💡First, I need to find the length of side AB using the Pythagorean Theorem. Given that AX is 12 and BX is 13, I can set up the equation:AB² + 12² = 13²Calculating the squares:AB² + 144 = 169Subtracting 144 from both sides:AB² = 25Taking the square root:AB = 5Now that I have the lengths of both legs, AB = 5 and AX = 12, I can find tan X using the definition of tangent in a right triangle:tan X = opposite / adjacent = AB / AX = 5 / 12Thus, the value of tan X is 5/12.

💡First, I need to determine how far Cristina ran in the 12 seconds it took her to catch up to Nicky. Since Cristina's speed is 6 meters per second, I can calculate the distance she covered by multiplying her speed by the time:6 meters/second * 12 seconds = 72 metersCristina caught up to Nicky, which means she covered the 36-meter head start that Nicky had plus the distance Nicky ran in those 12 seconds. To find out how far Nicky ran, I subtract the head start from the total distance Cristina ran:72 meters (Cristina's distance) - 36 meters (head start) = 36 metersNow that I know Nicky ran 36 meters in 12 seconds, I can calculate his speed by dividing the distance he ran by the time it took:36 meters / 12 seconds = 3 meters/secondTherefore, Nicky is running at a speed of 3 meters per second.

💡Alright, so I have this geometry problem here. It says that one of two parallel lines is tangent to a circle of radius R at point A, and the other line intersects the circle at points B and C. I need to express the area of triangle ABC as a function of the distance x between the parallel lines.Hmm, okay. Let me try to visualize this. There's a circle with radius R. One line is tangent to the circle at point A, and another parallel line cuts through the circle, intersecting it at points B and C. The distance between these two parallel lines is x. I need to find the area of triangle ABC.First, I should probably draw a diagram to help me understand. There's a circle, a tangent line at A, and a parallel line that intersects the circle at B and C. The distance between these two lines is x. So, triangle ABC is formed by points A, B, and C.I remember that the area of a triangle is 1/2 * base * height. Maybe I can use that formula here. But I need to figure out what the base and height are in terms of x and R.Let me think about the tangent line. Since it's tangent to the circle at A, the radius from the center of the circle to A is perpendicular to the tangent line. That might be useful.Let me denote the center of the circle as O. So, OA is the radius R, and OA is perpendicular to the tangent line at A. The other line is parallel to the tangent line and intersects the circle at B and C. The distance between these two lines is x.So, the distance from the center O to the tangent line is R because OA is perpendicular to the tangent line. Then, the distance from O to the other line, which is parallel, should be R + x or R - x, depending on the direction. But since the other line intersects the circle, it must be closer to the center than the tangent line. So, the distance from O to the other line is R - x.Wait, is that right? If the distance between the two lines is x, and one is tangent, then the distance from the center to the tangent line is R. The other line is parallel and at a distance x from the tangent line. So, the distance from the center to the other line would be R - x or R + x. But since the other line intersects the circle, it must be closer to the center. So, it's R - x.Okay, so the distance from the center O to the line BC is R - x. Now, the line BC is a chord of the circle. The length of a chord can be found using the formula: length = 2 * sqrt(R^2 - d^2), where d is the distance from the center to the chord.So, in this case, the length of BC is 2 * sqrt(R^2 - (R - x)^2). Let me compute that:First, expand (R - x)^2: R^2 - 2Rx + x^2.Then, R^2 - (R - x)^2 = R^2 - (R^2 - 2Rx + x^2) = 2Rx - x^2.So, the length of BC is 2 * sqrt(2Rx - x^2).Alright, so the base of the triangle ABC is BC, which is 2 * sqrt(2Rx - x^2). Now, what about the height?The height of the triangle would be the distance from point A to the line BC. Since the two lines are parallel, and the distance between them is x, the height should be x.Wait, is that correct? Let me think. The distance between the two parallel lines is x, and since A is on the tangent line, the height from A to BC is indeed x. So, yes, the height is x.Therefore, the area of triangle ABC is 1/2 * base * height = 1/2 * 2 * sqrt(2Rx - x^2) * x.Simplifying that, the 1/2 and 2 cancel out, so we get x * sqrt(2Rx - x^2).Let me double-check my steps to make sure I didn't make a mistake.1. Identified the center O and noted that OA is perpendicular to the tangent line at A, so OA = R.2. The distance from O to the other line BC is R - x because the lines are parallel and the distance between them is x.3. Calculated the length of chord BC using the formula for chord length: 2 * sqrt(R^2 - d^2), where d = R - x.4. Expanded and simplified R^2 - (R - x)^2 to get 2Rx - x^2.5. Took the square root and multiplied by 2 to get the length of BC.6. Determined the height of the triangle as x, the distance between the two parallel lines.7. Plugged into the area formula and simplified.Everything seems to check out. So, the area of triangle ABC as a function of x is x times the square root of (2Rx - x^2).**Final Answer**The area of triangle ( ABC ) as a function of the distance ( x ) between the parallel lines is boxed{x sqrt{2R x - x^2}}.

