💡First, I observe that the series alternates between subtraction and addition: (0 - 1 + 2 - 3 + 4 - 5 + dots + 10,002).To simplify the calculation, I can pair the terms:[(0 - 1) + (2 - 3) + (4 - 5) + dots + (10000 - 10001) + 10002]Each pair ((2k - (2k+1))) simplifies to (-1).Next, I determine the number of such pairs. Since the series ends at 10,002, which is an even number, there are:[frac{10002}{2} = 5001 text{ pairs}]However, the last term, 10,002, does not have a pair, so there are actually 5000 pairs and one unpaired term.Calculating the sum of the pairs:[5000 times (-1) = -5000]Adding the unpaired term:[-5000 + 10002 = 5002]

💡First, we calculate the area of the parallelogram (ABCD). The area of a parallelogram is given by the formula:[text{Area}_{text{parallelogram}} = text{base} times text{height}]Given the base (BC = 6) units and the height from (A) to line (BC) is 3 units, the area is:[text{Area}_{text{parallelogram}} = 6 times 3 = 18 text{ square units}]Next, we calculate the area of triangle (ACE). Since (ACE) is a right triangle at (C), we use the formula for the area of a triangle:[text{Area}_{text{triangle}} = frac{1}{2} times text{base} times text{height}]Here, the base is (AC) and the height is (CE). We need to find the length of (AC). Since (AC) is the diagonal of the parallelogram, we use the Pythagorean theorem:[AC = sqrt{BC^2 + text{height}^2} = sqrt{6^2 + 3^2} = sqrt{36 + 9} = sqrt{45} = 3sqrt{5} text{ units}]Now, the area of triangle (ACE) is:[text{Area}_{text{triangle}} = frac{1}{2} times 3sqrt{5} times 2 = 3sqrt{5} text{ square units}]Finally, we add the areas of the parallelogram and the triangle:[text{Total Area} = 18 + 3sqrt{5}]Thus, the combined area of the parallelogram and the triangle is (18 + 3sqrt{5}) square units.

💡Okay, so I have this problem here where I need to find the value of ( w ) given two cubic polynomials. The first polynomial is ( x^3 + 4x^2 + 5x - 13 = 0 ) with roots ( p ), ( q ), and ( r ). The second polynomial is ( x^3 + ux^2 + vx + w = 0 ) with roots ( p+q ), ( q+r ), and ( r+p ). I need to find ( w ).Hmm, I remember that Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. Maybe I can use that here. Let me recall Vieta's formulas for a cubic equation ( x^3 + ax^2 + bx + c = 0 ). The sum of the roots is ( -a ), the sum of the products of the roots taken two at a time is ( b ), and the product of the roots is ( -c ).So, for the first polynomial ( x^3 + 4x^2 + 5x - 13 = 0 ), the sum of the roots ( p + q + r ) should be ( -4 ). That's straightforward. The sum of the products two at a time, which is ( pq + pr + qr ), should be ( 5 ). And the product of the roots ( pqr ) is ( 13 ) because it's ( -(-13) ).Now, moving on to the second polynomial ( x^3 + ux^2 + vx + w = 0 ) with roots ( p+q ), ( q+r ), and ( r+p ). Again, using Vieta's formulas, the sum of the roots ( (p+q) + (q+r) + (r+p) ) should be ( -u ). Let me compute that:( (p+q) + (q+r) + (r+p) = 2p + 2q + 2r = 2(p + q + r) ).We already know ( p + q + r = -4 ), so this sum is ( 2(-4) = -8 ). Therefore, ( -u = -8 ), which means ( u = 8 ). Okay, that's one coefficient found.Next, the sum of the products of the roots taken two at a time for the second polynomial is ( v ). Let's compute that:( (p+q)(q+r) + (q+r)(r+p) + (r+p)(p+q) ).Hmm, that looks a bit complicated. Maybe I can expand each term and then combine like terms. Let me try that.First, expand ( (p+q)(q+r) ):( pq + pr + q^2 + qr ).Similarly, expand ( (q+r)(r+p) ):( qr + qp + r^2 + rp ).And expand ( (r+p)(p+q) ):( rp + rq + p^2 + pq ).Now, adding all these together:( (pq + pr + q^2 + qr) + (qr + qp + r^2 + rp) + (rp + rq + p^2 + pq) ).Let me collect like terms:- ( p^2 ): appears once- ( q^2 ): appears once- ( r^2 ): appears once- ( pq ): appears in the first term, third term, and second term? Wait, let me count:Wait, in the first expansion, ( pq ) is once, in the second expansion, ( qp ) is another ( pq ), and in the third expansion, ( pq ) is once more. So that's three ( pq ) terms.Similarly, ( pr ) appears in the first expansion, and then ( rp ) in the second and third expansions. So that's three ( pr ) terms.And ( qr ) appears in the first expansion, ( qr ) in the second, and ( rq ) in the third. So that's three ( qr ) terms.So putting it all together:( p^2 + q^2 + r^2 + 3pq + 3pr + 3qr ).Hmm, that's a bit messy. Maybe I can express this in terms of ( (p + q + r)^2 ) because I know ( p + q + r ) and ( pq + pr + qr ).Recall that ( (p + q + r)^2 = p^2 + q^2 + r^2 + 2(pq + pr + qr) ).So, ( p^2 + q^2 + r^2 = (p + q + r)^2 - 2(pq + pr + qr) ).We know ( p + q + r = -4 ) and ( pq + pr + qr = 5 ), so:( p^2 + q^2 + r^2 = (-4)^2 - 2(5) = 16 - 10 = 6 ).Okay, so ( p^2 + q^2 + r^2 = 6 ).Now, going back to the sum of the products two at a time for the second polynomial:( p^2 + q^2 + r^2 + 3(pq + pr + qr) = 6 + 3(5) = 6 + 15 = 21 ).So, ( v = 21 ). Got that.Now, the last part is the product of the roots of the second polynomial, which is ( (p+q)(q+r)(r+p) ). This product is equal to ( -w ) because in Vieta's formula for a cubic ( x^3 + ax^2 + bx + c = 0 ), the product of the roots is ( -c ). So, ( (p+q)(q+r)(r+p) = -w ). Therefore, ( w = - (p+q)(q+r)(r+p) ).So, I need to compute ( (p+q)(q+r)(r+p) ) and then take the negative of that to find ( w ).Hmm, how can I compute this product? Let me think. Maybe I can express this in terms of ( p + q + r ) and ( pq + pr + qr ) and ( pqr ), which I already know.Let me denote ( S = p + q + r = -4 ), ( T = pq + pr + qr = 5 ), and ( U = pqr = 13 ).So, ( (p+q)(q+r)(r+p) ). Let me expand this expression step by step.First, note that ( (p+q) = S - r ), ( (q+r) = S - p ), and ( (r+p) = S - q ).So, ( (p+q)(q+r)(r+p) = (S - r)(S - p)(S - q) ).So, this is the product ( (S - r)(S - p)(S - q) ). Hmm, that looks familiar. Isn't that the value of the polynomial evaluated at ( x = S )?Wait, if I have a polynomial with roots ( p ), ( q ), ( r ), then ( (x - p)(x - q)(x - r) = x^3 + 4x^2 + 5x - 13 ). So, evaluating this polynomial at ( x = S = -4 ) would give ( (-4 - p)(-4 - q)(-4 - r) ), which is exactly ( (S - p)(S - q)(S - r) ).So, ( (S - p)(S - q)(S - r) = (-4 - p)(-4 - q)(-4 - r) ).Therefore, ( (p+q)(q+r)(r+p) = (-4 - p)(-4 - q)(-4 - r) ).But wait, the polynomial evaluated at ( x = -4 ) is:( (-4)^3 + 4(-4)^2 + 5(-4) - 13 ).Let me compute that:First, ( (-4)^3 = -64 ).Then, ( 4(-4)^2 = 4*16 = 64 ).Then, ( 5(-4) = -20 ).And the constant term is ( -13 ).So, adding all together:( -64 + 64 - 20 - 13 = (-64 + 64) + (-20 -13) = 0 - 33 = -33 ).So, ( (-4 - p)(-4 - q)(-4 - r) = -33 ).Therefore, ( (p+q)(q+r)(r+p) = -33 ).But wait, earlier I said that ( (p+q)(q+r)(r+p) = (S - r)(S - p)(S - q) = (-4 - p)(-4 - q)(-4 - r) = -33 ).So, ( (p+q)(q+r)(r+p) = -33 ).But ( w = - (p+q)(q+r)(r+p) ), so ( w = -(-33) = 33 ).Therefore, ( w = 33 ).Wait, let me double-check my steps to make sure I didn't make a mistake.1. I used Vieta's formulas correctly for the first polynomial, so ( S = -4 ), ( T = 5 ), ( U = 13 ).2. For the second polynomial, I found ( u = 8 ) and ( v = 21 ) correctly.3. For ( w ), I recognized that ( (p+q)(q+r)(r+p) = (S - r)(S - p)(S - q) ), which is the polynomial evaluated at ( x = S ).4. Plugging ( x = -4 ) into the first polynomial gives ( (-4)^3 + 4(-4)^2 + 5(-4) - 13 = -64 + 64 - 20 -13 = -33 ).5. Therefore, ( (p+q)(q+r)(r+p) = -33 ), so ( w = -(-33) = 33 ).Yes, that seems correct. So, the value of ( w ) is 33.**Final Answer**The value of ( w ) is boxed{33}.

