🧠:[<think>Okay, so I have this problem here: For 0 < x < π/2, the expression (cot³x)/(cot3x) takes all real numbers except those in the open interval (a, b). I need to find a + b. Hmm, okay. Let's break this down step by step.First, I remember that cotangent is the reciprocal of tangent, so cot x = cos x / sin x. Also, there's a trigonometric identity for cot 3x. Maybe I can express cot3x in terms of cotx to simplify the given expression. Let me recall the triple angle formula for cotangent. I think it's something like cot3x = (cot³x - 3cotx)/(3cot²x - 1). Let me check that. If that's correct, then maybe I can substitute that into the denominator.Yes, let me verify. The triple angle formula for cotangent. Let's derive it if I can't remember. Starting from tan3x, since tan3x = (3tanx - tan³x)/(1 - 3tan²x). Therefore, cot3x = 1/tan3x = (1 - 3tan²x)/(3tanx - tan³x). Let's express this in terms of cotx. Since tanx = 1/cotx, so tan²x = 1/cot²x, tan³x = 1/cot³x. Substitute these into the expression:cot3x = (1 - 3*(1/cot²x)) / (3*(1/cotx) - (1/cot³x)) = (1 - 3/cot²x) / (3/cotx - 1/cot³x) Multiply numerator and denominator by cot³x to eliminate the fractions:Numerator: cot³x*(1) - 3cot³x*(1/cot²x) = cot³x - 3cotx Denominator: 3cot³x*(1/cotx) - cot³x*(1/cot³x) = 3cot²x - 1 So indeed, cot3x = (cot³x - 3cotx)/(3cot²x - 1). Great, that matches the formula I remembered.So the given expression is (cot³x)/(cot3x). Let's substitute the expression for cot3x:(cot³x) / [(cot³x - 3cotx)/(3cot²x - 1)] = cot³x * (3cot²x - 1)/(cot³x - 3cotx) Factor numerator and denominator if possible. Let's look at the denominator: cot³x - 3cotx = cotx(cot²x - 3). The numerator is 3cot²x - 1. So substituting back:= cot³x * (3cot²x - 1) / [cotx(cot²x - 3)] Simplify: cot³x / cotx = cot²x. So:= cot²x * (3cot²x - 1) / (cot²x - 3) Let me let y = cot²x to make this expression simpler. Let y = cot²x. Since x is between 0 and π/2, cotx is positive, so y is positive. Then the expression becomes:y * (3y - 1) / (y - 3) So the original expression (cot³x)/(cot3x) simplifies to [y(3y - 1)] / (y - 3), where y = cot²x and y > 0.Our goal is to find the range of this function as x varies in (0, π/2), which corresponds to y varying in (0, ∞), since cotx goes from infinity to 0 as x goes from 0 to π/2. Wait, actually, when x approaches 0, cotx approaches infinity, so y = cot²x approaches infinity. When x approaches π/2, cotx approaches 0, so y approaches 0. Therefore, y ∈ (0, ∞). So we need to find the range of the function f(y) = [y(3y - 1)] / (y - 3) for y > 0, y ≠ 3 (since denominator can't be zero). Wait, but y = cot²x, so when is y = 3? That would be when cot²x = 3, so cotx = √3, so x = π/6. So at x = π/6, the denominator becomes zero, but x = π/6 is within (0, π/2), so we have to exclude y = 3 from the domain.So the function f(y) is defined for y ∈ (0, 3) ∪ (3, ∞). We need to find the range of f(y). The problem states that the original expression takes all real numbers except those in the open interval (a, b). Therefore, the range of f(y) is ℝ (a, b), and we need to find a + b.To find the range, let's analyze the function f(y) = [3y² - y] / (y - 3). Let's write this as f(y) = (3y² - y)/(y - 3). Let's perform polynomial division or simplify the expression.Divide 3y² - y by y - 3. Let's do polynomial long division:Divide 3y² by y to get 3y. Multiply (y - 3) by 3y: 3y² - 9y. Subtract this from 3y² - y:(3y² - y) - (3y² - 9y) = (0y²) + ( - y + 9y ) = 8y.So we have 3y² - y = (y - 3)(3y) + 8y. Now, bring down the remaining terms. Wait, actually, since the original numerator is 3y² - y, after subtracting 3y² - 9y, we have 8y. So then:f(y) = (3y² - y)/(y - 3) = 3y + (8y)/(y - 3). Wait, but let's check:Wait, (y - 3)(3y) + 8y = 3y² - 9y + 8y = 3y² - y, which matches the numerator. Therefore:f(y) = 3y + (8y)/(y - 3). Hmm, maybe that's not the simplest. Alternatively, maybe split it differently.Alternatively, we can write f(y) = [3y² - y]/(y - 3) = [3y² - 9y + 8y]/(y - 3) = [3y(y - 3) + 8y]/(y - 3) = 3y + 8y/(y - 3). So yes, same result.Alternatively, write it as f(y) = 3y + 8y/(y - 3). Let me see if I can simplify further. Let's factor the 8y/(y - 3):8y/(y - 3) = 8(y - 3 + 3)/(y - 3) = 8[1 + 3/(y - 3)] = 8 + 24/(y - 3). Therefore:f(y) = 3y + 8 + 24/(y - 3) = 3y + 8 + 24/(y - 3). Hmm, not sure if that helps.Alternatively, maybe consider f(y) as a function and analyze its behavior. Let's think of f(y) as (3y² - y)/(y - 3). Let's analyze the limits as y approaches critical points and find the extrema.First, note that y ∈ (0, 3) ∪ (3, ∞). Let's split it into two intervals: (0, 3) and (3, ∞).First, let's consider y approaching 3 from the right (y → 3⁺). Then the denominator approaches 0 from the positive side, and the numerator approaches 3*(9) - 3 = 27 - 3 = 24. So 24 / 0⁺ → +∞. So as y approaches 3 from above, f(y) approaches +∞.Similarly, y approaching 3 from the left (y → 3⁻). Denominator approaches 0 from the negative side, numerator approaches 24. So 24 / 0⁻ → -∞. Therefore, as y approaches 3 from below, f(y) approaches -∞.Now, let's check the behavior as y approaches 0⁺. Then y is approaching 0 from the positive side. So f(y) = (0 - 0)/(0 - 3) = 0 / (-3) = 0. Wait, but substituting y = 0 into f(y): numerator is 0 - 0 = 0, denominator is -3. So f(y) approaches 0 as y approaches 0⁺.Wait, but actually, when y approaches 0, cot²x approaches 0, so x approaches π/2. Let me confirm:Yes, as x → π/2⁻, cotx → 0, so y = cot²x → 0. Therefore, f(y) approaches 0.Next, as y approaches ∞, which is when x approaches 0⁺. Let's compute the limit of f(y) as y → ∞. f(y) = (3y² - y)/(y - 3). Divide numerator and denominator by y:= (3y - 1)/(1 - 3/y). As y → ∞, this approaches (3y)/1 = 3y, which tends to ∞. So f(y) → ∞ as y → ∞.So overall, as y ranges over (0, 3) and (3, ∞), f(y) ranges from 0 to -∞ (on (0,3)) and from +∞ to +∞ (on (3, ∞))? Wait, let me check again.Wait, when y is in (0, 3):- At y approaching 0⁺, f(y) approaches 0.- As y approaches 3⁻, f(y) approaches -∞.Therefore, on (0, 3), f(y) decreases from 0 to -∞. So the range on (0, 3) is (-∞, 0).On the other interval, y ∈ (3, ∞):- As y approaches 3⁺, f(y) approaches +∞.- As y approaches ∞, f(y) approaches ∞. But does it have a minimum somewhere?Wait, we need to check if there's a local minimum or maximum on (3, ∞). Let's take the derivative of f(y) to find critical points.f(y) = (3y² - y)/(y - 3). Let's compute f'(y):Using the quotient rule: [(6y - 1)(y - 3) - (3y² - y)(1)] / (y - 3)²Compute numerator:(6y - 1)(y - 3) - (3y² - y) First expand (6y - 1)(y - 3):= 6y(y - 3) - 1(y - 3) = 6y² - 18y - y + 3 = 6y² - 19y + 3Subtract (3y² - y):= (6y² - 19y + 3) - 3y² + y = 3y² - 18y + 3Therefore, f'(y) = (3y² - 18y + 3)/(y - 3)²Set numerator equal to zero to find critical points:3y² - 18y + 3 = 0 Divide by 3: y² - 6y + 1 = 0 Solve quadratic equation: y = [6 ± √(36 - 4)]/2 = [6 ± √32]/2 = [6 ± 4√2]/2 = 3 ± 2√2So critical points at y = 3 + 2√2 ≈ 3 + 2.828 ≈ 5.828 and y = 3 - 2√2 ≈ 3 - 2.828 ≈ 0.172. However, since y is in (0, 3) ∪ (3, ∞), so y = 3 - 2√2 is approximately 0.172, which is in (0, 3), and y = 3 + 2√2 is in (3, ∞).Therefore, f(y) has critical points at y = 3 ± 2√2. Let's analyze these.First, in the interval (0, 3), the critical point is at y = 3 - 2√2 ≈ 0.172. Since on (0, 3), f(y) decreases from 0 to -∞, a critical point here might be a local minimum or maximum. Let's check the sign of f'(y) around y = 3 - 2√2.Wait, but let's compute the derivative's sign. Since f'(y) = (3y² - 18y + 3)/(y - 3)². The denominator is always positive except at y = 3. The numerator is 3y² - 18y + 3, which we can write as 3(y² - 6y + 1). The roots are at y = 3 ± 2√2. So for y < 3 - 2√2, the quadratic y² - 6y + 1 is positive (since parabola opens upwards). Between the roots, it's negative, and for y > 3 + 2√2, positive again.So in the interval (0, 3 - 2√2), the numerator is positive, so f'(y) is positive. In (3 - 2√2, 3), the numerator is negative, so f'(y) is negative. Therefore, at y = 3 - 2√2, f(y) has a local maximum. Similarly, in (3, ∞), the numerator is positive (for y > 3 + 2√2), so f'(y) is positive, and between (3, 3 + 2√2), the numerator is negative, so f'(y) is negative. Therefore, at y = 3 + 2√2, f(y) has a local minimum.Therefore, the function f(y) has a local maximum at y = 3 - 2√2 in (0, 3), and a local minimum at y = 3 + 2√2 in (3, ∞).So let's compute f(y) at these critical points.First, compute f(3 - 2√2):f(y) = (3y² - y)/(y - 3)Let y = 3 - 2√2. Let's compute numerator and denominator.First, compute y - 3 = (3 - 2√2) - 3 = -2√2.Compute numerator:3y² - y. Let's compute y² first.y = 3 - 2√2, so y² = (3)^2 + (2√2)^2 - 2*3*2√2 = 9 + 8 - 12√2 = 17 - 12√2.Therefore, 3y² = 3*(17 - 12√2) = 51 - 36√2.Subtract y: 51 - 36√2 - (3 - 2√2) = 51 - 36√2 - 3 + 2√2 = 48 - 34√2.Therefore, numerator = 48 - 34√2, denominator = -2√2.Thus, f(y) = (48 - 34√2)/(-2√2) = -(48 - 34√2)/(2√2) = (-48 + 34√2)/(2√2).Let's rationalize the denominator:Multiply numerator and denominator by √2:= [(-48 + 34√2)√2]/(2*2) = [(-48√2 + 34*2)]/4 = (-48√2 + 68)/4 = (68 - 48√2)/4 = 17 - 12√2.Similarly, compute f(y) at y = 3 + 2√2.First, y - 3 = 2√2.Compute numerator: 3y² - y.Compute y²:y = 3 + 2√2, so y² = 9 + 12√2 + 8 = 17 + 12√2.3y² = 3*(17 + 12√2) = 51 + 36√2.Subtract y: 51 + 36√2 - (3 + 2√2) = 51 + 36√2 - 3 - 2√2 = 48 + 34√2.Denominator: y - 3 = 2√2.Thus, f(y) = (48 + 34√2)/(2√2).Rationalize:Multiply numerator and denominator by √2:= [(48 + 34√2)√2]/(2*2) = [48√2 + 34*2]/4 = (48√2 + 68)/4 = (68 + 48√2)/4 = 17 + 12√2.So, at the critical points:- At y = 3 - 2√2, f(y) = 17 - 12√2 ≈ 17 - 12*1.414 ≈ 17 - 16.968 ≈ 0.032- At y = 3 + 2√2, f(y) = 17 + 12√2 ≈ 17 + 16.968 ≈ 33.968Wait, but this seems conflicting with our earlier analysis. Wait, when y = 3 - 2√2 ≈ 0.172, which is in (0, 3). We found that f(y) at this point is 17 - 12√2 ≈ 0.032, which is a positive value. However, earlier we thought that on the interval (0, 3), f(y) decreases from 0 to -∞, but here there's a local maximum at y ≈ 0.172 with f(y) ≈ 0.032, which is slightly positive. Then, after that, it decreases to -∞ as y approaches 3 from below.Wait a second, that changes things. So the function on (0, 3) actually starts at 0 when y approaches 0, increases to a local maximum of 17 - 12√2 (≈0.032), and then decreases to -∞ as y approaches 3. Therefore, the range on (0, 3) is (-∞, 17 - 12√2].But 17 - 12√2 is a positive number, very small, approximately 0.032. So from 0 up to 0.032, then decreasing to -infty.Wait, but how does that work? Let's see:At y = 0, f(y) = 0.At y = 3 - 2√2 ≈ 0.172, f(y) ≈ 0.032.Then as y approaches 3, f(y) approaches -∞. So the function f(y) on (0, 3) goes from 0 up to ≈0.032, then plummets to -∞. Therefore, the range on (0, 3) is (-∞, 17 - 12√2] ∪ [0, 0.032]? Wait, no, that can't be. Wait, if it starts at 0 when y is 0, increases to 17 - 12√2, then decreases to -∞, then the range is (-∞, 17 - 12√2]. But wait, how does it include 0? When y approaches 0, f(y) approaches 0. So the maximum is 17 - 12√2, which is slightly above 0, and then it goes down to -∞. So the range on (0, 3) is (-∞, 17 - 12√2], including 0 because as y approaches 0, f(y) approaches 0. Wait, but 17 - 12√2 is approximately 0.032, so it's a very small positive number. So the interval (-∞, 0.032], but including 0. Wait, but how does it cover from -infty up to 0.032? Because when y increases from 0 to 3 - 2√2, f(y) goes from 0 up to 0.032, then as y increases from 3 - 2√2 to 3, f(y) decreases from 0.032 down to -infty. Therefore, the range on (0, 3) is (-infty, 17 - 12√2], since it goes from 0 up to 17 - 12√2 and then down to -infty. Wait, but 0 is included because as y approaches 0, f(y) approaches 0. So 0 is part of the range. But the maximum is 17 - 12√2. Therefore, the range on (0, 3) is (-infty, 17 - 12√2], including 0? Wait, but as y approaches 0, f(y) approaches 0, but 0 is not in the interval (0, 3) for y, but since y approaches 0, the limit is 0. So the range on (0, 3) is (-infty, 17 - 12√2] ∪ {0}? No, that can't be, because when y approaches 0, f(y) approaches 0, but y never actually reaches 0. So since y is in (0, 3), the range would be (-infty, 17 - 12√2], but not including 0, right? Wait, because y approaches 0, so f(y) approaches 0, but doesn't actually reach 0. Wait, but wait, y ∈ (0,3). When y approaches 0⁺, f(y) approaches 0. So the infimum on (0,3) is -infty, and the supremum is 17 - 12√2. So the range on (0, 3) is (-infty, 17 - 12√2). But does it include 17 - 12√2? Since that value is achieved at y = 3 - 2√2, which is in (0, 3). So yes, the range on (0, 3) is (-infty, 17 - 12√2].Similarly, on the interval (3, ∞), f(y) approaches +∞ as y approaches 3 from above, and approaches +∞ as y approaches ∞. But there's a local minimum at y = 3 + 2√2, which is 17 + 12√2. Therefore, the function decreases from +∞ to 17 + 12√2 and then increases back to +∞. Therefore, the range on (3, ∞) is [17 + 12√2, ∞).Therefore, combining both intervals, the total range of f(y) is (-∞, 17 - 12√2] ∪ [17 + 12√2, ∞). Therefore, the expression (cot³x)/(cot3x) takes all real numbers except those in the open interval (17 - 12√2, 17 + 12√2). Therefore, a = 17 - 12√2 and b = 17 + 12√2. So a + b = (17 - 12√2) + (17 + 12√2) = 34.Wait, but let's verify this again. So the problem states that the expression takes all real numbers except those in the open interval (a, b). From our analysis, the range is (-∞, 17 - 12√2] ∪ [17 + 12√2, ∞). Therefore, the excluded interval is (17 - 12√2, 17 + 12√2). Therefore, a = 17 - 12√2 and b = 17 + 12√2. Then a + b = 34. Hmm, but 17 - 12√2 is approximately 0.03 and 17 + 12√2 is approximately 33.97, so their sum is 34. That seems reasonable.But let me check once again for errors. So the critical points were at y = 3 ± 2√2, and substituting those into f(y) gave 17 ∓ 12√2. Wait, hold on, when we computed f(y) at y = 3 - 2√2, we got 17 - 12√2, and at y = 3 + 2√2, we got 17 + 12√2. But wait, these results are positive. However, when y is in (0, 3), f(y) reaches a maximum of 17 - 12√2, which is a small positive number, then decreases to -∞. Therefore, the range on (0, 3) is (-∞, 17 - 12√2], including that maximum. And on (3, ∞), the range is [17 + 12√2, ∞). Therefore, combining these two intervals, the real numbers not covered are (17 - 12√2, 17 + 12√2). Therefore, the answer is a + b = (17 - 12√2) + (17 + 12√2) = 34.Yes, that seems correct. Therefore, the answer should be 34. Let me check if there is any mistake in computation.Wait, when I calculated f(y) at y = 3 - 2√2, let me recheck:y = 3 - 2√2.Compute f(y) = (3y² - y)/(y - 3)First compute y²:(3 - 2√2)^2 = 9 - 12√2 + 8 = 17 - 12√2.Therefore, 3y² = 51 - 36√2.Then subtract y: 51 - 36√2 - (3 - 2√2) = 51 - 3 - 36√2 + 2√2 = 48 - 34√2.Denominator: y - 3 = -2√2.Therefore, f(y) = (48 - 34√2)/(-2√2) = -(48 - 34√2)/(2√2).Let's compute numerator: -(48 - 34√2) = -48 + 34√2.Divide by 2√2:= (-48 + 34√2)/(2√2) = (-48)/(2√2) + (34√2)/(2√2) = (-24)/√2 + 17.Simplify:-24/√2 = -12√2.So f(y) = -12√2 + 17 = 17 - 12√2. Correct.Similarly, for y = 3 + 2√2:y² = (3 + 2√2)^2 = 9 + 12√2 + 8 = 17 + 12√2.3y² = 51 + 36√2.Subtract y: 51 + 36√2 - (3 + 2√2) = 48 + 34√2.Denominator: y - 3 = 2√2.Therefore, f(y) = (48 + 34√2)/(2√2) = (48)/(2√2) + (34√2)/(2√2) = 24/√2 + 17 = 12√2 + 17. Correct.Thus, indeed, the critical values are 17 ± 12√2. Therefore, the excluded interval is between these two values, so a = 17 - 12√2 and b = 17 + 12√2. Summing them gives 34.Therefore, the answer is 34.Final Answerboxed{34}</think>To solve the problem, we start by expressing the given function (frac{cot^3 x}{cot 3x}) in terms of (cot x). Using the triple angle formula for cotangent, (cot 3x = frac{cot^3 x - 3cot x}{3cot^2 x - 1}), we substitute this into the function:[frac{cot^3 x}{cot 3x} = frac{cot^3 x}{frac{cot^3 x - 3cot x}{3cot^2 x - 1}} = cot^3 x cdot frac{3cot^2 x - 1}{cot^3 x - 3cot x}]Simplifying this expression, we factor out (cot x) from the denominator:[= cot^2 x cdot frac{3cot^2 x - 1}{cot^2 x - 3}]Let (y = cot^2 x). The expression becomes:[frac{y(3y - 1)}{y - 3}]We need to analyze the range of this function for (y > 0), excluding (y = 3). We find the critical points by taking the derivative and setting it to zero. The critical points are (y = 3 pm 2sqrt{2}). Evaluating the function at these points:- For (y = 3 - 2sqrt{2}), the value is (17 - 12sqrt{2}).