answer：First, identify the number of people who visited the zoo on Friday, which is 3,575.Next, determine how many times more people visited on Saturday compared to Friday. In this case, it's five times as many.To find the number of people who visited on Saturday, multiply the number of Friday visitors by 5.Perform the multiplication: 3,575 multiplied by 5 equals 17,875.Therefore, 17,875 people visited the zoo on Saturday.

answer：Okay, so I have this problem about a line given by the equation ( l: (m+2)x - y + m - 2 = 0 ). There are four statements, A to D, and I need to figure out which ones are correct. Let me take it step by step.Starting with **Option A**: It says that if ( m = -3 ), then the slope of line ( l ) is ( 135^circ ). Hmm, I remember that the slope of a line is related to the angle it makes with the positive direction of the x-axis. Specifically, the slope ( m ) is equal to ( tan(theta) ), where ( theta ) is that angle. So, if the slope is ( 135^circ ), then ( tan(135^circ) = -1 ). That means the slope should be -1.Let me plug ( m = -3 ) into the equation of the line:[(-3 + 2)x - y + (-3) - 2 = 0]Simplifying that:[(-1)x - y - 5 = 0 implies -x - y - 5 = 0 implies x + y + 5 = 0]So, the equation becomes ( x + y + 5 = 0 ). To find the slope, I can rewrite this in slope-intercept form ( y = mx + b ):[y = -x - 5]So, the slope ( m ) is -1. Since ( tan(135^circ) = -1 ), that means the angle is indeed ( 135^circ ). So, **Option A is correct**.Moving on to **Option B**: It states that if the line ( l ) intersects the two coordinate axes at the same point, then ( m = pm 3 ). Hmm, intersecting both axes at the same point would mean that the x-intercept and y-intercept are equal. Let me recall how to find intercepts.For the x-intercept, set ( y = 0 ):[(m + 2)x + m - 2 = 0 implies (m + 2)x = -m + 2 implies x = frac{-m + 2}{m + 2}]For the y-intercept, set ( x = 0 ):[- y + m - 2 = 0 implies y = m - 2]So, the x-intercept is ( frac{-m + 2}{m + 2} ) and the y-intercept is ( m - 2 ). If they are the same point, then these two values must be equal:[frac{-m + 2}{m + 2} = m - 2]Let me solve this equation:Multiply both sides by ( m + 2 ):[-m + 2 = (m - 2)(m + 2)]Expanding the right side:[-m + 2 = m^2 - 4]Bring all terms to one side:[m^2 - 4 + m - 2 = 0 implies m^2 + m - 6 = 0]Factor the quadratic:Looking for two numbers that multiply to -6 and add to 1. Those are 3 and -2.[(m + 3)(m - 2) = 0]So, ( m = -3 ) or ( m = 2 ). Wait, the option says ( m = pm 3 ), which would be ( m = 3 ) or ( m = -3 ). But according to my solution, ( m = -3 ) or ( m = 2 ). So, ( m = 3 ) is not a solution. Therefore, **Option B is incorrect**.Next, **Option C**: It says that if line ( l ) is perpendicular to the line ( x + y = 0 ), then ( m = -1 ). I remember that two lines are perpendicular if the product of their slopes is -1.First, let me find the slope of ( x + y = 0 ). Rewriting it:[y = -x]So, the slope is -1. Therefore, the slope of a line perpendicular to this would be the negative reciprocal, which is 1.Now, let me find the slope of line ( l ). Starting with the equation:[(m + 2)x - y + m - 2 = 0]Rewriting in slope-intercept form:[-y = -(m + 2)x - m + 2 implies y = (m + 2)x + m - 2]So, the slope of line ( l ) is ( m + 2 ). For it to be perpendicular to ( x + y = 0 ), we need:[(m + 2) times (-1) = -1]Wait, actually, the condition is that the product of the slopes is -1. The slope of ( l ) is ( m + 2 ), and the slope of the other line is -1. So:[(m + 2) times (-1) = -1]Simplify:[-(m + 2) = -1 implies m + 2 = 1 implies m = -1]So, yes, ( m = -1 ). Therefore, **Option C is correct**.Finally, **Option D**: It states that there exists an ( m in mathbb{R} ) such that the distance from the origin ( (0,0) ) to line ( l ) is 5. I need to recall the formula for the distance from a point ( (x_0, y_0) ) to the line ( ax + by + c = 0 ):[text{Distance} = frac{|a x_0 + b y_0 + c|}{sqrt{a^2 + b^2}}]In our case, the line is ( (m + 2)x - y + (m - 2) = 0 ), so ( a = m + 2 ), ( b = -1 ), and ( c = m - 2 ). The point is ( (0,0) ), so plugging into the distance formula:[text{Distance} = frac{|(m + 2)(0) + (-1)(0) + (m - 2)|}{sqrt{(m + 2)^2 + (-1)^2}} = frac{|m - 2|}{sqrt{(m + 2)^2 + 1}}]We want this distance to be 5:[frac{|m - 2|}{sqrt{(m + 2)^2 + 1}} = 5]Let me square both sides to eliminate the square root:[frac{(m - 2)^2}{(m + 2)^2 + 1} = 25]Multiply both sides by the denominator:[(m - 2)^2 = 25[(m + 2)^2 + 1]]Expand both sides:Left side: ( (m - 2)^2 = m^2 - 4m + 4 )Right side: ( 25[(m + 2)^2 + 1] = 25[m^2 + 4m + 4 + 1] = 25[m^2 + 4m + 5] = 25m^2 + 100m + 125 )So, the equation becomes:[m^2 - 4m + 4 = 25m^2 + 100m + 125]Bring all terms to one side:[m^2 - 4m + 4 - 25m^2 - 100m - 125 = 0 implies -24m^2 - 104m - 121 = 0]Multiply both sides by -1 to make it positive:[24m^2 + 104m + 121 = 0]Now, let's compute the discriminant ( D ):[D = 104^2 - 4 times 24 times 121]Calculate ( 104^2 ):( 100^2 = 10000 ), ( 4^2 = 16 ), and ( 2 times 100 times 4 = 800 ), so ( (100 + 4)^2 = 10000 + 800 + 16 = 10816 ).Calculate ( 4 times 24 times 121 ):First, ( 4 times 24 = 96 ), then ( 96 times 121 ). Let's compute that:( 96 times 100 = 9600 ), ( 96 times 20 = 1920 ), ( 96 times 1 = 96 ). So, ( 9600 + 1920 + 96 = 11616 ).So, ( D = 10816 - 11616 = -800 ).Since the discriminant is negative, there are no real solutions for ( m ). Therefore, there does not exist a real number ( m ) such that the distance from the origin to line ( l ) is 5. So, **Option D is incorrect**.To recap:- **A**: Correct- **B**: Incorrect- **C**: Correct- **D**: IncorrectSo, the correct options are A and C.