💡Alright, so I have this function f(x) = 2cos²(x/2) + √3 sin x. I need to find its maximum value and the set of x where this maximum occurs. Then, in part II, given that tan(α/2) = 1/2, I have to find f(α). Hmm, okay, let's tackle part I first.Starting with f(x) = 2cos²(x/2) + √3 sin x. I remember there are some trigonometric identities that might help simplify this expression. The first term is 2cos²(x/2). I recall that cos²θ can be written using the double-angle identity: cos²θ = (1 + cos2θ)/2. So, if I let θ = x/2, then cos²(x/2) = (1 + cosx)/2. Therefore, 2cos²(x/2) becomes 2*(1 + cosx)/2, which simplifies to 1 + cosx. So, f(x) simplifies to 1 + cosx + √3 sinx.Now, f(x) = 1 + cosx + √3 sinx. This looks like a combination of sine and cosine terms. I remember that expressions of the form a cosx + b sinx can be rewritten as R sin(x + φ) or R cos(x + φ), where R is the amplitude and φ is the phase shift. Let me try that.First, let's consider the terms cosx and √3 sinx. Let me write them as √3 sinx + cosx. To combine these into a single sine function, I can use the identity: a sinx + b cosx = R sin(x + φ), where R = √(a² + b²) and tanφ = b/a.Here, a is √3 and b is 1. So, R = √[(√3)² + 1²] = √(3 + 1) = √4 = 2. Then, tanφ = b/a = 1/√3, which means φ = π/6 because tan(π/6) = 1/√3.So, √3 sinx + cosx can be written as 2 sin(x + π/6). Therefore, f(x) becomes 1 + 2 sin(x + π/6).Now, the function is f(x) = 1 + 2 sin(x + π/6). The maximum value of sinθ is 1, so the maximum value of f(x) is 1 + 2*1 = 3. That's straightforward.Next, I need to find all x where this maximum occurs. The maximum of sin(x + π/6) is 1 when x + π/6 = π/2 + 2πk, where k is any integer. Solving for x, we get x = π/2 - π/6 + 2πk = (3π/6 - π/6) + 2πk = (2π/6) + 2πk = π/3 + 2πk. So, the set of x where f(x) attains its maximum is x = π/3 + 2πk for all integers k.Alright, that seems solid for part I. Now, moving on to part II: If tan(α/2) = 1/2, find f(α). So, I need to compute f(α) where α is such that tan(α/2) = 1/2.Given f(x) = 2cos²(x/2) + √3 sinx, so f(α) = 2cos²(α/2) + √3 sinα. Hmm, I need to express this in terms of tan(α/2) = 1/2.I remember that there are identities that relate tan(α/2) to sinα and cosα. Specifically, the double-angle formulas. Let me recall:tan(α/2) = t, then sinα = 2t/(1 + t²) and cosα = (1 - t²)/(1 + t²). Also, cos²(α/2) can be expressed in terms of cosα. Wait, cos²(α/2) is (1 + cosα)/2, right?Yes, from the double-angle identity: cos²θ = (1 + cos2θ)/2. So, cos²(α/2) = (1 + cosα)/2.So, let's compute each term step by step.First, let me denote t = tan(α/2) = 1/2.Compute cosα: cosα = (1 - t²)/(1 + t²) = (1 - (1/2)²)/(1 + (1/2)²) = (1 - 1/4)/(1 + 1/4) = (3/4)/(5/4) = 3/5.Compute sinα: sinα = 2t/(1 + t²) = 2*(1/2)/(1 + (1/2)²) = (1)/(5/4) = 4/5.Now, compute cos²(α/2): cos²(α/2) = (1 + cosα)/2 = (1 + 3/5)/2 = (8/5)/2 = 4/5.So, 2cos²(α/2) = 2*(4/5) = 8/5.Next, compute √3 sinα: √3*(4/5) = (4√3)/5.Therefore, f(α) = 8/5 + (4√3)/5 = (8 + 4√3)/5.Hmm, that seems correct. Let me double-check the steps.1. tan(α/2) = 1/2, so t = 1/2.2. Using the identities: - sinα = 2t/(1 + t²) = 2*(1/2)/(1 + 1/4) = 1/(5/4) = 4/5. - cosα = (1 - t²)/(1 + t²) = (1 - 1/4)/(1 + 1/4) = (3/4)/(5/4) = 3/5.3. Compute cos²(α/2) = (1 + cosα)/2 = (1 + 3/5)/2 = (8/5)/2 = 4/5.4. Then, 2cos²(α/2) = 8/5.5. √3 sinα = √3*(4/5) = 4√3/5.6. Adding them together: 8/5 + 4√3/5 = (8 + 4√3)/5.Yes, that seems consistent. I don't see any errors in the calculations.Alternatively, another approach could be to express f(x) in terms of tan(α/2) directly, but I think the method I used is straightforward and less error-prone.So, summarizing:I. The maximum value of f(x) is 3, achieved when x = π/3 + 2πk for any integer k.II. f(α) = (8 + 4√3)/5.**Final Answer**(I) The maximum value is boxed{3}, attained at ( x = frac{pi}{3} + 2kpi ) for any integer ( k ).(II) The value of ( f(alpha) ) is boxed{dfrac{8 + 4sqrt{3}}{5}}.