- For (y = 3 + 2sqrt{2}), the value is (17 + 12sqrt{2}).Analyzing the behavior of the function as (y) approaches critical points and infinity, we determine the ranges for (y) in the intervals ((0, 3)) and ((3, infty)). The function ranges from (-infty) to (17 - 12sqrt{2}) on ((0, 3)) and from (17 + 12sqrt{2}) to (infty) on ((3, infty)).Thus, the function takes all real numbers except those in the open interval ((17 - 12sqrt{2}, 17 + 12sqrt{2})). Summing the endpoints (a) and (b):[a + b = (17 - 12sqrt{2}) + (17 + 12sqrt{2}) = 34][boxed{34}],<think>Okay, so I need to figure out for the function f(x) = cot³x / cot3x, where 0 < x < π/2, all real numbers are taken except those in the open interval (a, b). Then find a + b. Hmm, alright. Let me break this down step by step.First, I know that cotangent is the reciprocal of tangent, so cot x = cos x / sin x. But maybe there's a better way to approach this using trigonometric identities? Let me recall that cot3x can be expressed in terms of cot x. There's a formula for cotangent of multiple angles. Let me check that.Yes, the formula for cot3x is (cot³x - 3cotx) / (3cot²x - 1). Let me verify that. Starting from the tangent triple angle formula: tan3x = (3tanx - tan³x) / (1 - 3tan²x). Taking reciprocal for cotangent, we get cot3x = (1 - 3tan²x) / (3tanx - tan³x). Let's convert tan to cot. Since tanx = 1/cotx, so tan²x = 1/cot²x. Let me substitute that in.So, cot3x = (1 - 3*(1/cot²x)) / (3*(1/cotx) - (1/cot³x)) = (1 - 3/cot²x) / (3/cotx - 1/cot³x). Let's multiply numerator and denominator by cot³x to eliminate denominators. That gives:Numerator: cot³x - 3cotxDenominator: 3cot²x - 1So yes, cot3x = (cot³x - 3cotx)/(3cot²x - 1). Got that.So now, our function f(x) = cot³x / cot3x. Substitute the expression for cot3x:f(x) = cot³x / [(cot³x - 3cotx)/(3cot²x - 1)] = cot³x * (3cot²x - 1)/(cot³x - 3cotx)Let me simplify this. Let's factor out cotx in the denominator:Denominator: cotx(cot²x - 3)So f(x) = cot³x * (3cot²x - 1) / [cotx(cot²x - 3)] = [cot³x / cotx] * (3cot²x - 1)/(cot²x - 3) = cot²x * (3cot²x - 1)/(cot²x - 3)Let me let y = cot²x. Since 0 < x < π/2, cotx is positive, so y > 0. So substituting y, f(x) becomes:f(x) = y * (3y - 1)/(y - 3)So f(y) = y*(3y - 1)/(y - 3). Let me simplify this expression. Let's perform the multiplication:f(y) = [y(3y - 1)] / (y - 3) = (3y² - y)/(y - 3)Now, to simplify this further, perhaps perform polynomial division. Let's divide 3y² - y by y - 3.Divide 3y² - y by y - 3:First term: 3y² / y = 3y. Multiply divisor by 3y: 3y*(y - 3) = 3y² - 9ySubtract this from dividend:(3y² - y) - (3y² - 9y) = (-y + 9y) = 8ySo now, remainder is 8y. So the division gives 3y + 8y/(y - 3). Wait, no, that's not correct. Wait, after subtracting, we have:Dividend: 3y² - ySubtract 3y² - 9y: 0 + 8ySo then, bring down the next term, but there's no constant term. So we have:3y² - y divided by y - 3 is 3y + (8y)/(y - 3). Wait, but 8y divided by (y - 3) can be further broken down. Let me check:Wait, actually, the remainder is 8y, so we can express:(3y² - y)/(y - 3) = 3y + 8y/(y - 3). Hmm, but that seems incomplete. Wait, perhaps I made a mistake in the division.Wait, let's redo the division:Divide 3y² - y by y - 3.First term: 3y² divided by y is 3y. Multiply 3y by (y - 3): 3y² - 9y. Subtract this from 3y² - y:(3y² - y) - (3y² - 9y) = 0 + 8y. So the remainder is 8y.Therefore, the division gives 3y + (8y)/(y - 3). So f(y) = 3y + 8y/(y - 3)Alternatively, factor 8y/(y - 3) as 8*(y)/(y - 3). Let me write it as:f(y) = 3y + 8y/(y - 3). Hmm, maybe we can combine these terms?Alternatively, write f(y) as 3y + 8y/(y - 3) = 3y + 8 + 24/(y - 3). Wait, how?Wait, 8y/(y - 3) can be written as 8*(y - 3 + 3)/(y - 3) = 8*(1 + 3/(y - 3)) = 8 + 24/(y - 3). So then:f(y) = 3y + 8 + 24/(y - 3) = 3y + 8 + 24/(y - 3)But I'm not sure if that helps. Alternatively, perhaps set t = y - 3. Let me think.Alternatively, let's consider f(y) = (3y² - y)/(y - 3). Let me write this as a function in terms of y, and analyze its range.Since y = cot²x and x ∈ (0, π/2), so y ∈ (0, ∞). So we need to find the range of f(y) = (3y² - y)/(y - 3) for y ∈ (0, ∞) {3}, since y = 3 would make the denominator zero. But we need to check if y = 3 is attainable. Since y = cot²x, so cotx = sqrt(3) or -sqrt(3), but since x is in (0, π/2), cotx is positive, so y = 3 when cotx = sqrt(3), which occurs at x = π/6. So x = π/6 is in (0, π/2), so y = 3 is attainable, but at that point, the original function f(x) would have a denominator cot3x. Let me check if cot3x is defined at x = π/6.Wait, x = π/6, so 3x = π/2. Cotangent of π/2 is zero, so cot3x = 0. Therefore, f(x) = cot³x / 0, which is undefined. Therefore, y = 3 corresponds to x = π/6, but at that point, f(x) is undefined. Therefore, in the domain 0 < x < π/2, x ≠ π/6, so y ∈ (0, ∞) {3}.Therefore, y can be any positive real number except 3.Therefore, f(y) = (3y² - y)/(y - 3) with y ∈ (0, 3) ∪ (3, ∞). We need to find the range of this function. The problem states that f(x) takes all real numbers except those in the open interval (a, b). So we need to find a and b such that the range of f is ℝ (a, b).Therefore, the key is to analyze f(y) = (3y² - y)/(y - 3) over y ∈ (0, 3) and y ∈ (3, ∞), find the ranges for each interval, and then combine them to see the total range.So let's split into two cases:Case 1: y ∈ (0, 3)Case 2: y ∈ (3, ∞)For each case, compute the range of f(y).Let me first compute the derivative of f(y) to find critical points.But since this is a rational function, maybe we can analyze its behavior.Alternatively, since f(y) is a rational function, we can parametrize it and find its extrema.Alternatively, set f(y) = k and solve for y, then find for which k the equation has real solutions in y ∈ (0,3) ∪ (3, ∞).Let me try this approach.Let f(y) = k, so (3y² - y)/(y - 3) = kMultiply both sides by (y - 3):3y² - y = k(y - 3)Bring all terms to left-hand side:3y² - y - ky + 3k = 0Combine like terms:3y² - (1 + k)y + 3k = 0So quadratic equation in y: 3y² - (k + 1)y + 3k = 0For real solutions y, discriminant must be non-negative.Discriminant D = [-(k + 1)]² - 4*3*3k = (k + 1)² - 36kCompute D:D = k² + 2k + 1 - 36k = k² - 34k + 1For real solutions, D ≥ 0:k² - 34k + 1 ≥ 0Solve quadratic inequality:k² - 34k + 1 ≥ 0Find roots:k = [34 ± √(34² - 4*1*1)] / 2 = [34 ± √(1156 - 4)] / 2 = [34 ± √1152] / 2Simplify √1152: 1152 = 1024 + 128 = 1024 + 128 = 32² * 36? Wait, 1152 ÷ 64 = 18, so √1152 = √(64*18) = 8√18 = 8*3√2 = 24√2. Wait, let me check:Wait, 34² = 1156, so 1156 - 4 = 1152. √1152: 1152 = 1024 + 128 = 1024 + 128. Wait, 1152 ÷ 16 = 72. 72 = 36*2. So √1152 = √(16*72) = 4√72 = 4*6√2 = 24√2. Yes. Therefore,k = [34 ± 24√2]/2 = 17 ± 12√2Therefore, the quadratic is non-negative when k ≤ 17 - 12√2 or k ≥ 17 + 12√2So, discriminant is non-negative only when k is in (-∞, 17 - 12√2] ∪ [17 + 12√2, ∞). Therefore, these are the values of k for which the equation 3y² - (k + 1)y + 3k = 0 has real solutions.But even if discriminant is non-negative, we need to check whether the solutions y are in (0,3) or (3, ∞). Therefore, even if k is in (-∞, 17 - 12√2] ∪ [17 + 12√2, ∞), the solutions y must lie in (0,3) or (3, ∞). Therefore, the actual range of f(y) may be different.Alternatively, perhaps the function f(y) maps (0,3) to some interval and (3, ∞) to another interval, and their union is ℝ except (a,b). So we need to analyze the behavior of f(y) on both intervals.Let me analyze Case 1: y ∈ (0, 3)Compute the limit as y approaches 3 from the left:lim_{y→3^-} f(y) = lim_{y→3^-} (3y² - y)/(y - 3). The denominator approaches 0 from the negative side (since y < 3), numerator approaches 3*(9) - 3 = 27 - 3 = 24. So numerator approaches 24, denominator approaches 0 from negative side. Therefore, limit is -∞.Now, as y approaches 0 from the right:lim_{y→0^+} f(y) = (0 - 0)/(0 - 3) = 0 / (-3) = 0So as y goes from 0 to 3, f(y) goes from 0 to -∞. But does it cover all negative numbers? Wait, but maybe there's a critical point where the function attains a maximum or minimum.Similarly, for Case 2: y ∈ (3, ∞)Compute limit as y approaches 3 from the right:lim_{y→3^+} f(y) = (3*9 - 3)/(3 - 3) = 24/0, which is +∞.As y approaches ∞:lim_{y→∞} f(y) = (3y²)/(y) = 3y, so it goes to +∞.But again, between 3 and ∞, is the function increasing the entire time? Let's check derivative.Alternatively, compute derivative of f(y):f(y) = (3y² - y)/(y - 3)Using quotient rule: f’(y) = [ (6y - 1)(y - 3) - (3y² - y)(1) ] / (y - 3)^2Compute numerator:(6y - 1)(y - 3) - (3y² - y) = [6y(y) - 18y - y + 3] - 3y² + y= [6y² - 19y + 3] - 3y² + y= 3y² - 18y + 3Therefore, f’(y) = (3y² - 18y + 3)/(y - 3)^2Set numerator equal to zero to find critical points:3y² - 18y + 3 = 0Divide by 3:y² - 6y + 1 = 0Solutions:y = [6 ± √(36 - 4)]/2 = [6 ± √32]/2 = [6 ± 4√2]/2 = 3 ± 2√2So critical points at y = 3 + 2√2 ≈ 3 + 2.828 ≈ 5.828 and y = 3 - 2√2 ≈ 3 - 2.828 ≈ 0.172So for y ∈ (0,3), the critical point is at y ≈ 0.172. Let's check whether that's in (0,3). Yes, 3 - 2√2 ≈ 0.172 is in (0,3). Similarly, 3 + 2√2 ≈ 5.828 is in (3, ∞). Therefore, for Case 1: y ∈ (0,3), the critical point is at y = 3 - 2√2 ≈ 0.172. Let's compute f(y) at this critical point.Compute f(3 - 2√2):First, substitute y = 3 - 2√2 into f(y) = (3y² - y)/(y - 3)First compute y - 3 = (3 - 2√2) - 3 = -2√2Compute 3y² - y:First compute y² = (3 - 2√2)^2 = 9 - 12√2 + 8 = 17 - 12√2So 3y² = 3*(17 - 12√2) = 51 - 36√2Then subtract y: 51 - 36√2 - (3 - 2√2) = 51 - 36√2 - 3 + 2√2 = 48 - 34√2Therefore, f(y) = (48 - 34√2)/(-2√2) = (-48 + 34√2)/(2√2) = [34√2 - 48]/(2√2)Factor numerator: 34√2 - 48 = 2*(17√2 - 24)So f(y) = 2*(17√2 -24)/(2√2) = (17√2 -24)/√2Multiply numerator and denominator by √2 to rationalize:= (17√2 -24)√2 / 2 = (17*2 -24√2)/2 = (34 -24√2)/2 = 17 -12√2So f(y) at y = 3 - 2√2 is 17 -12√2 ≈ 17 - 16.97 ≈ 0.03. Wait, that seems low. Wait, but 17 -12√2 is approximately 17 - 16.97056 ≈ 0.02944. Hmm, but when y approaches 0, f(y) approaches 0, and at y ≈0.172, it's ≈0.029. Then as y approaches 3 from the left, f(y) approaches -∞. So in the interval (0,3), the function starts at 0, reaches a maximum of 17 -12√2 ≈0.029, then decreases to -∞. Therefore, the range for y ∈ (0,3) is (-∞, 17 -12√2].Wait, but at y approaching 0, f(y) approaches 0, then increases to 17 -12√2 ≈0.029, then decreases to -∞. Wait, that seems contradictory. Wait, if the critical point is a maximum, then the function increases from 0 to 17 -12√2, then decreases to -∞. Therefore, the range for y ∈ (0,3) is (-∞, 17 -12√2].Wait, but 17 -12√2 is approximately 0.029, so positive. So from 0, goes up to ~0.029, then plummets to -∞. Therefore, the range is (-∞, 17 -12√2] ∪ [0, something]. Wait, but no, wait, when y approaches 0, f(y) approaches 0. Then it goes up to 17 -12√2, which is a positive number, then goes down to -∞. Therefore, the range on (0,3) is (-∞, 17 -12√2] ∪ [0, 0.029]. Wait, no, that can't be. Wait, when y approaches 0, f(y) approaches 0. Then as y increases to 3 -2√2 ≈0.172, f(y) increases to 17 -12√2 ≈0.029. Then, as y continues to increase from 0.172 to 3, f(y) decreases from 0.029 to -∞. Therefore, the maximum value in (0,3) is 17 -12√2, and the function takes all values from -∞ up to 17 -12√2, and also from 0 to 17 -12√2? Wait, no. Wait, the function starts at 0 when y approaches 0, increases to a maximum of 17 -12√2, then decreases to -∞ as y approaches 3 from the left. Therefore, the range is (-∞, 17 -12√2]. But wait, when y is between 0 and 3 -2√2, the function is increasing from 0 to 17 -12√2. Then when y is between 3 -2√2 and 3, the function decreases from 17 -12√2 to -∞. Therefore, the range on (0,3) is (-∞, 17 -12√2]. Wait, but the function also passes through 0 at y approaching 0. So how does that work? Wait, if the function starts at 0, goes up to 17 -12√2, then down to -∞, then the range would be all real numbers less than or equal to 17 -12√2. But the function doesn't take any values between 0 and 17 -12√2 except near the peak. Wait, no, actually, when y is in (0, 3 -2√2), f(y) is increasing from 0 to 17 -12√2. Then when y is in (3 -2√2, 3), f(y) is decreasing from 17 -12√2 to -∞. Therefore, the range on (0,3) is (-∞, 17 -12√2] ∪ [0, 17 -12√2]. Wait, but that would suggest overlapping ranges, but since y is split into two intervals, (0, 3 -2√2) and (3 -2√2, 3). Hmm, perhaps I need to better analyze.Wait, actually, when y is in (0, 3 -2√2), the function f(y) increases from 0 to 17 -12√2. Then, when y is in (3 -2√2, 3), the function decreases from 17 -12√2 to -∞. Therefore, the range on y ∈ (0,3) is (-∞, 17 -12√2] ∪ [0, 17 -12√2]. Wait, but 17 -12√2 is approximately 0.029, so [0, 0.029] and (-∞, 0.029]. So combined, it's (-∞, 0.029]. But that can't be, because when y approaches 0 from the right, f(y) approaches 0. But when y increases from 0 to 3 -2√2, f(y) increases to 17 -12√2. Then, when y increases further towards 3, f(y) decreases to -∞. Therefore, the function covers all values from -∞ up to 17 -12√2, and also from 0 to 17 -12√2. Therefore, combining these, since as y moves from 0 to 3, f(y) goes 0 → 17 -12√2 → -∞. Therefore, the entire range for y ∈ (0,3) is (-∞, 17 -12√2] ∪ {0}? Wait, but that doesn't include the interval from 0 to 17 -12√2. Wait, this is getting confusing. Let me plot f(y) mentally.When y is very close to 0, f(y) is approximately (0 - 0)/( -3) = 0. Then, as y increases a little, say y = 0.1, f(y) = (0.03 - 0.1)/(0.1 -3) = (-0.07)/(-2.9) ≈0.0241. Then, at y =3 -2√2 ≈0.172, f(y) ≈0.029. Then, as y approaches 3 from below, f(y) tends to -∞. Wait, but how? If y approaches 3 from below, denominator (y -3) approaches 0 from the negative side, and numerator approaches 24. So 24 divided by a very small negative number is a large negative number. Therefore, as y approaches 3 from the left, f(y) approaches -∞.Wait, so from y=0 to y≈0.172, f(y) increases from 0 to ≈0.029. Then, from y≈0.172 to y=3, f(y) decreases from ≈0.029 to -∞. Therefore, in the interval y ∈ (0,3), f(y) takes all values from -∞ up to 17 -12√2 (≈0.029). But when y is moving from 0 to 0.172, it takes values from 0 to 0.029, and when moving from 0.172 to 3, it takes values from 0.029 down to -∞. Therefore, the entire range for y ∈ (0,3) is (-∞, 17 -12√2]. Because once it passes the maximum, it goes down to -∞, covering all values from -∞ up to 17 -12√2. But the part from 0 to 17 -12√2 is only covered once, but since it starts at 0, goes up to 17 -12√2, then down to -∞, the range is actually (-∞, 17 -12√2] union [0, 17 -12√2]. But since 17 -12√2 is positive, and the function reaches 17 -12√2 and then goes to -infty, so all values less than or equal to 17 -12√2 are covered. Wait, but how?Wait, when y increases from 0 to 0.172, f(y) increases from 0 to 0.029. Then, when y increases beyond 0.172 to 3, f(y) decreases from 0.029 to -infty. So, the maximum value is 0.029, and the function can take any value from -infty up to 0.029. So, the range on (0,3) is (-infty, 17 -12√2].But in that case, how does 0 fit in? When y approaches 0, f(y) approaches 0, but once y is greater than 0, f(y) starts increasing. Wait, but as y approaches 0 from the right, f(y) approaches 0. So 0 is a limit but not attained. Therefore, the range on (0,3) is (-infty, 17 -12√2]. Because even though the function starts near 0, it's never actually 0 in the interval (0,3), since y=0 is not included. Wait, but wait, the original domain is 0 < x < π/2, so y = cot²x is in (0, ∞). When x approaches 0, cotx approaches infinity, so y approaches infinity? Wait, no. Wait, when x approaches 0, cotx = cosx/sinx. As x approaches 0, sinx approaches 0, so cotx approaches infinity, so y = cot²x approaches infinity. Wait, hold on. Wait, there's a confusion here.Wait, original substitution: y = cot²x. When x approaches 0, cotx approaches infinity, so y approaches infinity. When x approaches π/2, cotx approaches 0, so y approaches 0. Therefore, as x increases from 0 to π/2, y decreases from infinity to 0. Therefore, y ∈ (0, ∞). But in the problem statement, x ∈ (0, π/2) except x = π/6 (where y = 3). So y ∈ (0,3) ∪ (3, ∞). Wait, but when x approaches 0, y approaches infinity. When x approaches π/2, y approaches 0. So y ∈ (0, ∞). But the point y = 3 corresponds to x = π/6, where f(x) is undefined. Therefore, y ∈ (0,3) ∪ (3, ∞). However, as x moves from 0 to π/6, y moves from infinity down to 3. Then, as x moves from π/6 to π/2, y moves from 3 down to 0.Therefore, y ∈ (0,3) corresponds to x ∈ (π/6, π/2), and y ∈ (3, ∞) corresponds to x ∈ (0, π/6). Therefore, in terms of y, when analyzing y ∈ (0,3), x is in (π/6, π/2), and y ∈ (3, ∞) corresponds to x ∈ (0, π/6).Therefore, when we look at f(y) for y ∈ (3, ∞), that corresponds to x ∈ (0, π/6). Let's analyze that.Case 2: y ∈ (3, ∞)In this interval, let's compute limits and behavior.First, as y approaches 3 from above:lim_{y→3^+} f(y) = (3*9 -3)/(3 -3) = 24/0⁺ = +∞As y approaches ∞:lim_{y→∞} f(y) = (3y² - y)/(y -3) ≈ 3y²/y = 3y → ∞So as y increases from 3 to ∞, f(y) starts at +∞ and goes to +∞. But there must be a minimum in between. From the derivative, we found critical points at y =3 +2√2 ≈5.828. So let's compute f(y) at y =3 +2√2.Compute f(3 +2√2):Again, f(y) = (3y² - y)/(y -3). Let's compute numerator and denominator.First, y =3 +2√2Compute y -3 =2√2Compute 3y² - y:First compute y² = (3 +2√2)^2 =9 +12√2 +8=17 +12√2So 3y² =3*(17 +12√2)=51 +36√2Subtract y: 51 +36√2 - (3 +2√2)=51 +36√2 -3 -2√2=48 +34√2Therefore, f(y)=(48 +34√2)/(2√2)=[48 +34√2]/(2√2)Factor numerator: 2*(24 +17√2)Thus, f(y)=2*(24 +17√2)/(2√2)=(24 +17√2)/√2Multiply numerator and denominator by √2:=(24√2 +17*2)/2=(24√2 +34)/2=17 +12√2So f(y) at y=3 +2√2 is 17 +12√2 ≈17 +16.97≈33.97Therefore, in the interval y ∈ (3, ∞), f(y) has a critical point at y=3 +2√2 where it attains a minimum of 17 +12√2. Then, as y increases beyond that, f(y) tends to infinity.Therefore, for y ∈ (3, ∞), the function f(y) decreases from +∞ to 17 +12√2 as y increases from 3 to 3 +2√2, then increases back to +∞ as y goes from 3 +2√2 to ∞. Therefore, the range on y ∈ (3, ∞) is [17 +12√2, ∞)Therefore, combining both cases:Case 1 (y ∈ (0,3)): f(y) ∈ (-∞, 17 -12√2]Case 2 (y ∈ (3, ∞)): f(y) ∈ [17 +12√2, ∞)But wait, in the original problem, it says the function takes all real numbers except those in the open interval (a, b). So the combined range is (-∞, 17 -12√2] ∪ [17 +12√2, ∞). Therefore, the missing interval is (17 -12√2, 17 +12√2). But the problem states "except those in the open interval (a, b)", so a =17 -12√2 and b=17 +12√2. Therefore, a + b = (17 -12√2) + (17 +12√2) =34.But wait, the answer is 34? Let me check the calculations again.Wait, when we combined the two ranges: (-∞, 17 -12√2] and [17 +12√2, ∞). Therefore, the function does not take any values in (17 -12√2, 17 +12√2). But since it's an open interval, (a,b) where a=17 -12√2 and b=17 +12√2. Therefore, a + b=34.Yes, this seems correct. Let me confirm once again:- For y ∈ (0,3), f(y) ranges from -∞ up to 17 -12√2.