_n = a_n + a_n². That's the key equation here.So, the first step is to write down the condition given by the arithmetic sequence. For any n in N*, 2S_n = a_n + a_n². Let me note that down:2S_n = a_n + (a_n)^2. --- (1)Our goal is to find a_{2009}. To do that, we need to find a general formula for a_n. Let's see if we can express S_n in terms of a_n and perhaps previous terms.We know that S_n is the sum of the first n terms, so S_n = a_1 + a_2 + ... + a_n. Also, S_n = S_{n-1} + a_n for n >= 2, with S_1 = a_1.So, let's try to use equation (1) for n and n-1. Let's write equation (1) for n:2S_n = a_n + (a_n)^2. --- (1)And for n-1:2S_{n-1} = a_{n-1} + (a_{n-1})^2. --- (2)But we also know that S_n = S_{n-1} + a_n. So, substituting S_{n-1} from equation (2) into this relation might help. Let's see.From equation (2), S_{n-1} = [a_{n-1} + (a_{n-1})^2]/2.Therefore, S_n = [a_{n-1} + (a_{n-1})^2]/2 + a_n.But from equation (1), S_n = [a_n + (a_n)^2]/2.Therefore, equating these two expressions for S_n:[a_{n-1} + (a_{n-1})^2]/2 + a_n = [a_n + (a_n)^2]/2.Multiply both sides by 2 to eliminate the denominators:a_{n-1} + (a_{n-1})^2 + 2a_n = a_n + (a_n)^2.Simplify both sides:Left side: a_{n-1} + (a_{n-1})^2 + 2a_n.Right side: a_n + (a_n)^2.Subtract the right side from both sides:Left side - Right side: [a_{n-1} + (a_{n-1})^2 + 2a_n] - [a_n + (a_n)^2] = 0.Simplify:a_{n-1} + (a_{n-1})^2 + 2a_n - a_n - (a_n)^2 = 0.Combine like terms:a_{n-1} + (a_{n-1})^2 + a_n - (a_n)^2 = 0.Rearranged:(a_{n-1} + (a_{n-1})^2) + (a_n - (a_n)^2) = 0.Hmm. Let's factor terms:Looking at a_{n-1} + (a_{n-1})^2 = a_{n-1}(1 + a_{n-1}).Similarly, a_n - (a_n)^2 = a_n(1 - a_n).So, equation becomes:a_{n-1}(1 + a_{n-1}) + a_n(1 - a_n) = 0. --- (3)This seems a bit complicated, but maybe we can find a pattern or recursion here. Let's see.Alternatively, maybe instead of subtracting S_{n} expressions for n and n-1, we can look for a recurrence relation between a_n and a_{n-1}.Alternatively, let's try to write the difference between S_n and S_{n-1}, which is a_n.But perhaps we can express the equation (1) for both n and n-1, then relate them.Wait, let me try to write equation (1) for n:2S_n = a_n + (a_n)^2.And equation (1) for n-1:2S_{n-1} = a_{n-1} + (a_{n-1})^2.Subtracting these two equations gives:2(S_n - S_{n-1}) = a_n + (a_n)^2 - a_{n-1} - (a_{n-1})^2.But S_n - S_{n-1} is a_n, so:2a_n = a_n + (a_n)^2 - a_{n-1} - (a_{n-1})^2.Simplify:Left side: 2a_n.Right side: a_n + (a_n)^2 - a_{n-1} - (a_{n-1})^2.Subtract the right side from both sides:2a_n - [a_n + (a_n)^2 - a_{n-1} - (a_{n-1})^2] = 0.Simplify:2a_n - a_n - (a_n)^2 + a_{n-1} + (a_{n-1})^2 = 0.Which becomes:a_n - (a_n)^2 + a_{n-1} + (a_{n-1})^2 = 0.Which is the same as equation (3). So, same result.So, we have:a_{n-1} + (a_{n-1})^2 + a_n - (a_n)^2 = 0.Let me rearrange terms:a_{n-1} + (a_{n-1})^2 = (a_n)^2 - a_n.So, left side is a quadratic in a_{n-1}, and the right side is a quadratic in a_n.Perhaps factor each side.Left side: a_{n-1}(1 + a_{n-1}).Right side: a_n(a_n - 1).So:a_{n-1}(1 + a_{n-1}) = a_n(a_n - 1).This is interesting. Let me write this as:a_{n} (a_n - 1) = a_{n-1}(a_{n-1} + 1).So, this seems like a recursive relation. Let's denote b_n = a_n(a_n - 1). Then, the equation becomes:b_n = a_{n-1}(a_{n-1} + 1).But since b_n = a_n(a_n - 1), then:a_n(a_n - 1) = a_{n-1}(a_{n-1} + 1).So, this is a recurrence relation. Let me write it again:a_n(a_n - 1) = a_{n-1}(a_{n-1} + 1). --- (4)This seems a bit non-linear. Maybe we can manipulate this to find a pattern.Alternatively, let's try to find a telescoping product or something. Let's see.Let me rearrange equation (4):a_n(a_n - 1) - a_{n-1}(a_{n-1} + 1) = 0.But not sure if that helps. Alternatively, consider that both sides are products of terms. Let's compute this for small n to see if we can spot a pattern.First, let's consider n=1. For n=1, S_1 = a_1. Then, according to equation (1):2S_1 = a_1 + (a_1)^2.But S_1 = a_1, so 2a_1 = a_1 + a_1².Simplify: 2a_1 - a_1 - a_1² = 0 → a_1 - a_1² = 0 → a_1(1 - a_1) = 0.But the terms are positive, so a_1 ≠ 0. Hence, 1 - a_1 = 0 ⇒ a_1 = 1.So, the first term is 1. That's a start.Now, let's compute a_2.Using equation (4):a_2(a_2 - 1) = a_1(a_1 + 1).We know a_1 = 1, so RHS is 1*(1 + 1) = 2.Thus, a_2(a_2 - 1) = 2.So, let's solve for a_2:a_2² - a_2 - 2 = 0.Quadratic equation: a_2 = [1 ± sqrt(1 + 8)] / 2 = [1 ± 3]/2.Since a_2 is positive, we take the positive root: (1 + 3)/2 = 2. So, a_2 = 2.Now, let's check if that works with the original condition. For n=2, S_2 = a_1 + a_2 = 1 + 2 = 3.Check if 2S_2 = a_2 + (a_2)^2. 2*3 = 6, and a_2 + a_2² = 2 + 4 = 6. Yes, that works.Now, let's compute a_3.Again, using equation (4):a_3(a_3 - 1) = a_2(a_2 + 1) = 2*(2 + 1) = 6.So, equation: a_3² - a_3 - 6 = 0.Solutions: [1 ± sqrt(1 + 24)] / 2 = [1 ± 5]/2. Taking the positive root: (1 + 5)/2 = 3. So, a_3 = 3.Check with original condition. S_3 = 1 + 2 + 3 = 6.Check 2S_3 = 12, and a_3 + a_3² = 3 + 9 = 12. Correct.Continuing, a_4:a_4(a_4 - 1) = a_3(a_3 + 1) = 3*4 = 12.Equation: a_4² - a_4 - 12 = 0.Solutions: [1 ± sqrt(1 + 48)] / 2 = [1 ± 7]/2. Positive root: (1 + 7)/2 = 4. So, a_4 = 4.Check S_4 = 1 + 2 + 3 + 4 = 10. 2S_4 = 20. a_4 + a_4² = 4 + 16 = 20. Correct.So, the pattern seems that a_n = n. Let's verify with a_5.a_5(a_5 -1 ) = a_4(a_4 +1 ) = 4*5 =20.Equation: a_5² - a_5 -20=0. Solutions: [1 ± sqrt(81)]/2 = [1 ±9]/2. Positive root:10/2=5. So, a_5=5.Indeed, the pattern holds. So, it seems that a_n = n for all n.If this is the case, then a_{2009} = 2009, which is option D.But wait, let's confirm this with induction to make sure.Base case: n=1, a_1=1. True.Assume that a_k =k for all k <=n. Then, for a_{n+1}, using equation (4):a_{n+1}(a_{n+1} -1) = a_n(a_n +1 ) = n(n +1).So, equation becomes a_{n+1}^2 -a_{n+1} -n(n +1)=0.Let’s solve this quadratic equation:a_{n+1} = [1 ± sqrt(1 +4n(n +1))]/2.Compute discriminant: 1 +4n(n +1) =4n² +4n +1= (2n +1)^2.Thus, sqrt(discriminant)=2n +1.Therefore, a_{n+1}= [1 + (2n +1)]/2 = (2n +2)/2= n +1. (We discard the negative root because a_{n+1} is positive.)Thus, by induction, a_n =n for all n. Therefore, the sequence is a_n =n.Hence, a_{2009}=2009, which is option D.But let me check with n=1 again to make sure. For n=1, S_1=1. Then, 2S_1=2, and a_1 + a_1²=1 +1=2. Correct. Similarly, all other terms check out as above. So, this seems to hold.Therefore, the answer should be D:2009.Final Answerboxed{D}</think>Given the sequence ({a_n}) with positive terms, where (S_n) denotes the sum of the first (n) terms, and the condition that (a_n), (S_n), (a_n^2) form an arithmetic sequence for any (n in mathbb{N}^*), we need to find (a_{2009}).First, we use the property of an arithmetic sequence: the middle term is the average of the other two terms. Therefore, we have:[2S_n = a_n + a_n^2]This can be rewritten as:[S_n = frac{a_n + a_n^2}{2}]We also know that (S_n = S_{n-1} + a_n). Substituting this into the equation, we get:[S_{n-1} + a_n = frac{a_n + a_n^2}{2}]Substituting (S_{n-1}) from the previous term's condition:[S_{n-1} = frac{a_{n-1} + a_{n-1}^2}{2}]Combining these, we have:[frac{a_{n-1} + a_{n-1}^2}{2} + a_n = frac{a_n + a_n^2}{2}]Multiplying through by 2 to clear the denominators:[a_{n-1} + a_{n-1}^2 + 2a_n = a_n + a_n^2]Rearranging terms, we get:[a_{n-1} + a_{n-1}^2 = a_n^2 - a_n]This can be factored as:[a_{n-1}(a_{n-1} + 1) = a_n(a_n - 1)]We solve this recurrence relation by induction. For (n=1), we find (a_1 = 1). Assuming (a_k = k) for all (k leq n), we solve for (a_{n+1}):[a_{n+1}(a_{n+1} - 1) = n(n + 1)]Solving the quadratic equation (a_{n+1}^2 - a_{n+1} - n(n + 1) = 0), we find:[a_{n+1} = n + 1]By induction, (a_n = n) for all (n). Therefore, (a_{2009} = 2009).The final answer is (boxed{D}).]

