answer：Alright, so I've got this problem here about a sequence {a_n} where the sum of the first n terms is denoted as S_n. They give me that a_1 is 1, and for any positive integer n, S_n equals n squared times a_n. I need to figure out what S_n is, and then choose the correct expression from the options given.Okay, let's start by understanding what's given. The first term, a_1, is 1. That's straightforward. Then, for any n, the sum of the first n terms, S_n, is equal to n squared times the nth term, a_n. So, S_n = n² * a_n.Hmm, that seems interesting. So, the sum of the first n terms is related directly to the nth term itself. I wonder how that relationship can help me find a general formula for S_n.Let me write down what I know:1. a_1 = 12. S_n = a_1 + a_2 + a_3 + ... + a_n3. S_n = n² * a_nSo, for n=1, S_1 = a_1 = 1. Plugging into the formula, S_1 = 1² * a_1 = 1*1 = 1. That checks out.Now, let's try n=2. S_2 = a_1 + a_2. According to the formula, S_2 = 2² * a_2 = 4 * a_2. But S_2 is also equal to a_1 + a_2, which is 1 + a_2. So, setting them equal:1 + a_2 = 4 * a_2Subtracting a_2 from both sides:1 = 3 * a_2So, a_2 = 1/3.Okay, that's good. Now, let's move on to n=3. S_3 = a_1 + a_2 + a_3. According to the formula, S_3 = 3² * a_3 = 9 * a_3. But S_3 is also 1 + 1/3 + a_3. So:1 + 1/3 + a_3 = 9 * a_3Combining the constants:4/3 + a_3 = 9 * a_3Subtracting a_3 from both sides:4/3 = 8 * a_3So, a_3 = (4/3) / 8 = (4/3) * (1/8) = 1/6.Alright, so a_3 is 1/6. Let's see if we can spot a pattern here. So far, we have:a_1 = 1a_2 = 1/3a_3 = 1/6Hmm, that looks familiar. 1, 1/3, 1/6... It seems like each term is half of the previous term's denominator. Wait, 1 is 1/1, then 1/3, then 1/6. The denominators are 1, 3, 6. Those are triangular numbers. 1, 3, 6, 10, 15,... which are sums of natural numbers.So, maybe a_n is 1 divided by the nth triangular number. The nth triangular number is n(n+1)/2. So, a_n = 2 / [n(n+1)].Let me test that. For n=1: 2 / [1*2] = 1, which matches a_1=1.For n=2: 2 / [2*3] = 2/6 = 1/3, which matches a_2=1/3.For n=3: 2 / [3*4] = 2/12 = 1/6, which matches a_3=1/6.Great, so it seems like a_n = 2 / [n(n+1)].Now, since S_n is the sum of the first n terms, let's write that out:S_n = a_1 + a_2 + a_3 + ... + a_nGiven that a_k = 2 / [k(k+1)], we can write:S_n = sum_{k=1}^n [2 / (k(k+1))]This looks like a telescoping series. Remember, telescoping series are those where terms cancel out when you write them all out. Let's see if we can express 2 / [k(k+1)] as a difference of two fractions.Yes, partial fractions. Let's decompose 2 / [k(k+1)].Let me set 2 / [k(k+1)] = A/k + B/(k+1). Multiplying both sides by k(k+1):2 = A(k+1) + BkExpanding:2 = A k + A + B kCombine like terms:2 = (A + B)k + ANow, this must hold for all k, so the coefficients of like terms must be equal on both sides. On the left side, the coefficient of k is 0, and the constant term is 2. On the right side, the coefficient of k is (A + B), and the constant term is A.So, setting up equations:A + B = 0A = 2From the second equation, A = 2. Plugging into the first equation:2 + B = 0 => B = -2Therefore, 