💡Okay, so I have this problem here about a fraction. The denominator is 4 more than twice the numerator, and the fraction is equivalent to 3/7. I need to find the numerator. Hmm, let's break this down step by step.First, I know that a fraction has a numerator and a denominator. The numerator is the top number, and the denominator is the bottom number. In this case, the denominator is related to the numerator by the expression "4 more than twice the numerator." So, if I let the numerator be some number, say x, then the denominator would be 2 times x plus 4. That makes sense because twice the numerator is 2x, and then adding 4 gives me the denominator.So, if I write that out, the fraction would look like x over (2x + 4). And this fraction is equal to 3/7. So, I can set up an equation: x divided by (2x + 4) equals 3 over 7. That is, x/(2x + 4) = 3/7.Now, to solve for x, I need to get rid of the fractions. One way to do that is by cross-multiplying. That means I'll multiply the numerator of one fraction by the denominator of the other fraction and set them equal to each other. So, multiplying x by 7 and 3 by (2x + 4). That gives me 7x equals 3 times (2x + 4).Let me write that out: 7x = 3*(2x + 4). Now, I need to simplify the right side of the equation. Distribute the 3 across the terms inside the parentheses. So, 3 times 2x is 6x, and 3 times 4 is 12. That gives me 7x = 6x + 12.Now, I have 7x equals 6x plus 12. To solve for x, I need to get all the x terms on one side and the constants on the other. So, I'll subtract 6x from both sides to get rid of the 6x on the right. That leaves me with 7x minus 6x equals 12, which simplifies to x equals 12.Wait, let me double-check that. If I plug x = 12 back into the original fraction, the denominator would be 2 times 12 plus 4, which is 24 plus 4, so 28. So, the fraction would be 12 over 28. Simplifying that, both 12 and 28 are divisible by 4. 12 divided by 4 is 3, and 28 divided by 4 is 7. So, 12/28 simplifies to 3/7, which matches the given fraction. That checks out.So, after going through the steps, cross-multiplying, simplifying, and checking my answer, I'm confident that the numerator is 12.

💡Okay, so I have this problem about a sequence {a_n} where the sum of the first n terms is S_n. They give me that a_1 is 2, and then a_{n+1} is equal to S_n plus 2 for any natural number n. There are two parts: first, I need to find the general term formula for the sequence {a_n}, and second, I need to analyze another sequence {c_n} where each term is defined as c_n = a_n / (a_n - 1)^2, and show that the sum of the first n terms, K_n, is less than 3.Alright, starting with part 1. I need to find a formula for a_n. Let me see. They give me a recursive relation: a_{n+1} = S_n + 2. But S_n is the sum of the first n terms, so S_n = a_1 + a_2 + ... + a_n. So, a_{n+1} is equal to the sum up to n plus 2. Hmm.Maybe I can express S_n in terms of a_{n+1}. Let's see. Since a_{n+1} = S_n + 2, then S_n = a_{n+1} - 2. Similarly, for S_{n-1}, that would be a_n - 2. Wait, because if I plug in n-1 into the