🔑:Since a_n, S_n, a_n^2 form an arithmetic sequence, and a_n > 0, we have 2S_n = a_n + a_n^2. When n=1, we get 2a_1 = a_1 + a_1^2, thus a_1 = 1. When n=2, 2(a_2 + 1) = a_2 + a_2^2, thus a_2 = 2. Therefore, the common difference d = a_2 - a_1 = 1. By the formula of the general term of an arithmetic sequence, we have a_{2009} = a_1 + 2008d = 1 + 2008 = 2009. Hence, the correct choice is: boxed{D}.

🧠:[<think>Okay, so the problem is to allocate 24 volunteer slots to 3 schools, with each school getting at least one slot and each school having a different number of slots. We need to find the number of ways to do this. Hmm, let me think step by step.First, since each school must get at least one slot, the minimum number of slots each school can have is 1. But also, the number of slots for each school has to be different. So we need three distinct positive integers that add up to 24. And then, once we find all such triplets, we have to consider the permutations of these allocations among the three schools.Wait, so maybe the problem can be broken down into two parts: first, find all possible sets of three distinct positive integers that sum to 24, and then for each such set, multiply by the number of ways to assign these numbers to the three schools, which would be 3! = 6 for each set. But we need to make sure that each set is unique and doesn't have duplicate numbers.But let me confirm. Let's start by thinking about how to partition 24 into three distinct positive integers. Let's denote the three numbers as a, b, c, such that a + b + c = 24, and a, b, c are distinct positive integers. Without loss of generality, let's assume that a < b < c. Then, each such ordered triplet (a, b, c) can be permuted in 6 ways to assign to the three schools. So first, find all triplets where a < b < c and a + b + c = 24, and then multiply the count by 6.So the key is to find the number of such triplets. How do we do that? Let's think. Since a, b, c are distinct positive integers with a < b < c, we can set up inequalities. The smallest possible values for a, b, c would be 1, 2, 3, which sum to 6. But we need them to sum to 24, which is much larger.Let me think of a systematic way. Let's start with a = 1. Then b must be at least 2, and c must be at least b + 1. So a =1, b =2, then c =24 -1 -2 =21. Then check if 1,2,21 are distinct and ordered. Yes, 1 < 2 <21. So that's one triplet.Next, keep a=1, increase b to 3. Then c =24 -1 -3=20. So triplet (1,3,20). Good.Similarly, a=1, b=4, c=19.Continuing this way, a=1, b can go up until a point where c is still greater than b.Wait, so for a=1, b can range from 2 up to such that c =24 -1 -b > b. So 24 -1 -b > b => 23 -b > b => 23 > 2b => b <11.5. Since b must be integer, b ≤11.So for a=1, b can be from 2 to 11. That gives 10 triplets: (1,2,21), (1,3,20), ..., (1,11,12).Wait, check when a=1, b=11: c=24 -1 -11=12. So 11 <12, so that's okay. So 10 triplets when a=1.Then, move to a=2. Now, a=2, so b must be at least 3, and c must be at least b+1. Let's see. So a=2, b=3, then c=24 -2 -3=19. So (2,3,19).Similarly, a=2, b=4, c=18. Continue this until c > b.Again, find the upper limit for b. For a=2, c=24 -a -b=22 -b. So c >b =>22 -b >b =>22 >2b =>b <11. So b can be from 3 to 10. Thus, when a=2, b ranges from 3 to 10. That's 8 values. So triplets (2,3,19), (2,4,18), ..., (2,10,12). Check when b=10: c=24 -2 -10=12. 10 <12, so okay. So 8 triplets for a=2.Next, a=3. Then b must be at least 4, c=24 -3 -b=21 -b. Then c >b =>21 -b >b =>21 >2b =>b <10.5. So b can go from 4 to 10. So b=4 to 10. But wait, when a=3 and b=10, c=21 -10=11. But c must be greater than b, so 11 >10, so okay. Wait, but 3,10,11. However, 10 <11, but does 3,10,11 sum to 24? 3+10+11=24. Yes. So that's okay.But wait, when a=3, b=4, c=17; a=3, b=5, c=16; ... up to b=10, c=11. So the number of triplets here is from b=4 to b=10, which is 7 triplets.Wait, 4 to10 inclusive is 7 numbers (4,5,6,7,8,9,10). So 7 triplets for a=3.Proceeding similarly, a=4. Then b must be at least 5, c=24 -4 -b=20 -b. Then c >b =>20 -b >b =>20>2b =>b <10. So b can go from 5 to9. Because when b=10, c=20 -10=10, which is equal to b, but they need to be distinct, so b must be up to 9. So b=5,6,7,8,9. That's 5 values. So triplets (4,5,15), (4,6,14), (4,7,13), (4,8,12), (4,9,11). Check sum: 4+5+15=24, etc. So 5 triplets.Next, a=5. Then b must be at least6, c=24 -5 -b=19 -b. c >b =>19 -b >b =>19>2b =>b <9.5. So b can be from6 to9. Let's check: b=6, c=13; b=7, c=12; b=8, c=11; b=9, c=10. So triplets (5,6,13), (5,7,12), (5,8,11), (5,9,10). That's 4 triplets.Moving to a=6. Then b must be at least7, c=24 -6 -b=18 -b. c >b =>18 -b >b =>18>2b =>b <9. So b=7,8. Let's check: b=7, c=11; b=8, c=10. So triplets (6,7,11), (6,8,10). So 2 triplets.a=7. Then b must be at least8, c=24 -7 -b=17 -b. c >b =>17 -b >b =>17>2b =>b <8.5. Since b must be at least8, the only possible value is b=8. Then c=17 -8=9. So triplet (7,8,9). Check sum:7+8+9=24. So 1 triplet.a=8. Then b must be at least9. c=24 -8 -b=16 -b. c >b =>16 -b >b =>16>2b =>b <8. But b must be at least9, which is impossible. So no triplets here.Similarly, a=9 or higher: a=9, then b must be at least10, c=24 -9 -b=15 -b. c >b =>15 -b >b =>15>2b =>b <7.5. But b must be at least10, so impossible. So no more triplets.So total number of triplets where a < b < c and a + b + c =24 is:For a=1:10a=2:8a=3:7a=4:5a=5:4a=6:2a=7:1Total:10+8=18, 18+7=25, 25+5=30, 30+4=34, 34+2=36, 36+1=37.Wait, so 37 triplets? Let me check again.Wait, when a=1, b from 2 to11: that's 11-2+1=10. Correct.a=2, b from3 to10:10-3+1=8. Correct.a=3, b from4 to10:10-4+1=7. Correct.a=4, b from5 to9:9-5+1=5. Correct.a=5, b from6 to9:9-6+1=4. Correct.Wait, no: for a=5, b=6 to9. 9-6 +1=4. Correct.a=6, b=7 to8:8-7 +1=2. Correct.a=7, b=8:1. Correct.So total triplets:10+8+7+5+4+2+1=37? Wait, 10+8 is 18, +7 is25, +5 is30, +4 is34, +2 is36, +1 is37. So 37.But that seems a lot. Let me verify with a different approach. The number of solutions in positive integers of a + b + c =24, with a,b,c distinct and ordered a < b < c.Alternatively, maybe we can use stars and bars with constraints. But stars and bars counts all solutions, but here we need distinct integers. Hmm, perhaps it's complicated. But maybe an alternative way.Alternatively, since a, b, c are distinct positive integers with a < b < c, we can write variables as x = a -1, y = b - a -1, z = c - b -1, w = ... Wait, perhaps that's overcomplicating.Alternatively, think of the problem as equivalent to partitioning 24 into three distinct parts, each at least 1. So similar to partitions, but here order matters? Wait, no, in partitions order doesn't matter, but here since we will assign them to different schools, so the order matters. Wait, no. Wait, in the triplet (a,b,c) we have a < b < c, so each unique combination is counted once. Then, when we assign these numbers to schools, each permutation is a different way, so multiply by 6.But first, how many triplets are there with a < b < c and a + b + c =24.Alternatively, let me check for some smaller number. Let's say if the total was 6. Then the only triplet is 1,2,3. So one triplet. If total is 9, triplets like 1,2,6;1,3,5;2,3,4. So three triplets. Then 3*6=18 ways. Wait, but how do we calculate that?Alternatively, perhaps use generating functions. The generating function for three distinct positive integers is x^(a + b + c) where a, b, c are distinct. So the generating function is (x + x^2 + x^3 +...)^3 minus the cases where two are equal or all are equal. Hmm, but this might be complicated.Alternatively, perhaps inclusion-exclusion. Total number of positive integer solutions to a + b + c =24 is C(24-1,3-1)=C(23,2)=253. Then subtract the cases where two are equal or all are equal.But since we need all distinct, first subtract the cases where two are equal. Let's compute that.Number of solutions where a = b: Let a = b =k, then 2k + c =24. So c=24 -2k. k can be from1 to11 (since 2k <=24 -1 =>k <=11.5). So k=1 to11. Each such k gives a solution (k,k,24-2k). Similarly for a = c and b =c. So three cases.But we need to subtract these but then add back the cases where all three are equal, since they were subtracted three times. But in our problem, all three equal would be a case where a = b =c. For sum 24, a = b =c =8. So there is one such solution: (8,8,8). So by inclusion-exclusion:Number of solutions with all distinct = total solutions - 3*number with two equal + 2*number with all equal.So total solutions:253Number with two equal: For each pair, a = b, a =c, b =c. For each pair, as above, for a = b: k from1 to11, 11 solutions. Similarly for a =c and b =c, each 11 solutions. So total 3*11=33.Number with all equal:1.Therefore, number of solutions with all distinct:253 -33 +2*1=253 -33=220 +2=222.Wait, but this counts all ordered triples where a, b, c are distinct positive integers. However, in our problem, we want unordered triples where a < b <c. So each such unordered triple is counted 6 times in the total count (since there are 6 permutations). So the number of unordered triples would be 222 /6=37. Which matches the previous count of37.Therefore, the number of ordered triples where a, b, c are distinct positive integers summing to24 is222, and the number of unordered triples is37. Then, since