answer：Alright, so I have this problem here where I need to find the sum from n equals 3 to infinity of g(n), where g(n) is defined as 1 over 3 to the n plus 1 over 4 to the n and so on. Hmm, okay, let me try to unpack this step by step.First, let me write down what g(n) is. It's a series starting from 3^n, so for each n, g(n) is the sum from k equals 3 to infinity of 1 over k to the n. So, mathematically, that's:g(n) = Σ (from k=3 to ∞) 1/k^nAnd I need to find the sum from n=3 to ∞ of g(n). So, putting it all together, the entire expression I need to compute is:Σ (from n=3 to ∞) [Σ (from k=3 to ∞) 1/k^n]Hmm, that looks like a double summation. Maybe I can switch the order of summation? I remember that sometimes swapping the order of summation can simplify things, especially if the terms are positive, which they are here since all terms are positive.So, let me try swapping the order. Instead of summing over n first and then k, I'll sum over k first and then n. That would give me:Σ (from k=3 to ∞) [Σ (from n=3 to ∞) 1/k^n]Okay, that seems manageable. Now, for each fixed k, the inner sum is a geometric series in terms of n. Let me recall that the sum from n=0 to ∞ of r^n is 1/(1 - r) for |r| < 1. But here, my series starts at n=3, not n=0. So, I need to adjust for that.The sum from n=3 to ∞ of 1/k^n is the same as the sum from n=0 to ∞ of 1/k^n minus the first three terms (n=0,1,2). So, let's write that out:Σ (from n=3 to ∞) 1/k^n = Σ (from n=0 to ∞) 1/k^n - Σ (from n=0 to 2) 1/k^nCalculating the first part, which is the full geometric series:Σ (from n=0 to ∞) 1/k^n = 1 / (1 - 1/k) = k / (k - 1)Now, subtracting the first three terms:Σ (from n=0 to 2) 1/k^n = 1 + 1/k + 1/k^2So, putting it all together:Σ (from n=3 to ∞) 1/k^n = [k / (k - 1)] - [1 + 1/k + 1/k^2]Let me simplify that expression. First, let's write k / (k - 1) as 1 + 1/(k - 1). Wait, actually, let me compute it step by step.Compute [k / (k - 1)] - [1 + 1/k + 1/k^2]:First, k / (k - 1) can be written as 1 + 1/(k - 1). So, subtracting 1 gives us 1/(k - 1). Then, subtracting the remaining terms:1/(k - 1) - 1/k - 1/k^2Hmm, that seems a bit messy. Maybe instead, let's get a common denominator for all terms. Let's see:k / (k - 1) is equal to [k^2] / [k(k - 1)]Similarly, 1 is [k^2(k - 1)] / [k^2(k - 1)], but that might complicate things. Maybe another approach.Alternatively, let's compute [k / (k - 1)] - [1 + 1/k + 1/k^2] directly.First, expand k / (k - 1):k / (k - 1) = 1 + 1/(k - 1)So, subtracting 1 gives 1/(k - 1). Then, subtract 1/k and 1/k^2:1/(k - 1) - 1/k - 1/k^2To combine these, let's find a common denominator. The denominators are (k - 1), k, and k^2. The least common denominator would be k^2(k - 1). Let's rewrite each term:1/(k - 1) = k^2 / [k^2(k - 1)]1/k = k(k - 1) / [k^2(k - 1)]1/k^2 = (k - 1) / [k^2(k - 1)]So, putting it all together:[k^2 - k(k - 1) - (k - 1)] / [k^2(k - 1)]Simplify the numerator:k^2 - [k^2 - k] - [k - 1] = k^2 - k^2 + k - k + 1 = 1So, the entire expression simplifies to 1 / [k^2(k - 1)]Therefore, Σ (from n=3 to ∞) 1/k^n = 1 / [k^2(k - 1)]Hmm, that's a nice simplification. So, going back to the double summation, we have:Σ (from k=3 to ∞) [1 / (k^2(k - 1))]So, now the problem reduces to computing this sum. Let me write that down:S = Σ (from k=3 to ∞) 1 / [k^2(k - 1)]I need to find a way to compute this sum. It might be helpful to use partial fractions to break this into simpler terms that telescope or can be summed more easily.Let me consider the term 1 / [k^2(k - 1)]. Let's try to express this as a combination of simpler fractions. Let's suppose:1 / [k^2(k - 1)] = A / (k - 1) + B / k + C / k^2Where A, B, and C are constants to be determined. Let's solve for A, B, and C.Multiplying both sides by k^2(k - 1):1 = A k^2 + B k(k - 1) + C(k - 1)Now, let's expand the right-hand side:A k^2 + B k(k - 1) + C(k - 1) = A k^2 + B k^2 - B k + C k - CCombine like terms:(A + B) k^2 + (-B + C) k - CSet this equal to the left-hand side, which is 1. So, we have:(A + B) k^2 + (-B + C) k - C = 1Since this must hold for all k, the coefficients of k^2, k, and the constant term must each be equal on both sides. Therefore, we can set up the following system of equations:1. Coefficient of k^2: A + B = 02. Coefficient of k: -B + C = 03. Constant term: -C = 1Let's solve this system step by step.From equation 3: -C = 1 ⇒ C = -1From equation 2: -B + C = 0 ⇒ -B + (-1) = 0 ⇒ -B -1 = 0 ⇒ -B = 1 ⇒ B = -1From equation 1: A + B = 0 ⇒ A + (-1) = 0 ⇒ A = 1So, we have A = 1, B = -1, C = -1.Therefore, the partial fraction decomposition is:1 / [k^2(k - 1)] = 1 / (k - 1) - 1 / k - 1 / k^2So, substituting back into the sum S:S = Σ (from k=3 to ∞) [1 / (k - 1) - 1 / k - 1 / k^2]Let me split this into three separate sums:S = Σ (from k=3 to ∞) 1 / (k - 1) - Σ (from k=3 to ∞) 1 / k - Σ (from k=3 to ∞) 1 / k^2Let me adjust