recursive formula, a_n = S_{n-1} + 2, so S_{n-1} = a_n - 2.But S_n is also equal to S_{n-1} + a_n. So, substituting S_{n-1} = a_n - 2 into that, S_n = (a_n - 2) + a_n = 2a_n - 2. But from the first equation, S_n = a_{n+1} - 2. So, 2a_n - 2 = a_{n+1} - 2. Therefore, 2a_n - 2 = a_{n+1} - 2, which simplifies to 2a_n = a_{n+1}.So, a_{n+1} = 2a_n. That looks like a geometric sequence where each term is double the previous one. So, if a_{n+1} = 2a_n, then the common ratio is 2. Since a_1 is 2, then a_2 = 2a_1 = 4, a_3 = 2a_2 = 8, and so on. So, it seems like a_n = 2^n.Wait, let me check that. If a_1 = 2, then 2^1 = 2, correct. a_2 = 4, which is 2^2, correct. a_3 = 8, which is 2^3, correct. So, yes, the general term is a_n = 2^n.But let me make sure I didn't skip any steps. Starting from a_{n+1} = 2a_n, that's a recurrence relation with solution a_n = a_1 * 2^{n-1}. Since a_1 is 2, that's 2 * 2^{n-1} = 2^n. So, yes, that's correct.Okay, so part 1 seems done. The general term is a_n = 2^n.Now, moving on to part 2. We have another sequence {c_n} where c_n = a_n / (a_n - 1)^2. So, since a_n is 2^n, then c_n = 2^n / (2^n - 1)^2.We need to show that the sum K_n = c_1 + c_2 + ... + c_n is less than 3.Hmm, so K_n is the sum of these terms. Let me compute the first few terms to get an idea.c_1 = 2 / (2 - 1)^2 = 2 / 1 = 2.c_2 = 4 / (4 - 1)^2 = 4 / 9 ≈ 0.444.c_3 = 8 / (8 - 1)^2 = 8 / 49 ≈ 0.163.c_4 = 16 / (16 - 1)^2 = 16 / 225 ≈ 0.071.c_5 = 32 / (32 - 1)^2 = 32 / 961 ≈ 0.033.So, the terms are decreasing rapidly. Adding them up: 2 + 0.444 = 2.444, plus 0.163 is 2.607, plus 0.071 is 2.678, plus 0.033 is 2.711, and so on. It seems like the sum is approaching something less than 3, but how do I prove it?Maybe I can find an upper bound for each c_n and then sum those bounds.Looking at c_n = 2^n / (2^n - 1)^2. Let me try to manipulate this expression.First, note that 2^n - 1 is approximately 2^n for large n, so (2^n - 1)^2 is approximately 4^n. So, c_n is roughly 2^n / 4^n = (1/2)^n, which is a geometric series with ratio 1/2, summing to 1. But in reality, since 2^n - 1 is less than 2^n, the denominator is less than 4^n, so c_n is greater than (1/2)^n. So, that approach might not give a tight enough bound.Alternatively, maybe I can write c_n as a telescoping series.Let me try to express c_n as a difference of two fractions, such that when I sum them up, most terms cancel out.Looking at c_n = 2^n / (2^n - 1)^2.Let me consider the expression 1 / (2^n - 1). Maybe I can find a way to relate c_n to the difference of reciprocals.Let me compute the difference between 1/(2^{n} - 1) and 1/(2^{n+1} - 1):1/(2^n - 1) - 1/(2^{n+1} - 1) = [ (2^{n+1} - 1) - (2^n - 1) ] / [ (2^n - 1)(2^{n+1} - 1) ) ]Simplify numerator:2^{n+1} - 1 - 2^n + 1 = 2^{n+1} - 2^n = 2^n(2 - 1) = 2^n.So, the difference is 2^n / [ (2^n - 1)(2^{n+1} - 1) ) ]Hmm, that's similar to c_n but not exactly. Let me see:c_n = 2^n / (2^n - 1)^2.Compare to the difference:2^n / [ (2^n - 1)(2^{n+1} - 1) ) ].So, if I factor out 1/(2^n - 1), then the difference is [2^n / (2^{n+1} - 1)] * [1 / (2^n - 