💡Okay, so I have this problem about a sequence {a_n} where the sum of the first n terms is S_n. They gave me that a₁ = 1, and S_n = n²a_n for any positive integer n. The problem has two parts: first, to find S₁, S₂, S₃, S₄, and guess what S_n is; second, to prove my guess and find a_n.Alright, starting with part (1). They want me to compute the first few S_n terms. Let me think about how to approach this.Given that S_n is the sum of the first n terms, so S₁ = a₁. Since a₁ is given as 1, that's straightforward. So S₁ = 1.Now, for S₂. S₂ is the sum of the first two terms, so S₂ = a₁ + a₂. But they also told me that S₂ = 2²a₂, which is 4a₂. So I can set up the equation:S₂ = a₁ + a₂ = 4a₂But a₁ is 1, so substituting that in:1 + a₂ = 4a₂Subtracting a₂ from both sides:1 = 3a₂So, a₂ = 1/3. Therefore, S₂ = 4a₂ = 4*(1/3) = 4/3.Alright, moving on to S₃. Similarly, S₃ is the sum of the first three terms: S₃ = a₁ + a₂ + a₃. But also, S₃ = 3²a₃ = 9a₃.So, setting up the equation:S₃ = a₁ + a₂ + a₃ = 9a₃We already know a₁ = 1 and a₂ = 1/3, so substituting those in:1 + 1/3 + a₃ = 9a₃Adding 1 and 1/3 gives 4/3, so:4/3 + a₃ = 9a₃Subtracting a₃ from both sides:4/3 = 8a₃So, a₃ = (4/3)/8 = (4/3)*(1/8) = 1/6. Therefore, S₃ = 9a₃ = 9*(1/6) = 3/2.Next, S₄. S₄ = a₁ + a₂ + a₃ + a₄ = 4²a₄ = 16a₄.So, let's plug in the known values:1 + 1/3 + 1/6 + a₄ = 16a₄First, let me compute the sum of the first three terms:1 + 1/3 + 1/6. Let's convert them to sixths:1 = 6/6, 1/3 = 2/6, 1/6 = 1/6. So adding them together: 6/6 + 2/6 + 1/6 = 9/6 = 3/2.So, 3/2 + a₄ = 16a₄Subtracting a₄ from both sides:3/2 = 15a₄Therefore, a₄ = (3/2)/15 = (3/2)*(1/15) = 1/10. So S₄ = 16a₄ = 16*(1/10) = 8/5.Alright, so now I have:S₁ = 1S₂ = 4/3S₃ = 3/2S₄ = 8/5Hmm, trying to see a pattern here. Let me write them as fractions:S₁ = 1 = 2/2S₂ = 4/3S₃ = 3/2S₄ = 8/5Wait, that might not be the most helpful. Alternatively, let's write them as:S₁ = 1 = 2*1 / (1 + 1) = 2/2S₂ = 4/3 = 2*2 / (2 + 1) = 4/3S₃ = 3/2 = 2*3 / (3 + 1) = 6/4 = 3/2S₄ = 8/5 = 2*4 / (4 + 1) = 8/5Oh! So it seems like S_n = 2n / (n + 1). Let me check that:For n=1: 2*1 / (1 + 1) = 2/2 = 1. Correct.n=2: 4/3. Correct.n=3: 6/4 = 3/2. Correct.n=4: 8/5. Correct.So, the conjecture is that S_n = 2n / (n + 1). That seems to fit.Alright, so part (1) is done. Now, moving on to part (2): proving this conjecture and finding a_n.To prove that S_n = 2n / (n + 1) for all positive integers n, given that S_n = n²a_n and a₁ = 1.I think induction is the way to go here. Let me recall how mathematical induction works. First, prove the base case, then assume it's true for n=k, and prove it for n=k+1.Base case: n=1.S₁ = a₁ = 1. According to the formula, S₁ = 2*1 / (1 + 1) = 2/2 = 1. So, it holds for n=1.Inductive step: Assume that for some integer k ≥ 1, S_k = 2k / (k + 1). We need to show that S_{k+1} = 2(k+1)/(k+2).Given that S_{k+1} = (k+1)²a_{k+1}, and also S_{k+1} = S_k + a_{k+1}.So, let's write that:S_{k+1} = S_k + a_{k+1} = (k+1)²a_{k+1}But from the inductive hypothesis, S_k = 2k / (k + 1). So substituting that in:2k / (k + 1) + a_{k+1} = (k+1)²a_{k+1}Let me solve for a_{k+1}. Let's subtract a_{k+1} from both sides:2k / (k + 1) = (k+1)²a_{k+1} - a_{k+1}Factor out a_{k+1} on the right:2k / (k + 1) = a_{k+1}[(k+1)² - 1]Compute (k+1)² - 1:(k+1)² - 1 = k² + 2k + 1 - 1 = k² + 2k = k(k + 2)So, we have:2k / (k + 1) = a_{k+1} * k(k + 2)Solving for a_{k+1}:a_{k+1} = [2k / (k + 1)] / [k(k + 2)] = [2k] / [(k + 1) * k(k + 2)] = 2 / [(k + 1)(k + 2)]So, a_{k+1} = 2 / [(k + 1)(k + 2)]Therefore, S_{k+1} = (k+1)²a_{k+1} = (k+1)² * [2 / ((k + 1)(k + 2))] = (k+1) * 2 / (k + 2) = 2(k + 1)/(k + 2)Which is exactly the formula we wanted for S_{k+1}. Therefore, by induction, the formula S_n = 2n / (n + 1) holds for all positive integers n.Now, to