the problem asks for the number of ways to allocate the slots such that each school gets at least one and the numbers are different. So each unordered triplet corresponds to6 ordered allocations. Therefore, the total number of ways is37*6=222. Wait, but that can't be, since in the problem statement it says "the number of slots for each school is different". So each school has a distinct number, so each allocation is a permutation of the triplet. Therefore, the answer should be37*6=222. But wait, this conflicts with the previous inclusion-exclusion approach. Wait, but no. Wait, according to the inclusion-exclusion, the number of ordered triples with distinct numbers is222, which is exactly37*6. So the answer would be222.But the problem is in Chinese, and the translation says "the number of ways to allocate such that each school gets at least one slot and the number of slots for each school is different". So that is, assign 24 slots to3 schools, each school at least1, all numbers different. So the answer should be the number of ordered triples (a,b,c) of positive integers, distinct, summing to24. Which is222. So why did we do the whole a <b <c thing?Wait, because in the first approach, we considered a <b <c and found37 triplets, then multiplied by6 to get222. While the inclusion-exclusion directly gives222 as the number of ordered triples. So both agree.But the problem is asking for "kinds". Depending on translation, "kinds" might mean the number of different allocations, considering that each school is distinct. So for example, giving school A 1, school B 2, school C21 is different from school A2, school B1, school C21. So in that case, the answer is222. However, maybe in Chinese, "kinds" might refer to the unordered partitions, but given the problem mentions "allocate to3 schools", which are distinguishable (since they are different schools), the answer should be222.But wait, let's check the problem again: "the number of ways to allocate such that each school gets at least one slot and the number of slots for each school is different". So "ways to allocate" would typically consider different assignments to different schools as different ways. Therefore, the answer should be222.But the initial approach of enumerating triplets with a <b <c and then multiplying by6 gives37*6=222. So that's correct.But wait, the answer expected here is maybe written as a number in boxed notation. So either222 or37*6=222. But I need to confirm.Wait, but let me check with an example. For instance, if total was6, then the triplets are1,2,3. The number of allocations would be6, since each school gets a different number. So1,2,3 can be assigned to schools in6 ways. Similarly, here, each triplet a <b <c can be assigned in6 ways. So the total is37*6=222.Alternatively, if the problem had said that the schools are indistinct, then the answer would be37. But since schools are distinct, it's222.But let me check the problem statement again: "allocate 24 volunteer slots to3 schools". Since schools are distinct entities, the allocation is an ordered triple. Therefore, the answer should be222.But wait, in the original Chinese problem, the term used is "kinds". Depending on translation, if "kinds" refers to the unique distributions regardless of school order, then it's37. If it refers to the number of assignments considering school order, then it's222. So need to figure out which is intended.But the problem says "allocate to3 schools", which usually implies that the schools are distinguishable. For example, if you have different schools, the way you distribute slots matters based on which school gets which number. Therefore, the answer should be222.But I need to confirm with my initial calculation. Since we found37 triplets with a <b <c, each can be permuted in6 ways, hence222.Alternatively, maybe the problem is expecting partitions with distinct parts, which is37, but considering the schools are different, it should be multiplied by6. So yes, the answer is222.But let's verify with a small example. Suppose we have6 slots,3 schools, each at least1, numbers different. Then the only triplet is1,2,3, which can be allocated in6 ways. So answer is6. Similarly, in this problem, it's the same reasoning, just scaled up.Therefore, the answer is37*6=222.But let me cross-verify with inclusion-exclusion. As we computed earlier, the number of ordered triples with distinct positive integers summing to24 is222. So that's the same as37*6=222.Therefore, the answer should be222. Hence, the number of ways is222, so boxed{222}.Final Answerboxed{222}</think>To allocate 24 volunteer slots to 3 schools such that each school gets at least one slot and the number of slots for each school is different, we need to find the number of ways to partition 24 into three distinct positive integers and then consider the permutations of these partitions among the schools.1. Finding distinct triplets: - We need to find all sets of three distinct positive integers (a, b, c) such that (a + b + c = 24) and (a < b < c). - By systematically checking possible values for (a) and calculating corresponding (b) and (c), we find the following triplets: - For (a = 1), (b) ranges from 2 to 11, giving 10 triplets. - For (a = 2), (b) ranges from 3 to 10, giving 8 triplets. - For (a = 3), (b) ranges from 4 to 10, giving 7 triplets. - For (a = 4), (b) ranges from 5 to 9, giving 5 triplets. - For (a = 5), (b) ranges from 6 to 9, giving 4 triplets. - For (a = 6), (b) ranges from 7 to 8, giving 2 triplets. - For (a = 7), (b = 8), giving 1 triplet. - Summing these, we get (10 + 8 + 7 + 5 + 4 + 2 + 1 = 37) distinct triplets.2. Considering permutations: - Each triplet ((a, b, c)) can be permuted in (3! = 6) ways to assign the slots to the three schools. - Therefore, the total number of ways to allocate the slots is (37 times 6 = 222).Thus, the number of ways to allocate the 24 volunteer slots to 3 schools such that each school gets at least one slot and the number of slots for each school is different is (boxed{222}).,<think>Okay, let's see. The problem is to allocate 24 volunteer slots to 3 schools, with each school getting at least one slot and the number of slots for each school being different. I need to find the number of ways to do this. Hmm, okay, let's break this down step by step.First, the basic idea here is that we have to distribute 24 slots among 3 schools. Normally, if there were no restrictions, the number of ways to allocate these slots would be calculated using the stars and bars method. But here, there are two conditions: each school must get at least one slot, and each school must have a different number of slots. So, we need to adjust for those constraints.Let me recall that for distributing n identical items into k distinct groups with each group getting at least one item, the formula is (n-1 choose k-1). In this case, without considering the different numbers condition, it would be (24-1 choose 3-1) = (23 choose 2). But since we have the added condition that each school must have a distinct number of slots, we need to figure out how that affects the count.So, first, each school must get at least one slot. Second, each school must have a different number of slots. So, the three schools must have three distinct positive integers that add up to 24. So, essentially, we need the number of ordered triples (a, b, c) such that a, b, c are positive integers, a + b + c = 24, and a, b, c are all distinct. Then, since the schools are distinct, each such triple would correspond to a different allocation.But wait, the problem says "the number of ways to allocate such that each school gets at least one slot and the number of slots for each school is different". So, perhaps we need to find the number of ordered triples (a, b, c) with a, b, c ≥ 1, a ≠ b ≠ c ≠ a, and a + b + c = 24. Then, each such ordered triple is a different allocation. So, the answer would be equal to the number of such ordered triples.Alternatively, maybe the problem is considering the schools as distinct, so each permutation of a set of numbers counts as a different allocation. For example, if one allocation is (1, 2, 21), then another allocation like (2, 1, 21) would be different because the schools are different. So, order matters here. So, we need to count all ordered triples where the numbers are distinct and add up to 24.But to find that, perhaps it's easier to first count the number of unordered triples (i.e., partitions of 24 into 3 distinct positive integers) and then multiply by the number of permutations (which is 6, since there are 3! ways to arrange three distinct numbers). But we need to make sure whether the problem wants ordered or unordered. But since the schools are distinct, the allocations are different if the numbers are assigned to different schools. So, order does matter here. So, each permutation counts as a different allocation.Therefore, the total number of allocations would be equal to the number of ordered triples (a, b, c) with a, b, c ≥ 1, a + b + c = 24, and a, b, c distinct. So, the problem reduces to finding the number of such ordered triples.Alternatively, another approach is to think of this as a partition problem. First, find all partitions of 24 into 3 distinct positive integers, and then