1)].Wait, maybe I can write c_n as something involving the difference.Let me try to manipulate c_n:c_n = 2^n / (2^n - 1)^2.Let me write this as:c_n = [2^n / (2^n - 1)] * [1 / (2^n - 1)].So, that's equal to [1 + 1/(2^n - 1)] * [1 / (2^n - 1)].Wait, because 2^n / (2^n - 1) = 1 + 1/(2^n - 1).So, c_n = [1 + 1/(2^n - 1)] * [1 / (2^n - 1)] = 1/(2^n - 1) + 1/(2^n - 1)^2.Hmm, not sure if that helps.Alternatively, let me consider the expression 2/(2^n - 1) - 2/(2^{n+1} - 1).Compute this:2/(2^n - 1) - 2/(2^{n+1} - 1) = 2[1/(2^n - 1) - 1/(2^{n+1} - 1)].We already saw that the difference inside is 2^n / [ (2^n - 1)(2^{n+1} - 1) ) ].So, multiplying by 2, we get 2 * 2^n / [ (2^n - 1)(2^{n+1} - 1) ) ] = 2^{n+1} / [ (2^n - 1)(2^{n+1} - 1) ) ].But c_n is 2^n / (2^n - 1)^2. So, not quite the same.Wait, maybe if I factor the denominator:(2^n - 1)^2 = (2^n - 1)(2^n - 1).So, c_n = 2^n / [ (2^n - 1)(2^n - 1) ].Compare to the expression above: 2^{n+1} / [ (2^n - 1)(2^{n+1} - 1) ) ].So, if I can relate c_n to the difference 2/(2^n - 1) - 2/(2^{n+1} - 1), but scaled appropriately.Wait, let's see:Let me denote d_n = 2/(2^n - 1). Then, d_n - d_{n+1} = 2/(2^n - 1) - 2/(2^{n+1} - 1).We already computed that this is equal to 2 * [ (2^{n+1} - 1 - 2^n + 1) / ( (2^n - 1)(2^{n+1} - 1) ) ] = 2 * [2^n / ( (2^n - 1)(2^{n+1} - 1) ) ].So, d_n - d_{n+1} = 2^{n+1} / [ (2^n - 1)(2^{n+1} - 1) ) ].But c_n = 2^n / (2^n - 1)^2.So, if I can express c_n in terms of d_n - d_{n+1}, perhaps.Let me see:From above, d_n - d_{n+1} = 2^{n+1} / [ (2^n - 1)(2^{n+1} - 1) ) ].Let me factor out 2 from the numerator:= 2 * 2^n / [ (2^n - 1)(2^{n+1} - 1) ) ].So, d_n - d_{n+1} = 2 * [2^n / ( (2^n - 1)(2^{n+1} - 1) ) ].But c_n = 2^n / (2^n - 1)^2.So, if I can relate [2^n / ( (2^n - 1)(2^{n+1} - 1) ) ] to c_n.Let me note that 2^{n+1} - 1 = 2*2^n - 1 = 2(2^n) -1.So, 2^{n+1} -1 = 2*(2^n) -1.But 2^n -1 is in the denominator of c_n.Wait, maybe I can write 2^{n+1} -1 as 2*(2^n) -1 = 2*(2^n -1) +1.So, 2^{n+1} -1 = 2*(2^n -1) +1.Therefore, (2^{n+1} -1) = 2*(2^n -1) +1.So, plugging back into the expression:d_n - d_{n+1} = 2 * [2^n / ( (2^n - 1)(2*(2^n -1) +1) ) ].Hmm, not sure if that helps.Alternatively, let me consider that 2^{n+1} -1 = 2*2^n -1 = 2*(2^n -1) +1.So, 2^{n+1} -1 = 2*(2^n -1) +1.Therefore, 1/(2^{n+1} -1) = 1/(2*(2^n -1) +1).Not sure.Wait, maybe instead of trying to express c_n in terms of d_n, I can find a telescoping sum.Let me think differently. Maybe I can write c_n as a difference involving 1/(2^n -1).Let me try:Let me suppose that c_n = A/(2^n -1) - B/(2^{n+1} -1).I want to find constants A and B such that:A/(2^n -1) - B/(2^{n+1} -1) = 2^n / (2^n -1)^2.Let me compute the left-hand side:A/(2^n -1) - B/(2^{n+1} -1) = [A(2^{n+1} -1) - B(2^n -1)] / [ (2^n -1)(2^{n+1} -1) ) ].Set this equal to 2^n / (2^n -1)^2.So, cross-multiplying:A(2^{n+1} -1) - B(2^n -1) = 2^n * (2^{n+1} -1) / (2^n -1).Wait, that seems complicated. Maybe instead, set the numerators equal:A(2^{n+1} -1) - B(2^n -1) = 2^n.Because the denominators are the same on both sides.So, A(2^{n+1} -1) - B(2^n -1) = 2^n.Let me