2 / [k(k+1)] = 2/k - 2/(k+1)So, our sum S_n becomes:S_n = sum_{k=1}^n [2/k - 2/(k+1)]Let's write out the terms:For k=1: 2/1 - 2/2For k=2: 2/2 - 2/3For k=3: 2/3 - 2/4...For k=n: 2/n - 2/(n+1)Now, let's add them all up:(2/1 - 2/2) + (2/2 - 2/3) + (2/3 - 2/4) + ... + (2/n - 2/(n+1))Notice how the -2/2 cancels with +2/2, the -2/3 cancels with +2/3, and so on, all the way up to -2/n cancels with +2/n. What's left is the first term of the first fraction and the second term of the last fraction:2/1 - 2/(n+1) = 2 - 2/(n+1)So, S_n = 2 - 2/(n+1)Let's simplify that:S_n = 2 - 2/(n+1) = [2(n+1) - 2] / (n+1) = [2n + 2 - 2] / (n+1) = 2n / (n+1)So, S_n = 2n / (n+1)Looking back at the options given:A: 2n/(n+1)B: (3n-1)/(n+1)C: (2n+1)/(n+2)D: 2n/(n+2)Our derived expression matches option A: 2n/(n+1)Just to double-check, let's verify with n=1, 2, 3.For n=1: S_1 = 2*1/(1+1) = 2/2 = 1, which matches a_1=1.For n=2: S_2 = 2*2/(2+1) = 4/3, which is 1 + 1/3 = 4/3. Correct.For n=3: S_3 = 2*3/(3+1) = 6/4 = 3/2, which is 1 + 1/3 + 1/6 = 1 + 1/3 + 1/6 = (6/6 + 2/6 + 1/6) = 9/6 = 3/2. Correct.So, everything checks out. The expression for S_n is indeed 2n/(n+1), which is option A.

answer：Okay, so I've got these three trigonometric expressions to evaluate. Let me take them one by one and see if I can figure them out. I remember some trigonometric identities, but I might need to look some up or think through them carefully.Starting with the first expression: cos²15° - sin²15°. Hmm, that looks familiar. I think there's an identity that relates cos²θ - sin²θ to something else. Let me recall... Oh yeah, it's the double-angle formula for cosine! So, cos²θ - sin²θ equals cos(2θ). That means cos²15° - sin²15° should be equal to cos(30°). Okay, cos(30°) is a standard angle. I remember that cos(30°) is √3/2. So, that should be the value for the first expression. That wasn't too bad.Moving on to the second expression: (1 + tan15°)/(1 - tan15°). Hmm, this looks like a fraction involving tangent. I wonder if there's a way to simplify this using tangent addition formulas. I recall that tan(A + B) = (tanA + tanB)/(1 - tanA tanB). If I set A = 45° and B = 15°, then tan(45° + 15°) would be tan60°, which is √3. Let me check: tan45° is 1, so plugging into the formula, (1 + tan15°)/(1 - tan15°) is indeed tan60°, which is √3. So, that simplifies nicely to √3. Cool, that worked out.Now, the third expression is a bit more complicated: sin²10° + cos²55° + √2 sin10° cos55°. Hmm, okay, so we have squares of sine and cosine, and a cross term with a square root. I need to see if I can combine these terms somehow or use some identities to simplify.First, let's recall that sin²θ + cos²θ = 1, but here the angles are different: 10° and 55°. Maybe I can express one in terms of the other? Wait, 55° is 45° + 10°, right? So, cos55° is cos(45° + 10°). Maybe I can expand that using the cosine addition formula: cos(A + B) = cosA cosB - sinA sinB. So, cos55° = cos45° cos10° - sin45° sin10°. Since cos45° and sin45° are both √2/2, that becomes (√2/2)(cos10° - sin10°). Hmm, not sure if that helps yet.Looking back at