the indices for the first two sums to make them easier to handle.For the first sum, let me let m = k - 1. When k=3, m=2, and as k approaches infinity, m approaches infinity. So, the first sum becomes:Σ (from m=2 to ∞) 1 / mSimilarly, the second sum is:Σ (from k=3 to ∞) 1 / kSo, S can be written as:S = [Σ (from m=2 to ∞) 1/m] - [Σ (from k=3 to ∞) 1/k] - [Σ (from k=3 to ∞) 1/k^2]Notice that the first two sums are almost the same, except the first starts at m=2 and the second starts at k=3. So, let's subtract them:[Σ (from m=2 to ∞) 1/m] - [Σ (from k=3 to ∞) 1/k] = 1/2 + [Σ (from m=3 to ∞) 1/m] - [Σ (from k=3 to ∞) 1/k] = 1/2Because the sums from m=3 to ∞ and k=3 to ∞ cancel each other out, leaving just 1/2.So, now S simplifies to:S = 1/2 - Σ (from k=3 to ∞) 1/k^2So, now I need to compute Σ (from k=3 to ∞) 1/k^2. I know that the sum from k=1 to ∞ of 1/k^2 is a well-known series, equal to π^2/6. So, let's use that.Σ (from k=1 to ∞) 1/k^2 = π^2 / 6Therefore, Σ (from k=3 to ∞) 1/k^2 = Σ (from k=1 to ∞) 1/k^2 - 1/1^2 - 1/2^2 = π^2 / 6 - 1 - 1/4Compute that:π^2 / 6 - 1 - 1/4 = π^2 / 6 - 5/4So, substituting back into S:S = 1/2 - (π^2 / 6 - 5/4) = 1/2 - π^2 / 6 + 5/4Combine the constants:1/2 + 5/4 = (2/4 + 5/4) = 7/4So, S = 7/4 - π^2 / 6Wait, hold on. That can't be right because earlier steps might have miscalculations. Let me double-check.Wait, actually, when I split the sums, I had:S = [Σ (from m=2 to ∞) 1/m] - [Σ (from k=3 to ∞) 1/k] - [Σ (from k=3 to ∞) 1/k^2]Which simplifies to:[Σ (from m=2 to ∞) 1/m - Σ (from k=3 to ∞) 1/k] - Σ (from k=3 to ∞) 1/k^2As I correctly noted, the first two sums cancel out except for the first term of the first sum, which is 1/2. So, that part is correct.Then, S = 1/2 - Σ (from k=3 to ∞) 1/k^2But Σ (from k=3 to ∞) 1/k^2 = Σ (from k=1 to ∞) 1/k^2 - 1 - 1/4 = π^2 / 6 - 5/4So, substituting back:S = 1/2 - (π^2 / 6 - 5/4) = 1/2 - π^2 / 6 + 5/4Combine constants:1/2 + 5/4 = (2/4 + 5/4) = 7/4So, S = 7/4 - π^2 / 6Wait, but that seems a bit odd because I was expecting a simpler answer, maybe a fraction without π. Did I make a mistake somewhere?Let me go back through my steps.Starting from the double summation:Σ (from n=3 to ∞) Σ (from k=3 to ∞) 1/k^n = Σ (from k=3 to ∞) Σ (from n=3 to ∞) 1/k^nWhich is correct.Then, for each k, Σ (from n=3 to ∞) 1/k^n is a geometric series with first term 1/k^3 and ratio 1/k.Sum of a geometric series is a / (1 - r), so here a = 1/k^3, r = 1/k, so sum is (1/k^3) / (1 - 1/k) = 1 / [k^2(k - 1)]That seems correct.Then, partial fraction decomposition:1 / [k^2(k - 1)] = 1/(k - 1) - 1/k - 1/k^2Which we verified by solving for A, B, C.Then, splitting the sum:Σ [1/(k - 1) - 1/k - 1/k^2] from k=3 to ∞Which becomes:Σ 1/(k - 1) from k=3 to ∞ - Σ 1/k from k=3 to ∞ - Σ 1/k^2 from k=3 to ∞Changing variable in the first sum: m = k - 1, so m starts at 2.So, Σ 1/m from m=2 to ∞ - Σ 1/k from k=3 to ∞ - Σ 1/k^2 from k=3 to ∞Which simplifies to:[Σ 1/m from m=2 to ∞ - Σ 1/k from k=3 to ∞] - Σ 1/k^2 from k=3 to ∞The first two sums cancel except for the first term of the first sum, which is 1/2.So, S = 1/2 - Σ 1/k^2 from k=3 to ∞Then, Σ 1/k^2 from k=3 to ∞ = Σ 1/k^2 from k=1 to ∞ - 1 - 1/4 = π^2 / 6 - 5/4Thus, S = 1/2 - (π^2 / 6 - 5/4) = 1/2 - π^2 / 6 + 5/4 = 7/4 - π^2 / 6Hmm, so that seems to be the result. But wait, the original problem didn't involve π^2, so maybe I made a mistake in the partial fraction decomposition or in the setup.Wait, let me double-check the partial fraction decomposition.We had:1 / [k^2(k - 1)] = A / (k - 1) + B / k + C / k^2Multiplying both sides by k^2(k - 1):1 = A k^2 + B k(k - 1) + C(k - 1)Expanding:1 = A k^2 + B k^2 - B k + C k - CCombine like terms:(A + B) k^2 + (-B + C) k + (-C) = 1So, setting coefficients equal:A + B = 0-B + C = 0-C = 1Solving:From third equation: C = -1From second equation: -B + (-1) = 0 ⇒ -B = 1 ⇒ B = -1From first equation: A + (-1) = 0 ⇒ A = 1So, the partial fractions are correct: 1/(k - 1) - 1/k - 1/k^2So, that part is correct.Then, the sum S = Σ [1/(k - 1) - 1/k - 1/k^2] from k=3 to ∞Which is:Σ 1/(k - 1) from k=3 to ∞ - Σ 1/k from k=3 to ∞ - Σ 1/k^2 from k=3 to ∞Changing variables in the first sum: m = k - 1, so m starts at 2.Thus, Σ 1/m from m=2 to ∞ - Σ 1/k from k=3 to ∞ - Σ 1/k^2 from k=3 to ∞Which is:[Σ 1/m from m=2 to ∞ - Σ 1/k from k=3 to ∞] - Σ 1/k^2 from k=3 to ∞The first two sums: Σ 1/m from m=2 to ∞ is 1/2 + 1/3 + 1/4 + ... and Σ 1/k from k=3 to ∞ is 1/3 + 1/4 + ... So, subtracting them gives 1/2.Thus, S = 1/2 - Σ 1/k^2 from k=3 to ∞Now, Σ 1/k^2 from k=3 to ∞ is indeed Σ 1/k^2 from k=1 to ∞ minus 1 - 1/4, which is π^2/6 - 5/4.So, S = 1/2 - (π^2/6 - 5/4) = 1/2 - π^2/6 + 5/4 = (1/2 + 5/4) - π^2/6 = 7/4 - π^2/6Hmm, that's approximately 1.75 - (1.6449) ≈ 0.1051, which seems plausible, but I was expecting a simpler answer. Maybe I made a mistake in the initial steps.Wait, let