find a_n. Since S_n = n²a_n, we can solve for a_n:a_n = S_n / n² = [2n / (n + 1)] / n² = 2 / [n(n + 1)]So, a_n = 2 / [n(n + 1)]. Let me verify this with the earlier terms.For n=1: a₁ = 2 / (1*2) = 1. Correct.n=2: a₂ = 2 / (2*3) = 1/3. Correct.n=3: a₃ = 2 / (3*4) = 1/6. Correct.n=4: a₄ = 2 / (4*5) = 1/10. Correct.Looks good. So, the general term a_n is 2 divided by n(n + 1).I think that wraps up both parts of the problem. I used induction to prove the conjecture for S_n and then derived a_n from that. All the initial terms check out, so I feel confident about the solution.**Final Answer**(1) The conjectured expression for ( S_n ) is ( boxed{dfrac{2n}{n + 1}} ).(2) The expression for ( a_n ) is ( boxed{dfrac{2}{n(n + 1)}} ).

💡Okay, so I have this function f(x) which is equal to x squared plus 2x minus 3. And the question is asking me to find the value of f(2). Hmm, okay, that sounds like I need to plug in 2 into the function and see what I get. Let me think about how to approach this step by step.First, I remember that when you have a function like f(x), and you want to find f(a) where a is some number, you just substitute a into every place where there's an x in the function. So in this case, since I need to find f(2), I should replace every x in the function with 2. That makes sense.So, starting with the function f(x) = x² + 2x - 3, I need to substitute x with 2. Let me write that out:f(2) = (2)² + 2*(2) - 3Okay, now I need to calculate each part step by step. Let's break it down:1. Calculate (2)²: That's 2 multiplied by itself, which is 4.2. Calculate 2*(2): That's 2 times 2, which is also 4.3. The last term is just -3, so I'll keep that as it is.Now, putting those together:f(2) = 4 + 4 - 3Next, I need to add and subtract these numbers in the correct order. I should do the addition first before the subtraction. So, 4 plus 4 is 8, and then subtracting 3 from that gives me:f(2) = 8 - 3 = 5Wait, let me double-check that to make sure I didn't make any mistakes. So, (2)² is definitely 4, 2 times 2 is 4, and then minus 3. Adding 4 and 4 gives 8, and 8 minus 3 is indeed 5. Okay, that seems right.I was a little confused at first about the order of operations, but since all the operations here are addition and subtraction, I can go from left to right without worrying about multiplication or division taking precedence. So, it's straightforward.Just to be extra sure, maybe I can visualize the function and see if plugging in 2 makes sense. If I think of the function f(x) = x² + 2x - 3, it's a quadratic function, which graphs as a parabola. The vertex form might help me understand it better, but for this problem, I just need the value at x=2.Alternatively, I could also think about the function in terms of its components. The x² term will dominate as x gets larger, but since we're only evaluating at x=2, it's manageable.Another way to check my work is to plug in x=2 into the original function again:f(2) = (2)² + 2*(2) - 3= 4 + 4 - 3= 8 - 3= 5Yep, same result. So, I feel confident that f(2) is indeed 5.I guess the key takeaway here is that substituting a value into a function is pretty straightforward. Just replace every instance of x with the given value and perform the arithmetic carefully. It's important to keep track of the signs and operations to avoid mistakes, especially with negative numbers or more complex expressions.Also, double-checking my work by recalculating or approaching the problem from a different angle helps ensure that I didn't make any errors. It's a good habit to verify my answers, especially when dealing with mathematical problems.In summary, substituting x=2 into the function f(x) = x² + 2x - 3 gives me f(2) = 5 after performing the necessary calculations step by step and verifying my result.