the expression: sin²10° + cos²55° + √2 sin10° cos55°. Let me see if I can rewrite cos²55° using the identity cos²θ = (1 + cos2θ)/2. So, cos²55° = (1 + cos110°)/2. Similarly, sin²10° can be written as (1 - cos20°)/2. So, substituting these into the expression:sin²10° + cos²55° + √2 sin10° cos55° = (1 - cos20°)/2 + (1 + cos110°)/2 + √2 sin10° cos55°.Simplifying the first two terms: (1 - cos20° + 1 + cos110°)/2 = (2 - cos20° + cos110°)/2 = 1 - (cos20° - cos110°)/2.Now, cos110° is equal to cos(90° + 20°), which is -sin20°. So, cos110° = -sin20°. Therefore, cos20° - cos110° = cos20° - (-sin20°) = cos20° + sin20°. So, the expression becomes 1 - (cos20° + sin20°)/2.Now, let's look at the remaining term: √2 sin10° cos55°. Earlier, I expressed cos55° as (√2/2)(cos10° - sin10°). So, substituting that in:√2 sin10° cos55° = √2 sin10° * (√2/2)(cos10° - sin10°) = (√2 * √2)/2 sin10° (cos10° - sin10°) = (2)/2 sin10° (cos10° - sin10°) = sin10° (cos10° - sin10°).Expanding that: sin10° cos10° - sin²10°.So, putting it all together, the entire expression is:1 - (cos20° + sin20°)/2 + sin10° cos10° - sin²10°.Hmm, let's see if we can simplify this further. Let's look at the term sin10° cos10°. I remember that sin2θ = 2 sinθ cosθ, so sin10° cos10° = (1/2) sin20°. Similarly, sin²10° can be written as (1 - cos20°)/2.So, substituting these:1 - (cos20° + sin20°)/2 + (1/2) sin20° - (1 - cos20°)/2.Let's simplify term by term:1 - (cos20° + sin20°)/2 + (1/2) sin20° - (1 - cos20°)/2.First, let's distribute the negative signs:1 - (cos20°)/2 - (sin20°)/2 + (sin20°)/2 - (1)/2 + (cos20°)/2.Now, let's combine like terms:- (cos20°)/2 + (cos20°)/2 cancels out.- (sin20°)/2 + (sin20°)/2 cancels out.So, we're left with 1 - 1/2 = 1/2.Wow, that simplified down to 1/2. So, the third expression equals 1/2.Let me just recap to make sure I didn't make any mistakes:1. For the first expression, used the double-angle identity for cosine, which worked out nicely to √3/2.2. For the second expression, recognized the tangent addition formula and saw that it simplifies to tan60°, which is √3.3. For the third expression, it was a bit more involved. I used the Pythagorean identity to rewrite the squares, then used the cosine addition formula to express cos55°, and then carefully expanded and simplified each term. After a lot of algebra, it all canceled out except for 1/2.I think that's correct. I didn't see any errors in my steps, and each identity I used seems appropriate. So, I feel confident about these answers.

answer：To find the value of ( x ) such that the arithmetic mean of ( x ), ( 3x ), 1000, and 3000 is 2018, follow these steps:1. **Understand the formula for arithmetic mean**: The arithmetic mean of a set of numbers is the sum of the numbers divided by the count of numbers. In this case, there are four numbers: ( x ), ( 3x ), 1000, and 3000.2. **Set up the equation**: [ frac{x + 3x + 1000 + 3000}{4} = 2018 ]3. **Combine like terms**: [ frac{4x + 4000}{4} = 2018 ]4. **Simplify the equation**: [ x + 1000 = 2018 ]5. **Solve for ( x )**: [ x = 2018 - 1000 ] [ x = 1018 ]The value of ( x ) is **1018**.