multiply by 6 (since each partition can be arranged in 3! ways). However, partitions usually consider unordered collections, so if we find the number of partitions of 24 into 3 distinct parts, each part at least 1, then each such partition corresponds to 6 ordered triples. So, if I can find the number of such partitions, multiplying by 6 would give the answer.So, the key question is: how many partitions of 24 into three distinct positive integers are there?To compute this, let's denote the three distinct positive integers as x, y, z, such that x < y < z and x + y + z = 24. Then, the number of such triplets (x, y, z) would be the number of partitions we need, and then we multiply by 6 to get the number of ordered allocations.So, first, let's find the number of triplets where x < y < z and x + y + z = 24.To find this, we can use the concept of generating partitions with distinct parts.Alternatively, we can use substitution. Let me set variables such that x, y, z are positive integers with x < y < z. Let's set x = a, y = a + b, z = a + b + c, where a, b, c are positive integers. Then, the sum x + y + z = a + (a + b) + (a + b + c) = 3a + 2b + c = 24. So, the equation is 3a + 2b + c = 24, where a, b, c ≥ 1.But this substitution might complicate things. Alternatively, another approach is to use the standard method for counting partitions with distinct parts.Alternatively, since x, y, z are distinct positive integers with x < y < z and x + y + z = 24, we can start by finding the minimum possible values for x, y, z.The smallest possible values for x, y, z would be 1, 2, 3, which sum to 6. But we need them to sum to 24, which is much larger. So, how can we adjust?Let me consider that for x < y < z, we can let x be as small as possible and then y and z adjust accordingly. But maybe a better way is to use substitution variables.Let me set x = a, y = a + d, z = a + d + e, where d and e are positive integers. Then, x + y + z = a + (a + d) + (a + d + e) = 3a + 2d + e = 24. So, we have 3a + 2d + e = 24, with a ≥ 1, d ≥ 1, e ≥ 1. Then, we need to find the number of solutions in positive integers to this equation.Alternatively, perhaps it's easier to think in terms of equations. Since x < y < z, then x ≥ 1, y ≥ x + 1 ≥ 2, z ≥ y + 1 ≥ 3. So, the minimum sum is 1 + 2 + 3 = 6. Our sum is 24, so the difference is 18. So, we need to distribute this difference among the three variables while maintaining the order.Alternatively, we can use the concept of partitions with constraints.Alternatively, perhaps the standard formula for the number of partitions of n into k distinct parts is a known formula, but I might not remember it. So, perhaps it's better to approach this with generating functions or by systematic counting.Alternatively, let's consider the problem of finding the number of triplets (x, y, z) with x < y < z and x + y + z = 24.We can use substitution variables. Let’s set x = a, y = a + b, z = a + b + c, where a, b, c are positive integers. Then, x + y + z = a + (a + b) + (a + b + c) = 3a + 2b + c = 24. So, the equation is 3a + 2b + c = 24, with a, b, c ≥ 1.So, we need to find the number of positive integer solutions (a, b, c) to this equation.Let’s adjust variables to convert this into non-negative integers. Let a’ = a - 1, b’ = b - 1, c’ = c - 1. Then, a’, b’, c’ ≥ 0. Substituting into the equation:3(a’ + 1) + 2(b’ + 1) + (c’ + 1) = 24Expanding:3a’ + 3 + 2b’ + 2 + c’ + 1 = 24Combine constants:3a’ + 2b’ + c’ + 6 = 24Therefore:3a’ + 2b’ + c’ = 18Now, we need to find the number of non-negative integer solutions (a’, b’, c’) to 3a’ + 2b’ + c’ = 18.This seems a bit complicated, but maybe we can fix a’ and b’ and solve for c’.For each possible value of a’ and b’, c’ is determined as c’ = 18 - 3a’ - 2b’. We need c’ ≥ 0.So, the number of solutions is equal to the number of pairs (a’, b’) such that 3a’ + 2b’ ≤ 18.So, let's find all non-negative integers a’ and b’ where 3a’ + 2b’ ≤ 18.To count this, perhaps iterate over possible a’ values.Let’s note that a’ can be from 0 up to floor(18/3) = 6.For each a’ from 0 to 6, compute the maximum possible b’ such that 2b’ ≤ 18 - 3a’.So, for each a’:max b’ = floor( (18 - 3a’) / 2 )Therefore, for each a’, the number of possible b’ is floor( (18 - 3a’)/2 ) + 1 (since b’ can start from 0 up to max b’).Let me compute this for each a’:a’ = 0:max b’ = floor(18 / 2) = 9. So, number of b’ = 9 + 1 = 10a’ = 1:max b’ = floor( (18 - 3)/2 ) = floor(15/2) = 7. So, number of b’ = 7 + 1 = 8a’ = 2:max b’ = floor( (18 - 6)/2 ) = floor(12/2) = 6. So, number of b’ = 6 + 1 = 7a’ = 3:max b’ = floor( (18 - 9)/2 ) = floor(9/2) = 4. So, number of b’ = 4 + 1 = 5a’ = 4:max b’ = floor( (18 - 12)/2 ) = floor(6/2) = 3. So, number of b’ = 3 + 1 = 4a’ = 5:max b’ = floor( (18 - 15)/2 ) = floor(3/2) = 1. So, number of b’ = 1 + 1 = 2a’ = 6:max b’ = floor( (18 - 18)/2 ) = floor(0/2) = 0. So, number of b’ = 0 + 1 = 1Now, sum up these numbers:10 + 8 + 7 + 5 + 4 + 2 + 1 = 37Therefore, there are 37 solutions (a’, b’, c’) to the equation 3a’ + 2b’ + c’ = 18, which correspond to 37 solutions (a, b, c) in positive integers.Therefore, the number of triplets (x, y, z) with x < y < z and x + y + z = 24 is 37.But wait, this seems a bit high. Let me check the calculation again.Wait, the sum for a’ from 0 to 6:a’=0:10a’=1:8a’=2:7a’=3:5a’=4:4a’=5:2a’=6:1Total: 10+8=18, 18+7=25, 25+5=30, 30+4=34, 34+2=36, 36+1=37. Yes, 37.But does that mean there are 37 ordered triples (a, b, c)? Wait no, this is the number of solutions (a', b', c') which correspond to (a, b, c) in the original variables. Each solution gives a unique triplet (x, y, z) = (a, a + b, a + b + c). So, each solution corresponds to one unique increasing triplet. Therefore, there are 37 such triplets.Therefore, the number of unordered triplets is 37. Then, since each allocation to schools is ordered (since schools are distinct), we need to multiply by 6 (the number of permutations of each triplet). Therefore, the total number of allocations would be 37 * 6 = 222.But wait, that seems like a possible answer, but let's verify.Alternatively, maybe there's a mistake in my substitution. Let me take a small example to check.Suppose we have n = 6, k = 3. The triplets with x < y < z and x + y + z = 6.The only triplet is (1, 2, 3). So, number of partitions is 1. Then, total allocations would be 6. Let's check with the formula.Using the same method:3a + 2b + c = 6 - 3 - 2 -1 = 0? Wait, no, wait.Wait, for n=6, x + y + z =6, x < y < z, so x >=1, y >=2, z >=3. So, x=1, y=2, z=3. That's the only one.So, according to our formula above, let's check.Original equation: 3a + 2b + c = 6 (since n=6 here). Then, substituting a’=a-1, b’=b-1, c’=c-1:3a’ + 2b’ + c’ = 6 - 3 - 2 -1 + 3 + 2 + 1 = Wait, no. Wait, perhaps my substitution was different. Wait, no, in the previous problem, the original equation was 3a + 2b + c = 24, but for n=6, perhaps the equation would be different. Wait, maybe my substitution is wrong. Wait, no, in the original problem, the substitution was for x, y, z such that x < y < z. So, x = a, y = a + b, z = a + b + c. Then, x + y + z = 3a + 2b + c. So, for n=6, 3a + 2b + c =6. Then, a, b, c >=1. Then, substituting a’=a-1, etc.:3(a’+1) + 2(b’+1) + (c’+1) =63a’ +3 + 2b’ +2 + c’ +1 =63a’ +2b’ +c’ +6=6So, 3a’ +2b’ +c’=0. The only solution is a’=b’=c’=0, which gives a=1, b=1, c=1. Then, x=1, y=2, z=3. So, that works. So, in that case, the number of solutions is 1, which is correct. So, the formula works here.But according to the previous calculation for n=6, the number of solutions would be 1. So, the formula seems to work. Therefore, perhaps in the original problem, the number of increasing triplets is indeed 37, leading to 37 *6=222 allocations.But let me test another example. Take n=7.Possible triplets with distinct positive integers adding to 7: (1,2,4). So, only one triplet. Then, allocations would be 6. Let's check with the formula.3a + 2b + c =7. Wait, no, the equation is 3a + 2b + c = sum. Wait, in this case, sum is 7. Wait, but in our substitution:x =a, y =a + b, z =a + b + c. Then, x + y + z =3a + 2b + c =7. Then, a, b, c >=1. So, substituting a’=a-1, etc.:3a’ +2b’ +c’ +6=7 => 3a’ +2b’ +c’=1.Solutions where a', b', c' >=0.Possible solutions: a’=0, b’=0, c’=1; a’=0, b’=0, c’=1. So, only one solution. Then, a=1, b=1, c=2. Then, x=1, y=2, z=1+1+2=4. Which is correct. So, only one triplet, which gives 6 allocations. So, formula works here.Another example: n=9.Possible triplets (1,2,6), (1,3,5), (2,3,4). So, three triplets. Therefore, 3*6=18 allocations.Using our formula:Original equation: 3a + 2b + c =9.Substituting variables:3(a’+1) + 2(b’+1) + (c’+1) =93a’ +3 + 2b’ +2 + c’ +1 =93a’ +2b’ +c’ +6=93a’ +2b’ +c’=3Now, find the number of non-negative integer solutions.Possible a’:a’ can be 0, 1 (since 3a’ <=3)a’=0:3a’=0, so 2b’ +c’=3. Possible b’:b’=0: c’=3b’=1: c’=1b’=2: c’= -1 (invalid)So, two solutions: (0,0,3), (0,1,1)a’=1:3a’=3, so 2b’ +c’=0. Only solution is b’=0, c’=0Thus, total solutions: 2 +1=3. Therefore, three triplets, which matches. So, the formula works here.Therefore, the formula seems to hold. Thus, in our original problem, with n=24, we found 37 triplets. Therefore, multiplying by 6 gives 222 allocations. So, the answer is 222.But wait, let me check another way. Let's think of the