💡Alright, so I've got this problem here: I need to find the equation of a circle. They've given me the center of the circle, which is at point C(2, -1). That's straightforward enough. But then there's this line, l: x - y - 1 = 0, which intercepts a chord on the circle, and the length of that chord is 2√2. Hmm, okay, so I need to figure out the radius of the circle to write its equation.First, I remember that the general equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius. Since the center is given as (2, -1), I can plug those into the equation right away. So, the equation becomes (x - 2)^2 + (y + 1)^2 = r^2. Now, I just need to find r.To find the radius, I think I need to use the information about the chord. The chord is intercepted by the line l, and its length is 2√2. I recall that the length of a chord in a circle can be related to the radius and the distance from the center of the circle to the chord. There's a formula that connects these three: if d is the distance from the center to the chord, and L is the length of the chord, then L = 2√(r^2 - d^2). So, in this case, L is 2√2. That means 2√2 = 2√(r^2 - d^2). If I divide both sides by 2, I get √2 = √(r^2 - d^2). Squaring both sides, that gives me 2 = r^2 - d^2. So, r^2 = d^2 + 2.Now, I need to find d, the distance from the center C(2, -1) to the line l: x - y - 1 = 0. I remember the formula for the distance from a point (x0, y0) to the line ax + by + c = 0 is |ax0 + by0 + c| / √(a^2 + b^2). Let me rewrite the equation of line l in the standard form: x - y - 1 = 0. So, a = 1, b = -1, and c = -1. Plugging in the coordinates of the center, x0 = 2 and y0 = -1, the distance d is |1*2 + (-1)*(-1) + (-1)| / √(1^2 + (-1)^2). Calculating the numerator: 1*2 is 2, (-1)*(-1) is 1, and then + (-1) is -1. So, 2 + 1 - 1 = 2. The absolute value of 2 is 2. The denominator is √(1 + 1) which is √2. So, d = 2 / √2. Simplifying that, 2 divided by √2 is √2. So, d = √2.Now, going back to the equation r^2 = d^2 + 2. Since d is √2, d^2 is (√2)^2 = 2. So, r^2 = 2 + 2 = 4. Therefore, r = √4 = 2.Now that I have the radius, I can plug it back into the equation of the circle. So, (x - 2)^2 + (y + 1)^2 = 4. Wait, let me double-check my steps to make sure I didn't make any mistakes. I found the distance from the center to the line correctly, right? Using the formula, I got √2. Then, using the chord length formula, I found r^2 = 4, so r = 2. That seems right.Just to visualize, the circle is centered at (2, -1) with a radius of 2. The line l: x - y - 1 = 0 cuts through the circle, creating a chord of length 2√2. The distance from the center to the line is √2, which is less than the radius, so the line does indeed intersect the circle at two points, forming a chord. I think that all adds up. So, the equation of the circle should be (x - 2)^2 + (y + 1)^2 = 4.**Final Answer**The equation of the circle is boxed{(x-2)^2+(y+1)^2=4}.

💡First, I need to understand the given condition: the square root of ( x ) is greater than 5 and less than 6. This can be written as:[5 < sqrt{x} < 6]To find the range of ( x ), I'll square all parts of the inequality. Squaring 5 gives 25, and squaring 6 gives 36. Therefore:[25 < x < 36]Next, I'll identify the integer values of ( x ) within this range. Starting from 26 and ending at 35, the integers are:[26, 27, 28, 29, 30, 31, 32, 33, 34, 35]Counting these values, there are 10 integers that satisfy the condition.