answer：Okay, so I have this problem here: there's a point (x₀, y₀) on the parabola y² = 8x, and it's at a distance of 3 from the focus of the parabola. I need to find the absolute value of y₀, which is |y₀|. Hmm, alright, let me think about how to approach this.First, I remember that the standard form of a parabola that opens to the right is y² = 4px, where p is the distance from the vertex to the focus. Comparing that to the given equation y² = 8x, it looks like 4p = 8, so p must be 2. That means the focus of this parabola is at (p, 0), which would be (2, 0). Got that down.Now, the problem says the point (x₀, y₀) is on the parabola and is 3 units away from the focus. So, I need to use the distance formula between the point (x₀, y₀) and the focus (2, 0). The distance formula is sqrt[(x₀ - 2)² + (y₀ - 0)²] = 3. Squaring both sides to eliminate the square root gives (x₀ - 2)² + y₀² = 9. That's one equation.But since the point (x₀, y₀) is on the parabola y² = 8x, I can substitute y₀² with 8x₀. So, replacing y₀² in the distance equation, I get (x₀ - 2)² + 8x₀ = 9. Let me expand (x₀ - 2)²: that's x₀² - 4x₀ + 4. So plugging that back in, the equation becomes x₀² - 4x₀ + 4 + 8x₀ = 9.Combining like terms: x₀² + 4x₀ + 4 = 9. Wait, that simplifies to x₀² + 4x₀ - 5 = 0. Hmm, okay, so I have a quadratic equation in terms of x₀. Let me try to factor this. Looking for two numbers that multiply to -5 and add to 4. Hmm, 5 and -1? Yeah, because 5 * (-1) = -5 and 5 + (-1) = 4. So, the equation factors to (x₀ + 5)(x₀ - 1) = 0. Therefore, x₀ = -5 or x₀ = 1.But wait a second, the parabola is y² = 8x, which means x must be non-negative because y² is always non-negative. So, x₀ can't be negative. Therefore, x₀ = -5 is not a valid solution. So, the only valid solution is x₀ = 1.Now that I have x₀, I can find y₀ by plugging it back into the equation of the parabola. So, y₀² = 8x₀ = 8*1 = 8. Therefore, y₀ = sqrt(8) or y₀ = -sqrt(8). Simplifying sqrt(8), that's 2*sqrt(2). So, y₀ is either 2√2 or -2√2. Since the problem asks for |y₀|, the absolute value, it doesn't matter if it's positive or negative. So, |y₀| = 2√2.Let me just double-check my steps to make sure I didn't make any mistakes. I found the focus correctly at (2, 0). Then I set up the distance equation correctly, substituted y₀² with 8x₀, expanded and simplified the equation, solved the quadratic, discarded the negative x solution because it's not valid for the parabola, and then found y₀ from the valid x₀. Everything seems to check out.I think that's the correct answer. |y₀| is 2√2.**Final Answer**The absolute value of ( y_0 ) is boxed{2sqrt{2}}.

answer：Okay, so I have this problem here about finding the length of the median to side BC in triangle ABC. The coordinates of the vertices are given: A is at (3, 3, 2), B is at (4, -3, 7), and C is at (0, 5, 1). The options are A: 2, B: 3, C: 4, D: 5. I need to figure out which one is the correct answer.First, let me recall what a median of a triangle is. A median is a line segment joining a vertex to the midpoint of the opposite side. So, in this case, the median to side BC would be the line segment from vertex A to the midpoint of BC.Alright, so the first step is to find the midpoint of BC. To find the midpoint, I can use the midpoint formula, which is the average of the corresponding coordinates of the two points. So, if I have points B(4, -3, 7) and C(0, 5, 1), the midpoint M will have coordinates:M_x = (B_x + C_x)/2 = (4 + 0)/2 = 2 M_y = (B_y + C_y)/2 = (-3 + 5)/2 = 1 M_z = (B_z + C_z)/2 = (7 + 1)/2 = 4So, the midpoint M is at (2, 1, 4). Got that.Now, the median is the line segment from point A(3, 3, 2) to point M(2, 1, 4). To find the length of this median, I need to calculate the distance between points A and M.The distance formula in three dimensions is similar to the two-dimensional case, but with an extra term for the z-coordinate. The formula is:Distance = sqrt[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]So, plugging in the coordinates of A and M:Distance = sqrt[(2 - 