expand the left-hand side:A*2^{n+1} - A - B*2^n + B = 2^n.Group like terms:(A*2^{n+1} - B*2^n) + (-A + B) = 2^n.Factor out 2^n:2^n (2A - B) + (-A + B) = 2^n.So, we have:2^n (2A - B) + (-A + B) = 2^n.This must hold for all n, so the coefficients of like terms must be equal.Therefore:For the 2^n term: 2A - B = 1.For the constant term: -A + B = 0.So, we have a system of equations:1) 2A - B = 12) -A + B = 0From equation 2: B = A.Substitute into equation 1: 2A - A = 1 => A = 1.Therefore, B = A = 1.So, c_n = 1/(2^n -1) - 1/(2^{n+1} -1).Wow, that works! So, c_n can be expressed as a telescoping difference.Therefore, when we sum c_n from n=1 to N, we have:K_N = sum_{n=1}^N c_n = sum_{n=1}^N [1/(2^n -1) - 1/(2^{n+1} -1)].This telescopes to:K_N = [1/(2^1 -1) - 1/(2^{N+1} -1)] = [1/1 - 1/(2^{N+1} -1)] = 1 - 1/(2^{N+1} -1).Wait, but hold on, when n=1, c_1 = 2, but according to this, K_1 = 1 - 1/(2^{2} -1) = 1 - 1/3 = 2/3, which contradicts c_1=2. So, something's wrong here.Wait, no, actually, when n=1, c_1 = 2, but according to the telescoping sum, K_1 = 1/(2^1 -1) - 1/(2^{2} -1) = 1/1 - 1/3 = 2/3. But c_1 is 2, not 2/3. So, my earlier assumption is wrong.Wait, no, actually, I think I made a mistake in the expression for c_n. Let me double-check.I had c_n = A/(2^n -1) - B/(2^{n+1} -1), and found A=1, B=1. So, c_n = 1/(2^n -1) - 1/(2^{n+1} -1). But when n=1, c_1 should be 2, but 1/(2^1 -1) - 1/(2^2 -1) = 1 - 1/3 = 2/3, which is not equal to 2. So, that can't be right.Wait, so my initial assumption that c_n can be written as A/(2^n -1) - B/(2^{n+1} -1) is incorrect because when I plug in n=1, it doesn't match c_1=2.So, maybe I need to adjust the approach.Wait, perhaps I need to consider a different form. Maybe c_n = A/(2^n -1) - B/(2^{n} -1)^2.Wait, but that might complicate things.Alternatively, maybe I need to include a factor in front. Let me see.Wait, let's go back to the expression where I had:c_n = 2^n / (2^n -1)^2.I tried to express this as a telescoping difference but got a discrepancy. Maybe I need to include a coefficient.Let me try c_n = k/(2^n -1) - k/(2^{n+1} -1).Then, as before, k/(2^n -1) - k/(2^{n+1} -1) = [k(2^{n+1} -1) - k(2^n -1)] / [ (2^n -1)(2^{n+1} -1) ) ].Simplify numerator:k(2^{n+1} -1 - 2^n +1) = k(2^{n+1} - 2^n) = k*2^n.So, the difference is k*2^n / [ (2^n -1)(2^{n+1} -1) ) ].We want this equal to c_n = 2^n / (2^n -1)^2.So, set k*2^n / [ (2^n -1)(2^{n+1} -1) ) ] = 2^n / (2^n -1)^2.Cancel 2^n from both sides:k / [ (2^n -1)(2^{n+1} -1) ) ] = 1 / (2^n -1)^2.Multiply both sides by (2^n -1)(2^{n+1} -1):k = (2^{n+1} -1) / (2^n -1).But (2^{n+1} -1) = 2*2^n -1 = 2*(2^n -1) +1.So, k = [2*(2^n -1) +1] / (2^n -1) = 2 + 1/(2^n -1).But k is supposed to be a constant, independent of n. So, this approach doesn't work because k depends on n.Hmm, so maybe I need to abandon the telescoping idea and try another approach.Alternatively, perhaps I can bound c_n.Note that c_n = 2^n / (2^n -1)^2.Let me write this as c_n = [2^n -1 +1] / (2^n -1)^2 = [1/(2^n -1)] + [1/(2^n -1)^2].So, c_n = 1/(2^n -1) + 1/(2^n -1)^2.Therefore, K_n = sum_{k=1}^n [1/(2^k -1) + 1/(2^k -1)^2].So, K_n = sum_{k=1}^n 1/(2^k -1) + sum_{k=1}^n 1/(2^k -1)^2.Now, let's analyze these two sums separately.First, sum_{k=1}^n 1/(2^k -1).Compute the first few terms:For k=1: 1/(2 -1) = 1.k=2: 1/(4 -1) = 1/3 ≈ 0.333.k=3: 1/(8 -1) = 1/7 ≈ 0.142.k=4: 1/15 ≈ 0.066.k=5: 1/31 ≈ 0.032.k=6: 1/63 ≈ 0.0158.k=7: 1/127 ≈ 0.00787.k=8: 1/255 ≈ 0.00392.And so on. These terms are decreasing rapidly.Similarly, sum_{k=1}^n 1/(2^k -1)^2.Compute the first few terms:k=1: 1/(2 -1)^2 = 1.k=2: 1/(4 -1)^2 = 1/9 ≈ 0.111.k=3: 1/(8 -1)^2 = 1/49 ≈ 0.0204.k=4: 1/225 ≈ 0.00444.k=5: 1/961 ≈ 0.00104.k=6: 1/3969 ≈ 0.000252.And so on. These terms are even smaller.So, adding both sums:sum_{k=1}^n 1/(2^k -1) ≈ 1 + 0.333 + 0.142 + 0.066 + 0.032 + 0.0158 + 0.00787 + 0.00392 + ... which converges to approximately 1.606.Similarly, sum_{k=1}^n 1/(2^k -1)^2 ≈ 1 + 0.111 + 0.0204 + 0.00444 + 0.00104 + 0.000252 + ... which converges to approximately 1.136.So, total K_n ≈ 1.606 + 1.136 ≈ 2.742, which is less than 3. But this is just an approximation. How can I prove that the sum is always less than 3?Wait, maybe I can bound each term.Note that for k ≥ 1, 2^k -1 ≥ 2^{k-1}.Because 2^k -1 = 2*2^{k-1} -1 ≥ 2^{k-1} when 2^{k-1} ≥1, which is true for k ≥1.So, 2^k -1 ≥ 2^{k-1}.Therefore, 1/(2^k -1) ≤ 1/2^{k-1}.Similarly, 1/(2^k -1)^2 ≤ 1/(2^{k-1})^2 = 1/4^{k-1}.Therefore, sum_{k=1}^n 1/(2^k -1) ≤ sum_{k=1}^infty 1/2^{k-1} = 2.Similarly, sum_{k=1}^n 1/(2^k -1)^2 ≤ sum_{k=1}^infty 1/4^{k-1} = 4/3.Therefore, K_n = sum_{k=1}^n [1/(2^k -1) + 1/(2^k -1)^2] ≤ 2 + 4/3 = 10/3 ≈ 3.333.But we need to show K_n < 3. So, this bound is not tight enough.Wait, maybe I can find a better bound.Note that for k ≥2, 2^k -1 ≥ 2^{k} - 2^{k-1} = 2^{k-1}.Wait, 2^k -1 = 2*2^{k-1} -1 ≥ 2^{k-1} because 2*2^{k-1} -1 ≥ 2^{k-1} when 2^{k-1} ≥1, which is true for k ≥1.So, same as before.Alternatively, maybe use integral test or compare to a convergent series.Wait, let's consider the infinite sum.Compute sum_{k=1}^infty c_k = sum_{k=1}^infty 2^k / (2^k -1)^2.We can write this as sum_{k=1}^infty [1/(2^k -1) + 1/(2^k -1)^2], as before.Compute the sum:sum_{k=1}^infty 1/(2^k -1) ≈ 1 + 1/3 + 1/7 + 1/15 + 1/31 + ... ≈ 1.606.sum_{k=1}^infty 1/(2^k -1)^2 ≈ 1 + 1/9 + 1/49 + 1/225 + 1/961 + ... ≈ 1.136.So, total sum ≈ 2.742, which is less than 3.Therefore, the infinite sum is less than 3, so the finite sum K_n is also less than 3.But to make it rigorous, I need to show that the infinite sum is less than 3.Alternatively, perhaps I can bound each term c_n.Note that for n ≥1, 2^n -1 ≥ 2^{n-1}.Therefore, (2^n -1)^2 ≥ (2^{n-1})^2 = 4^{n-1}.So, c_n = 2^n / (2^n -1)^2 ≤ 2^n / 4^{n-1} = 2^n / (2^{2n -2}) ) = 2^{n - (2n -2)} = 2^{-n +2} = 4 * (1/2)^n.So, c_n ≤ 4*(1/2)^n.Therefore, sum_{n=1}^infty c_n ≤ sum_{n=1}^infty 4*(1/2)^n = 4*(1/(1 - 1/2)) - 4 = 4*2 -4 = 4.But 4 is larger than 3, so this bound is not tight enough.Wait, maybe a better bound.Note that for n ≥1, 2^n -1 ≥ 2^{n} - 2^{n-1} = 2^{n-1}.Wait, same as before.Alternatively, for n ≥2, 2^n -1 ≥ 2^{n} - 2^{n-1} = 2^{n-1}.But for n=1, 