- For y ∈ (3, ∞), f(y) ranges from 17 +12√2 up to ∞.Therefore, the function skips the interval between 17 -12√2 and 17 +12√2. Therefore, the open interval (a,b) is (17 -12√2,17 +12√2), so a +b=34.Therefore, the answer is 34.Final Answerboxed{34}</think>To solve the problem, we need to analyze the function ( f(x) = frac{cot^3 x}{cot 3x} ) for ( 0 < x < frac{pi}{2} ). We start by expressing (cot 3x) using trigonometric identities:[cot 3x = frac{cot^3 x - 3cot x}{3cot^2 x - 1}]Substituting this into ( f(x) ), we get:[f(x) = frac{cot^3 x}{frac{cot^3 x - 3cot x}{3cot^2 x - 1}} = cot^2 x cdot frac{3cot^2 x - 1}{cot^2 x - 3}]Let ( y = cot^2 x ). Since ( 0 < x < frac{pi}{2} ), ( y ) ranges over ( (0, infty) ) except ( y = 3 ) (where ( x = frac{pi}{6} ) is excluded). Thus, ( f(x) ) becomes:[f(y) = frac{3y^2 - y}{y - 3}]We analyze the behavior of ( f(y) ) on the intervals ( (0, 3) ) and ( (3, infty) ).1. For ( y in (0, 3) ): - As ( y to 3^- ), ( f(y) to -infty ). - As ( y to 0^+ ), ( f(y) to 0 ). - The critical point ( y = 3 - 2sqrt{2} ) gives the maximum value ( f(3 - 2sqrt{2}) = 17 - 12sqrt{2} ).2. For ( y in (3, infty) ): - As ( y to 3^+ ), ( f(y) to +infty ). - As ( y to infty ), ( f(y) to infty ). - The critical point ( y = 3 + 2sqrt{2} ) gives the minimum value ( f(3 + 2sqrt{2}) = 17 + 12sqrt{2} ).Combining these results, the function ( f(y) ) takes all real values except those in the open interval ( (17 - 12sqrt{2}, 17 + 12sqrt{2}) ). Therefore, the values ( a ) and ( b ) are ( 17 - 12sqrt{2} ) and ( 17 + 12sqrt{2} ), respectively. Summing these values:[a + b = (17 - 12sqrt{2}) + (17 + 12sqrt{2}) = 34]Thus, the final answer is:[boxed{34}]]

🔑:19. Notice that operatorname{ctg} 3 x=frac{operatorname{ctg}^{3} x-3 operatorname{ctg} x}{3 operatorname{ctg}^{2} x-1}. Let y=frac{operatorname{ctg}^{3} x}{operatorname{ctg} 3 x}, t=operatorname{ctg}^{2} x, then3 t^{2}-(y+1)t+3 y=0 text {. }Thus this equation has real roots, henceDelta=(y+1)^{2}-36 y=y^{2}-34 y+1must be non-negative. And the roots of y^{2}-34 y+1=0 area=17-12 sqrt{2} text { and } b=17+12 sqrt{2} text {. }Therefore, Delta geqslant 0 if and only if y leqslant a or y geqslant b. Every y value outside of (a, b) is attainable, and thus, we havea+b=34 text {. }

🧠:[<think>Okay, so I need to find the value of the expression (frac{sin x}{2 - 3 cos x}) given that (cot frac{x}{2} = frac{1}{2}). Hmm, cotangent of half-angle... I remember there are some trigonometric identities related to half-angles. Let me recall them.First, I know that (cot frac{theta}{2}) can be expressed in terms of sine and cosine. The formula for cotangent of half-angle is (cot frac{theta}{2} = frac{sin theta}{1 - cos theta}) or maybe (frac{1 + cos theta}{sin theta})? Wait, let me check.Yes, actually, the cotangent of a half-angle can be written as (cot frac{theta}{2} = frac{1 + cos theta}{sin theta}). Let me verify that. If I start with the double-angle formulas, for example, (cos theta = 1 - 2 sin^2 frac{theta}{2}) and (sin theta = 2 sin frac{theta}{2} cos frac{theta}{2}). So, if I take (cot frac{theta}{2} = frac{cos frac{theta}{2}}{sin frac{theta}{2}}), then multiplying numerator and denominator by (2 sin frac{theta}{2}), I get (frac{2 cos frac{theta}{2} sin frac{theta}{2}}{2 sin^2 frac{theta}{2}} = frac{sin theta}{1 - cos theta}). Oh, so there are two expressions for (cot frac{theta}{2}): (frac{1 + cos theta}{sin theta}) and (frac{sin theta}{1 - cos theta}). Right, because (1 - cos theta = 2 sin^2 frac{theta}{2}) and (1 + cos theta = 2 cos^2 frac{theta}{2}), so substituting those in, the expressions make sense.So, given that (cot frac{x}{2} = frac{1}{2}), I can use either of these identities. Let me pick one that might help me relate (sin x) and (cos x) so I can compute the expression (frac{sin x}{2 - 3 cos x}).Let me use (cot frac{x}{2} = frac{1 + cos x}{sin x} = frac{1}{2}). So, (frac{1 + cos x}{sin x} = frac{1}{2}). That gives me an equation relating (sin x) and (cos x). Maybe I can solve for one variable in terms of the other.Let's denote (s = sin x) and (c = cos x). Then, the equation becomes (frac{1 + c}{s} = frac{1}{2}), which implies (2(1 + c) = s). So, (s = 2 + 2c). But we also know that (s^2 + c^2 = 1) from the Pythagorean identity. So, substituting (s = 2 + 2c) into that equation, we can solve for (c).Let me do that:(s = 2 + 2c)So,((2 + 2c)^2 + c^2 = 1)Expanding the square:(4 + 8c + 4c^2 + c^2 = 1)Combine like terms:(5c^2 + 8c + 4 = 1)Subtract 1:(5c^2 + 8c + 3 = 0)Now, solving this quadratic equation for (c):Quadratic equation (5c^2 + 8c + 3 = 0).Using the quadratic formula:(c = frac{-8 pm sqrt{64 - 60}}{10})Calculating the discriminant:(sqrt{64 - 60} = sqrt{4} = 2)So,(c = frac{-8 pm 2}{10})Thus, two solutions:1. (c = frac{-8 + 2}{10} = frac{-6}{10} = -frac{3}{5})2. (c = frac{-8 - 2}{10} = frac{-10}{10} = -1)Now, check if these solutions are valid. Since (c = cos x), cosine can be between -1 and 1, so both solutions are valid in that sense. But let's check back with the original equation.First, if (c = -1), then from (s = 2 + 2c = 2 + 2(-1) = 0). But if (sin x = 0), then the original expression (frac{sin x}{2 - 3 cos x}) would be 0, but also (cot frac{x}{2}) would involve division by zero since (sin x = 0). Wait, because if (sin x = 0), then in the equation (cot frac{x}{2} = frac{1 + cos x}{sin x}), we would have division by zero. Therefore, (c = -1) is not a valid solution because it leads to an undefined cotangent. Therefore, only (c = -frac{3}{5}) is acceptable.So, (c = -frac{3}{5}), then (s = 2 + 2c = 2 + 2(-frac{3}{5}) = 2 - frac{6}{5} = frac{10}{5} - frac{6}{5} = frac{4}{5}). Therefore, (sin x = frac{4}{5}) and (cos x = -frac{3}{5}).Now, plug these into the original expression (frac{sin x}{2 - 3 cos x}):Numerator: (sin x = frac{4}{5})Denominator: (2 - 3 cos x = 2 - 3(-frac{3}{5}) = 2 + frac{9}{5} = frac{10}{5} + frac{9}{5} = frac{19}{5})Therefore, the expression becomes (frac{frac{4}{5}}{frac{19}{5}} = frac{4}{19}).So, the value is (frac{4}{19}).Wait, let me double-check my steps.Starting from (cot frac{x}{2} = frac{1}{2}). Used identity (cot frac{x}{2} = frac{1 + cos x}{sin x}). So (frac{1 + cos x}{sin x} = frac{1}{2}) leading to (2(1 + cos x) = sin x). Then substituted into Pythagorean identity. Got quadratic equation, solved for (cos x = -3/5) and -1, rejected -1 because it leads to (sin x = 0) which would make cotangent undefined. Then found (sin x = 4/5) and (cos x = -3/5). Plugged into expression, got 4/19. That seems right.Alternatively, maybe use another identity to verify. Let me see. Another approach.Given (cot frac{x}{2} = frac{1}{2}). Then, (tan frac{x}{2} = 2). There's an identity that relates (sin x) and (tan frac{x}{2}). Recall that (sin x = 2 tan frac{x}{2} / (1 + tan^2 frac{x}{2})). Similarly, (cos x = (1 - tan^2 frac{x}{2}) / (1 + tan^2 frac{x}{2})).Let me try that.Given (tan frac{x}{2} = 2). Then:(sin x = frac{2 tan frac{x}{2}}{1 + tan^2 frac{x}{2}} = frac{2 times 2}{1 + 4} = frac{4}{5})(cos x = frac{1 - tan^2 frac{x}{2}}{1 + tan^2 frac{x}{2}} = frac{1 - 4}{1 + 4} = frac{-3}{5})So same results! Therefore, (sin x = frac{4}{5}) and (cos x = -frac{3}{5}). Then, substituting into the expression:(frac{sin x}{2 - 3 cos x} = frac{frac{4}{5}}{2 - 3(-frac{3}{5})} = frac{frac{4}{5}}{2 + frac{9}{5}} = frac{frac{4}{5}}{frac{19}{5}} = frac{4}{19}).Same answer. So, this confirms that the value is (frac{4}{19}). Therefore, the answer should be (boxed{dfrac{4}{19}}).Final Answerboxed{dfrac{4}{19}}</think>Given (cot frac{x}{2} = frac{1}{2}), we use the identity (cot frac{x}{2} = frac{1 + cos x}{sin x}). This gives us the equation:[frac{1 + cos x}{sin x} = frac{1}{2}]Solving for (sin x) and (cos x), we get:[2(1 + cos x) = sin x]Using the Pythagorean identity (sin^2 x + cos^2 x = 1), we substitute (sin x = 2(1 + cos x)) into the identity:[(2(1 + cos x))^2 + cos^2 x = 1]Expanding and simplifying:[4(1 + 2cos x + cos^2 x) + cos^2 x = 1][4 + 8cos x + 4cos^2 x + cos^2 x = 1][5cos^2 x + 8cos x + 3 = 0]Solving this quadratic equation using the quadratic formula:[cos x = frac{-8 pm sqrt{64 - 60}}{10} = frac{-8 pm 2}{10}]This gives two solutions:1. (cos x = -frac{3}{5})2. (cos x = -1)Rejecting (cos x = -1) because it leads to (sin x = 0) (which is invalid for cotangent), we use (cos x = -frac{3}{5}). Substituting back, we find:[sin x = 2(1 + (-frac{3}{5})) = 2 left(frac{2}{5}right) = frac{4}{5}]Finally, substituting (sin x = frac{4}{5}) and (cos x = -frac{3}{5}) into the expression:[frac{sin x}{2 - 3 cos x} = frac{frac{4}{5}}{2 - 3(-frac{3}{5})} = frac{frac{4}{5}}{2 + frac{9}{5}} = frac{frac{4}{5}}{frac{19}{5}} = frac{4}{19}]Thus, the value of the expression is (boxed{dfrac{4}{19}}).,<think>Okay, so I need to find the value of the expression sin x divided by (2 minus 3 cos x) when the cotangent of x over 2 is 1/2. Hmm, let me start by recalling some trigonometric identities that might help here.First, since we know cot(x/2) = 1/2, maybe I can find sin x and cos x in terms of cot(x/2). I remember there are formulas for sin x and cos x in terms of cot(x/2). Let me try to recall those.I think one of the half-angle identities relates cotangent to sine and cosine. The formula for cotangent of half-angle is cot(x/2) = (1 + cos x)/sin x. Wait, is that right? Let me check.Yes, the identity is cot(θ) = (1 + cos 2θ)/sin 2θ, so if θ is x/2, then 2θ is x. So substituting, cot(x/2) = (1 + cos x)/sin x. So given that cot(x/2) = 1/2, we can set up the equation:(1 + cos x)/sin x = 1/2.That gives us 2(1 + cos x) = sin x. Hmm, so maybe we can use this equation to express either sin x or cos x in terms of the other and then substitute into the original expression. Let me try that.Let me write the equation again: 2(1 + cos x) = sin x. Let me call this Equation (1).Now, we need to find sin x / (2 - 3 cos x). Let me denote the denominator as D = 2 - 3 cos x.If I can express sin x in terms of cos x using Equation (1), then substitute into the expression. Let's see.From Equation (1): sin x = 2(1 + cos x). So, sin x = 2 + 2 cos x.Wait, but sin x can't be more than 1 in absolute value. Let me check if this is possible. Wait, if sin x = 2 + 2 cos x, then depending on the value of cos x, this might be possible. Let me see. For example, if cos x is negative, maybe 2 + 2 cos x could be within [-1, 1]. Let's check:If cos x = -0.5, then sin x = 2 + 2*(-0.5) = 2 - 1 = 1. So sin x = 1, which is okay. If cos x = -1, sin x = 2 + 2*(-1) = 0. But sin x = 0 when cos x = -1? Wait, that's not possible. If cos x = -1, then x is π, and sin x = 0. So that is possible. Wait, but 2 + 2 cos x must be between -1 and 1. Let's solve for cos x.So 2 + 2 cos x must be between -1 and 1. So:-1 ≤ 2 + 2 cos x ≤ 1Subtract 2:-3 ≤ 2 cos x ≤ -1Divide by 2:-1.5 ≤ cos x ≤ -0.5But the range of cos x is [-1, 1], so cos x must be between -1 and -0.5. So that's possible.So, in Equation (1), we have sin x expressed in terms of cos x. So sin x = 2(1 + cos x). So now, the original expression is sin x divided by (2 - 3 cos x). Let's substitute sin x here:Expression = [2(1 + cos x)] / [2 - 3 cos x]So now we need to find the value of this expression. But we still have cos x in both numerator and denominator. So we need to find the value of cos x first. Let's see if we can find cos x from the given information.We have from Equation (1): sin x = 2 + 2 cos x.But we also know that sin^2 x + cos^2 x = 1. So substitute sin x:(2 + 2 cos x)^2 + cos^2 x = 1Let me expand that:(4 + 8 cos x + 4 cos^2 x) + cos^2 x = 1Combine like terms:4 + 8 cos x + 4 cos^2 x + cos^2 x = 1So 5 cos^2 x + 8 cos x + 4 = 1Subtract 1:5 cos^2 x + 8 cos x + 3 = 0Now we have a quadratic equation in terms of cos x. Let's solve for cos x.Let me denote y = cos x. Then the equation becomes:5y^2 + 8y + 3 = 0Use quadratic formula:y = [-8 ± sqrt(64 - 4*5*3)] / (2*5)Calculate discriminant:sqrt(64 - 60) = sqrt(4) = 2So solutions:y = [-8 ± 2]/10First solution:[-8 + 2]/10 = (-6)/10 = -3/5Second solution:[-8 - 2]/10 = (-10)/10 = -1So cos x = -3/5 or cos x = -1But earlier, we saw that cos x must be between -1 and -0.5 (since from sin x = 2 + 2 cos x, which requires that sin x is between -1 and 1, leading to cos x between -1.5 and -0.5, but cos x can't be less than -1). Therefore, cos x is between -1 and -0.5.So cos x = -3/5 is -0.6, which is between -1 and -0.5. And cos x = -1 is at the boundary. So both solutions are valid? Wait, but if cos x = -1, then sin x would be 2 + 2*(-1) = 0. Which is okay, sin x = 0 when x is π. Let's check if x = π satisfies cot(x/2) = 1/2.If x = π, then x/2 = π/2, and cot(π/2) = 0. But we were given that cot(x/2) = 1/2. So x cannot be π. Therefore, cos x = -1 is extraneous. Therefore, the valid solution is cos x = -3/5.Therefore, cos x = -3/5. Then sin x = 2 + 2 cos x = 2 + 2*(-3/5) = 2 - 6/5 = 4/5.Wait, but sin x should be positive? Let me check. If cos x is -3/5, then x is in the second or third quadrant. In the second quadrant, sin x is positive, and in the third quadrant, sin x is negative. But according to sin x = 2 + 2 cos x, if cos x = -3/5, then sin x = 2 + 2*(-3/5) = 2 - 6/5 = 4/5, which is positive. Therefore, x must be in the second quadrant.So, sin x = 4/5 and cos x = -3/5.Therefore, substituting into the expression:sin x / (2 - 3 cos x) = (4/5) / [2 - 3*(-3/5)] = (4/5) / [2 + 9/5] = (4/5) / [(10/5 + 9/5)] = (4/5) / (19/5) = (4/5) * (5/19) = 4/19.Therefore, the value is 4/19.Wait, let me double-check the calculations to make sure I didn't make an arithmetic mistake.Starting with cos x = -3/5. Then 2 - 3 cos x becomes 2 - 3*(-3/5) = 2 + 9/5. Convert 2 to fifths: 10/5 + 9/5 = 19/5. Then sin x is 4/5. So (4/5) divided by (19/5) is indeed (4/5)*(5/19) = 4/19. Yes, that's correct.Also, when solving the quadratic, the solutions were y = -3/5 and y = -1. And we rejected y = -1 because it leads to x = π, which doesn't satisfy cot(x/2) = 1/2. Therefore, the only valid solution is cos x = -3/5, leading to sin x = 4/5. Then substituting into the expression gives 4/19. So that's the answer.Alternatively, another way to approach this is using the identities for t = tan(x/2). Since we know cot(x/2) = 1/2, then tan(x/2) = 2. Then, using the identities:sin x = 2t/(1 + t^2) = 2*2/(1 + 4) = 4/5cos x = (1 - t^2)/(1 + t^2) = (1 - 4)/5 = -3/5So that's another way to get sin x and cos x. Then substituting into the expression gives the same result.Yes, that seems much quicker. So using the t = tan(x/2) substitution. Since cot(x/2) = 1/2, so tan(x/2) = 2. Then, sin x = 2t/(1 + t^2) = 4/5, cos x = (1 - t^2)/(1 + t^2) = -3/5. Therefore, sin x/(2 - 3 cos x) = (4/5)/(2 - 3*(-3/5)) = (4/5)/(2 + 9/5) = (4/5)/(19/5) = 4/19.Therefore, same answer. So either way, the result is 4/19.I think that's the correct answer.Final AnswerThe value of the expression is boxed{dfrac{4}{19}}.</think>Given the problem, we need to determine the value of the expression (frac{sin x}{2 - 3 cos x}) when (cot frac{x}{2} = frac{1}{2}).First, we use the identity for cotangent of a half-angle:[cot frac{x}{2} = frac{1 + cos x}{sin x}]Given (cot frac{x}{2} = frac{1}{2}), we set up the equation:[frac{1 + cos x}{sin x} = frac{1}{2}]This simplifies to:[2(1 + cos x) = sin x]Next, we use the Pythagorean identity (sin^2 x + cos^2 x = 1). Substituting (sin x = 2(1 + cos x)) into the identity, we get:[[2(1 + cos x)]^2 + cos^2 x = 1]Expanding and simplifying:[4(1 + 2cos x + cos^2 x) + cos^2 x = 1 4 + 8cos x + 4cos^2 x + cos^2 x = 1 5cos^2 x + 8cos x + 3 = 0]Solving this quadratic equation using the quadratic formula (y = cos x):[y = frac{-8 pm sqrt{64 - 60}}{10} = frac{-8 pm 2}{10}]This gives the solutions:[y = frac{-6}{10} = -frac{3}{5} quad text{or} quad y = frac{-10}{10} = -1]Since (cos x = -1) leads to (sin x = 0) which does not satisfy (cot frac{x}{2} = frac{1}{2}), we discard it. Thus, (cos x = -frac{3}{5}).Substituting (cos x = -frac{3}{5}) into (sin x = 2(1 + cos x)):[sin x = 2left(1 - frac{3}{5}right) = 2 left(frac{2}{5}right) = frac{4}{5}]Finally, substituting (sin x = frac{4}{5}) and (cos x = -frac{3}{5}) into the given expression:[frac{sin x}{2 - 3 cos x} = frac{frac{4}{5}}{2 - 3left(-frac{3}{5}right)} = frac{frac{4}{5}}{2 + frac{9}{5}} = frac{frac{4}{5}}{frac{19}{5}} = frac{4}{19}]Thus, the value of the expression is (boxed{dfrac{4}{19}}).]