3)^2 + (1 - 3)^2 + (4 - 2)^2]Let me compute each part step by step.First, (2 - 3) is -1, and squaring that gives 1.Next, (1 - 3) is -2, and squaring that gives 4.Then, (4 - 2) is 2, and squaring that gives 4.Adding these up: 1 + 4 + 4 = 9.Taking the square root of 9 gives 3.So, the length of the median from A to BC is 3 units. Looking back at the options, that's option B.Wait, let me double-check my calculations to make sure I didn't make any mistakes.Midpoint M: (4+0)/2=2, (-3+5)/2=1, (7+1)/2=4. That seems correct.Distance from A(3,3,2) to M(2,1,4):x difference: 2-3=-1, squared is 1 y difference: 1-3=-2, squared is 4 z difference: 4-2=2, squared is 4 Total: 1+4+4=9 Square root of 9 is 3. Yep, that's correct.Hmm, just to be thorough, maybe I should visualize this or think about it another way. Since it's a median, it should be shorter than the sides of the triangle, but in this case, the sides might be longer. Let me quickly compute the lengths of sides AB, BC, and AC to get a sense.Length of AB: distance between A(3,3,2) and B(4,-3,7)x difference: 4-3=1, squared is 1 y difference: -3-3=-6, squared is 36 z difference:7-2=5, squared is 25 Total:1+36+25=62 Distance: sqrt(62) ≈ 7.87Length of BC: distance between B(4,-3,7) and C(0,5,1)x difference:0-4=-4, squared is 16 y difference:5-(-3)=8, squared is 64 z difference:1-7=-6, squared is 36 Total:16+64+36=116 Distance: sqrt(116) ≈ 10.77Length of AC: distance between A(3,3,2) and C(0,5,1)x difference:0-3=-3, squared is 9 y difference:5-3=2, squared is 4 z difference:1-2=-1, squared is 1 Total:9+4+1=14 Distance: sqrt(14) ≈ 3.74So, sides are approximately 7.87, 10.77, and 3.74. The median is 3, which is shorter than all sides, which makes sense because a median is always shorter than the sides it connects.Wait, but in a triangle, a median can sometimes be longer than some sides, depending on the triangle. But in this case, 3 is shorter than all sides, which is fine.Alternatively, maybe I can use vectors to confirm. The vector from A to M is M - A, which is (2-3, 1-3, 4-2) = (-1, -2, 2). The length of this vector is sqrt[(-1)^2 + (-2)^2 + (2)^2] = sqrt[1 + 4 + 4] = sqrt[9] = 3. Yep, same result.Another way: maybe using coordinates to plot, but since it's 3D, plotting is a bit more complex. But since all calculations give me 3, I think that's solid.So, I feel confident that the length of the median is 3, which is option B.**Final Answer**boxed{B}

answer：Okay, so I'm trying to figure out how Wanda the Worm can eat 100,000 numbers in the first 2011 rows of Pascal's triangle without ever eating three numbers a, b, and c such that a + b = c. That seems tricky, but let me break it down step by step.First, I know that Pascal's triangle is constructed such that each entry is the sum of the two directly above it. So, each number is a combination, like C(n, k) which is the number of ways to choose k elements from a set of n elements. The top of the triangle is C(0,0) = 1.Wanda starts at the top, eats 1, and then moves to an adjacent number. She can't go back to a number she's already eaten, and she can't eat three numbers where one is the sum of the other two. So, I need to find a path for her that avoids such triples.Hmm, the problem mentions that she isn't restricted to the first 2011 rows, but she needs to eat 100,000 numbers within those rows. So, maybe she can move beyond the 2011 rows to find a path that allows her to collect 100,000 numbers without violating the sum condition.I remember that in Pascal's triangle, numbers can be even or odd. Maybe focusing on odd numbers could help because they don't add up to another odd number. Wait, let me think: if you add two odd numbers, you get an even number. So, if Wanda only eats odd numbers, she can't have a situation where a + b = c because c would have to be even, and she's only eating odd numbers. That seems promising.But how many odd numbers are there in the first 2011 rows? I know that the number of odd numbers in the nth row of Pascal's triangle corresponds to 2 raised to the