💡Okay, so I have this sequence {a_n} where a_1 is given as 1. The rule for the subsequent terms is a_n = a_{n-1} + 1/(n(n-1)) for n ≥ 2. I need to find the first five terms and then figure out a general formula for a_n.Starting with the first part, writing out the first five terms. I know a_1 is 1. Then, for a_2, I plug in n=2 into the formula: a_2 = a_1 + 1/(2*(2-1)) = 1 + 1/2 = 3/2. That makes sense.Next, a_3 would be a_2 + 1/(3*(3-1)) = 3/2 + 1/6. Let me calculate that: 3/2 is 9/6, plus 1/6 is 10/6, which simplifies to 5/3. Okay, so a_3 is 5/3.Moving on to a_4: a_3 + 1/(4*(4-1)) = 5/3 + 1/12. Converting 5/3 to twelfths, that's 20/12, plus 1/12 is 21/12, which simplifies to 7/4. So a_4 is 7/4.Finally, a_5: a_4 + 1/(5*(5-1)) = 7/4 + 1/20. Converting 7/4 to twentieths, that's 35/20, plus 1/20 is 36/20, which simplifies to 9/5. So a_5 is 9/5.Alright, so the first five terms are 1, 3/2, 5/3, 7/4, and 9/5. That seems to follow a pattern where the numerator is increasing by 2 each time and the denominator is increasing by 1 each time. So, for a_n, it looks like the numerator is 2n - 1 and the denominator is n. So, a_n = (2n - 1)/n.But let me verify that. If I use the recursive formula, can I derive this general term?Starting from a_1 = 1, which is (2*1 - 1)/1 = 1, that checks out.Assuming a_{n-1} = (2(n-1) - 1)/(n-1) = (2n - 3)/(n - 1). Then, a_n = a_{n-1} + 1/(n(n-1)) = (2n - 3)/(n - 1) + 1/(n(n-1)).To add these fractions, I need a common denominator, which would be n(n - 1). So, (2n - 3)/(n - 1) becomes (2n - 3)*n / (n(n - 1)) = (2n^2 - 3n)/(n(n - 1)). Adding 1/(n(n - 1)) gives (2n^2 - 3n + 1)/(n(n - 1)).Now, let's factor the numerator: 2n^2 - 3n + 1. Looking for two numbers that multiply to 2*1=2 and add to -3. Hmm, -1 and -2. So, 2n^2 - 2n - n + 1 = 2n(n - 1) -1(n - 1) = (2n - 1)(n - 1). So, the numerator factors as (2n - 1)(n - 1).Thus, a_n = (2n - 1)(n - 1)/(n(n - 1)) = (2n - 1)/n. The (n - 1) terms cancel out. So, yes, that works. Therefore, the general term is a_n = (2n - 1)/n.Alternatively, I can think of this as a telescoping series. Since each term a_n - a_{n-1} = 1/(n(n - 1)) = 1/(n - 1) - 1/n. So, if I sum from k=2 to n, the differences a_k - a_{k-1} will telescope, leaving a_n - a_1 = (1 - 1/2) + (1/2 - 1/3) + ... + (1/(n-1) - 1/n) = 1 - 1/n. Therefore, a_n = a_1 + 1 - 1/n = 1 + 1 - 1/n = 2 - 1/n, which is the same as (2n - 1)/n.So, both methods confirm that the general term is a_n = (2n - 1)/n.