🔑:## First Solution.begin{aligned}frac{sin x}{2-3 cos x} & =frac{2 sin frac{x}{2} cos frac{x}{2}}{2left(cos ^{2} frac{x}{2}+sin ^{2} frac{x}{2}right)-3left(cos ^{2} frac{x}{2}-sin ^{2} frac{x}{2}right)} & =frac{2 sin frac{x}{2} cos frac{x}{2}}{-cos ^{2} frac{x}{2}+5 sin ^{2} frac{x}{2}}end{aligned}If we divide the numerator and the denominator in the last equality by sin ^{2} frac{x}{2}, we getfrac{sin x}{2-3 cos x}=frac{2 operatorname{ctg} frac{x}{2}}{-operatorname{ctg}^{2} frac{x}{2}+5}=frac{1}{-frac{1}{4}+5}=frac{4}{19}

🧠:[<think>Okay, let's see. I need to find the sum of the sequence a₁ + a₂ + ... + a₉₉ where each term aₙ is given by 1 divided by [(n+1)√n + n√(n+1)]. Hmm, that denominator looks a bit complicated. Maybe I can simplify it somehow?First, let me write down the general term again to make sure I have it right:aₙ = 1 / [ (n + 1)√n + n√(n + 1) ]I wonder if there's a way to rationalize the denominator here. When there's a sum of square roots in the denominator, rationalizing might help. The standard method is to multiply numerator and denominator by the conjugate. Let's try that.The denominator is (n + 1)√n + n√(n + 1). The conjugate would be (n + 1)√n - n√(n + 1). So, multiply numerator and denominator by that:aₙ = [1 * ((n + 1)√n - n√(n + 1))] / [ ((n + 1)√n + n√(n + 1)) * ((n + 1)√n - n√(n + 1)) ]Let me compute the denominator first. It's a difference of squares:[( (n + 1)√n )² - (n√(n + 1))² ]Calculating each term:First term: (n + 1)² * (√n)² = (n² + 2n + 1) * n = n³ + 2n² + nSecond term: n² * (√(n + 1))² = n² * (n + 1) = n³ + n²So the denominator becomes:(n³ + 2n² + n) - (n³ + n²) = (n³ - n³) + (2n² - n²) + n = n² + nSo, denominator simplifies to n(n + 1).Now, the numerator after multiplying by the conjugate is (n + 1)√n - n√(n + 1). Therefore, the term aₙ becomes:[(n + 1)√n - n√(n + 1)] / [n(n + 1)]Let me split this fraction into two separate terms:= [(n + 1)√n / (n(n + 1))] - [n√(n + 1) / (n(n + 1))]Simplify each part:First term: (n + 1) cancels out in numerator and denominator, so √n / nSecond term: n cancels out in numerator and denominator, so √(n + 1) / (n + 1)Thus, aₙ simplifies to:(√n / n) - (√(n + 1) / (n + 1))Wait a minute, that looks like it could telescope when we sum over n. Let's check:√n / n is equal to 1 / √n, since √n divided by n is 1 / n^(1 - 1/2) = 1 / n^(1/2) = 1 / √n. Similarly, √(n + 1) / (n + 1) is 1 / √(n + 1). So, actually, aₙ = 1/√n - 1/√(n + 1). So that's even better.Wait, let me confirm:√n / n = n^(1/2) / n = n^(-1/2) = 1 / √n. Yes.Similarly, √(n + 1) / (n + 1) = (n + 1)^(-1/2) = 1 / √(n + 1). So indeed, aₙ = 1/√n - 1/√(n + 1). That's a telescoping series!Therefore, when we sum a₁ + a₂ + ... + a₉₉, each term cancels out the next one. Let me write out the first few terms to see:For n=1: 1/√1 - 1/√2For n=2: 1/√2 - 1/√3For n=3: 1/√3 - 1/√4...For n=99: 1/√99 - 1/√100So when we add them all up, the -1/√2 from the first term cancels with the +1/√2 from the second term. Similarly, the -1/√3 cancels with the +1/√3 from the third term, and so on, up until the last term, which is -1/√100. All intermediate terms cancel, leaving us with the first term of the first expression and the last term of the last expression.Therefore, the total sum is:1/√1 - 1/√100Simplify that:1/1 - 1/10 = 1 - 0.1 = 0.9But wait, 0.9 is 9/10. But let me write it as fractions to be precise.1 is 1/1, and 1/√100 is 1/10, so 1 - 1/10 = 9/10. So the sum is 9/10.Therefore, the answer should be 9/10, which is 0.9. But since the problem probably expects a fraction, boxed{dfrac{9}{10}}.Wait, but let me double-check. Did I make any mistake in the algebra?Original term: aₙ = 1 / [ (n + 1)√n + n√(n + 1) ]After rationalizing, we got to [ (n + 1)√n - n√(n + 1) ] / [n(n + 1)]Then splitting into two terms:(n + 1)√n / [n(n + 1)] - n√(n + 1) / [n(n + 1)]Simplifies to √n / n - √(n + 1)/(n + 1) = 1/√n - 1/√(n + 1)Yes, that seems correct.Sum from n=1 to 99: (1/1 - 1/√2) + (1/√2 - 1/√3) + ... + (1/√99 - 1/√100)All the middle terms cancel, leaving 1 - 1/√100. Since √100 is 10, that's 1 - 1/10 = 9/10. So yes, 9/10 is correct.Therefore, the final answer is 9/10.Final Answerboxed{dfrac{9}{10}}</think>Given the sequence ( a_n = frac{1}{(n+1)sqrt{n} + nsqrt{n+1}} ), we need to find the sum ( a_1 + a_2 + cdots + a_{99} ).First, we rationalize the denominator of ( a_n ). Multiply the numerator and denominator by the conjugate of the denominator:[a_n = frac{1}{(n+1)sqrt{n} + nsqrt{n+1}} cdot frac{(n+1)sqrt{n} - nsqrt{n+1}}{(n+1)sqrt{n} - nsqrt{n+1}}]The denominator becomes a difference of squares:[[(n+1)sqrt{n}]^2 - [nsqrt{n+1}]^2 = (n+1)^2 n - n^2 (n+1) = n(n+1)^2 - n^2(n+1) = n(n+1)(n+1 - n) = n(n+1)]The numerator simplifies to:[(n+1)sqrt{n} - nsqrt{n+1}]Thus, the term ( a_n ) becomes:[a_n = frac{(n+1)sqrt{n} - nsqrt{n+1}}{n(n+1)} = frac{sqrt{n}}{n} - frac{sqrt{n+1}}{n+1} = frac{1}{sqrt{n}} - frac{1}{sqrt{n+1}}]This is a telescoping series. When summing from ( n = 1 ) to ( n = 99 ), most terms cancel out:[sum_{n=1}^{99} left( frac{1}{sqrt{n}} - frac{1}{sqrt{n+1}} right) = left( frac{1}{sqrt{1}} - frac{1}{sqrt{2}} right) + left( frac{1}{sqrt{2}} - frac{1}{sqrt{3}} right) + cdots + left( frac{1}{sqrt{99}} - frac{1}{sqrt{100}} right)]All intermediate terms cancel, leaving:[frac{1}{sqrt{1}} - frac{1}{sqrt{100}} = 1 - frac{1}{10} = frac{9}{10}]Thus, the final answer is:[boxed{dfrac{9}{10}}],<think>Okay, so I need to find the sum of the series a₁ + a₂ + ... + a₉₉ where each term aₙ is given by 1 divided by [(n+1)√n + n√(n+1)]. Hmm, that looks a bit complicated at first glance. Let me see if I can simplify the expression for aₙ somehow. Maybe rationalizing the denominator or finding a telescoping pattern?First, let me write down the general term again to get a clear view:aₙ = 1 / [(n+1)√n + n√(n+1)]The denominator has two terms: (n+1)√n and n√(n+1). Both terms involve square roots and coefficients that are linear in n. Maybe if I factor something out? Let's see.Alternatively, rationalizing the denominator might help. Since the denominator is of the form A + B, where A and B are terms with square roots, multiplying numerator and denominator by (A - B) would rationalize it. Let me try that.So, let me set A = (n+1)√n and B = n√(n+1). Then, multiplying numerator and denominator by (A - B):aₙ = [1 * (A - B)] / [(A + B)(A - B)] = (A - B) / (A² - B²)Let me compute A² and B².First, A² = [(n+1)√n]² = (n+1)² * n = n(n+1)²Similarly, B² = [n√(n+1)]² = n²(n+1)So, A² - B² = n(n+1)² - n²(n+1) = n(n+1)[(n+1) - n] = n(n+1)(1) = n(n+1)Therefore, the denominator simplifies to n(n+1).So, now aₙ becomes (A - B) / [n(n+1)] = [(n+1)√n - n√(n+1)] / [n(n+1)]Let me split this into two fractions:= [(n+1)√n / (n(n+1))] - [n√(n+1) / (n(n+1))]Simplifying each term:First term: [(n+1)√n / (n(n+1))] = √n / n = 1/√nSecond term: [n√(n+1) / (n(n+1))] = √(n+1)/(n+1) = 1/√(n+1)Therefore, aₙ simplifies to 1/√n - 1/√(n+1)Oh! That's a telescoping series! That makes the sum much easier. So the sum from n=1 to 99 of aₙ is the sum of [1/√n - 1/√(n+1)] from n=1 to 99.When we expand this sum, most terms will cancel out:(1/√1 - 1/√2) + (1/√2 - 1/√3) + (1/√3 - 1/√4) + ... + (1/√99 - 1/√100)Each -1/√k cancels with the next term's +1/√k, leaving only the first term of the first expression and the last term of the last expression. Therefore, the total sum is 1/√1 - 1/√100.Since √1 is 1 and √100 is 10, this simplifies to 1 - 1/10 = 9/10.So, the answer should be 9/10. Let me double-check to make sure I didn't make a mistake in simplifying aₙ.Starting again, aₙ = 1 / [(n+1)√n + n√(n+1)]Multiply numerator and denominator by (n+1)√n - n√(n+1):Denominator becomes [(n+1)√n]^2 - [n√(n+1)]^2 = (n+1)^2 * n - n^2 * (n+1) = n(n+1)[n+1 - n] = n(n+1)(1) = n(n+1)Numerator becomes (n+1)√n - n√(n+1)So, aₙ = [(n+1)√n - n√(n+1)] / [n(n+1)] = [(n+1)√n / (n(n+1))] - [n√(n+1) / (n(n+1))] = (√n / n) - (√(n+1) / (n+1)) = 1/√n - 1/√(n+1)Yes, that's correct. Therefore, the telescoping sum indeed works out, and the sum from n=1 to 99 is 1 - 1/10 = 9/10. So, the answer is 9/10.Final Answerboxed{dfrac{9}{10}}</think>To find the sum (a_1 + a_2 + cdots + a_{99}) where (a_n = frac{1}{(n+1)sqrt{n} + nsqrt{n+1}}), we start by simplifying the general term (a_n).First, we rationalize the denominator by multiplying the numerator and denominator by ((n+1)sqrt{n} - nsqrt{n+1}):[a_n = frac{1}{(n+1)sqrt{n} + nsqrt{n+1}} cdot frac{(n+1)sqrt{n} - nsqrt{n+1}}{(n+1)sqrt{n} - nsqrt{n+1}}]The denominator becomes:[[(n+1)sqrt{n}]^2 - [nsqrt{n+1}]^2 = n(n+1)^2 - n^2(n+1) = n(n+1)((n+1) - n) = n(n+1)]The numerator becomes:[(n+1)sqrt{n} - nsqrt{n+1}]Thus, the simplified form of (a_n) is:[a_n = frac{(n+1)sqrt{n} - nsqrt{n+1}}{n(n+1)} = frac{sqrt{n}}{n} - frac{sqrt{n+1}}{n+1} = frac{1}{sqrt{n}} - frac{1}{sqrt{n+1}}]This shows that (a_n) is a telescoping series. When we sum the terms from (n=1) to (n=99):[sum_{n=1}^{99} a_n = sum_{n=1}^{99} left( frac{1}{sqrt{n}} - frac{1}{sqrt{n+1}} right)]Expanding the series, we see that most terms cancel out:[left( frac{1}{sqrt{1}} - frac{1}{sqrt{2}} right) + left( frac{1}{sqrt{2}} - frac{1}{sqrt{3}} right) + cdots + left( frac{1}{sqrt{99}} - frac{1}{sqrt{100}} right)]This leaves only the first term of the first expression and the last term of the last expression:[frac{1}{sqrt{1}} - frac{1}{sqrt{100}} = 1 - frac{1}{10} = frac{9}{10}]Thus, the sum (a_1 + a_2 + cdots + a_{99}) is:[boxed{dfrac{9}{10}}]]

🔑:[Solution] Sincebegin{aligned}a_{n} & =frac{1}{(n+1) sqrt{n}+n sqrt{n+1}}=frac{(n+1) sqrt{n}-n sqrt{n+1}}{(n+1)^{2} n-n^{2}(n+1)} & =frac{(n+1) sqrt{n}-n sqrt{n+1}}{n(n+1)}=frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}}end{aligned}Therefore, (a_{1}+a_{2}+cdots+a_{99}=sum_{n=1}^{99}left(frac{1}{sqrt{n}}-frac{1}{sqrt{n+1}}right)=1-frac{1}{sqrt{100}}=frac{9}{10}).