problem as finding all ordered triples (a, b, c) where a, b, c are distinct positive integers, a + b + c =24. The number of such ordered triples is equal to the number of permutations of the partitions, so 6 times the number of partitions into distinct parts.Alternatively, another formula: the number of ordered triples (a, b, c) with a, b, c ≥1, a, b, c distinct, and a + b + c =24.Alternatively, the standard approach is to subtract the cases where two are equal or all are equal from the total number of solutions with a, b, c ≥1.So, first, total number of solutions with a, b, c ≥1: This is the stars and bars problem, which is C(24 -1, 3 -1)=C(23, 2)=253.Then, subtract the cases where two or more numbers are equal.So, using inclusion-exclusion:Total = all solutions - solutions with at least two equal + solutions with all three equal.Wait, inclusion-exclusion formula for distinct elements:Number of solutions with all distinct = Total solutions - (number with at least two equal) + ... Wait, inclusion-exclusion is:Number of solutions with all distinct = Total solutions - C(3,2)*number of solutions with a specific pair equal + C(3,3)*number of solutions with all three equal.Wait, yes, that's correct.So, formula:N(distinct) = N(total) - 3*N(a = b) + 3*N(a = b = c)Wait, because there are C(3,2)=3 pairs, each pair can be equal, and then subtract the overcounted cases where all three are equal.So, first, compute N(total)=C(23,2)=253.Then, compute N(a = b). If a = b, then the problem reduces to 2a + c =24, with a ≥1, c ≥1. So, a can be from 1 to 11 (since 2a ≤23). Therefore, number of solutions is 11. But wait, let's see:If a = b, then 2a + c =24. Since a ≥1 and c ≥1, then 2a can be from 2 to 22, so c =24 -2a. To have c ≥1, 24 -2a ≥1 => 2a ≤23 => a ≤11.5. Since a is integer, a ≤11. Therefore, a=1 to 11, which gives 11 solutions. Similarly, for each pair (a,b), (a,c), (b,c), there are 11 solutions each. So, 3*11=33.But wait, N(a = b) is 11, and similarly for N(a = c) and N(b = c). So, total overcount is 3*11=33.Then, the number of solutions where all three are equal: a = b = c. Then, 3a =24 => a=8. So, only one solution: a=8, b=8, c=8.Therefore, applying inclusion-exclusion:N(distinct) = N(total) - 3*N(a = b) + 3*N(a = b = c)Wait, inclusion-exclusion formula is:Number of solutions with all distinct = Total solutions - sum of solutions with each pair equal + sum of solutions with all three equal.But in our case, the formula is:N(distinct) = N(total) - 3*N(a = b) + 1*N(a = b = c)Because when we subtract the pairs, we subtract the cases where all three are equal three times, so we need to add them back two times? Wait, let's recall the inclusion-exclusion principle.When computing the number of elements not having any of the properties (here, not having any equal pairs), the formula is:|A| = Total - Σ|A_i| + Σ|A_i ∩ A_j| - Σ|A_i ∩ A_j ∩ A_k| + ... But in our case, the properties are A_i: the ith and jth elements are equal. But since we have three variables, the pairs are (a,b), (a,c), (b,c). Each A_i is the set of solutions where a particular pair is equal.So, the formula is:Number of solutions with all distinct = Total - |A_ab ∪ A_ac ∪ A_bc|By inclusion-exclusion:|A_ab ∪ A_ac ∪ A_bc| = |A_ab| + |A_ac| + |A_bc| - |A_ab ∩ A_ac| - |A_ab ∩ A_bc| - |A_ac ∩ A_bc| + |A_ab ∩ A_ac ∩ A_bc|But |A_ab ∩ A_ac| is the set of solutions where a = b and a = c, i.e., a = b = c. Similarly for other intersections. So, each pairwise intersection is |A_abc|, the set where all three are equal. And the triple intersection is also |A_abc|.Therefore,|A_ab ∪ A_ac ∪ A_bc| = 3*|A_ab| - 3*|A_abc| + |A_abc|= 3*|A_ab| - 2*|A_abc|Therefore,Number of solutions with all distinct = Total - [3*|A_ab| - 2*|A_abc|] = Total - 3*|A_ab| + 2*|A_abc|But in our case, how many are |A_ab| and |A_abc|.As computed, |A_ab| =11, |A_abc|=1.Therefore,Number of distinct solutions = 253 - 3*11 + 2*1 =253 -33 +2=222.So, this matches the previous result of 37*6=222.Therefore, the number of ordered triples (a, b, c) with a, b, c ≥1, a + b + c=24, and all distinct is 222.Therefore, the answer is 222.But wait, let's cross-verify this with the partition approach.Earlier, we found that the number of triplets (x, y, z) with x < y < z and x + y + z=24 is 37. Each such triplet can be arranged in 6 ways, so 37*6=222. So, same result.Therefore, both methods give 222. So, the answer is 222.Therefore, the number of ways is 222.Final AnswerThe number of ways is boxed{222}.</think>To allocate 24 volunteer slots to 3 schools such that each school gets at least one slot and the number of slots for each school is different, we need to find the number of ordered triples ((a, b, c)) where (a, b, c) are distinct positive integers summing to 24.First, we use the principle of inclusion-exclusion to calculate the number of such ordered triples:1. Total number of solutions without restrictions: The number of ways to distribute 24 slots into 3 schools with each school getting at least one slot is given by the stars and bars method: [ binom{24-1}{3-1} = binom{23}{2} = 253 ]2. Subtract cases where two schools have the same number of slots: - For each pair (e.g., (a = b)), the equation becomes (2a + c = 24). Here, (a) ranges from 1 to 11, giving 11 solutions per pair. - There are 3 pairs, so we subtract (3 times 11 = 33).3. Add back cases where all three schools have the same number of slots: - The equation (3a = 24) has only one solution ((a = 8)), so we add back (2 times 1 = 2) (since we subtracted these cases three times, we need to add them back twice).Applying inclusion-exclusion:[253 - 33 + 2 = 222]Thus, the number of ways to allocate the 24 volunteer slots such that each school gets at least one slot and the number of slots for each school is different is (boxed{222}).]

🔑:The number of ways to allocate 24 volunteer slots to 3 schools, with each school getting at least one slot, is C_{23}^{2}= 253; the number of ways to allocate the slots such that all 3 schools receive the same number of slots is 1; the number of ways to allocate the slots such that exactly 2 schools receive the same number of slots (i.e., satisfying 2 x+z=24 and x neq z with positive integer solutions) is 10 times 3=30. Therefore, the number of ways to allocate the slots such that all 3 schools receive different numbers of slots is 253-1-30=222.

🧠:[<think>Okay, so I need to find the general formula for the nth term of the sequence {a_n} where the sum of the first n terms, S_n, is given by S_n = 2a_n - n. Then, for the second part, I have to define another sequence b_n = 2^n / (a_n a_{n+1}), find the sum T_n of the first n terms of this sequence, and determine the smallest positive integer m such that T_n < m/20 for all natural numbers n. Alright, let's start with the first part.First, the problem gives me that the sum of the first n terms, S_n, is equal to 2a_n - n. So S_n = 2a_n - n. I need to find a formula for a_n. Let's recall that the sum of the first n terms S_n is related to the nth term a_n by the relation S_n = S_{n-1} + a_n. That's the recursive formula for the sum. So maybe I can use that to find a recursive relation for a_n.Given S_n = 2a_n - n and S_{n} = S_{n-1} + a_n. So substituting the first equation into the second, we get:2a_n - n = S_{n-1} + a_nBut S_{n-1} can also be written using the first equation, replacing n with n-1:S_{n-1} = 2a_{n-1} - (n - 1)So substituting back into the previous equation:2a_n - n = (2a_{n-1} - (n - 1)) + a_nSimplify the right-hand side:2a_n - n = 2a_{n-1} - n + 1 + a_nCombine like terms:2a_n - n = a_n + 2a_{n-1} - n + 1Subtract a_n and add n to both sides:2a_n - n - a_n + n = 2a_{n-1} + 1Simplify left side:a_n = 2a_{n-1} + 1So we have a recurrence relation: a_n = 2a_{n-1} + 1. This is a linear nonhomogeneous recurrence relation. To solve this, we can find the homogeneous solution and a particular solution.First, solve the homogeneous equation: a_n - 2a_{n-1} = 0The characteristic equation is r - 2 = 0, so r = 2. Therefore, the homogeneous solution is A*2^n.Now, find a particular solution for the nonhomogeneous equation a_n = 2a_{n-1} + 1. Since the nonhomogeneous term is a constant (1), we can try a constant particular solution. Let’s assume a particular solution is a constant C.Substitute into the equation:C = 2C + 1Solve for C:C - 2C = 1 => -C = 1 => C = -1Therefore, the general solution is a_n = homogeneous + particular = A*2^n + (-1). So:a_n = A*2^n - 1Now we need to determine the constant A using an initial condition. Let's find the initial term a_1.We can use the given formula S_n = 2a_n - n. For n = 1, S_1 = a_1 = 2a_1 - 1So:a_1 = 2a_1 - 1Subtract 2a_1 from both sides:-a_1 = -1 => a_1 = 1Now plug n = 1 into the general solution:1 = A*2^1 - 1 => 1 = 2A - 1 => 2A = 2 => A = 1Therefore, the general formula for a_n is:a_n = 2^n - 1Wait, let me check that. If a_n = 2^n - 1, then let's verify for n=1, a_1 = 2^1 -1 = 1, which matches. Let's check the recurrence relation.a_n = 2a_{n-1} + 1Compute 2a_{n-1} +1 = 2*(2^{n-1} -1) +1 = 2^n - 2 +1 = 2^n -1 = a_n. Perfect, it satisfies the recurrence. Also, let's check S_n.Given S_n = 2a_n -n, let's compute S_n using the formula for a_n.S_n = 2*(2^n -1) -n = 2^{n+1} -2 -nBut also, S_n is the sum of the first n terms of a_n. Let's compute the sum