number of ones in the binary representation of n. So, for example, row 0 has 1 odd number, row 1 has 2, row 2 has 2, row 3 has 4, and so on. This is related to Sierpiński triangles and fractal patterns in Pascal's triangle.If I can find a way to traverse through these odd numbers without getting stuck, that would solve the problem. But how do I ensure that Wanda can move through them without forming a triple a, b, c where a + b = c? Since she's only eating odd numbers, c would have to be even, so she can't form such a triple because she's not eating even numbers. Wait, is that right?Let me double-check: if a and b are odd, then a + b is even. Since Wanda is only eating odd numbers, she won't eat an even number c, so she can't have a situation where a + b = c because c isn't in her path. So, as long as she sticks to eating only odd numbers, she won't violate the condition. That seems correct.Now, how many odd numbers are there in the first 2011 rows? Each row n has 2^{number of ones in binary representation of n} odd numbers. So, for n from 0 to 2010, we need to sum 2^{number of ones in binary(n)}.But calculating that directly seems complicated. Maybe there's a pattern or a way to estimate it. I remember that the total number of odd numbers in the first N rows is roughly proportional to N * log N, but I'm not sure. Alternatively, maybe it's exponential in the number of bits of N.Wait, 2011 in binary is 11111011011, which has 8 ones. So, the number of odd numbers in row 2011 is 2^8 = 256. But that's just one row. The total number of odd numbers up to row 2011 would be the sum from n=0 to 2010 of 2^{number of ones in binary(n)}.I think this sum is actually equal to 3^k where k is the number of bits in N. Wait, 2011 is less than 2048, which is 2^11. So, maybe the total number of odd numbers up to row 2047 is 3^11. Let me check: 3^11 is 177147, which is more than 100,000. So, up to row 2047, there are 177,147 odd numbers. Since 2011 is less than 2047, the total number of odd numbers up to row 2011 is slightly less, but still more than 100,000.So, if Wanda can traverse all the odd numbers in the first 2011 rows, she can eat more than 100,000 numbers without violating the sum condition. But how does she traverse them without getting stuck?I remember that in Pascal's triangle, the odd numbers form a fractal pattern known as the Sierpiński triangle. This pattern is self-similar and has a lot of connectivity. So, maybe there's a Hamiltonian path through all the odd numbers, allowing Wanda to eat them one by one without getting stuck.But how do I prove that such a path exists? Maybe by induction. Let's try that. For the base case, when n=1, the first two rows have three odd numbers: 1, 1, 1. Wanda can eat them in order: 1 (top), then 1 (left), then 1 (right). So, it works for n=1.Assume that for some k, the first 2^k rows have a path that allows Wanda to eat all the odd numbers. Then, for k+1, the first 2^{k+1} rows can be divided into three sections: the first 2^k rows, the next 2^k rows, and the last row. Each of these sections has a similar structure, and by the inductive hypothesis, there's a path through each section. So, Wanda can traverse the first section, then move to the second, then to the third, and finally to the last row, connecting them all together.This seems a bit hand-wavy, but I think the idea is that the structure of Pascal's triangle allows for such a path due to its recursive nature. Each larger triangle contains smaller triangles, and by connecting them appropriately, Wanda can traverse all the odd numbers.So, putting it all together: since the number of odd numbers in the first 2011 rows is more than 100,000, and there exists a path that allows Wanda to eat all these odd numbers without forming a forbidden triple, it's possible for her to eat 100,000 numbers in the first 2011 rows.I think that's the gist of it. Maybe I missed some details, but the key idea is focusing on odd numbers and using the structure of Pascal's triangle to ensure a valid path exists.