🧠:[<think>Okay, so I need to figure out how many ways the smaller cubes can be painted such that when they're assembled into a 2cm cube, the entire surface is white. The original cube is made up of eight 1cm cubes, right? Each edge is 2cm, so it's a 2x2x2 cube. First, let me visualize this. A 2x2x2 cube has 8 smaller cubes, each at the corners. Each small cube has three faces exposed to the outside of the larger cube. Wait, no, in a 2x2x2 cube, every small cube is at a corner, and each corner cube has three faces exposed. That's correct. So each of the eight small cubes has three faces that are part of the surface of the larger cube. Now, the problem says that eight faces of the smaller cubes are painted black, and the rest are white. Then, from these painted smaller cubes, we can assemble a cube whose entire surface is white. So when we put them together, all the outer faces of the large cube must be white. But each small cube has three faces, and when they're put into the larger cube, some of their faces are on the outside, some are internal. The internal faces don't matter because they're not visible. So the key is that all the faces of the larger cube must be white, which means that for each small cube, the three faces that are on the outside must not be painted black. However, the internal faces can be either black or white, but since we are only painting eight faces in total, those eight black faces must be distributed among the internal faces or the non-visible faces of the small cubes. Wait, no. Wait, actually, the total number of black faces across all small cubes is eight. Each small cube can have multiple black faces, but in total, there are eight black faces. But when we assemble the cube, all the outer faces must be white. Therefore, any black face on a small cube must be on an internal face (i.e., a face that is glued to another small cube) or on a face that's not part of the outer surface. But since in the 2x2x2 cube, each small cube is in a corner, each small cube has three faces exposed. Therefore, each small cube has three faces that are part of the outer surface, and three faces that are internal (but in reality, in a 2x2x2 cube, each small cube is adjacent to three other small cubes, so each of its three internal faces is adjacent to another small cube). Therefore, if any of those internal faces are painted black, it's okay because they won't be visible once the cube is assembled. However, the outer faces of the small cubes (the ones that become the outer surface of the large cube) must not be painted black. But the problem says that eight faces of the smaller cubes are painted black. So in total, across all eight small cubes, there are eight black faces. However, when we assemble the cube, all the outer faces (which are 6 faces of the large cube, each consisting of 4 small faces, so 24 small faces) must be white. Therefore, none of the 24 outer small faces can be black. Thus, the eight black faces must be among the internal faces. But each small cube has three internal faces. Since there are eight small cubes, each with three internal faces, that's 24 internal faces. So the total number of internal faces is 24, and the total number of outer faces is 24. But each small cube has three outer faces and three internal faces. Wait, no. Wait, each face of the large cube is 2x2=4 small faces. So the large cube has 6 faces, each with 4 small faces, so 24 small faces. Therefore, each small cube contributes three faces to the outer surface. Therefore, each small cube has three outer faces and three inner faces. Thus, the total number of inner faces across all small cubes is 8 small cubes * 3 inner faces each = 24 inner faces. Similarly, the total number of outer faces is 24. But we have to paint eight of the small cube's faces black. But all the outer faces must be white. Therefore, the eight black faces must be among the 24 inner faces. So the problem reduces to choosing eight inner faces (from the 24 total inner faces) to paint black, such that when the cube is assembled, the arrangement of black faces doesn't affect the outer whiteness. However, the cubes are indistinct except for their painting. So the problem is asking for the number of ways to paint eight inner faces black, considering that the small cubes can be interchanged if they are identically painted. Wait, but the problem says: "Assume that identically painted smaller cubes cannot be distinguished and can therefore be interchanged." So if two small cubes have the same pattern of painted faces, they are considered identical and can be swapped, so the overall arrangement is considered the same. Therefore, the problem is equivalent to counting the number of colorings up to the symmetry of the cube. That is, considering that two colorings are the same if one can be transformed into the other by permuting the small cubes (since they are indistinct when identically painted). But how does that affect the count? Let me think. We need to count the number of distinct colorings under the action of the cube's automorphism group. Because the small cubes can be interchanged, any two colorings that are related by a symmetry of the cube are considered the same. Therefore, we need to use Burnside's lemma or similar combinatorial methods to account for symmetries. But maybe there's a different approach here. Let me try to parse the problem again. The problem states: "In how many ways can the smaller cubes be painted? Assume that identically painted smaller cubes cannot be distinguished and can therefore be interchanged." So we need to count colorings of the eight small cubes with some number of black faces (total eight black faces across all small cubes) such that the outer surface of the assembled cube is entirely white. Moreover, colorings are considered the same if they can be transformed into each other by permuting identically painted small cubes. Wait, perhaps the key is that when we assemble the cube, the positions of the small cubes don't matter as long as the overall cube's surface is white. So essentially, the problem is to color eight small cubes (each with three inner faces) with black, such that exactly eight inner faces are black in total, and considering that two colorings are the same if you can permute the small cubes (since they are indistinct when painted identically). But maybe even more precise: since the small cubes can be interchanged if they are identically painted, the number of distinct colorings is equal to the number of colorings where the colorings are considered up to permutation of the small cubes. So the colorings are multisets of painted small cubes. That is, the order doesn't matter, only the counts of each type of painted small cube. Therefore, the problem reduces to finding the number of multisets of eight small cubes, each small cube having some number of black inner faces (each small cube has three inner faces), such that the total number of black inner faces is eight, and when assembling the cube, all outer faces are white (which is already guaranteed because we are only coloring inner faces). Wait, but no. Wait, the outer faces are white because we didn't paint them. The problem says eight faces of the smaller cubes are painted black, and the rest are white. So the outer faces (which are 24 in total) are white, and the eight black faces are among the inner faces. But each small cube has three inner faces, so we need to choose eight black faces among these 24 inner faces. However, each small cube's inner faces are adjacent to other small cubes. But when assembling the cube, we can arrange the small cubes in different positions. However, since the small cubes can be interchanged if they are identically painted, the actual positions don't matter. Therefore, the problem is equivalent to: Count the number of colorings of eight small cubes (each with three inner faces) such that exactly eight inner faces are black in total, considering that two colorings are equivalent if one can be obtained from the other by permuting the small cubes. Therefore, the problem is equivalent to finding the number of equivalence classes under the action of the cube's symmetry group on the set of colorings with eight black inner faces. But Burnside's lemma is the tool here. Let me recall Burnside's lemma: the number of equivalence classes is equal to the average number of fixed points of the group's elements. Therefore, we need to consider all the symmetries of the cube (the automorphism group), which has 24 elements: 6 faces times 4 rotations each, but actually, the cube's symmetry group has 24 elements: 6 rotations fixing a face, 8 rotations fixing a vertex, 6 rotations fixing an edge, and the identity. Wait, actually, the cube's rotation group has 24 elements. If we include reflections, it's 48, but since the problem doesn't mention reflections as distinct, but in standard cube puzzles, usually rotations are considered. But the problem might be considering all symmetries, including reflections. Wait, the problem says "can be assembled", so perhaps we can flip the cube, so reflections are allowed. Hmm. But let's check. The problem states: "a cube can be assembled whose surface is white. In how many ways can the smaller cubes be painted? Assume that identically painted smaller cubes cannot be distinguished and can therefore be interchanged." So "interchanged" probably includes all cube symmetries, both rotations and reflections. Therefore, the group is the full octahedral group of 48 elements. But perhaps the problem is in 3D, so the symmetry group is 48. Hmm. But maybe in the original problem statement, since it's a math competition problem, they might consider the rotation group only. But without explicit mention, it's safer to assume the rotation group. But this is a crucial point. Hmm. Wait, in the problem statement, it's mentioned that the small cubes can be interchanged if they are identically painted. So if there's a symmetry of the cube (rotation or reflection) that maps one small cube to another, then if their colorings are the same, they can be interchanged. Therefore, the group is the full symmetry group including reflections. So 48 elements. But to use Burnside's lemma, we need to calculate for each symmetry of the cube, the number of colorings fixed by that symmetry, then average over all symmetries. However, this seems complicated, but perhaps manageable. Alternatively, since the problem is about painting inner faces, perhaps the symmetries correspond to the symmetries of the cube acting on the small cubes. Each small cube is a corner of the larger cube, so the group acts on the 8 corners. Therefore, the group is the octahedral group acting on the 8 corners. So the number of symmetries is 48. Therefore, we need to compute the number of colorings of the 8 small cubes (each small cube has three inner faces) with eight black inner faces in total, under the action of the octahedral group. Each coloring is a function assigning to each small cube a subset of its three inner faces, such that the total number of black faces is eight. But considering that the small cubes are indistinct except for their coloring, so two colorings are equivalent if there is a symmetry of the cube that maps one coloring to the other. This is classic Burnside's lemma application. But to apply Burnside's lemma, I need to:1. Determine all the conjugacy classes of the octahedral group.2. For each conjugacy class, determine the number of colorings fixed by any element of that conjugacy class.3. Compute the average over all group elements.But this seems complicated. Let me recall that the octahedral group (including reflections) has the following conjugacy classes:- Identity (1 element)- 90-degree rotations about axes through centers of opposite faces (6 elements: 3 axes, each with two directions)- 180-degree rotations about face axes (3 elements)- 120-degree rotations about axes through opposite vertices (8 elements: 4 axes, each two directions)- 180-degree rotations about axes through midpoints of opposite edges (6 elements)- Reflections in planes through opposite edges (6 elements)- Reflections in planes through centers of opposite faces (3 elements)- Rotoreflections: combination of a rotation and reflection (possibly equivalent to other elements)Wait, actually, the full octahedral group has 48 elements, and the conjugacy classes are:1. Identity: 12. 180-degree rotations about face axes: 33. 180-degree rotations about edge axes: 64. 120-degree rotations (both directions) about vertex axes: 85. 90-degree and 270-degree rotations about face axes: 6 each (total 12)6. Reflections: 9 (3 through face centers, 6 through edge planes)Wait, actually, different sources might classify them differently, but perhaps for simplicity, let's refer to the standard conjugacy classes in the octahedral group. Alternatively, since this is getting too complicated, maybe focus on the rotation group first. The rotation group has 24 elements with the following conjugacy classes:- Identity: 1- 180-degree rotations about face axes: 3- 120-degree rotations about vertex axes: 8 (divided into two conjugacy classes of 4 each, but actually, in cube rotation group, rotations by 120° and 240° are in separate classes)Wait, I need to check:The rotation group of the cube (proper rotations) has conjugacy classes:1. Identity: 1 element.2. 90-degree rotations about face axes: 6 elements (3 axes, each with two directions: clockwise and counterclockwise).3. 180-degree rotations about face axes: 3 elements.4. 120-degree rotations about vertex axes: 8 elements (4 axes, each with two directions).Wait, no. Wait, the rotation group of the cube:- Identity: 1- 180-degree rotations about face axes: 3 (each axis through centers of opposite faces, rotated 180°)- 180-degree rotations about edge axes: 6 (each axis through midpoints of opposite edges, rotated 180°)- 120-degree rotations about body diagonals (vertex axes): 8 (each axis through opposite vertices, rotated 120° and 240°, so two rotations per axis, total 8)But wait, that sums up to 1 + 3 + 6 + 8 = 18 elements, which is not 24. So perhaps my breakdown is wrong.Actually, the correct conjugacy classes for the rotation group of the cube (order 24) are:1. Identity: 1 element.2. 180-degree rotations about axes through centers of opposite faces: 3 elements.3. 180-degree rotations about axes through midpoints of opposite edges: 6 elements.4. 120-degree rotations (both 120° and 240°) about axes through opposite vertices: 8 elements.Wait, 1 + 3 + 6 + 8 = 18, still not 24. Hmm.Wait, perhaps 90-degree rotations. Let me check another source.The rotation group of the cube has:- 1 identity,- 6 rotations of 90° around face axes,- 3 rotations of 180° around face axes,- 6 rotations of 180° around edge axes,- 8 rotations of 120° around vertex axes.Total: 1 + 6 + 3 + 6 + 8 = 24. Yes, that's correct.So conjugacy classes:1. Identity (order 1): 1 element.2. 90° rotations about face axes (order 4): 6 elements (3 axes, each with two directions).3. 180° rotations about face axes (order 2): 3 elements.4. 180° rotations about edge axes (order 2): 6 elements.5. 120° rotations about vertex axes (order 3): 8 elements (4 axes, each with two directions: 120° and 240°).Therefore, these are the conjugacy classes. Similarly, the full octahedral group (including reflections) has 48 elements. But since the problem allows interchanging cubes via reflections (as assembling the cube can involve flipping it), we need to consider the full group. However, perhaps the problem is intended to use rotations only. Hmm. But the problem doesn't specify, which is confusing. But given that it's a competition problem, it might be expecting a rotation-only approach. However, let's check both possibilities.But perhaps there is a smarter way to approach this problem without Burnside's lemma.Wait, here's another approach. Since the small cubes are indistinct except for their painting, and we need to assign to each small cube some number of black inner faces (each small cube has three inner faces), such that the total number of black inner faces is eight.But each small cube's inner faces are adjacent to three other small cubes. However, when considering the entire cube, the arrangement must be such that when the cube is assembled, the black faces are internal. But since we are allowed to permute the small cubes, we can arrange them in any order. Therefore, as long as we have a collection of small cubes with some number of black inner faces (each black inner face will be adjacent to another small cube's face), but since we don't care about the adjacency (the problem doesn't state any constraints on the internal black faces except that the total number is eight), the only constraint is that each small cube has three inner faces, and exactly eight of these across all small cubes are black. But since the cubes can be permuted, the problem reduces to counting the number of multisets of eight small cubes, each with 0, 1, 2, or 3 black inner faces, such that the total number of black inner faces is eight. But each small cube must have 0, 1, 2, or 3 black inner faces, and since they are indistinct, the order doesn't matter. Wait, but each small cube is a corner cube with three inner faces. However, the inner faces of each small cube are in specific orientations. For example, in a coordinate system, each corner cube's inner faces are along the negative x, negative y, negative z directions, depending on their position. However, if we can permute the small cubes, then the specific orientations of the black faces don't matter, because we can rotate the entire cube. Therefore, the coloring is up to rotation. But this is getting complicated. Let's think again.Suppose we model each small cube as having three "slots" for black paint on their inner faces. The total number of slots is 8 small cubes * 3 inner faces = 24. We need to choose eight slots to paint black. However, the small cubes are considered indistinct if they have the same number of black inner faces, regardless of the positions of the black faces. Because the cube can be rotated, so the positions of the black inner faces on a small cube don't matter—since we can rotate the small cube within the larger cube. Wait, but can we rotate the small cubes?Wait, actually, the small cubes are fixed in their orientations. When assembling the cube, each small cube is placed in a corner, and their inner faces are adjacent to other small cubes. However, the problem states that the smaller cubes can be interchanged if identically painted. Therefore, the key is that two small cubes are considered the same if they have the same number of black inner faces, regardless of which inner faces are black. Because if you can rotate the small cube, the positions of the black faces can be permuted. Wait, but the inner faces are adjacent to specific other small cubes. For example, if a small cube is in the (0,0,0) corner, its three inner faces are adjacent to the small cubes at (1,0,0), (0,1,0), and (0,0,1). If another small cube in a different corner has a black inner face, that face is adjacent to a different set of small cubes. However, since we can permute the small cubes (i.e., rearrange them into different corners), the adjacency relationships can be changed. Therefore, the specific positions of the black inner faces don't matter, because by permuting the small cubes, we can assign any small cube to any corner, and thus any black inner face can be adjacent to any other small cube. Therefore, the problem reduces to assigning to each small cube a number of black inner faces (0, 1, 2, or 3), such that the total is eight, and considering that small cubes with the same number of black inner faces are indistinct. Therefore, the problem is equivalent to finding the number of multisets of eight elements (the small cubes), where each element can be of a type 0, 1, 2, or 3 (representing the number of black inner faces on the small cube), and the sum of all elements is eight. However, this is not exactly correct, because the number of black inner faces per cube is limited by the number of inner faces. Each small cube has three inner faces, so the number of black inner faces on a single small cube can be 0, 1, 2, or 3. Therefore, the problem is equivalent to solving the equation:x0 + x1 + x2 + x3 = 8with the constraint that each xi is a non-negative integer, where x0 is the number of small cubes with 0 black inner faces, x1 with 1, etc., and the total number of black inner faces is:0*x0 + 1*x1 + 2*x2 + 3*x3 = 8.Wait, that's correct. So the problem is now to find the number of non-negative integer solutions (x0, x1, x2, x3) to the system:x0 + x1 + x2 + x3 = 8and1*x1 + 2*x2 + 3*x3 = 8.But since we are considering multisets, each solution corresponds to a distinct way of painting the cubes, considering that cubes with the same number of black inner faces are indistinct.Therefore, we need to compute the number of non-negative integer solutions to the above system.Let me write the equations again:1. x0 + x1 + x2 + x3 = 82. x1 + 2x2 + 3x3 = 8We need to find the number of non-negative integer solutions (x0, x1, x2, x3).To solve this, let's express x0 = 8 - x1 - x2 - x3. Then substitute into the second equation:x1 + 2x2 + 3x3 = 8We need to find the number of non-negative integers x1, x2, x3 such that x1 + 2x2 + 3x3 = 8.This is a classic Diophantine equation. Let's solve for variables x3, x2, x1.Let me consider possible values of x3:x3 can be 0, 1, 2 since 3*3=9 >8.Case 1: x3 = 0Then the equation becomes x1 + 2x2 = 8Possible x2: 0,1,2,3,4For each x2, x1 =8 - 2x2x2=0: x1=8x2=1: x1=6x2=2: x1=4x2=3: x1=2x2=4: x1=0So 5 solutions in this case.Case 2: x3 =1Equation becomes x1 + 2x2 =8 -3=5Possible x2: 0,1,2 (since 2*3=6>5)x2=0: x1=5x2=1: x1=3x2=2: x1=1So 3 solutions.Case3: x3=2Equation becomes x1 + 2x2 =8 -6=2Possible x2:0,1x2=0: x1=2x2=1: x1=0So 2 solutions.Case4: x3=3: 3*3=9>8, so invalid.Total solutions: 5+3+2=10.Therefore, there are 10 solutions. Each solution corresponds to a different way of distributing the black inner faces among the small cubes, considering that small cubes with the same number of black inner faces are indistinct.But wait, but this seems too low. The answer is 10? But let's check:For example, one solution is x3=0, x2=4, x1=0. Wait, no, in Case1, x3=0, x2=4, x1=0, x0=8 -0 -4 -0=4. So four small cubes with 0 black faces, four with 2 black faces. But each small cube can have up to 3 black faces. So is that acceptable? Yes, because 2 is allowed.But the problem states that eight faces of the smaller cubes are painted black. Each small cube can have multiple black faces. So if we have four small cubes each with two black inner faces, that's 4*2=8, which is allowed.But according to our count, there are 10 such solutions. So does that mean the answer is 10?But the problem is about the number of ways to paint the smaller cubes, considering that identically painted cubes can be interchanged. So if we have multiple small cubes with the same number of black inner faces, swapping them doesn't create a new coloring.Therefore, the answer is 10. But is this correct?Wait, but in the actual problem, the small cubes are in specific positions, but since we can interchange them, the positions don't matter. Therefore, the colorings are considered up to permutation. Therefore, the number of colorings is the number of distinct multisets, which is the number of solutions to the equations above, which is 10. So is the answer boxed{10}? But let me think again. For each solution (x0, x1, x2, x3), we have a certain number of small cubes with 0,1,2,3 black inner faces. But each small cube is a corner cube with three inner faces. However, each inner face is adjacent to another small cube. If we have a small cube with, say, two black inner faces, then those two black inner faces are adjacent to two different small cubes. But when we assemble the cube, those adjacent small cubes might have their own black inner faces. But since we can permute the small cubes, maybe the adjacency doesn't matter because we can rearrange the small cubes to avoid conflicts. Wait, but the problem doesn't specify any constraints on the adjacency of black faces. It only requires that the total number of black faces is eight, and the outer surfaces are white. Since the outer surfaces are already white (as we are only painting inner faces), the only constraint is the total number of black faces. Therefore, the problem reduces to painting eight inner faces black, with the cubes being indistinct when painted the same. Therefore, the count is indeed the number of multisets, which is 10. But wait, another thought. Each small cube's inner faces are in specific orientations. For example, if a small cube has one black inner face, that black face could be oriented in different directions. If the cube can be rotated, then different