of a_n = 2^k -1 from k=1 to n.Sum_{k=1}^n (2^k -1) = Sum_{k=1}^n 2^k - Sum_{k=1}^n 1 = (2^{n+1} -2) -nWhich matches 2^{n+1} -2 -n, so that's correct. Therefore, the general formula is a_n = 2^n -1.So part 1 is done. The nth term is 2^n -1.Now part 2: Let b_n = 2^n / (a_n a_{n+1}), and T_n is the sum of the first n terms of b_n. We need to find the smallest positive integer m such that T_n < m/20 for all n ∈ ℕ*.First, let's write down b_n. Since a_n = 2^n -1, so a_{n+1} = 2^{n+1} -1. Therefore:b_n = 2^n / [(2^n -1)(2^{n+1} -1)]We need to compute T_n = Sum_{k=1}^n b_k. Then find the smallest m such that T_n < m/20 for all n.To compute T_n, perhaps we can find a telescoping sum. Let's see if b_n can be expressed as the difference of two terms, so that when we sum them up, most terms cancel out.Looking at the expression for b_n:b_n = 2^n / [(2^n -1)(2^{n+1} -1)]Let’s denote denominator as (2^n -1)(2^{n+1} -1). Let's see if we can write this as a telescoping series. Maybe using partial fractions.Let’s suppose that:2^n / [(2^n -1)(2^{n+1} -1)] = C/(2^n -1) - D/(2^{n+1} -1)Find constants C and D such that:2^n = C(2^{n+1} -1) - D(2^n -1)Let’s expand the right-hand side:C*2^{n+1} - C - D*2^n + D= (2C*2^n - D*2^n) + (-C + D)Factor out 2^n:= 2^n*(2C - D) + (-C + D)This should equal the left-hand side, which is 2^n. Therefore, we can equate coefficients:For 2^n term: 2C - D = 1Constant term: -C + D = 0Solve the system:From the constant term: -C + D = 0 => D = CSubstitute into the first equation: 2C - C = 1 => C = 1Therefore, D = 1So we have:2^n / [(2^n -1)(2^{n+1} -1)] = 1/(2^n -1) - 1/(2^{n+1} -1)Therefore, b_n = 1/(2^n -1) - 1/(2^{n+1} -1)Hence, the sum T_n = Sum_{k=1}^n [1/(2^k -1) - 1/(2^{k+1} -1)].This is a telescoping series. Let's write out the terms:For k=1: 1/(2^1 -1) -1/(2^2 -1) = 1/1 -1/3For k=2: 1/(2^2 -1) -1/(2^3 -1) = 1/3 -1/7For k=3: 1/7 -1/15...For k=n: 1/(2^n -1) -1/(2^{n+1} -1)When we sum all these up, the intermediate terms cancel out:T_n = [1 -1/3] + [1/3 -1/7] + [1/7 -1/15] + ... + [1/(2^n -1) -1/(2^{n+1} -1)]All the inner terms cancel, so T_n = 1 - 1/(2^{n+1} -1)Therefore, T_n = 1 - 1/(2^{n+1} -1)So we have T_n = 1 - 1/(2^{n+1} -1). Now, we need to find the smallest positive integer m such that T_n < m/20 for all n ∈ ℕ^*.First, note that T_n is an increasing sequence since each term b_n is positive, so adding more terms increases T_n. However, as n approaches infinity, T_n approaches 1 - 0 = 1. So the supremum of T_n is 1. Therefore, 1 is the limit as n approaches infinity.But the problem says "the smallest positive integer m such that T_n < m/20 holds true for all n ∈ ℕ^*". So we need m/20 > T_n for all n. Since the supremum of T_n is 1, m/20 must be greater than 1. Hence, m > 20. But we need the smallest such m. However, we need to confirm if T_n actually reaches up to some value less than 1, but approaching 1. Wait, but since as n approaches infinity, T_n approaches 1. So for any epsilon >0, there exists an n such that T_n >1 - epsilon. Therefore, to have T_n < m/20 for all n, we must have m/20 ≥1, so m ≥20. But m has to be an integer, so m=20. However, since T_n approaches 1, but never actually reaches 1. So T_n is always less than 1. Therefore, if m/20 ≥1, then m=20 would give m/20=1, but since T_n <1 for all n, then 1 would be an upper bound. But the problem says "the smallest positive integer m such that T_n < m/20 holds true for all n ∈ ℕ^*". Wait, but if m=20, then m/20=1, and T_n <1 for all n, so indeed T_n <1 < 20/20=1? Wait, no. Wait, T_n is approaching 1 but is always less than 1, so T_n <1=20/20. But if m=20, then the inequality is T_n <1. But T_n approaches 1. So for m=20, the inequality would be T_n <1, but since T_n approaches 1, there is no n where T_n equals 1, but for any n, T_n is strictly less than 1. Therefore, T_n <1 holds for all n. Therefore, m=20. However, the problem says "the smallest positive integer m such that T_n < m/20 holds true for all n". So m=20 is the minimal m since if we take m=19, then 19/20=0.95, but as n increases, T_n approaches 1, which is greater than 0.95. Hence, for m=19, there exists some n (sufficiently large) where T_n >19/20. Therefore, m=20 is the smallest integer such that T_n <20/20=1 for all n. Therefore, the answer is m=20.Wait, but let's check for n=1,2,... what is T_n?For n=1:T_1 = b_1 =2^1 / (a_1 a_2) = 2 / (1 * 3) =2/3≈0.6667 <1For n=2:T_2 = 2/3 + 4/(3*7)= 2/3 +4/21=14/21 +4/21=18/21≈0.8571 <1n=3:T_3=18/21 +8/(7*15)=18/21 +8/105=90/105 +8/105=98/105≈0.9333 <1n=4:T_4≈98/105 +16/(15*31)=98/105 +16/465≈0.9333 +0.0344≈0.9677 <1n=5:T_5≈0.9677 +32/(31*63)=≈0.9677 +32/1953≈0.9677+0.0164≈0.9841 <1Continuing this, as n increases, T_n approaches 1. So T_n is always less than 1, but for any m less than 20, say m=19, then 19/20=0.95. But when n=4, T_4≈0.9677>0.95. So m=19 would not satisfy T_n <0.95 for all n. Similarly, m=20 gives 1, which is an upper bound. Therefore, the minimal m is 20.Wait, but the question says "the smallest positive integer m such that T_n < m/20 holds true for all n ∈ ℕ^*". Since T_n approaches 1, m/20 must be greater than 1. But 1=20/20, so m=20 is the minimal integer such that m/20=1, and since T_n <1 for all n, then m=20 is the answer.But wait, the problem says "the smallest positive integer m such that T_n < m/20 holds true for all n ∈ ℕ^*". So since T_n approaches 1, for any m such that m/20 >1, i.e., m>20, it would hold, but the minimal m is 20. However, if we check T_n <1, which is exactly m=20. But the inequality is strict: T_n < m/20. Since T_n approaches 1 from below, for m=20, T_n <1 for all n. Therefore, m=20 is the minimal such integer.Therefore, the answer is 20.Wait, but let me check once more. Suppose m=20. Then T_n <20/20=1. Since T_n approaches 1, but never reaches or exceeds 1, so T_n is always less than 1. Therefore, m=20 works. If we tried m=19, then m/20=0.95. But as we saw, T_4≈0.9677>0.95. So m=19 doesn't work. Therefore, m=20 is indeed the minimal.Therefore, the answer to part 2 is m=20.Final Answer1. The general formula for the (n^{th}) term is (boxed{2^n - 1}).2. The smallest positive integer (m) is (boxed{20}).</think>Given that the sum of the first (n) terms, (S_n), of the sequence ({a_n}) is represented by (S_n = 2a_n - n):1. To find the general formula for the (n^{th}) term of the sequence ({a_n}): - We start with the given relation (S_n = 2a_n - n) and the recursive formula for the sum (S_n = S_{n-1} + a_n). - Substituting (S_{n-1} = 2a_{n-1} - (n - 1)) into the recursive formula, we get: [ 2a_n - n = 2a_{n-1} - (n - 1) + a_n ] - Simplifying this, we find the recurrence relation (a_n = 2a_{n-1} + 1). - Solving this recurrence relation, we find the homogeneous solution (A cdot 2^n) and a particular solution (-1), leading to the general solution (a_n = 2^n - 1). - Verifying with the initial condition (a_1 = 1), we confirm the solution is correct.2. For (b_n = frac{2^n}{a_n a_{n+1}}) and the sum (T_n): - Substituting (a_n = 2^n - 1) and (a_{n+1} = 2^{n+1} - 1), we express (b_n) as: [ b_n = frac{2^n}{(2^n - 1)(2^{n+1} - 1)} ] - Using partial fractions, we decompose (b_n) into a telescoping series: [ b_n = frac{1}{2^n - 1} - frac{1}{2^{n+1} - 1} ] - Summing the first (n) terms, we find: [ T_n = 1 - frac{1}{2^{n+1} - 1} ] - Since (T_n) approaches 1 as (n) approaches infinity, the smallest integer (m) such that (T_n < frac{m}{20}) for all (n) is 20, because (T_n < 1 = frac{20}{20}). Final Answer1. The general formula for the (n^{th}) term is (boxed{2^n - 1}).2. The smallest positive integer (m) is (boxed{20}).,<think>Okay, so I need to solve this problem about the sequence {a_n} where the sum of the first n terms, S_n, is given by S_n = 2a_n - n. Then there's a second part involving another sequence {b_n} and finding the smallest m such that T_n < m/20 for all n. Let's start with the first part.First, the problem says that S_n = 2a_n - n. Since S_n is the sum of the first n terms, that means S_n = a_1 + a_2 + ... + a_n. So the formula given relates the sum of the first n terms to the nth term itself. Interesting. We need to find a general formula for a_n.I remember that for sequences, the nth term can be found by subtracting S_{n-1} from S_n. Specifically, a_n = S_n - S_{n-1} for n >= 2, and a_1 = S_1. Let's apply that here.First, let's find a_1. Since S_1 = a_1, plugging n=1 into the given formula: S_1 = 2a_1 - 1. But S_1 is also a_1, so:a_1 = 2a_1 - 1Solving for a_1:a_1 - 2a_1 = -1 => -a_1 = -1 => a_1 = 1.Okay, so the first term is 1. Now let's find a_n for n >= 2. Using the formula a_n = S_n - S_{n-1}.Given that S_n = 2a_n - n, and S_{n-1} = 2a_{n-1} - (n - 1). Therefore,a_n = S_n - S_{n-1} = [2a_n - n] - [2a_{n-1} - (n - 1)]Simplify:a_n = 2a_n - n - 2a_{n-1} + n - 1Wait, let's do the subtraction step by step:First term: 2a_n - nSecond term: - [2a_{n-1} - (n - 1)] = -2a_{n-1} + n - 1So adding those together:a_n = (2a_n - n) - 2a_{n-1} + n - 1Simplify the terms:The -n and +n cancel out, so:a_n = 2a_n - 2a_{n-1} - 1Now, let's bring all the a_n terms to one side:a_n - 2a_n = -2a_{n-1} - 1Which simplifies to:- a_n = -2a_{n-1} - 1Multiply both sides by -1:a_n = 2a_{n-1} + 1So we have a recurrence relation here: a_n = 2a_{n-1} + 1 for n >= 2, with a_1 = 1.Hmm, this is a linear nonhomogeneous recurrence