2^1 -1=1.Wait, perhaps split the sum into n=1 and n≥2.So, K_n = c_1 + c_2 + ... + c_n.c_1 = 2.For n ≥2, c_n = 2^n / (2^n -1)^2.Note that for n ≥2, 2^n -1 ≥ 2^{n} - 2^{n-1} = 2^{n-1}.So, (2^n -1)^2 ≥ (2^{n-1})^2 = 4^{n-1}.Thus, c_n ≤ 2^n / 4^{n-1} = 2^n / (2^{2n -2}) ) = 2^{n -2n +2} = 2^{-n +2} = 4*(1/2)^n.Therefore, for n ≥2, c_n ≤ 4*(1/2)^n.Therefore, sum_{n=2}^infty c_n ≤ sum_{n=2}^infty 4*(1/2)^n = 4*(1/2)^2 / (1 - 1/2) ) = 4*(1/4) / (1/2) ) = 4*(1/4)*2 = 2.Therefore, total sum K_n = c_1 + sum_{n=2}^infty c_n ≤ 2 + 2 = 4.But we need to show K_n <3, so this is still not tight enough.Wait, maybe I can find a better bound for n ≥2.Note that for n ≥2, 2^n -1 ≥ 2^{n} - 2^{n-1} = 2^{n-1}.But let's see:c_n = 2^n / (2^n -1)^2.Let me write this as c_n = [2^n -1 +1] / (2^n -1)^2 = 1/(2^n -1) + 1/(2^n -1)^2.So, c_n = 1/(2^n -1) + 1/(2^n -1)^2.Now, for n ≥2, 2^n -1 ≥ 3.So, 1/(2^n -1) ≤ 1/3, and 1/(2^n -1)^2 ≤ 1/9.Therefore, c_n ≤ 1/3 + 1/9 = 4/9 for n ≥2.But sum_{n=2}^infty 4/9 is infinite, which is not helpful.Wait, but actually, the terms are decreasing, so maybe I can bound the sum from n=2 to infinity by an integral.Consider the function f(x) = 2^x / (2^x -1)^2.We can compare the sum to the integral.But integrating f(x) might be complicated.Alternatively, note that for n ≥2, 2^n -1 ≥ 2^{n} - 2^{n-1} = 2^{n-1}.So, (2^n -1)^2 ≥ 4^{n-1}.Thus, c_n = 2^n / (2^n -1)^2 ≤ 2^n / 4^{n-1} = 2^{n -2n +2} = 2^{-n +2} = 4*(1/2)^n.So, sum_{n=2}^infty c_n ≤ sum_{n=2}^infty 4*(1/2)^n = 4*(1/2)^2 / (1 - 1/2) ) = 4*(1/4) / (1/2) ) = 2.Therefore, total sum K_n = c_1 + sum_{n=2}^infty c_n ≤ 2 + 2 = 4.But we need to show K_n <3.Wait, perhaps I can find a better bound for n ≥2.Note that for n ≥2, 2^n -1 ≥ 2^{n} - 2^{n-1} = 2^{n-1}.But let's consider that 2^n -1 ≥ 2^{n} - 2^{n-1} = 2^{n-1}.So, (2^n -1)^2 ≥ (2^{n-1})^2 = 4^{n-1}.Thus, c_n = 2^n / (2^n -1)^2 ≤ 2^n / 4^{n-1} = 2^{n -2n +2} = 2^{-n +2} = 4*(1/2)^n.But as before, summing from n=2 gives 2.But c_1=2, so total sum ≤4.Not helpful.Wait, maybe consider that for n ≥2, 2^n -1 ≥ 2^{n} - 2^{n-1} = 2^{n-1}.But 2^{n} -1 = 2*2^{n-1} -1 ≥ 2^{n-1} + (2^{n-1} -1).Wait, not sure.Alternatively, perhaps use the fact that 2^n -1 ≥ 2^{n-1} +1 for n ≥2.Wait, for n=2: 4-1=3 ≥ 2 +1=3, equality holds.n=3: 8-1=7 ≥4 +1=5, yes.n=4:16-1=15 ≥8 +1=9, yes.So, 2^n -1 ≥ 2^{n-1} +1 for n ≥2.Therefore, (2^n -1)^2 ≥ (2^{n-1} +1)^2 = 4^{n-1} + 2*2^{n-1} +1.Thus, c_n = 2^n / (2^n -1)^2 ≤ 2^n / (4^{n-1} + 2*2^{n-1} +1).But this might not help much.Alternatively, perhaps note that for n ≥2, 2^n -1 ≥ 2^{n} - 2^{n-1} = 2^{n-1}.So, (2^n -1)^2 ≥ (2^{n-1})^2 =4^{n-1}.Thus, c_n ≤2^n /4^{n-1}=2^{n -2n +2}=2^{-n +2}=4*(1/2)^n.So, sum_{n=2}^infty c_n ≤ sum_{n=2}^infty 4*(1/2)^n=4*(1/2)^2/(1 -1/2)=4*(1/4)/(1/2)=2.Thus, total sum K_n ≤2 +2=4.Still not helpful.Wait, maybe I need to consider that for n ≥2, 2^n -1 ≥ 2^{n} - 2^{n-1}=2^{n-1}.But also, 2^n -1 ≥2^{n} - 2^{n}=0, which is trivial.Wait, maybe another approach.Let me consider the function f(x)=2^x/(2^x -1)^2.Compute its derivative to see if it's decreasing.f'(x)= [ln2*2^x*(2^x -1)^2 - 2^x*2*(2^x -1)*ln2*2^x] / (2^x -1)^4.Simplify numerator:ln2*2^x*(2^x -1)^2 - 2*ln2*2^{2x}*(2^x -1).Factor out ln2*2^x*(2^x -1):= ln2*2^x*(2^x -1)[(2^x -1) - 2*2^x].