orientations of the black faces are considered the same. However, in our previous analysis, we assumed that the number of black inner faces per cube is all that matters, not their orientation. But in reality, two small cubes with one black inner face each can have their black faces in different directions, which might not be equivalent under cube symmetries. Wait, this complicates things. For example, a small cube with one black inner face could have that face in one of three directions. If we consider that these are different, even when permuting the cubes, then our previous count is invalid. But the problem says: "Assume that identically painted smaller cubes cannot be distinguished and can therefore be interchanged." So "identically painted" is the key. If two small cubes have the same number of black inner faces, regardless of their orientations, they are considered identical. But actually, no. If the small cubes have black faces in different positions, even if the count is the same, they are not identically painted. Wait, the problem says: "identically painted smaller cubes cannot be distinguished". So if two small cubes have the same painting, considering the positions of the black faces, they are indistinct. However, if the black faces are in different positions, they are different. But since we can rotate the small cubes themselves, the orientation of the black faces on a single small cube doesn't matter. Wait, but the problem does not mention rotating the small cubes themselves. It says "identically painted smaller cubes cannot be distinguished and can therefore be interchanged". So if two small cubes have the same number of black faces, but in different orientations, are they considered identical? If the small cubes are indistinct except for their painting, then their painting includes the positions of the black faces. Therefore, two small cubes with the same number of black faces but in different positions would be considered different. However, since we can interchange them (i.e., permute their positions in the larger cube), perhaps the overall cube can be rotated, so the specific positions of the black faces on the small cubes don't matter. This is a critical point. Let's consider two scenarios:1. The small cubes are considered identical if they have the same number of black inner faces, regardless of where those black faces are located. Then the number of colorings is the number of multisets, which is 10.2. The small cubes are considered distinct if their black inner faces are in different orientations, even if the count is the same. Then the number of colorings is higher, and we need to account for the different ways of assigning black faces to each small cube, considering the symmetries.But the problem states: "identically painted smaller cubes cannot be distinguished and can therefore be interchanged". So "identically painted" should mean that if two small cubes have the same painting (i.e., the same set of black faces in the same positions), then they are indistinct. However, if their black faces are in different positions, even with the same count, they are different. But since we can rotate the entire cube, perhaps different orientations of black faces on the small cubes can be rotated to match each other. Wait, but the small cubes are fixed in the larger cube. When we assemble the cube, each small cube is placed in a specific corner, and their black inner faces are adjacent to specific other small cubes. However, since we can permute the small cubes (i.e., place any small cube in any corner), the specific adjacency of black faces can be changed. Therefore, the only thing that matters is how many black inner faces each small cube has, not their specific positions. Because by permuting the small cubes, we can effectively move the black inner faces to different adjacencies. Therefore, the problem is equivalent to coloring each small cube with a certain number of black inner faces (0,1,2,3), and the total being eight. Since the small cubes can be permuted, two colorings are equivalent if they have the same counts of black inner faces per small cube, regardless of the specific distribution. Therefore, the answer is 10. But let me check with a simple example. Suppose all eight small cubes have one black inner face. Then the total black faces are 8. This is one solution. Another solution is seven small cubes with 1 black face and one small cube with 1 black face. Wait, no. Wait, if x1=8, x0=0, x2=0, x3=0. But x1 + 2x2 + 3x3=8, so x1=8 is allowed. But in reality, each small cube can have at most three black faces. So x1=8 would mean eight small cubes each with 1 black inner face. But there are only eight small cubes, so yes, that's possible. Similarly, x3=2, x2=1, etc.But according to our equation, there are 10 solutions. Therefore, the answer is 10. But in the original problem statement, it's possible that the answer is higher because the orientation of the black faces matters. However, according to the problem's instruction that identically painted cubes can be interchanged, which implies that as long as two small cubes have the same number of black inner faces, they are considered identical, regardless of the positions of the black faces. Therefore, the count is indeed 10.But wait, another perspective. Suppose that each small cube's inner faces are labeled (e.g., front, left, bottom), but since the cube can be rotated, these labels are relative. Therefore, the number of black inner faces on a small cube is what matters, not their specific orientation. Therefore, two small cubes with the same number of black inner faces are considered identical, even if the black faces are on different sides. Hence, the answer is 10.But to confirm, let me think of small cases. For example, if all eight black faces are on different small cubes (each small cube has one black face), then there's only one way, since all small cubes are identical. If seven small cubes have one black face, and one small cube has one black face, but that's not possible. Wait, no. Wait, if the total number of black faces is eight, and each small cube has at most three. So possible distributions are combinations where the sum is eight, with up to three per cube. The number of such distributions, considering that the order of the cubes doesn't matter, is indeed 10.Therefore, the answer should be 10. But the problem is from a competition, and 10 seems low. Maybe I made a mistake in assuming that the orientation of the black faces doesn't matter. Let's think again.Suppose two small cubes each have one black inner face. If their black faces are on different inner faces (e.g., one on the front face and one on the left face), but since the cube can be rotated, these two small cubes can be interchanged via a rotation that swaps front and left. Therefore, these two small cubes would be considered identical. Therefore, the orientation of the black inner faces on a small cube doesn't matter because the cube can be rotated. Therefore, when counting, we only need to consider the number of black inner faces per small cube, not their positions. Therefore, the answer is indeed 10.Hence, the number of ways is 10. So the final answer is boxed{10}. Final Answerboxed{10}</think>The problem involves determining the number of ways to paint eight smaller cubes such that the assembled 2 cm cube has a completely white surface. Each smaller cube has three inner faces, and exactly eight of these inner faces must be painted black. The key steps are as follows:1. Understanding the Cube Structure: The 2 cm cube is composed of eight 1 cm cubes, each with three inner faces. The total number of inner faces is (8 times 3 = 24). We need to paint eight of these inner faces black.2. Constraints: The outer faces of the assembled cube must be white, meaning all black faces must be internal. The smaller cubes are indistinct if they have the same number of black faces, allowing them to be interchanged.3. Counting Solutions: We need to find the number of ways to distribute eight black faces among the eight smaller cubes. This is equivalent to solving the system of equations: [ x_0 + x_1 + x_2 + x_3 = 8 quad text{(total number of smaller cubes)} ] [ 0 cdot x_0 + 1 cdot x_1 + 2 cdot x_2 + 3 cdot x_3 = 8 quad text{(total number of black faces)} ] where (x_i) represents the number of smaller cubes with (i) black faces.4. Solving the Equations: We consider possible values for (x_3), (x_2), and (x_1): - Case 1: (x_3 = 0) leads to (x_1 + 2x_2 = 8) with solutions ((x_2, x_1) = (0, 8), (1, 6), (2, 4), (3, 2), (4, 0)). - Case 2: (x_3 = 1) leads to (x_1 + 2x_2 = 5) with solutions ((x_2, x_1) = (0, 5), (1, 3), (2, 1)). - Case 3: (x_3 = 2) leads to (x_1 + 2x_2 = 2) with solutions ((x_2, x_1) = (0, 2), (1, 0)).5. Total Solutions: Summing the solutions from all cases gives (5 + 3 + 2 = 10) distinct ways to distribute the black faces.Thus, the number of ways to paint the smaller cubes is (boxed{10}).,<think>Okay, let's try to figure out this cube painting problem. So we have a cube that's 2 cm on each edge, made up of eight 1 cm smaller cubes. The task is to paint some of the faces of these smaller cubes black such that when we assemble them into the bigger cube, the entire outer surface is white. We need to find how many ways we can paint the smaller cubes, considering that identical smaller cubes can be interchanged, meaning they are indistinct.First, let me visualize the cube. A 2x2x2 cube has eight smaller cubes, each at the corners. Each small cube has three faces exposed in the big cube. Wait, no, actually, in a 2x2x2 cube, each small cube has three faces exposed only if they're on the corner. But actually, in a 2x2x2 cube, every small cube is a corner cube, right? Because there's no internal cubes; all are on the corners or edges. Wait, no, in a 2x2x2 cube, every small cube is a corner cube. Each has three faces exposed. So when you put them together, each of those three faces is part of the outer surface of the big cube. Therefore, if we want the entire outer surface of the big cube to be white, then all the faces of the small cubes that are on the exterior must be white.But the problem says that eight faces of the smaller cubes are painted black, and the rest are white. So out of all the small cube faces, eight are black. But when we assemble the big cube, all the outer faces must be white. So the eight black faces must be on the internal faces of the small cubes, the ones that are glued to another small cube when assembled. Wait, but in a 2x2x2 cube, how many internal faces are there?Let me think. Each face of the big cube is 2x2, so each face has four small faces. The total number of small faces on the big cube is 6 faces * 4 = 24. But each small cube has three faces exposed, and there are eight small cubes, so 8*3=24, which matches. So the total number of exterior faces is 24. The rest are internal. Each small cube has 6 faces total, so 8 small cubes have 48 faces. Subtract the 24 external ones, so 24 internal faces.The problem says that eight of the small cube faces are painted black, and the rest are white. So the eight black faces must be among the internal faces, since all external ones must be white. Therefore, the problem reduces to choosing eight internal faces to paint black, such that when the cubes are assembled into the big cube, the black faces are all internal.But wait, the problem mentions that "from them, a cube can be assembled whose surface is white." So maybe there's more to it? Like, the cubes can be arranged in different ways, but we have to count colorings such that there exists at least one assembly where the surface is all white. But the problem also says "the smaller cubes can be assembled whose surface is white," so we need to count colorings where it's possible to arrange the small cubes (with their painted faces) into the big cube with all outer faces white. Since the small cubes can be indistinct except for their coloring, so if two colorings can be turned into each other by permuting the small cubes, they are considered the same. Wait, actually the problem says "identically painted smaller cubes cannot be distinguished and can therefore be interchanged." So if two small cubes have the same painting, they can be swapped. Therefore, when counting colorings, we have to consider colorings up to the permutation of indistinct small cubes.Wait, this is getting complicated. Let me parse the problem statement again:"In the picture, there is a cube with an edge of 2 cm made up of eight smaller cubes with an edge of 1 cm. Eight faces of the smaller cubes are painted black, and the rest are white. From them, a cube can be assembled whose surface is white. In how many ways can the smaller cubes be painted? Assume that identically painted smaller cubes cannot be distinguished and can therefore be interchanged."So, rephrasing: we need to paint eight 1x1 faces black (on the small cubes), such that when we use these small cubes to build the 2x2 cube, the outer surfaces are entirely white. The question is how many such colorings are possible, considering that small cubes with identical painting are interchangeable.Therefore, two colorings are considered the same if one can be transformed into the other by permuting small cubes that have the same color pattern. So, for example, if two small cubes have exactly the same faces painted, swapping them doesn't create a new coloring.But perhaps more formally, the colorings are considered up to the equivalence relation where two colorings are equivalent if one can be obtained from the other by permuting small cubes of the same type.Therefore, the problem reduces to counting the number of equivalence classes under this relation, where each class corresponds to a coloring scheme that allows assembling the big cube with all white exterior.But let's step back.First, since the big cube's outer surface must be white, all the outer-facing faces of the small cubes must be white. Therefore, any black face must be on an internal face (i.e., a face that is adjacent to another small cube in the assembled big cube). Each internal face is shared between two small cubes. Therefore, when we paint a face black, we have to paint the corresponding face on the adjacent small cube as well? Wait, no. Wait, in the problem statement, it says "eight faces of the smaller cubes are painted black". Each face is a face of a small cube. Since internal faces are shared between two small cubes, painting an internal face on one small cube would mean that the adjacent small cube's face is also part of that internal face. But in the problem statement, it's eight faces of the small cubes, not eight internal faces of the big cube.Wait, hold on. If we have eight small cubes, each with 6 faces. So total 48 faces. The problem says eight of these are painted black. Each black face is a face of a small cube. However, in the assembled cube, the external faces (24 in total) must all be white. Therefore, the eight black faces must be among the internal faces. The internal faces are 24 total (as calculated before: each small cube has 3 internal faces, 8*3=24). So we need to choose 8 black faces among these 24 internal faces. However, we need to choose them in such a way that when the cube is assembled, the black faces are only on the internal sides. But since the small cubes can be arranged in different ways, but in the problem statement, it says "from them, a cube can be assembled whose surface is white." So the coloring must allow for at least one arrangement where all outer faces are white, which implies that all the black faces are on the internal sides of that particular arrangement.But in this problem, the cube is made up of eight smaller cubes with eight faces painted black. So the eight black faces are on the small cubes, but when assembled, those eight black faces must be entirely internal. Since in the assembled cube, each internal face is shared between two small cubes. So if we have an internal face between two small cubes, only one of them can have that face painted black? Wait, but each face is part of a single small cube.Wait, confusion here. Let's clarify:In the big cube, each internal face is an interface between two small cubes. Each such interface consists of two faces: one from each adjacent small cube. So, for each internal face of the big cube, there are two small cube faces involved.But in the problem statement, it's the small cubes' faces that are painted. So if we paint a face of a small cube that is internal (i.e., glued to another small cube), then in the assembled cube, that face is adjacent to another small cube's face. However, the problem allows us to paint eight such small cube faces (not pairs). So, for example, if we paint one face of a small cube that is internal, the adjacent small cube's face could be white or black. The problem doesn't require that the adjacent face is also black. Wait, but when assembling the cube, if we have two small cubes adjacent to each other, their adjacent faces are pressed together. So if one is black and the other is white, does that matter? The problem only requires that the outer surfaces are white. The internal faces can be any color, but in the problem, we are painting eight small cube faces black, regardless of their position. However, when assembling, those black faces must not be on the outside. Therefore, the eight black faces must be on the internal sides of the small cubes in whatever assembly we choose.But the problem says "from them, a cube can be assembled whose surface is white." So as long as there exists some way to arrange the small cubes (possibly rotating them) such that all the black faces are internal. So the same small cubes can be rotated such that their black faces are hidden inside.Wait, but small cubes can be rotated. So even if a small cube has a black face on what would be an external face in some orientation, we can rotate it so that the black face is on an internal face. Therefore, the key is that each black face can be positioned in an internal face when assembling the cube.But since the cube has to be assembled with all external faces white, the eight black faces must be arranged such that each black face is on an internal face (i.e., between two small cubes) in the assembled cube.But each black face is a face of a small cube. So when assembling the cube, for each black face, we need to have it adjacent to another small cube's face (which could be black or white). The problem doesn't restrict the color of the internal faces, only that the external ones are white. But the eight black faces are just eight faces of the small cubes, which are internal in the assembled cube.Wait, but the problem says "eight faces of the smaller cubes are painted black", not "eight internal faces". So we have eight faces painted black, but when assembled into the cube, those eight faces must not be on the exterior. So the eight black faces must be assigned to internal positions in the cube. But how does that work? Because when you assemble the cube, each small cube's position determines which of its faces are internal or external. So depending on how you place the small cubes, the same painted face could be internal or external.But the problem states that "from them, a cube can be assembled whose surface is white." So there must exist an arrangement where all eight black faces are internal. Therefore, the coloring must be such that for each small cube with a black face, that black face can be oriented towards an internal face in some assembly.But since we can rotate the small cubes, maybe the key is that each black face is on a face that can be internal in some orientation. However, in a 2x2x2 cube, each small cube is a corner cube, so each small cube has three faces that are external. The other three faces are internal, adjacent to the three neighboring small cubes.Therefore, each small cube in the assembled big cube has three external faces and three internal faces. So, for a small cube to have a black face, that black face must be on one of its three internal faces. Because if a black face were on an external face, then that would violate the requirement of the entire surface being white.Therefore, the problem reduces to selecting eight internal faces (three per small cube, but we need to select eight in total) such that each black face is on an internal face of its small cube in the assembled big cube.But since we can arrange the small cubes in different positions and orientations, the same coloring can be achieved by permuting the small cubes and rotating them.But the problem says that identically painted small cubes cannot be distinguished and can be interchanged. Therefore, two colorings are considered the same if they can be transformed into each other by permuting small cubes with the same color patterns.This seems like a problem that involves Burnside's lemma or some form of counting under group actions, but I need to be careful.Alternatively, perhaps we can model the problem as assigning the eight black faces to the internal faces of the small cubes in such a way that the arrangement is possible. However, each internal face in the big cube is shared between two small cubes, so painting an internal face on one small cube would leave the adjacent small cube's face white. Wait, but the problem allows us to paint eight small cube faces black, regardless of their adjacency.Wait, maybe not. Let's think again.In the big cube, each internal face is an interface between two small cubes. Each such interface is two faces: one from each small cube. If we paint one of them black and the other white, then in the assembled cube, the black face is adjacent to a white face. But the problem doesn't care about the internal colors, only that the external ones are white. Therefore, it's allowed for internal interfaces to have one black and one white face. The problem only restricts that the external faces are white.Therefore, the eight black faces can be any eight internal faces (i.e., faces that are adjacent to another small cube in the assembled cube), regardless of their pairing. So the total number of internal faces is 24 (as each small cube has three internal faces, 8*3=24). We need to choose eight of these to paint black. But the problem is that when we assemble the cube, we have to place the small cubes such that the black faces are indeed on the internal sides. However, since we can rotate the small cubes, the same small cube can be oriented in different ways. Therefore, the key is that for each small cube, the number of black faces it has must be possible to be oriented such that all its black faces are on internal sides when placed in the big cube.But each small cube in the big cube has three internal faces. Therefore, a small cube can have up to three black faces, as long as they are on the internal sides. Wait, but the problem says that "eight faces of the smaller cubes are painted black". So it's possible for a small cube to have multiple black faces.But here's the catch: if a small cube has a black face, in order for it to be placed in the big cube with that black face internal, the black face must be aligned with an internal face. Since each small cube in the big cube has three internal faces, which correspond to the three adjacent small cubes. Therefore, any black face on a small cube must be on one of those three internal faces when placed in the big cube.However, when considering the problem statement, we have to paint eight faces black, and then assemble the cube such that all external faces are white. Therefore, the black faces must be assigned to the internal faces of the assembled cube.But the challenge is that the small cubes can be rotated, so the same physical small cube (with a specific painting) can be placed in different orientations. Therefore, the number of black faces on a small cube can vary depending on how it's rotated. Wait, no. The painting is fixed; once the small cube is painted, the positions of the black faces are fixed. But we can rotate the cube when placing it into the big cube, so that different faces become the internal ones.Therefore, if a small cube has some black faces, we need to be able to rotate it such that all its black faces are on the internal sides of the big cube. But each small cube in the big cube has three internal faces. Therefore, if a small cube has k black faces, we must be able to rotate it so that those k black faces are among the three internal faces. Therefore, k must be at most 3, but since we only have eight black faces total, and eight small cubes, the average is one per cube. But some cubes can have more, some less.But the key is that for each small cube, the number of black faces it has must be less than or equal to 3, and those black faces must be on three different faces (since you can't have two black faces on the same