💡Okay, so I have this problem about two sequences, {a_n} and {b_n}. The sequence {a_n} is an arithmetic sequence, which I remember means that each term increases by a constant difference. The sum of the first n terms of {b_n} is given by S_n = (b_n - 1). Hmm, that's interesting. Also, I know that a_2 = b_1 and a_5 = b_2. Alright, part (I) asks for the general formula for {a_n}. Since it's an arithmetic sequence, the general formula is usually a_n = a_1 + (n-1)d, where a_1 is the first term and d is the common difference. So, I need to find a_1 and d.Given that a_2 = b_1 and a_5 = b_2, I need to find expressions for b_1 and b_2. But wait, the sum of the first n terms of {b_n} is S_n = (b_n - 1). Let me think about that. For n=1, S_1 should be just b_1, right? So S_1 = b_1 = (b_1 - 1). Wait, that seems strange. If S_1 = b_1, then according to the formula, S_1 = (b_1 - 1). So, setting them equal: b_1 = b_1 - 1. Hmm, that would imply 0 = -1, which doesn't make sense. Maybe I did something wrong.Wait, maybe I misinterpreted the formula. It says the sum of the first n terms is S_n = (b_n - 1). So, for n=1, S_1 = b_1 = (b_1 - 1). That still doesn't make sense. Maybe it's a typo or I'm misunderstanding. Alternatively, perhaps it's S_n = (b_n - 1), meaning the sum is equal to b_n minus 1. So, for n=1, S_1 = b_1 = (b_1 - 1). That still leads to 0 = -1, which is impossible. Maybe the formula is different? Or perhaps it's S_n = b_{n} - 1, meaning the sum is equal to the nth term minus 1. That still doesn't resolve the issue for n=1.Wait, perhaps the formula is S_n = (b_n - 1), meaning the sum is equal to b_n minus 1. So, for n=1, S_1 = b_1 = (b_1 - 1). That still implies 0 = -1, which is impossible. Maybe I need to consider that S_n is the sum up to n, so for n=1, S_1 = b_1, and according to the formula, S_1 = (b_1 - 1). So, b_1 = b_1 - 1, which is impossible. Maybe the formula is supposed to be S_n = b_{n+1} - 1? Or maybe S_n = b_n - 1, but starting from n=2?Wait, let me check the original problem again. It says, "the sum of the first n terms of the sequence {b_n} is S_n = (b_n - 1)." So, for n=1, S_1 = b_1 = (b_1 - 1). That still doesn't make sense. Maybe the formula is S_n = (b_{n} - 1), but for n ≥ 2? Or perhaps it's a different interpretation.Alternatively, maybe it's a recursive formula. Let me think. If S_n = (b_n - 1), then S_{n} = b_n - 1. But S_n is also equal to S_{n-1} + b_n. So, substituting, we have S_n = S_{n-1} + b_n. But S_n is also equal to b_n - 1. Therefore, S_{n-1} + b_n = b_n - 1. So, S_{n-1} = -1. That would mean that for all n ≥ 2, S_{n-1} = -1. So, S_1 = -1, S_2 = -1, S_3 = -1, etc. But S_1 is also b_1, so b_1 = -1. Then S_2 = b_1 + b_2 = -1 + b_2 = -1, so b_2 = 0. Similarly, S_3 = b_1 + b_2 + b_3 = -1 + 0 + b_3 = -1, so b_3 = 0. Wait, but then S_n = -1 for all n, which would mean that all terms after the first are zero. But let's check if that works with the given conditions.Given that a_2 = b_1 = -1 and a_5 = b_2 = 0. Since {a_n} is an arithmetic sequence, the common difference d can be found by a_5 - a_2 = 0 - (-1) = 1. The number of steps between a_2 and a_5 is 3 (from n=2 to n=5), so d = 1/3. Therefore, the general formula for {a_n} would be a_n = a_1 + (n-1)(1/3). But we know a_2 = -1, so a_2 = a_1 + (2-1)(1/3) = a_1 + 1/3 = -1. Therefore, a_1 = -1 - 1/3 = -4/3. So, a_n = -4/3 + (n-1)(1/3) = (-4/3) + (n/3 - 1/3) = (n - 5)/3. Hmm, that seems a bit messy, but let's see if it works.Wait, but if S_n = -1 for all n, then b_n = S_n + 1 = 0 for n ≥ 2, which makes b_2 = 0, b_3 = 0, etc. But then a_5 = b_2 = 0, which is consistent with the arithmetic sequence we found. However, the problem states that S_n = (b_n - 1), which for n=1 would imply S_1 = b_1 - 1, but S_1 is also b_1, so b_1 = b_1 - 1, which is impossible. Therefore, my assumption that S_n = (b_n - 1) for all n must be incorrect, or perhaps the formula is misinterpreted.Wait, maybe the formula is S_n = b_n - 1, but not for all n. Maybe it's a different relationship. Let me think again. If S_n is the sum of the first n terms of {b_n}, and it's given by S_n = (b_n - 1), then for n=1, S_1 = b_1 = (b_1 - 1), which is impossible. Therefore, perhaps the formula is S_n = b_{n+1} - 1? Let me check that.If S_n = b_{n+1} - 1, then for n=1, S_1 = b_2 - 1 = b_1. So, b_2 = b_1 + 1. For n=2, S_2 = b_3 - 1 = b_1 + b_2 = b_1 + (b_1 + 1) = 2b_1 + 1. Therefore, b_3 = 2b_1 + 2. Similarly, for n=3, S_3 = b_4 - 1 = b_1 + b_2 + b_3 = b_1 + (b_1 + 1) + (2b_1 + 2) = 4b_1 + 3. So, b_4 = 4b_1 + 4. Hmm, this seems to be forming a pattern where each b_n is related to b_1.But I also know that a_2 = b_1 and a_5 = b_2. Since {a_n} is arithmetic, a_5 = a_2 + 3d. So, a_5 - a_2 = 3d. Given that a_5 = b_2 and a_2 = b_1, then b_2 - b_1 = 3d. But from earlier, b_2 = b_1 + 1, so 1 = 3d, which means d = 1/3. Therefore, the common difference is 1/3.Now, let's try to find b_1. From the arithmetic sequence, a_2 = b_1 = a_1 + d. So, a_1 = b_1 - d = b_1 - 1/3. Similarly, a_5 = a_1 + 4d = (b_1 - 1/3) + 4*(1/3) = b_1 - 1/3 + 4/3 = b_1 + 1. But a_5 is also equal to b_2, which is b_1 + 1. So, that checks out.Now, going back to the sum S_n = b_{n+1} - 1. For n=1, S_1 = b_2 - 1 = b_1. So, b_2 = b_1 + 1. For n=2, S_2 = b_3 - 1 = b_1 + b_2 = b_1 + (b_1 + 1) = 2b_1 + 1. Therefore, b_3 = 2b_1 + 2. For n=3, S_3 = b_4 - 1 = b_1 + b_2 + b_3 = b_1 + (b_1 + 1) + (2b_1 + 2) = 4b_1 + 3. So, b_4 = 4b_1 + 4.Looking at the pattern, it seems that b_n = (2^{n-1})b_1 + (2^{n-1} - 1). Let me test this for n=1: b_1 = (2^{0})b_1 + (2^{0} - 1) = b_1 + 0, which is correct. For n=2: b_2 = (2^{1})b_1 + (2^{1} - 1) = 2b_1 + 1, but earlier we have b_2 = b_1 + 1. Wait, that doesn't match. Hmm, maybe my assumption is wrong.Alternatively, perhaps the sequence {b_n} is geometric. Let me check. If S_n = b_{n+1} - 1, and S_n = S_{n-1} + b_n, then substituting, we get S_n = S_{n-1} + b_n = (b_n - 1) + b_n = 2b_n - 1. But S_n is also equal to b_{n+1} - 1. Therefore, 2b_n - 1 = b_{n+1} - 1, which simplifies to b_{n+1} = 2b_n. So, the sequence {b_n} is geometric with a common ratio of 2.Wait, but earlier we had b_2 = b_1 + 1, and if it's geometric with ratio 2, then b_2 = 2b_1. Therefore, 2b_1 = b_1 + 1, which implies b_1 = 1. So, b_1 = 1, b_2 = 2, b_3 = 4, etc. But wait, earlier we had a_2 = b_1 = 1 and a_5 = b_2 = 2. Since {a_n} is arithmetic, a_5 = a_2 + 3d, so 2 = 1 + 3d, which gives d = 1/3. Therefore, the general formula for {a_n} is a_n = a_1 + (n-1)(1/3). We know a_2 = 1, so a_1 + (2-1)(1/3) = a_1 + 1/3 = 1. Therefore, a_1 = 1 - 1/3 = 2/3. So, a_n = 2/3 + (n-1)(1/3) = (2/3) + (n/3 - 1/3) = (n + 1)/3.Wait, but let's check if this works with the sum S_n = b_{n+1} - 1. For n=1, S_1 = b_2 - 1 = 2 - 1 = 1, which is equal to b_1 = 1. For n=2, S_2 = b_3 - 1 = 4 - 1 = 3, which should be equal to b_1 + b_2 = 1 + 2 = 3. That works. For n=3, S_3 = b_4 - 1 = 8 - 1 = 7, which should be equal to b_1 + b_2 + b_3 = 1 + 2 + 4 = 7. That works too. So, it seems that {b_n} is a geometric sequence with b_1 = 1 and ratio 2, so b_n = 2^{n-1}.But wait, earlier I thought S_n = b_{n+1} - 1, but according to the problem statement, S_n = (b_n - 1). So, there might be a confusion here. Let me clarify.If S_n = b_n - 1, then for n=1, S_1 = b_1 - 1 = b_1, which implies 0 = -1, which is impossible. Therefore, my initial assumption that S_n = b_{n+1} - 1 might be incorrect. Alternatively, perhaps the formula is S_n = b_{n} - 1, but starting from n=2. Let me check.If n=2, S_2 = b_2 - 1. But S_2 = b_1 + b_2. So, b_1 + b_2 = b_2 - 1, which implies b_1 = -1. So, b_1 = -1. Then, for n=1, S_1 = b_1 = -1, which according to the formula S_1 = b_1 - 1 would imply -1 = -1 -1, which is -1 = -2, which is false. Therefore, this approach doesn't work either.Wait, maybe the formula is S_n = (b_{n} - 1), but it's a different relationship. Let me consider that S_n = b_n - 1, so b_n = S_n + 1. But S_n is also equal to S_{n-1} + b_n. Therefore, S_n = S_{n-1} + b_n. Substituting b_n = S_n + 1, we get S_n = S_{n-1} + S_n + 1. Simplifying, 0 = S_{n-1} + 1, so S_{n-1} = -1. Therefore, for all n ≥ 2, S_{n-1} = -1. So, S_1 = -1, S_2 = -1, S_3 = -1, etc. But S_1 is also b_1, so b_1 = -1. Then S_2 = b_1 + b_2 = -1 + b_2 = -1, so b_2 = 0. Similarly, S_3 = b_1 + b_2 + b_3 = -1 + 0 + b_3 = -1, so b_3 = 0. This pattern continues, so all terms after b_1 are zero.But then, given that a_2 = b_1 = -1 and a_5 = b_2 = 0. Since {a_n} is arithmetic, the common difference d can be found by a_5 - a_2 = 0 - (-1) = 1. The number of steps between a_2 and a_5 is 3 (from n=2 to n=5), so d = 1/3. Therefore, the general formula for {a_n} is a_n = a_1 + (n-1)(1/3). We know a_2 = -1, so a_2 = a_1 + (2-1)(1/3) = a_1 + 1/3 = -1. Therefore, a_1 = -1 - 1/3 = -4/3. So, a_n = -4/3 + (n-1)(1/3) = (-4/3) + (n/3 - 1/3) = (n - 5)/3.Wait, but let's check if this works. For n=2, a_2 = (2 - 5)/3 = (-3)/3 = -1, which matches. For n=5, a_5 = (5 - 5)/3 = 0, which also matches. So, the general formula for {a_n} is a_n = (n - 5)/3.Now, for part (II), we need to find the sum of the first n terms of {b_n}, which is S_n. From earlier, we saw that S_n = -1 for all n, but that leads to a contradiction for n=1. However, if we consider that S_n = b_n - 1, and given that b_n = 0 for n ≥ 2, then S_n = 0 - 1 = -1 for n ≥ 2, but S_1 = b_1 = -1. So, S_n = -1 for all n. But that seems too trivial and contradicts the initial condition for n=1.Alternatively, perhaps the formula S_n = b_n - 1 is only valid for n ≥ 2, and S_1 is defined separately. If that's the case, then for n=1, S_1 = b_1 = -1, and for n ≥ 2, S_n = b_n - 1. But then, for n=2, S_2 = b_2 - 1 = 0 - 1 = -1, which is consistent with S_2 = b_1 + b_2 = -1 + 0 = -1. Similarly, for n=3, S_3 = b_3 - 1 = 0 - 1 = -1, which matches S_3 = b_1 + b_2 + b_3 = -1 + 0 + 0 = -1. So, in this case, S_n = -1 for all n ≥ 1.But that seems a bit strange because the sum of the first n terms is always -1, regardless of n. However, given the conditions, it seems to hold. Therefore, the sum S_n = -1 for all n.Wait, but let me double-check. If S_n = -1 for all n, then b_n = S_n - S_{n-1} = (-1) - (-1) = 0 for n ≥ 2, and b_1 = S_1 = -1. So, the sequence {b_n} is -1, 0, 0, 0, ..., which satisfies S_n = -1 for all n. Therefore, the sum of the first n terms is always -1.But the problem statement says "the sum of the first n terms of the sequence {b_n} is S_n = (b_n - 1)". If S_n = -1, then b_n - 1 = -1, so b_n = 0 for all n. But that contradicts b_1 = -1. Therefore, there must be a misunderstanding in the problem statement.Alternatively, perhaps the formula is S_n = b_{n} - 1, but it's a different interpretation. Maybe it's S_n = (b_n - 1), meaning the sum is equal to b_n minus 1, but not necessarily for all n. Or perhaps it's a misstatement, and the correct formula is S_n = b_{n+1} - 1, which would make more sense.Given the confusion, I think the correct approach is to assume that S_n = b_{n+1} - 1, which leads to a consistent solution where {b_n} is a geometric sequence with b_1 = 1 and ratio 2, so b_n = 2^{n-1}, and S_n = 2^n - 1. This would satisfy S_n = b_{n+1} - 1, as S_n = 2^{n} - 1 = b_{n+1} - 1, since b_{n+1} = 2^{n}.But wait, let's check this with the given conditions. If S_n = 2^n - 1, then b_n = S_n - S_{n-1} = (2^n - 1) - (2^{n-1} - 1) = 2^n - 2^{n-1} = 2^{n-1}. So, b_n = 2^{n-1}. Therefore, b_1 = 1, b_2 = 2, b_3 = 4, etc. Then, a_2 = b_1 = 1 and a_5 = b_2 = 2. Since {a_n} is arithmetic, a_5 = a_2 + 3d, so 2 = 1 + 3d, which gives d = 1/3. Therefore, a_n = a_1 + (n-1)(1/3). We know a_2 = 1, so a_1 + (2-1)(1/3) = a_1 + 1/3 = 1, so a_1 = 2/3. Therefore, a_n = 2/3 + (n-1)(1/3) = (2 + n - 1)/3 = (n + 1)/3.Wait, but earlier I thought S_n = -1, but that led to contradictions. So, perhaps the correct interpretation is S_n = b_{n+1} - 1, leading to S_n = 2^n - 1 and b_n = 2^{n-1}. This seems consistent with the given conditions and avoids the contradiction for n=1.Therefore, the general formula for {a_n} is a_n = (n + 1)/3, and the sum S_n = 2^n - 1.But let me verify this again. If S_n = 2^n - 1, then for n=1, S_1 = 2^1 - 1 = 1, which should be equal to b_1. So, b_1 = 1. For n=2, S_2 = 2^2 - 1 = 3, which should be b_1 + b_2 = 1 + b_2 = 3, so b_2 = 2. Similarly, S_3 = 2^3 - 1 = 7, which should be b_1 + b_2 + b_3 = 1 + 2 + b_3 = 7, so b_3 = 4. This pattern continues, so {b_n} is indeed a geometric sequence with b_n = 2^{n-1}.Given that a_2 = b_1 = 1 and a_5 = b_2 = 2, and {a_n} is arithmetic, we found that a_n = (n + 1)/3. Let's check a few terms:a_1 = (1 + 1)/3 = 2/3a_2 = (2 + 1)/3 = 1a_3 = (3 + 1)/3 = 4/3a_4 = (4 + 1)/3 = 5/3a_5 = (5 + 1)/3 = 2This seems correct, with a common difference of 1/3.Therefore, the answers are:(I) a_n = (n + 1)/3(II) S_n = 2^n - 1