relation. The standard method to solve this is to find the homogeneous solution and a particular solution.The homogeneous recurrence is a_n - 2a_{n-1} = 0. The characteristic equation is r - 2 = 0, so r = 2. Thus, the homogeneous solution is C * 2^n.For the particular solution, since the nonhomogeneous term is a constant (1), we can try a constant solution. Let's assume a particular solution is a constant K. Substitute into the recurrence:K = 2K + 1Solving for K:K - 2K = 1 => -K = 1 => K = -1.Therefore, the general solution is:a_n = homogeneous + particular = C * 2^n + (-1)Now apply the initial condition a_1 = 1:1 = C * 2^1 - 1 => 1 = 2C - 1 => 2C = 2 => C = 1.Therefore, the general formula is:a_n = 2^n - 1.Wait, let me check if this works for a_1: 2^1 -1 = 2 -1 =1. Correct. Then check the recurrence:a_n = 2a_{n-1} +1If a_{n-1} = 2^{n-1} -1, then 2a_{n-1} +1 = 2*(2^{n-1} -1) +1 = 2^n -2 +1 = 2^n -1 = a_n. Yes, that's correct. So the general formula for a_n is 2^n -1.Wait, but let me cross-validate this with the original S_n formula. Let's compute S_n using the formula S_n = 2a_n -n. If a_n = 2^n -1, then S_n = 2*(2^n -1) -n = 2^{n+1} -2 -n. On the other hand, the sum S_n = a_1 + a_2 + ... +a_n = sum_{k=1}^n (2^k -1) = sum_{k=1}^n 2^k - sum_{k=1}^n 1 = (2^{n+1} - 2) - n. Which is the same as 2^{n+1} -2 -n. So yes, that's consistent. Therefore, part 1's answer is a_n = 2^n -1.Okay, part 1 seems done. Now part 2: Let b_n = 2^n / (a_n a_{n+1}), and T_n is the sum of the first n terms of {b_n}. Find the smallest positive integer m such that T_n < m/20 for all n in N*.So first, let's write down b_n. Since a_n = 2^n -1, then a_{n+1} = 2^{n+1} -1. Therefore:b_n = 2^n / [ (2^n -1)(2^{n+1} -1) ]We need to find T_n = sum_{k=1}^n b_k, and find the smallest m such that T_n < m/20 for all n. So we need to find the supremum (least upper bound) of T_n over all n, multiply it by 20, and take the smallest integer m that is greater than or equal to 20*sup T_n.But first, let's try to compute T_n. Maybe the sum telescopes.Looking at the expression for b_n, it might be possible to express it as a telescoping series. Let's try to decompose it into partial fractions.Let’s suppose that b_n can be written as C_n - C_{n+1}, such that when we sum from 1 to n, most terms cancel out.So, let's assume:2^n / [ (2^n -1)(2^{n+1} -1) ) ] = C_n - C_{n+1}We need to find C_n such that:C_n - C_{n+1} = 2^n / [ (2^n -1)(2^{n+1} -1) ]Let’s see if we can express this as (A / (2^n -1)) + (B / (2^{n+1} -1)). Wait, partial fractions.Let’s set:2^n / [ (2^n -1)(2^{n+1} -1) ] = A / (2^n -1) + B / (2^{n+1} -1)Multiply both sides by (2^n -1)(2^{n+1} -1):2^n = A(2^{n+1} -1) + B(2^n -1)Let’s simplify the right-hand side:A*2^{n+1} - A + B*2^n - B = (2A + B)2^n - (A + B)This must equal 2^n. Therefore, equate coefficients:For 2^n term: 2A + B = 1Constant term: - (A + B) = 0So we have the system:1) 2A + B = 12) A + B = 0From equation 2: B = -A. Substitute into equation 1:2A - A = 1 => A = 1. Then B = -1.Therefore,2^n / [ (2^n -1)(2^{n+1} -1) ] = 1/(2^n -1) - 1/(2^{n+1} -1)So indeed, b_n = [1/(2^n -1)] - [1/(2^{n+1} -1)]Therefore, the sum T_n = sum_{k=1}^n [1/(2^k -1) - 1/(2^{k+1} -1)].This is a telescoping series. Let's write out the terms:When k=1: 1/(2^1 -1) - 1/(2^2 -1) = 1/1 - 1/3k=2: 1/3 - 1/7k=3: 1/7 - 1/15...k=n: 1/(2^n -1) - 1/(2^{n+1} -1)So adding all these up, most terms cancel:T_n = [1 - 1/3] + [1/3 -1/7] + [1/7 -1/15] + ... + [1/(2^n -1) -1/(2^{n+1} -1)]All intermediate terms cancel, leaving:T_n = 1 - 1/(2^{n+1} -1)Therefore, T_n = 1 - 1/(2^{n+1} -1)Now, we need to find the smallest positive integer m such that T_n < m/20 for all n in N^*.First, let's analyze the expression T_n = 1 - 1/(2^{n+1} -1). Since 1/(2^{n+1} -1) is positive and decreases as n increases, T_n is increasing and bounded above by 1. So the limit as n approaches infinity is 1. Therefore, the supremum of T_n is 1. Thus, we need 1 < m/20 => m > 20. Since m must be a positive integer, the smallest m is 21. But wait, hold on. The problem says "the smallest positive integer m such that T_n < m/20 holds true for all n ∈ N^*". However, T_n approaches 1 as n approaches infinity. Therefore, 1 < m/20 => m > 20. Since m must be an integer, the smallest m is 21. However, we must check if T_n is strictly less than 21/20 for all n. But 21/20 is 1.05, and T_n approaches 1, which is less than 1.05. Also, for all n, T_n = 1 - 1/(2^{n+1} -1) < 1 < 21/20. Wait, hold on. If T_n approaches 1, and 1 is less than 21/20, which is 1.05, then actually, T_n is always less than 1.05, but even m=20 would give m/20=1. But T_n approaches 1. So if m=20, then m/20=1. But T_n is strictly less than 1 for all n? Wait, let's check.Wait, when n approaches infinity, T_n approaches 1. So for any finite n, T_n is less than 1, but as n grows without bound, T_n approaches 1. Therefore, the supremum is 1, but it's never actually reached. So if we need T_n < m/20 for all n, then m/20 must be greater than 1. Since 1 is the supremum. Therefore, m must be greater than 20, so the smallest integer m is 21. Because 21/20 = 1.05, which is greater than 1. But wait, even m=21 gives 1.05, but since T_n approaches 1, which is less than 1.05. So m=21 would work, but perhaps even a smaller m would also work? Wait, but m must be a positive integer. Let's check the values.Wait, m has to be such that m/20 is greater than T_n for all n. Since T_n approaches 1, we need m/20 > 1. Therefore, m > 20. The smallest integer greater than 20 is 21. So m=21. However, maybe even m=20 is sufficient? Because T_n is strictly less than 1 for all n. Let's see. For m=20, m/20=1. But T_n approaches 1. So even though T_n never reaches 1, for m=20, 1 would be the limit. But the inequality is strict: T_n < 1. So for m=20, we have T_n < 1. But since T_n approaches 1, for any epsilon>0, there exists n such that T_n > 1 - epsilon. So for m=20, even though T_n never reaches 1, but since 1 - epsilon can be as close to 1 as we want, but the problem states that T_n < m/20 for all n. If m=20, then m/20=1, and T_n is always less than 1. Therefore, m=20 would satisfy T_n <1 for all n. Wait, but hold on. Wait, when n approaches infinity, T_n approaches 1, but never reaches it. So T_n is strictly less than 1 for all finite n. Therefore, if m=20, then T_n <1 holds for all n. So why is the answer 21? Wait, maybe there is a mistake here.Wait, let's check T_n for some specific n. Let's compute T_n for n=1,2,3,... and see.Given T_n = 1 - 1/(2^{n+1}-1). So:For n=1: T_1 = 1 - 1/(2^2 -1) = 1 -1/3 ≈ 0.6667For n=2: T_2 = 1 -1/(2^3 -1)=1 -1/7≈0.8571n=3: 1 -1/15≈0.9333n=4:1 -1/31≈0.9677n=5:1 -1/63≈0.9841n=6:1 -1/127≈0.9921n=7:1 -1/255≈0.9961...As n increases, T_n approaches 1, but each term is still less than 1. Therefore, T_n <1 for all n. Therefore, m=20 gives m/20=1, so T_n <1 <20/20=1 is not correct. Wait, m=20 gives m/20=1. But T_n <1. Therefore, T_n <1=20/20. So actually, m=20 would satisfy T_n < m/20 for all n. But wait, T_n approaches 1, so for any m=20, m/20=1, but T_n is strictly less than 1. So technically, m=20 is sufficient. But maybe the problem wants strict inequality, so even though T_n approaches 1, it never reaches 1, so T_n <1 holds for all n. Therefore, the smallest m is 20. But why would the problem say "smallest positive integer m such that T_n < m/20 holds true for all n"? If m=20 works, because 20/20=1 and T_n <1 for all n. But maybe there is a misunderstanding here.Wait, but m/20 must be greater than T_n for all n. If m=20, then m/20=1, which is greater than T_n for all n. So m=20 is sufficient, and it's the smallest such m. Because if m=19, then 19/20=0.95, but T_n approaches 1, which is 0.95 <1, but T_n approaches 1, which would be greater than 0.95. Wait, no. Wait, T_n approaches 1, so for m=19, 19/20=0.95. But as n increases, T_n approaches 1, which would eventually exceed 0.95. Therefore, there exists some n where T_n >0.95. Therefore, m=19 is insufficient. Similarly, m=20 gives 1, and since T_n <1 for all n, m=20 works. Therefore, the smallest m is 20.But wait, in the problem statement, it says "the smallest positive integer m such that T_n < m/20 holds true for all n ∈ N^*". If m=20, then T_n <1 for all n, which is true, as T_n approaches 1 from below. So m=20 is the minimal m. However, the answer might be 20, but the problem says "smallest positive integer m", so 20 is possible. But wait, wait. Wait, when n approaches infinity, T_n approaches 1. Therefore, for any m=20, T_n approaches 1, so the inequality is T_n <1. Since 1 is not less than 1, but strictly less. But in the limit, as n approaches infinity, T_n approaches 1, but never reaches it. So for all finite n, T_n is less than 1, hence T_n <1 holds for all n. Therefore, m=20 is acceptable.But maybe there's a catch here. Wait, let's check with n=1, T_1=1 -1/3=2/3≈0.6667, which is less than 1. For n=2, T_2=1-1/7≈0.8571, still less than 1. n=3, ≈0.9333, n=4≈0.9677, n=5≈0.9841, etc. All of these are less than 1. So T_n is always less than 1, so m=20 would work. So why would the answer be 21? Maybe I'm missing something here.Wait, maybe the problem is written in Chinese, and "m/20" is the upper bound, so T_n < m/20. Wait, if T_n approaches 1, then m/20 must be greater than 1. But 1 is the limit. So m must be greater