physical face of the small cube). Wait, actually, a small cube can have multiple black faces on different faces, and when placing it into the big cube, we can rotate it so that those black faces are aligned with internal faces.Therefore, the constraints are:1. Each small cube can have 0, 1, 2, or 3 black faces, but the total across all small cubes must be 8.2. For each small cube with k black faces, those k faces must be on different faces of the small cube (since each face is either black or white, and we can't have multiple black faces on the same physical face).Wait, actually, no. Each face of the small cube is either black or white. So a small cube can have any number of black faces from 0 to 6, but in our case, the total is 8. However, when assembling the big cube, each small cube is placed in a corner, with three faces exposed (external) and three faces internal. So to have all external faces white, the three external faces of the small cube in the big cube must be white. Therefore, the three internal faces of the small cube can be black or white. But the problem states that eight faces of the smaller cubes are painted black. So the total number of black faces is eight, and these can be on any faces of the small cubes, but when assembled, the three external faces of each small cube must be white. Therefore, the black faces must be among the three internal faces of each small cube in the assembled big cube.But here's the catch: the internal faces of a small cube in the assembled big cube correspond to the three faces adjacent to other small cubes. Therefore, if a small cube has a black face on one of its internal faces, then the adjacent small cube's face must be white (since the total black faces are only eight, but each internal face is shared between two small cubes). Wait, no. The problem says that eight faces of the smaller cubes are painted black. Each such face is part of one small cube. Therefore, if two small cubes are adjacent in the big cube, one could have a black face on the shared interface, and the other could have a white face. Since the problem doesn't require the internal interfaces to be any particular color, only the external ones. Therefore, it's allowed for one face of the interface to be black and the other white.But in that case, when we count the eight black faces, each is on a single small cube. Therefore, the problem reduces to choosing eight small cube faces that are internal in some assembly of the big cube.But since the small cubes can be rotated and arranged in different positions, how does this affect the count?Wait, perhaps the problem is similar to: how many colorings are there of eight small cubes, each with some faces painted black, such that the total number of black faces is eight, and there exists an arrangement (permutation and rotation) of the small cubes into a 2x2x2 cube where all external faces are white. Then, considering that two colorings are equivalent if they can be transformed into each other by permuting indistinct small cubes.But this seems complicated. Let's approach step by step.First, note that in the assembled big cube, each small cube has three external faces (which must be white) and three internal faces (which can be black or white). However, the problem states that exactly eight faces of the smaller cubes are painted black. Since all external faces must be white, all black faces must be internal. Therefore, the eight black faces are distributed among the internal faces of the small cubes.Each small cube has three internal faces. Therefore, the total number of internal faces across all small cubes is 8 * 3 = 24. But we are to choose eight of these to paint black.However, each internal face in the big cube is shared between two small cubes. Therefore, in the big cube, there are 12 internal faces (each internal face of the big cube is shared between two small cubes; the big cube has 6 faces, each divided into 4 small faces, so total 24 small faces, 12 internal interfaces). Wait, perhaps not. Let's think:In a 2x2x2 cube, how many internal faces are there? Each face of the big cube is 2x2, so 4 small faces. The internal structure: along each axis, there are layers. For example, in the x-axis, there are two layers: front and back. Between them, there's an internal layer where the faces are adjacent. Similarly for y and z axes. Each internal layer has 2x2=4 internal faces. There are three axes, each contributing 4 internal faces, so total 12 internal faces in the big cube. Each internal face is shared between two small cubes. Therefore, each internal face in the big cube corresponds to two small cube faces (one from each adjacent small cube). Therefore, in total, there are 12 internal faces in the big cube, each consisting of two small cube faces.But in the problem, we have eight small cube faces painted black. These must all be on the internal faces of the big cube. But each internal face of the big cube can have 0, 1, or 2 black small cube faces. However, since we only have eight black faces, and there are 12 internal faces (each with two small cube faces), the maximum number of black faces is 24, but we have only eight.Therefore, the problem is equivalent to selecting eight small cube faces that lie on the internal faces of the big cube, with the constraint that no two selected small cube faces are on the same big cube internal face. Wait, no. Wait, each big cube internal face is an interface between two small cubes, so if we paint one or both of the small cube faces on that interface, but in our problem, we can paint any number, as long as the total is eight. However, the problem does not restrict the number of black faces per internal interface.But here's the key: when we assemble the big cube, the small cubes can be arranged in different ways. However, the problem states that the small cubes are indistinct except for their coloring. So if two small cubes have the same coloring, they can be swapped. Therefore, the colorings are considered up to permutation of identical cubes.But this seems to complicate the counting. Alternatively, perhaps the problem is to count the number of colorings (assigning eight black faces to the 24 internal small cube faces) up to the symmetries of the cube, considering that small cubes can be permuted and rotated.But maybe not symmetries of the cube, but rather considering that two colorings are equivalent if one can be obtained from the other by permuting small cubes that are identical in coloring. Hmm.Alternatively, perhaps the problem is simpler: since the small cubes are indistinct except for their colorings, the number of colorings is equal to the number of distinct colorings when considering that two colorings are the same if they can be transformed into each other by permuting cubes with identical colorings. So it's similar to counting the number of colorings under the action of the symmetric group on the cubes, where the group action permutes cubes with the same color.This is similar to counting the number of colorings with eight black faces, where cubes are considered the same if they have the same number and arrangement of black faces, considering rotations.Wait, no. Because even if two cubes have the same number of black faces but in different positions, they can be distinguished if their colorings are not rotationally equivalent.But the problem states: "identically painted smaller cubes cannot be distinguished and can therefore be interchanged." So if two small cubes have the same coloring up to rotation, then they are considered identical and can be interchanged. Therefore, when counting the colorings, we need to consider two colorings equivalent if one can be obtained by permuting small cubes and rotating them such that the overall coloring is the same.This seems very abstract. Maybe an example would help.Suppose we have two small cubes, each with one black face. If their black faces are on different positions (e.g., one has a black face on the front, another on the top), but by rotating the cubes, they can be made to look the same. Then they are considered identical. So the key is that two small cubes are considered identical if their colorings are rotationally equivalent.Therefore, the problem reduces to counting the number of colorings with eight black faces (all on internal faces when assembled), where colorings are considered the same if they differ by permuting small cubes that are rotationally equivalent.Therefore, the problem is equivalent to counting the number of multisets of rotationally distinct small cube colorings, with a total of eight black faces, such that the colorings can be arranged into a big cube with all external faces white.But even this is not straightforward. Maybe we need to think in terms of orbits under the group of cube symmetries combined with permutations of identical small cubes.Alternatively, perhaps we can model this by considering the types of small cubes based on their black internal faces.Each small cube, when placed in the big cube, has three internal faces. Let's consider the possible configurations of black faces on these three internal faces.Each small cube can have 0, 1, 2, or 3 black faces on its internal faces. However, the total number of black faces across all small cubes must be eight.Moreover, since the small cubes are indistinct except for their colorings, we need to count the number of colorings up to permutation of small cubes with the same number and arrangement of black faces.But even more precisely, two small cubes are considered the same if their black faces can be rotated to match each other. Therefore, we need to consider the equivalence classes of small cube colorings under rotation.So first, let's determine how many distinct types of small cubes there are, based on the number and arrangement of black faces on their three internal faces.Since the three internal faces of a small cube are mutually adjacent (as in a corner), the possible configurations are:0 black faces: All three internal faces are white.1 black face: One of the three internal faces is black. However, due to rotational symmetry, all such configurations are equivalent. Because we can rotate the cube so that the black face is on any of the three internal faces.2 black faces: Two of the three internal faces are black. These can be adjacent or opposite. Wait, in three dimensions, on a cube, two faces can be adjacent or opposite. However, on three mutually adjacent faces (as in the internal faces of a corner cube), the three internal faces are all adjacent to each other (each pair meets at an edge). Therefore, any two of them are adjacent. So there's only one distinct configuration with two black faces: two adjacent faces.Wait, no. If we have three internal faces, each adjacent to the other two, then choosing any two of them will be adjacent. Therefore, there's only one rotational equivalence class for two black faces.Similarly, three black faces: all three internal faces are black. Only one configuration.Therefore, the distinct types of small cubes based on their internal black faces are:- Type 0: 0 black faces.- Type 1: 1 black face.- Type 2: 2 black faces.- Type 3: 3 black faces.Each of these types is distinct under rotation.Therefore, when considering the colorings of the eight small cubes, we can assign each small cube to one of these four types, with the total number of black faces being 8. However, since the small cubes of the same type are indistinct, the number of colorings is the number of multisets of these types that sum to 8 black faces.But we need to consider that the arrangement must allow for the cubes to be assembled into a big cube with all external faces white. However, if all black faces are on the internal faces (as per the types above), then assembling the cube with the black faces on the internal sides would satisfy the condition. Therefore, as long as the total number of black faces is eight, and each small cube's black faces are counted as Type 1, 2, or 3, then the cube can be assembled appropriately.But wait, no. Because if two adjacent small cubes in the big cube both have black faces on their shared internal face, that would count as two black faces on the same internal interface. However, the problem doesn't prohibit this; it only requires that the external faces are white. Therefore, even if two adjacent small cubes both have black faces on their shared interface, that's allowed. The problem doesn't restrict the internal colors.Therefore, as long as we have eight black faces distributed among the internal faces of the small cubes (i.e., each small cube's black faces are among their three internal faces), then we can assemble the big cube with all external faces white by appropriately rotating each small cube so that their black faces are internal.Therefore, the problem reduces to counting the number of multisets of the four types (0, 1, 2, 3) with eight small cubes, where the sum of black faces is 8, considering that cubes of the same type are indistinct.However, we need to account for rotational equivalence. Wait, no. Wait, the problem says "identically painted smaller cubes cannot be distinguished and can therefore be interchanged." So if two small cubes have the same type (i.e., same number of black faces up to rotation), they are considered identical and can be interchanged. Therefore, the count is the number of ways to assign the eight small cubes to the four types, such that the total number of black faces is eight, and considering that the order of cubes of the same type doesn't matter.This is equivalent to finding the number of non-negative integer solutions to the equation:a + b + c + d = 8,where:- a is the number of Type 0 cubes (0 black faces),- b is the number of Type 1 cubes (1 black face),- c is the number of Type 2 cubes (2 black faces),- d is the number of Type 3 cubes (3 black faces),and the total number of black faces is:0*a + 1*b + 2*c + 3*d = 8.So we have two equations:1. a + b + c + d = 8,2. b + 2c + 3d = 8.We need to find the number of non-negative integer solutions (a, b, c, d) to these equations.Since a = 8 - b - c - d, substituting into the second equation:b + 2c + 3d = 8,with b, c, d ≥ 0.So we need to find the number of non-negative integer solutions (b, c, d) to b + 2c + 3d = 8.Let's solve this equation.We can iterate over possible values of d (number of Type 3 cubes), since 3d ≤ 8.Possible d: 0, 1, 2.Case 1: d = 0Then the equation becomes b + 2c = 8.Possible c: 0 to 4 (since 2*4=8).For each c, b = 8 - 2c.So solutions:c=0, b=8c=1, b=6c=2, b=4c=3, b=2c=4, b=0That's 5 solutions.Case 2: d = 1Then the equation becomes b + 2c + 3 = 8 ⇒ b + 2c = 5.Possible c: 0 to 2 (since 2*3=6 >5)c=0: b=5c=1: b=3c=2: b=1That's 3 solutions.Case 3: d = 2Then the equation becomes b + 2c + 6 = 8 ⇒ b + 2c = 2.Possible c: 0 to 1c=0: b=2c=1: b=0That's 2 solutions.d cannot be 3, since 3*3=9 >8.Total solutions: 5 + 3 + 2 = 10.Therefore, there are 10 possible distributions of cube types (b, c, d) that satisfy the equations. Each of these corresponds to a different way of painting the cubes, considering that cubes of the same type are indistinct.However, the problem might have additional constraints. Specifically, we need to ensure that the cubes can actually be arranged in the big cube without overlapping black faces. Wait, but earlier we considered that the problem allows any distribution of black faces on internal sides, even if two adjacent small cubes have black faces on their shared interface. Since the problem doesn't restrict the internal colors, only that the external ones are white. Therefore, as long as the total number of black faces is eight, and each cube's black faces are on their internal sides (which we've already encoded in the types), then the arrangement is possible.Therefore, the answer should be 10. But wait, let me check.Wait, but is each solution (b, c, d) unique up to cube permutations? For example, two different distributions might result in the same multiset of cube types. But since we're already considering the number of solutions (i.e., different distributions of b, c, d), which correspond to different multisets, then yes, each solution is a distinct way of painting the cubes where cubes of the same type are indistinct.Therefore, the answer should be 10.But let me think again. Suppose we have different distributions. For example, one solution is d=2, c=1, b=0. Wait, no, in case d=2, we have:For d=2:b + 2c = 2. So c=0: b=2; c=1: b=0.So the solutions are (d=2, c=0, b=2) and (d=2, c=1, b=0). But in the case d=2, c=0, b=2, then total cubes: a=8 - b - c - d = 8 -2 -0 -2=4. So four Type 0 cubes, two Type 1, none Type 2, two Type 3.But in the equation, the variables are a, b, c, d, but we derived solutions for b, c, d. However, a is determined as 8 - b - c - d.Therefore, each solution (b, c, d) gives a unique multiset {a, b, c, d}, so the total number of such multisets is 10.Therefore, the answer is 10. Therefore, the number of ways is boxed{10}.Wait, but I need to confirm if there's a mistake here. The problem might have more complexity due to the fact that some cube types cannot coexist due to geometric constraints in the cube assembly.For example, suppose we have a cube with three black faces (Type 3). When placed in the big cube, its three internal faces are all black. However, each internal face is adjacent to another small cube. If another small cube adjacent to it also has a black face on the shared interface, then that's allowed, as the problem doesn't restrict internal colors. So in terms of feasibility, any combination is allowed.Therefore, as long as the total black faces sum to eight, and each small cube's black faces are on internal faces (which is ensured by the types), the assembly is possible. Therefore, the answer is 10.But let me check with an example.Take the case where all eight black faces are on Type 1 cubes (each with one black face). So eight Type 1 cubes, which is one of the solutions (d=0, c=0, b=8). But we have eight small cubes, each with one black face, so total eight black faces. However, each small cube must be placed in the big cube such that its single black face is internal. Since each small cube is a corner cube with three internal faces, we can rotate each Type 1 cube so that its black face is on any of the three internal faces. Therefore, this is possible.Another example: two Type 3 cubes (each with three black faces) and one Type 2 cube (two black faces), and the rest Type 0. Wait, but in the solutions above, for example, d=2, c=1, b=0. Total black faces: 3*2 + 2*1 = 8. So two Type 3, one Type 2, five Type 0. This should be possible. Place the two Type 3 cubes in opposite corners of the big cube. Each has three black faces on their internal sides, which are adjacent to three other cubes. The Type 2 cube can be placed somewhere else, with two black faces adjacent to two other cubes. Since there's no restriction on internal colors, this is allowed.Therefore, all 10 solutions are feasible. Therefore, the answer is 10.But wait, the problem is from a competition, and the answer is likely to be a small integer. However, 10 seems a bit low. Let me check again.Wait, maybe there's a mistake in assuming that the distinct types are only based on the number of black faces. Actually, two small cubes with the same number of black faces but in different configurations (which cannot be rotated to match) would be considered different. For example, a cube with two adjacent black faces versus two opposite black faces. Wait, but in our case, the black faces are on the three internal faces of the small cube, which are all mutually adjacent (as in a corner). Therefore, any two black faces on the internal faces are adjacent. Therefore, there is no such thing as two opposite black faces among the three internal faces. Since the three internal faces meet at a corner, any two of them share an edge. Therefore, in this specific case, any two black faces on the internal faces are adjacent, so there is only one rotational equivalence class for two black faces. Similarly, one black face can be rotated to any of the three internal faces, so there's only one equivalence class for one black face.Therefore, our initial assumption is correct: the distinct types are based solely on the number of black faces (0, 1, 2, 3), as the arrangements are rotationally equivalent.Therefore, the count of 10 is accurate.However, to be thorough, let's list all 10 solutions:Case d=0:c=0, b=8: 8 Type 1 cubes.c=1, b=6: 6 Type 1, 1 Type 2, 1 Type 0.c=2, b=4: 4 Type 1, 2 Type 2, 2 Type 0.c=3, b=2: 2 Type 1, 3 Type 2, 3 Type 0.c=4, b=0: 0 Type 1, 4 Type 2, 4 Type 0.Wait, no. Wait, a is 8 - b - c - d. Since d=0 in this case:For each solution (b, c):c=0, b=8: a=8 -8 -0 -0=0. So 0 Type 0, 8 Type 1, 0 Type 2, 0 Type 3.c=1, b=6: a=8 -6 -1 -0=1. So 1 Type 0, 6 Type 1, 1 Type 2.c=2, b=4: a=8 -4 -2 -0=2.c=3, b=2: a=8 -2 -3 -0=3.c=4, b=0: a=8 -0 -4 -0=4.Similarly for other cases.Case d=1:c=0, b=5: a=8 -5 -0 -1=2.c=1, b=3: a=8 -3 -1 -1=3.c=2, b=1: a=8 -1 -2 -1=4.Case d=2:c=0, b=2: a=8 -2 -0 -2=4.c=1, b=0: a=8 -0 -1 -2=5.So listing all 10 solutions:1. d=0, c=0, b=8, a=0: 8 Type 1.2. d=0, c=1, b=6, a=1: 1 Type 0, 6 Type 1, 1 Type 2.3. d=0, c=2, b=4, a=2: 2 Type 0, 4 Type 1, 2 Type 2.4. d=0, c=3, b=2, a=3: 3 Type 0, 2 Type 1, 3 Type 2.5. d=0, c=4, b=0, a=4: 4 Type 0, 0 Type 1, 4 Type 2.6. d=1, c=0, b=5, a=2: 2 Type 0, 5 Type 1, 0 Type 2, 1 Type 3.7. d=1, c=1, b=3, a=3: 3 Type 0, 3 Type 1, 1 Type 2, 1 Type 3.8. d=1, c=2, b=1, a=4: 4 Type 0, 1 Type 1, 2 Type 2, 1 Type 3.9. d=2, c=0, b=2, a=4: 4 Type 0, 2 Type 1, 0 Type 2, 2 Type 3.10. d=2, c=1, b=0, a=5: 5 Type 0, 0 Type 1, 1 Type 2, 2 Type 3.Each of these 10 solutions corresponds to a different way of distributing the black faces among the small cubes, considering that cubes of the same type are indistinct. Therefore, the answer is 10.However, I recall that in similar combinatorial cube problems, the answer sometimes is different due to overcounting or undercounting some symmetries. But in this problem, since we're considering the cubes up to permutation of identical ones, and not up to the cube's symmetries, the count is based purely on the multiset of cube types.Therefore, I think the answer is 10. So boxed{10} is the final answer.Final Answerboxed{10}</think>The problem involves a cube with an edge length of 2 cm, composed of eight smaller 1 cm cubes. We need to determine the number of ways to paint eight faces of the smaller cubes black such that the assembled cube has a completely white surface. Identically painted smaller cubes are indistinguishable and can be interchanged.Key steps:1. Understanding the Cube Structure: - Each small cube in the 2x2x2 cube is a corner cube with three faces exposed (external) and three faces internal. - All external faces must be white, so the eight black faces must be internal.2. Internal Faces Calculation: - Each small cube has three internal faces, leading to a total of (8 times 3 = 24) internal faces. - We need to choose 8 of these 24 internal faces to paint black.3. Cube Types and Symmetry: - Each small cube can have 0, 1, 2, or 3 black faces on its internal faces. - Due to rotational symmetry, small cubes with the same number of black faces are considered identical.4. Equation Setup: - Let (a), (b), (c), and (d) represent the number of small cubes with 0, 1, 2, and 3 black faces respectively. - We need to solve the system: [ a + b + c + d = 8 quad text{(total number of small cubes)} ] [ b + 2c + 3d = 8 quad text{(total number of black faces)} ]5. Solving the Equations: - Iterate over possible values of (d) (0, 1, 2) and find corresponding values of (b) and (c). - Total solutions: 10.6. Verification: - Each solution corresponds to a valid distribution of black faces, considering rotational symmetry and indistinguishability of identical cubes.The number of ways to paint the smaller cubes is (boxed{10}).]

🔑:From the task, it follows that each of the eight small cubes, from which the large cube is composed, has three white faces that share a common vertex. The remaining faces of each small cube are either black or white. In total, eight faces should be black, regardless of how we arrange the small cubes; what matters is only how many faces are colored. Moreover, it doesn't matter if we color, for example, the first and second faces or the first and third faces - we can rotate the small cube to convert one case to another and vice versa. Let's list the possible colorings of the small cubes. The rows record the number of black faces on each of the small cubes, always from the highest number mathrm{k} to the lowest:- 3,3,2,0,0,0,0,0,bullet 3,3,1,1,0,0,0,0,- 3,2,2,1,0,0,0,0,- 3,2,1,1,1,0,0,0,- 3,1,1,1,1,1,0,0,- 2,2,2,2,0,0,0,0,- 2,2,2,1,1,0,0,0- 2,2,1,1,1,1,0,0,- 2, 1, 1, 1, 1, 1, 1, 0,- 1,1,1,1,1,1,1,1.This gives us a total of 10 different colorings of the small cubes.