💡Okay, so I have this problem where I need to find the minimum value of the sum of distances from a moving point P on the parabola y² = 4x to two lines: l₁: 3x - 4y + 12 = 0 and l₂: x + 2 = 0. The options are A:1, B:2, C:3, D:4. Hmm, let me try to figure this out step by step.First, I remember that for a parabola, the definition involves a focus and a directrix. The standard form of a parabola that opens to the right is y² = 4ax, where the focus is at (a, 0) and the directrix is x = -a. Comparing this with our equation y² = 4x, it looks like 4a = 4, so a = 1. Therefore, the focus F is at (1, 0) and the directrix is x = -1.Wait, the line l₂ is x + 2 = 0, which simplifies to x = -2. That's different from the directrix x = -1. So, the distance from any point P on the parabola to l₂ is not the same as the distance to the directrix. Hmm, maybe I need another approach.I recall that for any point on the parabola, the distance to the focus is equal to the distance to the directrix. So, if I can express the distance to l₂ in terms of the distance to the focus, maybe that will help.Let me denote point P as (x, y) on the parabola. Then, the distance from P to l₂: x + 2 = 0 is |x + 2| / sqrt(1² + 0²) = |x + 2|. Since the parabola is y² = 4x, x is always non-negative (since y² is non-negative and 4x must be non-negative). So, x + 2 is always positive, so the distance simplifies to x + 2.Now, the distance from P to l₁: 3x - 4y + 12 = 0. The formula for the distance from a point (x, y) to the line Ax + By + C = 0 is |Ax + By + C| / sqrt(A² + B²). So, plugging in, the distance is |3x - 4y + 12| / sqrt(3² + (-4)²) = |3x - 4y + 12| / 5.So, the total distance we're trying to minimize is (x + 2) + |3x - 4y + 12| / 5.Since P is on the parabola y² = 4x, we can express x in terms of y: x = y² / 4. So, substituting x into the total distance expression, we get:Total distance = (y² / 4 + 2) + |3*(y² / 4) - 4y + 12| / 5Simplify that:Total distance = (y² / 4 + 2) + |(3y² / 4 - 4y + 12)| / 5Hmm, this seems a bit complicated. Maybe I can simplify the expression inside the absolute value:3y² / 4 - 4y + 12. Let's see if this quadratic in y can be factored or simplified. Alternatively, perhaps I can consider the expression without the absolute value and see if it's always positive or negative.Let me compute the discriminant of 3y² / 4 - 4y + 12 to see if it has real roots. The discriminant D = b² - 4ac = (-4)² - 4*(3/4)*12 = 16 - 36 = -20. Since the discriminant is negative, the quadratic doesn't cross the y-axis, meaning it's always positive or always negative. Since the coefficient of y² is positive (3/4), the quadratic is always positive. Therefore, the absolute value can be removed without changing the sign:Total distance = (y² / 4 + 2) + (3y² / 4 - 4y + 12) / 5Now, let's combine these terms:First, let's write both terms with a common denominator to make it easier:= (5*(y² / 4 + 2) + 3y² / 4 - 4y + 12) / 5Wait, actually, that might not be the best approach. Alternatively, let's compute each term separately:Total distance = y² / 4 + 2 + (3y² / 4 - 4y + 12) / 5Let me compute each part:First part: y² / 4 + 2Second part: (3y² / 4 - 4y + 12) / 5So, combining them:Total distance = y² / 4 + 2 + (3y² / 4)/5 - (4y)/5 + 12/5Simplify each term:= y² / 4 + 2 + (3y²) / 20 - (4y)/5 + 12/5Now, let's combine like terms:For y² terms: y² / 4 + 3y² / 20 = (5y² + 3y²) / 20 = 8y² / 20 = 2y² / 5For y terms: -4y / 5For constants: 2 + 12/5 = 10/5 + 12/5 = 22/5So, total distance becomes:2y² / 5 - 4y / 5 + 22/5Hmm, this is a quadratic in y: (2/5)y² - (4/5)y + 22/5To find the minimum value, since the coefficient of y² is positive, the parabola opens upwards, so the minimum occurs at the vertex.The vertex of a quadratic ay² + by + c is at y = -b/(2a). Here, a = 2/5, b = -4/5.So, y = -(-4/5) / (2*(2/5)) = (4/5) / (4/5) = 1.So, the minimum occurs at y = 1. Let's plug this back into the total distance expression:Total distance = 2*(1)² / 5 - 4*(1)/5 + 22/5 = 2/5 - 4/5 + 22/5 = (2 - 4 + 22)/5 = 20/5 = 4.Wait, so the minimum total distance is 4? But let me double-check because I might have made a mistake in simplifying.Wait, when I expressed the total distance, I had:Total distance = (y² / 4 + 2) + (3y² / 4 - 4y + 12) / 5But when I combined