💡Okay, so I have this problem about a rectangle folding. Let me try to visualize it. The rectangle has one side of length 5 and the other side less than 4. When it's folded so that two opposite corners coincide, the crease formed is √6. I need to find the length of the other side.First, let me sketch the rectangle in my mind. Let's label the rectangle ABCD, where AB is 5 units, and BC is the unknown length, let's call it 't', which is less than 4. So, AB = 5, BC = t, CD = 5, and DA = t.Now, when the rectangle is folded so that two opposite corners coincide, I assume it's folded along the diagonal. Wait, but the crease isn't the diagonal itself because the crease is given as √6, which is shorter than the diagonal. So, maybe it's not the diagonal. Hmm.Let me think. If we fold the rectangle such that two opposite corners coincide, the crease will be the perpendicular bisector of the diagonal. So, if we have rectangle ABCD, and we fold corner A onto corner C, the crease will be the line that perpendicularly bisects the diagonal AC.So, diagonal AC has length √(5² + t²) = √(25 + t²). The crease is the perpendicular bisector of this diagonal, so it will intersect AC at its midpoint. The length of the crease is given as √6.I remember that in a rectangle, the length of the crease formed by folding along the perpendicular bisector of the diagonal can be calculated using some geometric properties. Maybe I can use similar triangles or the Pythagorean theorem here.Let me denote the midpoint of AC as O. So, AO = OC = (√(25 + t²))/2. The crease, let's call it EF, is perpendicular to AC and passes through O. The length of EF is √6.Now, I need to find the relationship between the sides of the rectangle and the length of the crease. Maybe I can consider triangle AOE, where E is one end of the crease on side AB, and F is the other end on side AD.Wait, actually, since the crease is the perpendicular bisector, it will intersect the sides of the rectangle. Let me denote the points where the crease intersects the sides AB and AD as E and F respectively.So, triangle AOE is a right triangle with AO as one leg, OE as the other leg, and AE as the hypotenuse. But I'm not sure if that's the right approach.Alternatively, maybe I can use coordinate geometry. Let me place the rectangle on a coordinate system with point A at (0,0), B at (5,0), C at (5,t), and D at (0,t). The diagonal AC goes from (0,0) to (5,t). The midpoint O is at (2.5, t/2).The crease EF is the perpendicular bisector of AC. The slope of AC is (t - 0)/(5 - 0) = t/5. Therefore, the slope of the perpendicular bisector EF is -5/t.Since EF passes through O, its equation is y - t/2 = (-5/t)(x - 2.5).Now, I need to find where this crease intersects the sides of the rectangle. Let's find the intersection points with AB and AD.First, intersection with AB: AB is the bottom side from (0,0) to (5,0). On AB, y = 0. Plugging into the equation of EF:0 - t/2 = (-5/t)(x - 2.5)- t/2 = (-5/t)(x - 2.5)Multiply both sides by t:- t²/2 = -5(x - 2.5)Multiply both sides by -1:t²/2 = 5(x - 2.5)Divide both sides by 5:t²/10 = x - 2.5So, x = 2.5 + t²/10Similarly, intersection with AD: AD is the left side from (0,0) to (0,t). On AD, x = 0. Plugging into the equation of EF:y - t/2 = (-5/t)(0 - 2.5)y - t/2 = (-5/t)(-2.5)y - t/2 = (12.5)/tSo, y = t/2 + 12.5/tNow, the crease EF has endpoints at (2.5 + t²/10, 0) and (0, t/2 + 12.5/t). The length of EF is given as √6. So, let's calculate the distance between these two points.Distance formula:√[( (2.5 + t²/10 - 0)^2 + (0 - (t/2 + 12.5/t))^2 )] = √6Simplify the expression inside the square root:First, the x-coordinate difference: 2.5 + t²/10The y-coordinate difference: - (t/2 + 12.5/t)So, the distance squared is:(2.5 + t²/10)^2 + (t/2 + 12.5/t)^2 = 6Let me compute each term separately.First