🧠:[<think>Okay, so I have this problem here about an arithmetic sequence. Let me try to figure it out step by step. The problem says that in the arithmetic sequence {a_n}, a_2 is 1 and a_5 is 4. Then there are two parts: first, find the general formula for the sequence, and second, let b_n be 2 raised to the power of a_n, and find the sum of the first n terms of {b_n}. Hmm, okay. Let me start with part (1).First, arithmetic sequences have a common difference between consecutive terms. So, the general formula for an arithmetic sequence is usually a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference. But here, they give me a_2 and a_5. Let me write down what I know.Given:a_2 = 1a_5 = 4Since it's an arithmetic sequence, each term increases by d. So, a_2 is the first term plus d, right? Because a_1 + d = a_2. Similarly, a_5 would be a_1 + 4d. Wait, no: hold on, the nth term is a_1 + (n-1)d. So, a_2 is a_1 + (2-1)d = a_1 + d. Similarly, a_5 is a_1 + (5-1)d = a_1 + 4d. So, given that a_2 = 1 and a_5 = 4, I can set up two equations.So:1) a_1 + d = 12) a_1 + 4d = 4Now, I can solve these two equations to find a_1 and d. Let me subtract the first equation from the second to eliminate a_1.(a_1 + 4d) - (a_1 + d) = 4 - 1That simplifies to:3d = 3So, d = 1.Now plug back d = 1 into the first equation to find a_1.a_1 + 1 = 1 => a_1 = 0.So, the first term is 0 and the common difference is 1. Therefore, the general formula for the arithmetic sequence is:a_n = a_1 + (n - 1)d = 0 + (n - 1)(1) = n - 1.Wait, let me check that. If a_n = n - 1, then a_2 would be 2 - 1 = 1, which matches the given a_2 = 1. And a_5 would be 5 - 1 = 4, which matches a_5 = 4. Perfect, that works.So part (1) answer is a_n = n - 1. Let me just make sure I didn't make a mistake here. Let me list out the terms: a_1 = 0, a_2 = 1, a_3 = 2, a_4 = 3, a_5 = 4... Yep, each term increases by 1, starting at 0. That seems right.Okay, moving on to part (2). Let b_n = 2^{a_n}, so each term of {b_n} is 2 raised to the corresponding term in the arithmetic sequence. Since we found a_n = n - 1, then b_n = 2^{n - 1}. So, the sequence {b_n} is 2^{0}, 2^{1}, 2^{2}, 2^{3}, ..., 2^{n-1}. Wait a second, that looks like a geometric sequence where each term is multiplied by 2. Because 2^{n - 1} is the same as starting at 1 (when n=1) and multiplying by 2 each time. So the first term is 1, then 2, 4, 8, etc.Therefore, the sum of the first n terms of a geometric sequence is given by S_n = a_1*(r^n - 1)/(r - 1), where a_1 is the first term and r is the common ratio. In this case, a_1 is 2^{0} = 1, and r is 2. So plugging into the formula:S_n = 1*(2^n - 1)/(2 - 1) = (2^n - 1)/1 = 2^n - 1.Therefore, the sum S_n is 2^n - 1.Wait, let me verify with small n. Let's take n=1: b_1 = 2^{0} = 1. Sum S_1 =1. According to the formula, 2^1 -1=1. Correct.For n=2: b_1 + b_2 =1 +2=3. Formula gives 2^2 -1=4-1=3. Correct.n=3:1+2+4=7. Formula: 2^3 -1=8-1=7. Correct. So seems right.But wait, the problem says "the sum of the first n terms of the sequence {b_n}". But {b_n} is 2^{a_n}, and a_n is n-1. So, yeah, b_n is 2^{n-1}, which is a geometric sequence with first term 1 and ratio 2. Therefore, sum is 2^n -1. So the answer is 2^n -1.Wait, just to make sure I didn't confuse anything. Let me see. The problem is part (2): Let b_n =2^{a_n}, find S_n, the sum of the first n terms. So, given that a_n =n -1, so b_n=2^{n -1}. Then the sum S_n = sum_{k=1}^n b_k = sum_{k=1}^n 2^{k-1} }.Yes, that's a geometric series with n terms, starting from 2^0 up to 2^{n-1}. The sum is indeed (2^n -1)/(2 -1) =2^n -1. So, yes, that's correct.Therefore, part (2) answer is 2^n -1.Wait, but just to double-check, let me compute S_4: terms are 1, 2, 4, 8. Sum is 15. Formula gives 2^4 -1=16 -1=15. Correct. So all tests pass. Therefore, the answer is correct.So, summarizing:(1) The general formula for {a_n} is a_n = n -1.(2) The sum S_n for {b_n} is 2^n -1.Final Answer(1) The general formula is boxed{a_n = n - 1}. (2) The sum of the first ( n ) terms is boxed{2^n - 1}.</think>Given the arithmetic sequence ({a_n}) where (a_2 = 1) and (a_5 = 4):1. Finding the general formula for the sequence ({a_n}): The nth term of an arithmetic sequence is given by (a_n = a_1 + (n-1)d), where (a_1) is the first term and (d) is the common difference. Using the given terms: [ begin{cases} a_1 + d = 1 a_1 + 4d = 4 end{cases} ] Subtracting the first equation from the second: [ (a_1 + 4d) - (a_1 + d) = 4 - 1 implies 3d = 3 implies d = 1 ] Substituting (d = 1) back into the first equation: [ a_1 + 1 = 1 implies a_1 = 0 ] Therefore, the general formula for the sequence is: [ a_n = 0 + (n - 1) cdot 1 = n - 1 ] So, the answer is (boxed{a_n = n - 1}).2. Finding the sum of the first (n) terms of the sequence ({b_n}) where (b_n = 2^{a_n}): Given (a_n = n - 1), we have (b_n = 2^{n - 1}). The sequence ({b_n}) is a geometric sequence with the first term (b_1 = 2^0 = 1) and common ratio (r = 2). The sum of the first (n) terms of a geometric sequence is given by: [ S_n = frac{a_1 (r^n - 1)}{r - 1} ] Substituting (a_1 = 1) and (r = 2): [ S_n = frac{1 (2^n - 1)}{2 - 1} = 2^n - 1 ] Therefore, the sum of the first (n) terms is (boxed{2^n - 1}).,<think>Okay, so I have this arithmetic sequence {a_n}, and I know two specific terms: a_2 is 1, and a_5 is 4. The first part asks me to find the general formula for the sequence. Then the second part is about defining another sequence {b_n} where each term is 2 raised to the power of a_n, and finding the sum of the first n terms of this {b_n} sequence. Let me tackle part (1) first.Alright, arithmetic sequences have a common difference between consecutive terms. The general formula for an arithmetic sequence is usually given by a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference. But since I know a_2 and a_5, maybe I can set up equations based on that.Given that a_2 = 1, which would be a_1 + (2-1)d = a_1 + d = 1. Similarly, a_5 = 4, so that's a_1 + (5-1)d = a_1 + 4d = 4. So now I have two equations:1) a_1 + d = 12) a_1 + 4d = 4I need to solve these two equations for a_1 and d. Let me subtract the first equation from the second to eliminate a_1.(a_1 + 4d) - (a_1 + d) = 4 - 1This simplifies to 3d = 3, so d = 1. Then plugging back into the first equation: a_1 + 1 = 1, so a_1 = 0.So the common difference is 1, and the first term is 0. Therefore, the general formula should be a_n = 0 + (n - 1)*1 = n - 1. Let me check that against the given terms. For n=2, a_2 = 2 - 1 = 1, which matches. For n=5, a_5 = 5 - 1 = 4, which also matches. So that seems correct.So part (1) is solved: a_n = n - 1.Now moving on to part (2). They define b_n = 2^{a_n}, so substituting the general formula from part (1), b_n = 2^{n - 1}. Then we need to find the sum of the first n terms of {b_n}, which is S_n = b_1 + b_2 + ... + b_n.But wait, let's see what {b_n} looks like. Since b_n = 2^{n - 1}, then each term is 2 raised to the power of (n - 1). So:b_1 = 2^{1 - 1} = 2^0 = 1b_2 = 2^{2 - 1} = 2^1 = 2b_3 = 2^{3 - 1} = 4b_4 = 8, and so on. So this is a geometric sequence where the first term is 1 and the common ratio is 2. Because each term is multiplied by 2 to get the next term.Therefore, the sum of the first n terms of a geometric sequence is given by S_n = a_1*(r^n - 1)/(r - 1). Here, a_1 is the first term, which is 1, and r is the common ratio, which is 2. So plugging in:S_n = 1*(2^n - 1)/(2 - 1) = (2^n - 1)/1 = 2^n - 1.Wait, let me confirm. If n = 1, then S_1 = 1, which matches 2^1 - 1 = 1. For n = 2, S_2 = 1 + 2 = 3, and 2^2 - 1 = 3. For n = 3, S_3 = 1 + 2 + 4 = 7, and 2^3 - 1 = 7. That seems to work. So the sum S_n is 2^n - 1.Therefore, part (2) answer is 2^n - 1.But let me double-check everything again. For part (1), the arithmetic sequence starting at 0 with common difference 1 gives a_n = n - 1. Yes, because when n=1, a_1 = 0, which is correct. Then each subsequent term increases by 1. So that's right.For part (2), since each term of {b_n} is 2^(n - 1), which is a geometric sequence starting at 1 with ratio 2, the sum is indeed 2^n - 1. Because the formula for the sum of the first n terms of a geometric series is S_n = a_1*(r^n - 1)/(r - 1). Here, a_1 is 1, r is 2, so (2^n - 1)/(2 - 1) = 2^n - 1. Yep, that's correct.I think that's all. Both parts check out with the given terms and the formulas make sense.Final Answer(1) The general formula is boxed{a_n = n - 1}.(2) The sum of the first ( n ) terms is boxed{2^n - 1}.</think>Given the arithmetic sequence ({a_n}) where (a_2 = 1) and (a_5 = 4):1. To find the general formula for the sequence, we start with the general form of an arithmetic sequence (a_n = a_1 + (n-1)d). Using the given terms: [ begin{cases} a_1 + d = 1 a_1 + 4d = 4 end{cases} ] Subtracting the first equation from the second, we get: [ 3d = 3 implies d = 1 ] Substituting (d = 1) back into the first equation: [ a_1 + 1 = 1 implies a_1 = 0 ] Therefore, the general formula is: [ a_n = 0 + (n-1) cdot 1 = n - 1 ] [ boxed{a_n = n - 1} ]2. For the sequence (b_n = 2^{a_n}), substituting (a_n = n - 1) gives (b_n = 2^{n - 1}). This is a geometric sequence with the first term (b_1 = 1) and common ratio (r = 2). The sum of the first (n) terms of a geometric sequence is given by: [ S_n = frac{a_1 (r^n - 1)}{r - 1} ] Substituting (a_1 = 1) and (r = 2): [ S_n = frac{1 (2^n - 1)}{2 - 1} = 2^n - 1 ] [ boxed{2^n - 1} ]]

🔑:Solution: (1) According to the problem, a_5-a_2=3d=3, therefore d=1, therefore a_n=n-1 (ninmathbb{N}^*). Thus, the general formula for the sequence is a_n=n-1.(2) From (1), we have b_n=2^{n-1}, therefore the sequence {b_n} is a geometric sequence with the first term 1 and common ratio 2, therefore S_n= dfrac{1-2^n}{1-2}=2^n-1.Therefore, the answers are: (1) The general formula for the sequence {a_n} is boxed{a_n=n-1}. (2) The sum of the first n terms S_n of the sequence {b_n} is boxed{S_n=2^n-1}.

🧠:[<think>Alright, so I have two problems to solve here, both about conic sections: a hyperbola and an ellipse. Let's take them one at a time.Problem (1): Find the standard equation of the hyperbola that has the same foci as the ellipse x²/40 + y²/15 = 1 and an eccentricity of e = 5/4.Okay, first, I need to recall some properties of ellipses and hyperbolas. For an ellipse, the standard equation is x²/a² + y²/b² = 1, where a > b, and the foci are located at (±c, 0) where c² = a² - b². The eccentricity of an ellipse is e = c/a, which is less than 1.For a hyperbola, the standard equation is x²/A² - y²/B² = 1, and the foci are located at (±C, 0), where C² = A² + B². The eccentricity of a hyperbola is e = C/A, which is greater than 1.The problem states that the hyperbola shares the same foci as the given ellipse, so first I need to find the foci of the ellipse x²/40 + y²/15 = 1.Given the ellipse equation, a² = 40 and b² = 15. Then, c² = a² - b² = 40 - 15 = 25. So, c = 5. Therefore, the foci of the ellipse are at (±5, 0). So, the hyperbola also has foci at (±5, 0), meaning that for the hyperbola, C = 5.The eccentricity of the hyperbola is given as e = 5/4. For a hyperbola, e = C/A, so 5/4 = 5/A. Solving for A: A = 5 / (5/4) = 4. So, A = 4.Now, since for the hyperbola, C² = A² + B², we can solve for B². C is 5, so:5² = 4² + B² 25 = 16 + B² B² = 25 - 16 = 9 Therefore, B = 3.So, the standard equation of the hyperbola is x²/A² - y²/B² = 1, which becomes x²/16 - y²/9 = 1.Let me just check if that makes sense. The foci are at ±C = ±5, which for the hyperbola, C = 5. Since A = 4 and B = 3, C² = 16 + 9 = 25, which is correct. The eccentricity e = C/A = 5/4, which matches the given. So that seems right.Problem (2): Given that the center of an ellipse is at the origin, it passes through the point P(3,2), its foci are on the x-axis, and the length of its major axis is three times the length of its minor axis, find the equation of the ellipse.Alright, another ellipse problem. Let's parse the information.- Center at origin: So equation is x²/a² + y²/b² = 1.- Foci on x-axis: So major axis is along the x-axis, hence a > b. Therefore, the standard form is correct as written.- Length of major axis is three times the length of minor axis: The major axis length is 2a, minor axis length is 2b. So 2a = 3*(2b) => a = 3b.- Passes through point P(3,2): So substituting x=3 and y=2 into the ellipse equation should satisfy it.So, let's start writing down the given relationships.First, a = 3b. So, we can express a in terms of b.The equation of the ellipse is x²/(9b²) + y²/b² = 1.Now, since it passes through (3, 2), substitute x=3 and y=2:(3²)/(9b²) + (2²)/b² = 1 9/(9b²) + 4/b² = 1 (1/b²) + (4/b²) = 1 5/b² = 1 So, b² = 5. Therefore, b = √5.Since a = 3b, then a = 3√5, so a² = (3√5)² = 9*5 = 45.Therefore, the equation of the ellipse is x²/45 + y²/5 = 1.Let me verify this. If the major axis is three times the minor, then major length is 2a = 6√5, minor is 2b = 2√5. 6√5 is indeed three times 2√5. Also, the foci are on the x-axis, which they are because it's an ellipse with a > b. The foci would be at (±c, 0), where c² = a² - b² = 45 - 5 = 40, so c = 2√10. That's fine.Now, checking if the point (3, 2) lies on the ellipse:x²/45 + y²/5 = (9)/45 + (4)/5 = (1/5) + (4/5) = 5/5 = 1. Perfect, that works.So, that seems correct.Summary of Solutions:1. The hyperbola equation is x²/16 - y²/9 = 1.2. The ellipse equation is x²/45 + y²/5 = 1.Final Answer(1) boxed{dfrac{x^2}{16} - dfrac{y^2}{9} = 1}(2) boxed{dfrac{x^2}{45} + dfrac{y^2}{5} = 1}</think>(1) To find the standard equation of the hyperbola with the same foci as the ellipse (frac{x^2}{40} + frac{y^2}{15} = 1) and an eccentricity of (e = frac{5}{4}):- The foci of the ellipse are calculated using (c^2 = a^2 - b^2). Here, (a^2 = 40) and (b^2 = 15), so (c^2 = 40 - 15 = 25) giving (c = 5). The foci are at ((pm 5, 0)).- For the hyperbola, the foci are the same, so (C = 5). The eccentricity (e = frac{5}{4}) gives (C/A = frac{5}{4}). Solving for (A), we get (A = 4).- Using (C^2 = A^2 + B^2), we find (B^2 = C^2 - A^2 = 25 - 16 = 9). Thus, (B = 3).- The standard equation of the hyperbola is (frac{x^2}{16} - frac{y^2}{9} = 1).(2) To find the equation of the ellipse with center at the origin, passing through point (P(3,2)), foci on the x-axis, and major axis three times the minor axis:- The major axis length is three times the minor axis length, so (a = 3b). The equation is (frac{x^2}{(3b)^2} + frac{y^2}{b^2} = 1) or (frac{x^2}{9b^2} + frac{y^2}{b^2} = 1).- Substituting point (P(3,2)) into the equation: (frac{9}{9b^2} + frac{4}{b^2} = 1). Simplifying, (frac{1}{b^2} + frac{4}{b^2} = 1) gives (b^2 = 5).- Therefore, (a = 3b) leads to (a^2 = 9b^2 = 45).- The equation of the ellipse is (frac{x^2}{45} + frac{y^2}{5} = 1). Final Answer(1) boxed{dfrac{x^2}{16} - dfrac{y^2}{9} = 1}(2) boxed{dfrac{x^2}{45} + dfrac{y^2}{5} = 1},<think>Okay, so I need to solve two problems here. Let me start with the first one.Problem 1: Find the standard equation of the hyperbola that has the same foci as the ellipse (frac{x^{2}}{40} + frac{y^{2}}{15} = 1) and an eccentricity of (e = frac{5}{4}).Alright, first, I need to recall some properties of ellipses and hyperbolas. For an ellipse, the standard equation is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) where (a > b), and the foci are located at ((pm c, 0)) where (c = sqrt{a^2 - b^2}). The eccentricity of an ellipse is given by (e = frac{c}{a}).In this problem, the given ellipse is (frac{x^2}{40} + frac{y^2}{15} = 1). So here, (a^2 = 40) and (b^2 = 15). Therefore, (a = sqrt{40}) and (b = sqrt{15}). Then the distance to the foci (c) would be (sqrt{a^2 - b^2} = sqrt{40 - 15} = sqrt{25} = 5). So the foci of the ellipse are at ((pm 5, 0)).Now, the hyperbola we're looking for has the same foci, which are at ((pm 5, 0)). For a hyperbola, the standard equation is (frac{x^2}{A^2} - frac{y^2}{B^2} = 1), where the foci are at ((pm C, 0)) with (C = sqrt{A^2 + B^2}). The eccentricity of a hyperbola is (e = frac{C}{A}).We know the hyperbola's eccentricity (e = frac{5}{4}) and that its foci are at ((pm 5, 0)). Therefore, (C = 5) for the hyperbola. Using the eccentricity formula, (e = frac{C}{A}), so we can solve for (A):[frac{5}{4} = frac{5}{A} implies A = frac{5 times 4}{5} = 4]So, (A = 4). Then, since (C = sqrt{A^2 + B^2}), we can plug in the known values to find (B^2):[5 = sqrt{4^2 + B^2} implies 5 = sqrt{16 + B^2} implies 25 = 16 + B^2 implies B^2 = 9]Therefore, (B = 3). Now, plugging these into the standard equation of the hyperbola:[frac{x^2}{16} - frac{y^2}{9} = 1]Let me just double-check that. The foci are at (pm 5), which comes from (C = sqrt{16 + 9} = 5), which matches. The eccentricity is (5/4), which is correct because (C/A = 5/4). So that seems right.Problem 2: Given that the center of an ellipse is at the origin, it passes through the point (P(3,2)), its foci are on the (x)-axis, and the length of its major axis is three times the length of its minor axis, find the equation of the ellipse.Okay, let's start with the standard equation of an ellipse centered at the origin with foci on the (x)-axis: (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a > b). The major axis length is (2a) and the minor axis length is (2b). The problem states that the major axis is three times the minor axis, so:[2a = 3 times 2b implies a = 3b]So (a = 3b). Therefore, the equation of the ellipse becomes (frac{x^2}{(3b)^2} + frac{y^2}{b^2} = 1) or (frac{x^2}{9b^2} + frac{y^2}{b^2} = 1).We also know the ellipse passes through the point (P(3,2)). Plugging (x = 3) and (y = 2) into the equation:[frac{3^2}{9b^2} + frac{2^2}{b^2} = 1 implies frac{9}{9b^2} + frac{4}{b^2} = 1 implies frac{1}{b^2} + frac{4}{b^2} = 1 implies frac{5}{b^2} = 1 implies b^2 = 5]Therefore, (b^2 = 5) and since (a = 3b), (a^2 = 9b^2 = 9 times 5 = 45). Hence, the equation of the ellipse is:[frac{x^2}{45} + frac{y^2}{5} = 1]Wait, let me check if the foci are correctly located. The distance of foci from the center is (c = sqrt{a^2 - b^2} = sqrt{45 - 5} = sqrt{40} = 2sqrt{10}), so foci at ((pm 2sqrt{10}, 0)), which are on the x-axis, as required. The major axis length is (2a = 2sqrt{45} = 6sqrt{5}), and minor axis length is (2b = 2sqrt{5}). Indeed, (6sqrt{5} = 3 times 2sqrt{5}), so that checks out. The point (3,2) is on the ellipse:[frac{9}{45} + frac{4}{5} = frac{1}{5} + frac{4}{5} = 1]Perfect. So that seems correct.Final Answer(1) The standard equation of the hyperbola is boxed{dfrac{x^2}{16} - dfrac{y^2}{9} = 1}.(2) The equation of the ellipse is boxed{dfrac{x^2}{45} + dfrac{y^2}{5} = 1}.</think>(1) To find the standard equation of the hyperbola with the same foci as the ellipse (frac{x^2}{40} + frac{y^2}{15} = 1) and an eccentricity of (e = frac{5}{4}):- The given ellipse has (a^2 = 40) and (b^2 = 15). The distance to the foci is (c = sqrt{a^2 - b^2} = sqrt{40 - 15} = 5). Thus, the foci are at ((pm 5, 0)).- For the hyperbola, the foci are at ((pm 5, 0)), so (C = 5). The eccentricity (e = frac{5}{4}) gives (C = eA), leading to (A = 4).- Using (C = sqrt{A^2 + B^2}), we find (B^2 = C^2 - A^2 = 25 - 16 = 9).- The standard equation of the hyperbola is (frac{x^2}{16} - frac{y^2}{9} = 1).(2) To find the equation of the ellipse centered at the origin, passing through (P(3,2)), with foci on the (x)-axis, and major axis three times the minor axis:- The major axis length is three times the minor axis length, so (a = 3b).- The standard form is (frac{x^2}{9b^2} + frac{y^2}{b^2} = 1). Substituting (P(3,2)) gives (frac{9}{9b^2} + frac{4}{b^2} = 1), leading to (b^2 = 5).- Therefore, (a^2 = 9b^2 = 45). The equation of the ellipse is (frac{x^2}{45} + frac{y^2}{5} = 1). Final Answer(1) The standard equation of the hyperbola is boxed{dfrac{x^2}{16} - dfrac{y^2}{9} = 1}.(2) The equation of the ellipse is boxed{dfrac{x^2}{45} + dfrac{y^2}{5} = 1}.]

🔑:(1) Let the standard equation of the hyperbola that has the same foci as the ellipse frac{x^{2}}{40} + frac{y^{2}}{15} = 1 and an eccentricity of e = frac{5}{4} be frac{x^{2}}{m} - frac{y^{2}}{25-m} = 1 (m, 25-m > 0). Then, e = frac{5}{4} = frac{c}{a} = sqrt{1 + frac{25-m}{m}}, solving for m we get m = 16. therefore The required standard equation of the hyperbola is: boxed{frac{x^{2}}{16} - frac{y^{2}}{9} = 1}.(2) Let the required standard equation of the ellipse be: frac{x^{2}}{a^{2}} + frac{y^{2}}{b^{2}} = 1 (a > b > 0). Then, frac{9}{a^{2}} + frac{4}{b^{2}} = 1, 2a = 3 times 2b, solving for b^2 and a^2 we get b^{2} = 5, a^{2} = 45. therefore The standard equation of the ellipse is: boxed{frac{x^{2}}{45} + frac{y^{2}}{5} = 1}.