them, I think I might have miscalculated.Wait, let me re-express:Total distance = y² / 4 + 2 + (3y² / 4 - 4y + 12)/5Let me compute each term:y² / 4 is (1/4)y²2 is 2(3y² / 4 - 4y + 12)/5 is (3/20)y² - (4/5)y + 12/5So, adding them together:(1/4 + 3/20)y² + (-4/5)y + (2 + 12/5)Compute 1/4 + 3/20: 5/20 + 3/20 = 8/20 = 2/5Compute 2 + 12/5: 10/5 + 12/5 = 22/5So, total distance = (2/5)y² - (4/5)y + 22/5Yes, that's correct. Then, the vertex is at y = -b/(2a) = (4/5)/(2*(2/5)) = (4/5)/(4/5) = 1. So, y=1.Plugging back in:(2/5)(1)^2 - (4/5)(1) + 22/5 = 2/5 - 4/5 + 22/5 = (2 - 4 + 22)/5 = 20/5 = 4.So, the minimum total distance is 4. Therefore, the answer should be D:4.But wait, let me think again. Is there another way to approach this problem, maybe using reflections or properties of parabolas?I remember that in some optimization problems involving distances, reflecting a point across a line can help find the shortest path. Maybe I can use that idea here.Given that P is on the parabola y²=4x, and we need the sum of distances from P to l₁ and l₂. Since l₂ is x=-2, which is a vertical line, and the parabola's directrix is x=-1, which is different.Wait, but the distance from P to l₂ is x + 2, as I found earlier. And since x = y² / 4, it's (y² / 4) + 2.Alternatively, maybe I can think of the sum of distances as the distance from P to l₁ plus the distance from P to l₂, and try to minimize this sum.But I already did that by expressing everything in terms of y and found the minimum at y=1, giving a total distance of 4.Alternatively, maybe I can parametrize the parabola. Since y²=4x, I can write P as (t², 2t) where t is a parameter. Then, express the distances in terms of t.Let me try that.Let P = (t², 2t). Then, the distance to l₂: x + 2 = 0 is |t² + 2| / sqrt(1) = t² + 2.The distance to l₁: 3x - 4y + 12 = 0 is |3t² - 4*(2t) + 12| / 5 = |3t² - 8t + 12| / 5.So, total distance D(t) = t² + 2 + |3t² - 8t + 12| / 5.Again, since 3t² - 8t + 12 is a quadratic in t, let's check its discriminant: D = 64 - 144 = -80 < 0. So, it's always positive, so absolute value can be removed.Thus, D(t) = t² + 2 + (3t² - 8t + 12)/5.Combine terms:= (5t² + 10 + 3t² - 8t + 12)/5= (8t² - 8t + 22)/5Which is the same as (8/5)t² - (8/5)t + 22/5.To find the minimum, take derivative with respect to t and set to zero.dD/dt = (16/5)t - 8/5 = 0So, (16/5)t = 8/5 => t = (8/5)/(16/5) = 8/16 = 1/2.So, t = 1/2.Now, plug t = 1/2 into D(t):D(1/2) = (8*(1/2)^2 - 8*(1/2) + 22)/5 = (8*(1/4) - 4 + 22)/5 = (2 - 4 + 22)/5 = 20/5 = 4.Same result! So, the minimum total distance is indeed 4.Wait, but in the initial approach, I considered using the focus. Let me see if that approach was correct.I thought that the distance from P to l₂ is x + 2, and since x = y²/4, which is the same as the distance from P to the focus F(1,0) plus something. Wait, actually, for a parabola, the distance from P to the focus equals the distance from P to the directrix. The directrix is x = -1, so the distance from P to directrix is x + 1. But l₂ is x = -2, so the distance from P to l₂ is x + 2, which is (x + 1) + 1. So, distance to l₂ = distance to directrix + 1.Therefore, the sum of distances from P to l₁ and l₂ is equal to distance from P to l₁ plus (distance to directrix + 1). But since distance to directrix is equal to distance to focus, which is |PF|.So, the sum becomes distance from P to l₁ + |PF| + 1.But wait, is there a way to relate this to the distance from the focus to l₁?Hmm, maybe. If I can express the sum as |PF| + distance from P to l₁ + 1, then perhaps the minimum occurs when P lies on the line segment connecting F and l₁, but I'm not sure.Alternatively, maybe the minimal sum is the distance from F to l₁ plus 1, because |PF| + distance from P to l₁ is minimized when P is on the line connecting F and l₁, making the sum equal to the distance from F to l₁.Wait, that might be the case. Let me check.The distance from F(1,0) to l₁: 3x - 4y + 12 = 0 is |3*1 - 4*0 + 12| / 5 = |15| / 5 = 3.So, if the sum is |PF| + distance from P to l₁ + 1, and if |PF| + distance from P to l₁ is minimized when P is on the line connecting F and l₁, then the minimal sum would be 3 + 1 = 4. Which matches our previous result.Therefore, the minimal sum is indeed 4.So, the answer is D:4.