me go back to the original problem.We have g(n) = Σ (from k=3 to ∞) 1/k^nAnd we need to compute Σ (from n=3 to ∞) g(n) = Σ (from n=3 to ∞) Σ (from k=3 to ∞) 1/k^nWe swapped the order of summation to get Σ (from k=3 to ∞) Σ (from n=3 to ∞) 1/k^nWhich is Σ (from k=3 to ∞) [1/k^3 + 1/k^4 + 1/k^5 + ...] = Σ (from k=3 to ∞) [1/k^3 / (1 - 1/k)] = Σ (from k=3 to ∞) 1/[k^2(k - 1)]So, that seems correct.Then, partial fractions gave us 1/(k - 1) - 1/k - 1/k^2Thus, S = Σ (from k=3 to ∞) [1/(k - 1) - 1/k - 1/k^2] = 1/2 - Σ (from k=3 to ∞) 1/k^2Which is 1/2 - (π^2/6 - 1 - 1/4) = 1/2 - π^2/6 + 5/4 = 7/4 - π^2/6Wait, but the original problem didn't specify any constraints on the answer, so maybe this is the correct answer. However, I recall that in the initial problem statement, the user had a different approach and got 1/3. Let me check that.Wait, in the initial problem, the user wrote:"For a positive integer n, define g(n) = 1/3^n + 1/4^n + 1/5^n + ... Find the sum from n=3 to ∞ of g(n)."Then, the user's solution was:They considered the sum as a double summation, swapped the order, and for each k, summed over n starting from 3. They recognized it as a geometric series with first term 1/k^3 and ratio 1/k, so the sum is 1/(k^2(k - 1)).Then, they decomposed 1/(k^2(k - 1)) into partial fractions: 1/(k - 1) - 1/k - 1/k^2.Then, they summed over k from 3 to ∞, which gave them:Σ [1/(k - 1) - 1/k - 1/k^2] from k=3 to ∞They noticed that the first two terms telescope, leaving 1/2, and the last term is Σ 1/k^2 from k=3 to ∞, which they expressed as π^2/6 - 1 - 1/4.Thus, S = 1/2 - (π^2/6 - 5/4) = 7/4 - π^2/6But wait, in the initial problem, the user's solution ended up with 1/3, which contradicts this result. So, perhaps there was a mistake in their approach.Wait, let me check the initial problem again.The user wrote:"For a positive integer n, define g(n) = 1/3^n + 1/4^n + 1/5^n + ... Find the sum from n=3 to ∞ of g(n)."Then, their solution was:They considered the double summation, swapped the order, and for each k, summed over n starting from 3. They recognized it as a geometric series with first term 1/k^3 and ratio 1/k, so the sum is 1/(k^2(k - 1)).Then, they decomposed 1/(k^2(k - 1)) into partial fractions: 1/(k - 1) - 1/k - 1/k^2.Then, they summed over k from 3 to ∞, which gave them:Σ [1/(k - 1) - 1/k - 1/k^2] from k=3 to ∞They noticed that the first two terms telescope, leaving 1/2, and the last term is Σ 1/k^2 from k=3 to ∞, which they expressed as π^2/6 - 1 - 1/4.Thus, S = 1/2 - (π^2/6 - 5/4) = 7/4 - π^2/6But in their initial solution, they concluded the answer was 1/3, which seems inconsistent with this result.Wait, perhaps the user made a mistake in their initial solution. Let me check their steps.They wrote:"The sum of the numbers in the nth column (starting from n=3) is an infinite geometric series, with first term 1/(n+2)^3 and common ratio 1/(n+2). Hence, the sum of its terms is [1/(n+2)^3] / [1 - 1/(n+2)] = 1/[(n+2)^2(n+1)] = 1/[(n+1)(n+2)] = 1/(n+1) - 1/(n+2)."Wait, hold on. They seem to have misapplied the indices. Let me parse this.They said that the sum of the nth column is a geometric series with first term 1/(n+2)^3 and ratio 1/(n+2). But actually, for each fixed k, the inner sum over n is a geometric series starting at n=3, so the first term is 1/k^3, not 1/(n+2)^3.Wait, perhaps they confused the indices. Let me clarify.In the double summation, when swapping the order, for each k, we sum over n from 3 to ∞. So, the first term is 1/k^3, ratio is 1/k, so the sum is 1/k^3 / (1 - 1/k) = 1/[k^2(k - 1)].But the user wrote:"the sum of the numbers in the nth column (starting from n=3) is an infinite geometric series, with first term 1/(n+2)^3 and common ratio 1/(n+2)."Wait, that seems incorrect because for each column n, the terms are 1/3^n, 1/4^n, 1/5^n, etc., so the first term is 1/3^n, not 1/(n+2)^3.Wait, perhaps they misapplied the indices. Let me think.If we consider the double summation as rows and columns, where each row is fixed n and each column is fixed k. So, for column k=3, the terms are 1/3^3, 1/3^4, 1/3^5, etc., which is a geometric series with first term 1/3^3 and ratio 1/3.Similarly, for column k=4, it's 1/4^3, 1/4^4, etc., with first term 1/4^3 and ratio 1/4.So, for each column k, the sum is 1/k^3 + 1/k^4 + 1/k^5 + ... = 1/k^3 / (1 - 1/k) = 1/[k^2(k - 1)]So, the user's initial approach seems to have confused the indices, thinking that for column n, the first term is 1/(n+2)^3, which is not correct. Instead, for column k, the first term is 1/k^3.Therefore, their decomposition into partial fractions was incorrect because they used n instead of k. They ended up with 1/[(n+1)(n+2)] which is not the same as our 1/[k^2(k - 1)].So, their initial solution was flawed because they misapplied the indices, leading them to an incorrect telescoping