than 20. Therefore, m=21. Because m=20 gives m/20=1, but T_n approaches 1, which is not less than 1. But T_n is always less than 1, so even though it approaches 1, it never actually reaches or exceeds 1. Therefore, m=20 would suffice as 1 is the upper bound, and since T_n is strictly less than 1, m=20. However, this might be a point of contention. In real analysis, the supremum is 1, but since it's not attained, the inequality T_n <1 holds for all n. Therefore, m=20.But in some contexts, when you have a limit approaching a value, even if it doesn't reach it, sometimes people require the bound to be strictly greater than the supremum. However, mathematically, since T_n is always less than 1, then m=20 is the minimal integer such that m/20=1, which is an upper bound. But maybe the problem expects m=21 because 1 is not less than 1, but strictly less. Wait, but the problem states T_n < m/20 for all n. Since T_n <1 for all n, then m=20 is acceptable because 1 is not less than 1? Wait, no. Wait, m/20 has to be greater than T_n for all n. If m=20, then m/20=1. Then the inequality is T_n <1. But T_n approaches 1. So for any epsilon>0, there exists an n such that T_n >1 - epsilon. But does there exist an n for which T_n >=1? No. Because T_n =1 -1/(2^{n+1}-1), which is always less than 1. Therefore, T_n <1 for all n. Therefore, m=20 satisfies the inequality T_n <1=20/20 for all n. Therefore, the minimal m is 20.But I need to verify this. Let me take m=20. Then the required inequality is T_n <20/20=1. But T_n =1 -1/(2^{n+1}-1). Since 1/(2^{n+1}-1) is positive, T_n=1 - positive, so T_n <1. Therefore, for all n, T_n <1. Hence, m=20. However, the problem says "the smallest positive integer m such that T_n < m/20 holds true for all n". Therefore, if m=20 works, why would the answer be 21? Maybe the problem is in the translation, or maybe there is a miscalculation.Wait, perhaps there is a miscalculation in the expression for T_n. Let's rederive T_n.Given that b_n = [1/(2^n -1) -1/(2^{n+1}-1)]. Therefore, T_n = sum_{k=1}^n [1/(2^k -1) -1/(2^{k+1}-1)] = 1/(2^1 -1) -1/(2^{n+1}-1) =1 -1/(2^{n+1}-1). Therefore, yes, T_n=1 -1/(2^{n+1}-1). Therefore, T_n is always less than 1. Therefore, T_n <1. Hence, m=20 gives T_n <1, which is equivalent to T_n <20/20. Therefore, m=20 is the minimal m. Therefore, the answer is 20. But maybe there's a mistake here. Let me check the original problem again.Wait, the problem says "Find the smallest positive integer m such that T_n < m/20 holds true for all n ∈ N^*". If T_n <1 for all n, then m=20 is the smallest integer since 20/20=1, and m=19 would give 19/20=0.95, but T_n approaches 1, so for large n, T_n approaches 1, which is greater than 0.95. Therefore, m=19 is insufficient. Therefore, m=20 is indeed the smallest integer such that T_n < m/20 for all n. Hence, the answer is m=20.But in my initial thought, I thought m=21 because of the limit approaching 1, but since T_n never reaches 1, m=20 is sufficient. So the correct answer should be 20.Wait, but let's check n=1: T_1=2/3≈0.6667 <1n=2:≈0.8571 <1n=3:≈0.9333 <1n=4:≈0.9677 <1n=5:≈0.9841 <1As n increases, it approaches 1 but never reaches or exceeds it. Therefore, 1 is an upper bound, and T_n <1 for all n. Hence, m=20. Therefore, the answer is m=20. So the final answer for part 2 is 20.But maybe I made a mistake in telescoping the series? Let me check the steps again.We had:b_n =2^n / [(2^n -1)(2^{n+1} -1)].We assumed partial fractions decomposition:2^n / [(2^n -1)(2^{n+1} -1)] = A/(2^n -1) + B/(2^{n+1} -1)Then multiplying out:2^n = A(2^{n+1} -1) + B(2^n -1)Expanding:2^n = A*2^{n+1} -A + B*2^n - BGrouping terms:2^n = (2A + B)2^n - (A + B)Equating coefficients:2A + B =1 (for 2^n term)-(A + B)=0 (constant term)Solving gives A=1, B=-1. Therefore,b_n=1/(2^n -1) -1/(2^{n+1}-1). Therefore, sum T_n=1 -1/(2^{n+1}-1). So that's correct.Therefore, conclusion is T_n <1 for all n. Therefore, m=20.So why did I initially think 21? Probably because of confusion between closed and open intervals, but since the bound is strict inequality, m=20 is sufficient.Therefore, part 1: a_n=2^n -1, part 2: m=20.But to make sure, maybe the problem had a typo or I misinterpret it. The problem says "the smallest positive integer m such that T_n < m/20 holds true for all n ∈ N^*". If T_n approaches 1, then for the inequality to hold for all n, m must satisfy m/20 > sup{T_n}=1. Therefore, m>20. Since m must be an integer, m=21. Wait, but this contradicts with previous logic.Wait, this is the critical point. Is the supremum of T_n equal to 1? Yes, because as n approaches infinity, T_n approaches 1. However, T_n never actually reaches 1. Therefore, the supremum is 1, but it's not attained. Therefore, if the problem requires T_n < m/20 for all n, and since T_n can get arbitrarily close to 1, then m/20 must be at least 1. However, since T_n is always less than 1, m=20 works because 20/20=1. However, in real numbers, even though T_n approaches 1, but is always less than 1, but in terms of the inequality, T_n <1 for all n. Therefore, m=20 is acceptable.However, in some mathematical contexts, when dealing with limits, if you have a strict inequality that approaches a bound, you might need the bound to be strictly greater than the limit. But in this problem, it's specified that T_n < m/20 for all n. Since T_n is always less than 1, m=20 satisfies this. So unless there's a specific requirement that m/20 must be greater than the supremum, which is 1, then m must be greater than 20, hence 21. However, depending on the interpretation, the answer could be 20 or 21.To resolve this ambiguity, let's consider the difference between the supremum and the maximum. The supremum of T_n is 1, but it's not attained; there is no maximum. Therefore, T_n can get as close to 1 as desired but never reaches it. Therefore, for all n, T_n <1. Therefore, m=20 satisfies T_n <1=20/20. Hence, m=20 is acceptable.But if the problem had said "less than or equal to", then m=20 would not work because T_n never reaches 1. But since it's strict inequality, m=20 works. Therefore, the answer is m=20.However, to be thorough, let's check the problem statement again. It says "Find the smallest positive integer m such that T_n < m/20 holds true for all n ∈ N^*". Since T_n is strictly less than 1 for all n, and m is a positive integer, then m=20 gives m/20=1, and since T_n <1, the inequality holds. If we take m=19, then m/20=0.95, and when n is large enough, T_n approaches 1, so there exists an n such that T_n >0.95. For example, T_n=1 -1/(2^{n+1}-1). Let's solve for T_n >0.95:1 -1/(2^{n+1}-1) >0.95=> 1/(2^{n+1}-1) <0.05=> 2^{n+1}-1 >20=>2^{n+1} >21=>n+1 >log2(21)≈4.39=>n>3.39, so n>=4. Therefore, when n=4, T_4=1 -1/31≈0.9677>0.95. Therefore, m=19 is insufficient.Therefore, m=20 is the smallest integer where T_n=1 -1/(2^{n+1}-1) <1=20/20 for all n. Hence, the answer is 20.Therefore, part 1: a_n=2^n -1, part 2: m=20.Final Answer1. The general formula for the (n^{th}) term is (boxed{2^n - 1}). 2. The smallest positive integer (m) is (boxed{20}).</think>Given that the sum of the first (n) terms, (S_n), of the sequence ({a_n}) is represented by (S_n = 2a_n - n):1. To find the general formula for the (n^{th}) term of the sequence ({a_n}): - For (n = 1), we have (S_1 = a_1 = 2a_1 - 1), which gives (a_1 = 1). - For (n geq 2), using the recurrence relation derived from (a_n = S_n - S_{n-1}): [ a_n = 2a_{n-1} + 1 ] - Solving this linear nonhomogeneous recurrence relation with the initial condition (a_1 = 1), we find the general solution: [ a_n = 2^n - 1 ]2. To find the smallest positive integer (m) such that (T_n < frac{m}{20}) for all (n in mathbb{N}^*): - Given (b_n = frac{2^n}{a_n a_{n+1}}) where (a_n = 2^n - 1) and (a_{n+1} = 2^{n+1} - 1), we decompose (b_n) using partial fractions: [ b_n = frac{2^n}{(2^n - 1)(2^{n+1} - 1)} = frac{1}{2^n - 1} - frac{1}{2^{n+1} - 1} ] - The sum (T_n) of the first (n) terms of ({b_n}) is a telescoping series: [ T_n = sum_{k=1}^n left( frac{1}{2^k - 1} - frac{1}{2^{k+1} - 1} right) = 1 - frac{1}{2^{n+1} - 1} ] - Since (T_n) approaches 1 as (n) approaches infinity but is always less than 1 for all finite (n), the smallest positive integer (m) such that (T_n < frac{m}{20}) for all (n) is (20). Final Answer1. The general formula for the (n^{th}) term is (boxed{2^n - 1}). 2. The smallest positive integer (m) is (boxed{20}).]

🔑:(I) When n=1, S_1 = 2a_1 - 1, thus a_1 = 1.For n geq 2, S_n = 2a_n - n = 2a_{n-1} - (n-1).Subtracting these expressions yields a_n = 2a_{n-1} + 1.So, a_n + 1 = 2(a_{n-1} + 1).Hence, the sequence {a_n + 1} is a geometric sequence with the first term equal to 2 and the common ratio equal to 2.Thus, a_n + 1 = 2^n, and a_n = 2^n - 1.(II) From (I), we know a_n = 2^n - 1. Thus,b_n = frac{2^n}{a_n a_{n+1}} = frac{2^n}{(2^n - 1)(2^{n+1} - 1)} = frac{1}{2^n - 1} - frac{1}{2^{n+1} - 1}.Consequentially,begin{align}T_n &= left(1 - frac{1}{2^2 - 1}right) + left(frac{1}{2^2 - 1} - frac{1}{2^3 - 1}right) + dots + left(frac{1}{2^n - 1} - frac{1}{2^{n+1} - 1}right) &= 1 - frac{1}{2^{n+1} - 1}.end{align}As the sequence {T_n} is monotonically increasing, and lim_{n to +infty} T_n = 1, we have frac{m}{20} geq 1, implying m geq 20.Therefore, the smallest possible value for m is 20, i.e., boxed{m = 20}.