term: (2.5 + t²/10)^2= (5/2 + t²/10)^2= (25/4) + 2*(5/2)*(t²/10) + (t²/10)^2= 25/4 + (5t²)/10 + t^4/100= 25/4 + t²/2 + t^4/100Second term: (t/2 + 12.5/t)^2= (t/2 + 25/(2t))^2= (t/2)^2 + 2*(t/2)*(25/(2t)) + (25/(2t))^2= t²/4 + (25/2) + 625/(4t²)Now, add both terms together:25/4 + t²/2 + t^4/100 + t²/4 + 25/2 + 625/(4t²) = 6Combine like terms:25/4 + 25/2 = 25/4 + 50/4 = 75/4t²/2 + t²/4 = (2t² + t²)/4 = 3t²/4t^4/100 + 625/(4t²)So, the equation becomes:75/4 + 3t²/4 + t^4/100 + 625/(4t²) = 6Multiply both sides by 100t² to eliminate denominators:75/4 * 100t² + 3t²/4 * 100t² + t^4/100 * 100t² + 625/(4t²) * 100t² = 6 * 100t²Simplify each term:75/4 * 100t² = 75 * 25t² = 1875t²3t²/4 * 100t² = 3 * 25t^4 = 75t^4t^4/100 * 100t² = t^6625/(4t²) * 100t² = 625 * 25 = 15625Right side: 6 * 100t² = 600t²So, the equation becomes:1875t² + 75t^4 + t^6 + 15625 = 600t²Bring all terms to the left side:t^6 + 75t^4 + 1875t² + 15625 - 600t² = 0Combine like terms:t^6 + 75t^4 + (1875 - 600)t² + 15625 = 0t^6 + 75t^4 + 1275t² + 15625 = 0Hmm, this seems complicated. Maybe I made a mistake in the calculations. Let me double-check.Wait, when I multiplied 75/4 by 100t², that should be 75/4 * 100t² = 75 * 25t² = 1875t². That seems correct.Similarly, 3t²/4 * 100t² = 3 * 25t^4 = 75t^4. Correct.t^4/100 * 100t² = t^6. Correct.625/(4t²) * 100t² = 625 * 25 = 15625. Correct.Right side: 6 * 100t² = 600t². Correct.So, the equation is indeed:t^6 + 75t^4 + 1275t² + 15625 = 600t²Wait, but 1875t² - 600t² is 1275t², correct.So, moving 600t² to the left:t^6 + 75t^4 + 1275t² + 15625 = 0This is a sixth-degree equation, which is quite complex. Maybe there's a simpler way to approach this problem.Let me think again. Maybe instead of using coordinate geometry, I can use properties of rectangles and creases.When folding a rectangle along the perpendicular bisector of its diagonal, the crease length can be related to the sides of the rectangle. I recall that the length of the crease (EF) can be found using the formula:EF = (2 * AB * BC) / √(AB² + BC²)Wait, is that correct? Let me verify.Alternatively, I think the length of the crease can be found using the formula:EF = (2 * AB * BC) / √(AB² + BC²)But let me derive it properly.Consider the rectangle ABCD with AB = 5, BC = t. The diagonal AC has length √(25 + t²). The crease EF is the perpendicular bisector of AC, so it intersects AC at its midpoint O.The length of EF can be found by considering the right triangles formed by the crease and the sides of the rectangle.Let me denote the distance from O to E as 'a' and from O to F as 'b'. Since EF is perpendicular to AC, we can use similar triangles.In triangle AOE and triangle ABC, they are similar because both are right triangles and share an angle.So, AO / AB = OE / BCAO is half of AC, so AO = √(25 + t²)/2Thus,(√(25 + t²)/2) / 5 = OE / tSo,OE = (√(25 + t²)/2) * (t / 5) = (√(25 + t²) * t) / 10Similarly, the length of EF is twice OE because EF is the full length of the crease, so:EF = 2 * OE = 2 * (√(25 + t²) * t) / 10 = (√(25 + t²) * t) / 5But we are given that EF = √6. So,(√(25 + t²) * t) / 5 = √6Multiply both sides by 5:√(25 + t²) * t = 5√6Square both sides:(25 + t²) * t² = 25 * 6(25 + t²) * t² = 150Expand the left side:25t² + t^4 = 150Rearrange:t^4 + 25t² - 150 = 0Let me set u = t², so the equation becomes:u² + 25u - 150 = 0Now, solve for u using the quadratic formula:u = [-25 ± √(25² + 4 * 150)] / 2u = [-25 ± √(625 + 600)] / 2u = [-25 ± √1225] / 2u = [-25 ± 35] / 2We have two solutions:u = (-25 + 35)/2 = 10/2 = 5u = (-25 - 35)/2 = -60/2 = -30Since u = t² cannot be negative, we discard -30. So, u = 5.Thus, t² = 5, so t = √5.But wait, the problem states that the other side is less than 4. √5 is approximately 2.236, which is less than 4, so that's valid.Therefore, the length of the other side is √5.I think this makes sense. Let me just verify the steps quickly.1. Folded rectangle, crease is perpendicular bisector of diagonal.2. Used similar triangles to relate OE and EF.3. Derived equation EF = (√(25 + t²) * t)/5 = √64. Solved for t and got t = √5.Yes, that seems correct. So, the answer should be √5, which is option D.