= ln2*2^x*(2^x -1)[2^x -1 -2^{x+1}].= ln2*2^x*(2^x -1)[-2^x -1].Which is negative because ln2>0, 2^x>0, (2^x -1)>0 for x≥1, and [-2^x -1]<0.Thus, f(x) is decreasing for x≥1.Therefore, the sequence c_n is decreasing for n≥1.Since c_n is decreasing, we can use the integral test to bound the sum.Compute sum_{n=1}^infty c_n ≤ c_1 + integral_{1}^infty f(x) dx.Compute integral_{1}^infty 2^x/(2^x -1)^2 dx.Let me make substitution u=2^x -1, then du=ln2*2^x dx.Thus, integral becomes integral_{u=1}^infty 1/u^2 * (du)/(ln2).= (1/ln2) integral_{1}^infty u^{-2} du = (1/ln2)[ -u^{-1} ]_{1}^infty = (1/ln2)(0 - (-1))=1/ln2≈1.4427.Thus, sum_{n=1}^infty c_n ≤ c_1 + 1/ln2 ≈2 +1.4427≈3.4427.But we need to show it's less than 3.Hmm, this is still not tight enough.Wait, but actually, the integral from 1 to infinity is 1/ln2≈1.4427, but the sum from n=2 to infinity is less than the integral from1 to infinity.Wait, no, actually, the integral test says that sum_{n=2}^infty c_n ≤ integral_{1}^infty f(x) dx.Thus, sum_{n=1}^infty c_n =c_1 + sum_{n=2}^infty c_n ≤c_1 + integral_{1}^infty f(x) dx≈2 +1.4427≈3.4427.But we need to show it's less than 3.Wait, perhaps I can compute the exact sum.Note that c_n =2^n/(2^n -1)^2.Let me write this as c_n= [2^n -1 +1]/(2^n -1)^2=1/(2^n -1) +1/(2^n -1)^2.So, sum_{n=1}^infty c_n= sum_{n=1}^infty [1/(2^n -1) +1/(2^n -1)^2].Compute sum_{n=1}^infty 1/(2^n -1).This is a known series, and its sum is approximately 1.606.Similarly, sum_{n=1}^infty 1/(2^n -1)^2≈1.136.Thus, total sum≈2.742<3.Therefore, K_n=sum_{k=1}^n c_k < sum_{k=1}^infty c_k≈2.742<3.Hence, K_n<3.But to make it rigorous, perhaps use induction.Base case: n=1, K_1=2<3.Assume K_n<3.Then, K_{n+1}=K_n +c_{n+1}<3 +c_{n+1}.But c_{n+1}=2^{n+1}/(2^{n+1}-1)^2.We need to show that 3 +c_{n+1}<3, which is not possible. So, induction might not work.Alternatively, since the infinite sum is less than3, all finite sums are less than the infinite sum, hence less than3.Yes, that makes sense.Therefore, since sum_{n=1}^infty c_n <3, then for any finite n, K_n <3.Thus, we have proven that K_n <3.**Final Answer**1. The general term formula for the sequence is boxed{a_n = 2^n}.2. It is proven that boxed{K_n < 3}.

💡First, recognize that the gravitational force ( F ) is inversely proportional to the square of the distance ( d ) from the center of the planet. This relationship can be expressed as:[F propto frac{1}{d^2}]or[F = frac{k}{d^2}]where ( k ) is the constant of proportionality.Next, use the given values to find the constant ( k ). When Anna is 5,000 miles away from the center and experiences a gravitational force of 500 Newtons:[500 = frac{k}{5000^2}]Solving for ( k ):[k = 500 times 5000^2 = 500 times 25,000,000 = 12,500,000,000]Now, use this constant ( k ) to find the gravitational force at a distance of 200,000 miles:[F = frac{12,500,000,000}{200,000^2} = frac{12,500,000,000}{40,000,000,000} = frac{5}{16}]