series.Therefore, the correct approach is the one I followed, leading to S = 7/4 - π^2/6.But wait, the problem didn't specify whether to express the answer in terms of π or not. It just asked to find the sum. So, perhaps the answer is indeed 7/4 - π^2/6, which is approximately 0.105.But let me check if there's another way to approach this problem without involving π^2.Wait, another thought: maybe I can express the sum S = Σ (from k=3 to ∞) 1/[k^2(k - 1)] as a telescoping series without involving the Σ 1/k^2 term.Let me revisit the partial fraction decomposition:1/[k^2(k - 1)] = 1/(k - 1) - 1/k - 1/k^2So, S = Σ (from k=3 to ∞) [1/(k - 1) - 1/k - 1/k^2]Which is:Σ (from k=3 to ∞) [1/(k - 1) - 1/k] - Σ (from k=3 to ∞) 1/k^2The first sum is a telescoping series:Σ (from k=3 to ∞) [1/(k - 1) - 1/k] = lim (m→∞) [1/2 - 1/m] = 1/2Because the terms cancel out: 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - ... up to 1/(m-1) - 1/m, which leaves 1/2 - 1/m, and as m approaches infinity, 1/m approaches 0, so the sum is 1/2.Then, the second sum is Σ (from k=3 to ∞) 1/k^2, which is π^2/6 - 1 - 1/4 = π^2/6 - 5/4Thus, S = 1/2 - (π^2/6 - 5/4) = 1/2 - π^2/6 + 5/4 = 7/4 - π^2/6So, that seems to be the correct answer.But wait, the initial problem didn't involve π, so maybe I made a mistake in the partial fraction decomposition or in the setup.Alternatively, perhaps the problem expects a different approach.Wait, another idea: maybe instead of swapping the order of summation, I can express g(n) in terms of the Riemann zeta function, ζ(n), which is Σ (from k=1 to ∞) 1/k^n. Then, g(n) = ζ(n) - 1 - 1/2^nSo, the sum we need is Σ (from n=3 to ∞) [ζ(n) - 1 - 1/2^n]Which is Σ (from n=3 to ∞) ζ(n) - Σ (from n=3 to ∞) 1 - Σ (from n=3 to ∞) 1/2^nBut Σ (from n=3 to ∞) 1 is divergent, which suggests that this approach isn't valid, as the original sum converges.Wait, that can't be right because the original sum is convergent. So, perhaps expressing g(n) as ζ(n) - 1 - 1/2^n is correct, but then summing over n from 3 to ∞ would involve subtracting divergent series, which isn't helpful.Therefore, perhaps the initial approach of swapping the order of summation is the correct way, leading to S = 7/4 - π^2/6.Alternatively, maybe I can compute the sum numerically to check.Compute S = 7/4 - π^2/6 ≈ 1.75 - (1.6449) ≈ 0.1051Alternatively, compute the original double summation numerically for a few terms and see if it approaches 0.1051.Compute the first few terms:For n=3:g(3) = 1/3^3 + 1/4^3 + 1/5^3 + ... ≈ 1/27 + 1/64 + 1/125 + ... ≈ 0.037 + 0.0156 + 0.008 + ... ≈ 0.0606For n=4:g(4) = 1/3^4 + 1/4^4 + 1/5^4 + ... ≈ 1/81 + 1/256 + 1/625 + ... ≈ 0.0123 + 0.0039 + 0.0016 + ... ≈ 0.0178For n=5:g(5) ≈ 1/243 + 1/1024 + 1/3125 + ... ≈ 0.0041 + 0.00097 + 0.00032 + ... ≈ 0.0054Adding these up: 0.0606 + 0.0178 + 0.0054 ≈ 0.0838Adding n=6: g(6) ≈ 1/729 + 1/4096 + ... ≈ 0.00137 + 0.000244 ≈ 0.00161Total so far: ≈ 0.0838 + 0.00161 ≈ 0.0854n=7: g(7) ≈ 1/2187 + ... ≈ 0.000457Total: ≈ 0.0854 + 0.000457 ≈ 0.08585n=8: ≈ 0.000152Total: ≈ 0.08585 + 0.000152 ≈ 0.086And so on. So, the partial sums are approaching around 0.086, which is less than our earlier calculation of approximately 0.1051. Hmm, that suggests that perhaps my analytical result is incorrect.Wait, but the partial sums are only up to n=8, and the terms decrease rapidly, so maybe the total converges to around 0.1.Alternatively, perhaps my numerical estimation is too rough.Wait, let me compute more accurately.Compute g(3):1/27 ≈ 0.0370371/64 ≈ 0.0156251/125 ≈ 0.0081/216 ≈ 0.004631/343 ≈ 0.0029151/512 ≈ 0.0019531/729 ≈ 0.0013721/1000 ≈ 0.001Adding these: 0.037037 + 0.015625 ≈ 0.052662+0.008 ≈ 0.060662+0.00463 ≈ 0.065292+0.002915 ≈ 0.068207+0.001953 ≈ 0.07016+0.001372 ≈ 0.071532+0.001 ≈ 0.072532So, g(3) ≈ 0.0725Similarly, g(4):1/81 ≈ 0.01234571/256 ≈ 0.003906251/625 ≈ 0.00161/1296 ≈ 0.00077161/2401 ≈ 0.00041641/4096 ≈ 0.0002441Adding these: 0.0123457 + 0.00390625 ≈ 0.01625195+0.0016 ≈ 0.01785195+0.0007716 ≈ 0.01862355+0.0004164 ≈ 0.01903995+0.0002441 ≈ 0.01928405So, g(4) ≈ 0.019284g(5):1/243 ≈ 0.00411521/1024 ≈ 0.00097661/3125 ≈ 0.000321/7776 ≈ 0.00012861/16807 ≈ 0.0000595Adding these: 0.0041152 + 0.0009766 ≈ 0.0050918+0.00032 ≈ 0.0054118+0.0001286 ≈ 0.0055404+0.0000595 ≈ 0.0056So, g(5) ≈ 0.0056g(6):1/729 ≈ 0.00137171/4096 ≈ 0.00024411/15625 ≈ 0.0000641/46656 ≈ 0.0000214Adding these: 0.0013717 + 0.0002441 ≈ 0.0016158+0.000064 ≈ 0.0016798+0.0000214 ≈ 0.0017012So, g(6) ≈ 0.0017g(7):1/2187 ≈ 0.00045721/16384 ≈ 0.0000611/78125 ≈ 0.0000128Adding these: 0.0004572 + 0.000061 ≈ 0.0005182+0.0000128 ≈ 0.000531So, g(7) ≈ 0.000531g(8):1/6561 ≈ 0.00015241/65536 ≈ 0.00001526Adding these: ≈ 0.0001677So, g(8) ≈ 0.0001677Now, summing these up:g(3) ≈ 0.0725g(4) ≈ 0.019284g(5) ≈ 0.0056g(6) ≈ 0.0017g(7) ≈ 0.000531g(8) ≈ 0.0001677Adding these:0.0725 + 0.019284 ≈ 0.091784+0.0056 ≈ 0.097384+0.0017 ≈ 0.099084+0.000531 ≈ 0.099615+0.0001677 ≈ 0.0997827So, up to n=8, the sum is approximately 0.09978, which is about 0.1, which is close to our analytical result of approximately 0.1051.Given that the terms beyond n=8 are very small, the total sum is likely around 0.105, which matches our analytical result of 7/4 - π^2/6 ≈ 0.105.Therefore, despite the initial confusion and the user's incorrect approach, the correct answer is indeed 7/4 - π^2/6.But wait, let me check the numerical value of 7/4 - π^2/6:7/4 = 1.75π^2 ≈ 9.8696π^2/6 ≈ 1.6449So, 1.75 - 1.6449 ≈ 0.1051Which matches our numerical approximation.Therefore, the correct answer is 7/4 - π^2/6.But wait, the problem didn't specify to express the answer in terms of π, so perhaps it's acceptable to leave it as is.Alternatively, maybe there's a way to express this without π, but I don't see an obvious way.Wait, another thought: perhaps I made a mistake in the partial fraction decomposition.Let me double-check:We had:1 / [k^2(k - 1)] = A / (k - 1) + B / k + C / k^2Multiplying both sides by k^2(k - 1):1 = A k^2 + B k(k - 1) + C(k - 1)Expanding:1 = A k^2 + B k^2 - B k + C k - CCombine like terms:(A + B) k^2 + (-B + C) k + (-C) = 1So, setting coefficients:A + B = 0-B + C = 0-C = 1Solving:From third equation: C = -1From second equation: -B + (-1) = 0 ⇒ B = -1From first equation: A + (-1) = 0 ⇒ A = 1Thus, partial fractions are correct: 1/(k - 1) - 1/k - 1/k^2Therefore, the decomposition is correct.Thus, the sum S = 7/4 - π^2/6 is indeed the correct answer.But wait, the initial problem's user solution concluded 1/3, which is approximately 0.333, which is larger than our result of approximately 0.105. So, their approach was flawed.Therefore, the correct answer is 7/4 - π^2/6.But let me check if this can be simplified further or expressed differently.Alternatively, perhaps the problem expects a different approach, such as recognizing that the sum can be expressed in terms of the digamma function or other special functions, but I think 7/4 - π^2/6 is the simplest form.Therefore, the final answer is 7/4 - π^2/6.But wait, let me check if the problem expects a fractional answer without π. Maybe I made a mistake in the setup.Wait, another approach: perhaps instead of swapping the order of summation, I can consider the sum as a double summation and find a closed-form expression.But I think the approach I took is correct, leading to S = 7/4 - π^2/6.Alternatively, perhaps the problem expects a different answer, but given the steps, I think this is correct.Therefore, the sum is 7/4 - π^2/6.But wait, let me compute 7/4 - π^2/6 numerically:7/4 = 1.75π^2 ≈ 9.8696π^2/6 ≈ 1.6449So, 1.75 - 1.6449 ≈ 0.1051Which is approximately 0.105, which matches our earlier numerical approximation.Therefore, I think this is the correct answer.**Final Answer**boxed{dfrac{7}{4} - dfrac{pi^2}{6}}

answer：Okay, so I have this problem: I need to find a four-digit natural number with distinct non-zero digits. That means each digit is different and none of them are zero. The special property this number should have is that if I add it to its reverse, the result is divisible by 101. Hmm, interesting. Let me try to break this down step by step.First, let me understand what it means for a number to be divisible by 101. If a number is divisible by 101, then when you divide it by 101, there's no remainder. So, if I take the four-digit number, add it to its reverse, and the sum is divisible by 101, that means the sum is some multiple of 101.Let me represent the four-digit number in a general form. Let's say the number is ABCD, where A, B, C, D are digits from 1 to 9 (since they are non-zero and distinct). So, the number itself is 1000*A + 100*B + 10*C + D. Its reverse would be DCBA, which is 1000*D + 100*C + 10*B + A.Now, if I add these two numbers together, ABCD + DCBA, what do I get? Let's compute that:ABCD + DCBA = (1000*A + 100*B + 10*C + D) + (1000*D + 100*C + 10*B + A)Let me simplify this expression:= 1000*A + A + 100*B + 10*B + 10*C + 100*C + D + 1000*D= (1000*A + A) + (100*B + 10*B) + (10*C + 100*C) + (D + 1000*D)= 1001*A + 110*B + 110*C + 1001*DHmm, interesting. So, the sum is 1001*A + 110*B + 110*C + 1001*D.I notice that 1001 and 110 are both multiples of 11. Let me factor that out:= 11*(91*A + 10*B + 10*C + 91*D)So, the sum is 11 times (91*A + 10*B + 10*C + 91*D). Now, the problem states that this sum must be divisible by 101. So, 11*(91*A + 10*B + 10*C + 91*D) must be divisible by 101.Since 11 and 101 are both prime numbers, and they are distinct, the product 11*101 is 1111. So, for 11*(something) to be divisible by 101, that something must be divisible by 101. In other words, (91*A + 10*B + 10*C + 91*D) must be divisible by 101.So, now the problem reduces to finding digits A, B, C, D (distinct, non-zero) such that 91*A + 10*B + 10*C + 91*D is divisible by 101.Let me write that as:91*(A + D) + 10*(B + C) ≡ 0 mod 101So, 91*(A + D) + 10*(B + C) ≡ 0 mod 101.Hmm, okay. So, I need to find A, B, C, D such that this congruence holds.Let me think about how to approach this. Maybe I can express this equation in terms of modulo 101.First, note that 91 is congruent to -10 mod 101 because 91 + 10 = 101, which is 0 mod 101. So, 91 ≡ -10 mod 101.Similarly, 10 is just 10 mod 101.So, substituting 91 with -10 in the equation:-10*(A + D) + 10*(B + C) ≡ 0 mod 101Let me factor out the 10:10*(-(A + D) + (B + C)) ≡ 0 mod 101Since 10 and 101 are coprime (they have no common factors other than 1), I can divide both sides by 10:-(A + D) + (B + C) ≡ 0 mod 101Which simplifies to:(B + C) - (A + D) ≡ 0 mod 101Or,(B + C) ≡ (A + D) mod 101But since A, B, C, D are digits from 1 to 9, the maximum possible value for A + D is 9 + 8 = 17 (since digits are distinct), and the minimum is 1 + 2 = 3. Similarly, B + C can range from 3 to 17.So, the difference (B + C) - (A + D) can range from -14 to 14. Therefore, the only way for (B + C) - (A + D) to be congruent to 0 mod 101 is if it is actually equal to 0. Because the difference is between -14 and 14, and 101 is much larger, so the only multiple of 101 in that range is 0.Therefore, we have:(B + C) = (A + D)So, the sum of the middle two digits equals the sum of the first and last digits.That's a useful condition. So, now, I need to find a four-digit number ABCD with distinct non-zero digits such that B + C = A + D.Okay, so perhaps I can look for such numbers where the sum of the middle digits equals the sum of the first and last digits.Let me think of some examples.Take 1234. Let's check: A=1, B=2, C=3, D=4. Then B + C = 5, and A + D = 5. So, 2 + 3 = 1 + 4 = 5. So, this satisfies the condition.Let me check if 1234 + 4321 is divisible by 101.1234 + 4321 = 5555.Now, 5555 divided by 101 is 55. So, yes, 5555 is divisible by 101.So, 1234 is such a number.Wait, so 1234 works. But the problem says "is there a four-digit natural number...", so yes, 1234 is an example.But maybe I should check if there are others or if this is unique.Let me try another number. Let's say 1324.A=1, B=3, C=2, D=4.B + C = 5, A + D = 5. So, same sum.1324 + 4231 = 5555, same as before. So, same result.So, 1324 also works.Similarly, 2134.A=2, B=1, C=3, D=4.B + C = 4, A + D = 6. Not equal. So, doesn't satisfy.Wait, but 2134: B + C = 1 + 3 = 4, A + D = 2 + 4 = 6. Not equal. So, doesn't work.Another one: 1243.A=1, B=2, C=4, D=3.B + C = 6, A + D = 4. Not equal.Wait, 1243: 2 + 4 = 6, 1 + 3 = 4. Not equal.So, doesn't work.Another number: 1342.A=1, B=3, C=4, D=2.B + C = 7, A + D = 3. Not equal.Hmm.Wait, maybe I should think of numbers where the sum of the middle digits equals the sum of the first and last digits.So, for example, if I fix A and D such that their sum is S, then B and C must also sum to S.So, perhaps I can choose A and D such that their sum is S, and then choose B and C from the remaining digits such that their sum is also S.Since all digits must be distinct and non-zero, I need to ensure that A, B, C, D are all different.Let me try with A=1, D=4, so S=5.Then, B and C must be two distinct digits from 2,3,5,6,7,8,9 that sum to 5.But the only digits left are 2,3,5,6,7,8,9. The possible pairs that sum to 5 are 2 and 3, but 2 and 3 are available. So, B=2, C=3 or B=3, C=2.So, numbers like 1234 and 1324, as I had before.Similarly, if I take A=2, D=3, so S=5.Then, B and C must be digits from 1,4,5,6,7,8,9 that sum to 5. But the only pair is 1 and 4. So, B=1, C=4 or B=4, C=1.So, numbers like 2143 and 2413.Let me check 2143:2143 + 3412 = 5555, which is divisible by 101.Similarly, 2413 + 3142 = 5555.So, these also work.Similarly, if I take A=3, D=2, same thing.But wait, digits must be distinct, so A=3, D=2, then B and C must be 1 and 4.So, numbers like 3142 and 3412.3142 + 2413 = 5555, which works.Similarly, 3412 + 2143 = 5555.Okay, so there are multiple numbers that satisfy this condition.Another example: A=1, D=5, so S=6.Then, B and C must be digits from 2,3,4,6,7,8,9 that sum to 6.Possible pairs: 2 and 4.So, numbers like 1245 and 1425.Let me check 1245 + 5421 = 6666.6666 divided by 101 is 66, so yes, divisible by 101.Similarly, 1425 + 5241 = 6666.So, these also work.Similarly, A=2, D=4, S=6.B and C must be from 1,3,5,6,7,8,9 that sum to 6.Possible pairs: 1 and 5.So, numbers like 2154 and 2514.2154 + 4512 = 6666, which is divisible by 101.Similarly, 2514 + 4152 = 6666.Okay, so this pattern seems to hold.So, in general, for any four-digit number where the sum of the middle two digits equals the sum of the first and last digits, the number plus its reverse will be divisible by 101.Therefore, such numbers exist, and 1234 is one example.I think I've got it. So, the answer is yes, and one such number is 1234.

answer：First, I need to determine the weekly salaries of employees x, y, and z based on the given relationships and total payment.Let’s denote the weekly salary of employee y as Rs. Y.According to the problem, employee x is paid 120 percent of what y is paid, so x's salary is:X = 120% of YX = 1.2YEmployee z is paid 75 percent of what x is paid, so z's salary is:Z = 75% of XZ = 0.75XThe total weekly salary paid to all three employees is Rs. 1500, so we can write the equation:X + Y + Z = 1500Substituting the expressions for X and Z in terms of Y, we get:1.2Y + Y + 0.75(1.2Y) = 1500Now, let's solve for Y:1.2Y + Y + 0.9Y = 15003.1Y = 1500Y = 1500 / 3.1Y = 483.87 (approximately)Now we can find X and Z:X = 1.2Y = 1.2 * 483.87 = 580.65 (approximately)Z = 0.75X = 0.75 * 580.65 = 435.49 (approximately)So, the weekly salaries are approximately:x = Rs. 580.65y = Rs. 483.87z = Rs. 435.49Now, let's calculate the salary of x after 4 weeks with a 5% increase each week.The salary of x after the first week (W1) is:W1 = X * (1 + 5/100) = X * 1.05After the second week (W2), it will be:W2 = W1 * 1.05 = X * (1.05)^2After the third week (W3), it will be:W3 = W2 * 1.05 = X * (1.05)^3After the fourth week (W4), it will be:W4 = W3 * 1.05 = X * (1.05)^4Now, let's calculate W4:W4 = 580.65 * (1.05)^4W4 = 580.65 * 1.21550625W4 = 705.79 (approximately)So, after 4 weeks, the individual earnings are approximately:x = Rs. 705.79y = Rs. 483.87 (no change as there is no mention of an increase for y)z = Rs. 435.49 (no change as there is no mention of an increase for z)

answer：Okay, so I have this problem here: Given the function ( f(x) = |x + a| + |x - 2| ), and the solution set of ( f(x) leq |x - 4| ) contains the interval ([1, 2]). I need to find the range of values for ( a ).Hmm, let me try to understand what this means. The function ( f(x) ) is the sum of two absolute value expressions. The inequality ( f(x) leq |x - 4| ) must hold true for all ( x ) in the interval ([1, 2]). So, for every ( x ) between 1 and 2, the expression ( |x + a| + |x - 2| ) must be less than or equal to ( |x - 4| ).First, maybe I should simplify both sides of the inequality to see if I can find a relationship involving ( a ).Starting with the right-hand side, ( |x - 4| ). Since ( x ) is between 1 and 2, ( x - 4 ) will be negative because 4 is larger than both 1 and 2. So, ( |x - 4| = 4 - x ).Now, the left-hand side is ( |x + a| + |x - 2| ). Let's analyze each absolute value term separately.First, ( |x - 2| ). Since ( x ) is between 1 and 2, ( x - 2 ) will be negative or zero. So, ( |x - 2| = 2 - x ).So, substituting these into the left-hand side, we have:( |x + a| + (2 - x) ).Therefore, the inequality becomes:( |x + a| + (2 - x) leq 4 - x ).Hmm, let's simplify this. Subtract ( (2 - x) ) from both sides:( |x + a| leq (4 - x) - (2 - x) ).Simplify the right-hand side:( 4 - x - 2 + x = 2 ).So, the inequality reduces to:( |x + a| leq 2 ).That's simpler. So, for all ( x ) in ([1, 2]), ( |x + a| leq 2 ).This absolute value inequality can be rewritten as:( -2 leq x + a leq 2 ).Which can be split into two inequalities:1. ( x + a geq -2 )2. ( x + a leq 2 )Let me write these separately:1. ( a geq -2 - x )2. ( a leq 2 - x )So, for each ( x ) in ([1, 2]), ( a ) must satisfy both ( a geq -2 - x ) and ( a leq 2 - x ).Since this has to be true for all ( x ) in ([1, 2]), ( a ) must be greater than or equal to the maximum value of ( -2 - x ) over ([1, 2]), and less than or equal to the minimum value of ( 2 - x ) over ([1, 2]).Let me find these maximum and minimum values.First, consider ( -2 - x ) over ( x in [1, 2] ).As ( x ) increases from 1 to 2, ( -2 - x ) decreases because it's a linear function with a negative slope. So, the maximum value occurs at the smallest ( x ), which is 1.Calculating ( -2 - 1 = -3 ).Similarly, the minimum value occurs at the largest ( x ), which is 2.Calculating ( -2 - 2 = -4 ).But since we're looking for ( a geq -2 - x ) for all ( x ) in ([1, 2]), ( a ) must be greater than or equal to the maximum of ( -2 - x ), which is -3.Now, let's look at the other inequality: ( a leq 2 - x ).Again, ( x ) is in ([1, 2]). The function ( 2 - x ) is a linear function decreasing as ( x ) increases. So, the maximum value occurs at ( x = 1 ), which is ( 2 - 1 = 1 ), and the minimum value occurs at ( x = 2 ), which is ( 2 - 2 = 0 ).But since ( a ) has to be less than or equal to ( 2 - x ) for all ( x ) in ([1, 2]), ( a ) must be less than or equal to the minimum of ( 2 - x ), which is 0.Putting these together, ( a ) must satisfy:( -3 leq a leq 0 ).So, the range of values for ( a ) is ([-3, 0]).Let me double-check this to make sure I didn't make a mistake.Starting from the inequality ( |x + a| leq 2 ), which implies ( -2 leq x + a leq 2 ). Then, solving for ( a ):1. ( a geq -2 - x )2. ( a leq 2 - x )For all ( x ) in ([1, 2]), the first inequality requires ( a ) to be at least as big as the maximum of ( -2 - x ) over that interval. Since ( x ) is between 1 and 2, ( -2 - x ) is between -4 and -3. The maximum is -3, so ( a geq -3 ).The second inequality requires ( a ) to be at most as big as the minimum of ( 2 - x ) over ([1, 2]). ( 2 - x ) is between 0 and 1, so the minimum is 0, meaning ( a leq 0 ).Therefore, combining these, ( a ) must be between -3 and 0, inclusive. So, the range is ([-3, 0]).I think that makes sense. Let me test with ( a = -3 ) and ( a = 0 ) to see if the inequality holds.First, let's take ( a = -3 ). Then, ( f(x) = |x - 3| + |x - 2| ).For ( x ) in ([1, 2]), ( |x - 3| = 3 - x ) and ( |x - 2| = 2 - x ). So, ( f(x) = (3 - x) + (2 - x) = 5 - 2x ).The right-hand side is ( |x - 4| = 4 - x ).So, the inequality becomes ( 5 - 2x leq 4 - x ).Simplify: ( 5 - 2x leq 4 - x ) => ( 5 - 4 leq 2x - x ) => ( 1 leq x ).Which is true because ( x geq 1 ) in our interval. So, equality holds at ( x = 1 ), which is the boundary.Now, let's take ( a = 0 ). Then, ( f(x) = |x| + |x - 2| ).For ( x ) in ([1, 2]), ( |x| = x ) and ( |x - 2| = 2 - x ). So, ( f(x) = x + (2 - x) = 2 ).The right-hand side is ( |x - 4| = 4 - x ).So, the inequality becomes ( 2 leq 4 - x ).Simplify: ( 2 leq 4 - x ) => ( x leq 2 ).Which is true because ( x leq 2 ) in our interval. So, equality holds at ( x = 2 ), which is the other boundary.Therefore, both endpoints ( a = -3 ) and ( a = 0 ) satisfy the inequality for all ( x ) in ([1, 2]). Hence, the range of ( a ) is indeed ([-3, 0]).I think that's solid. I don't see any mistakes in my reasoning.**Final Answer**The range of values for ( a ) is boxed{[-3, 0]}.