answer：Okay, so I have this problem about two similar isosceles triangles, ABC and A₁B₁C₁. Both are isosceles, with AB equal to BC in triangle ABC, and A₁B₁ equal to B₁C₁ in triangle A₁B₁C₁. They are similar with a ratio of AB to A₁B₁ being 2:1. The vertices A₁, B₁, and C₁ are on the sides CA, AB, and BC respectively, and A₁B₁ is perpendicular to AC. I need to find angle B.Hmm, let me try to visualize this. Triangle ABC is isosceles with AB = BC, so that means angle A is equal to angle C. Wait, no, actually, in an isosceles triangle, the angles opposite the equal sides are equal. So if AB = BC, then angle C is equal to angle A. Wait, no, hold on. In triangle ABC, if AB = BC, then the sides opposite angles C and A are equal. So angle at A is equal to angle at C. So angle BAC = angle BCA. That makes sense.Similarly, triangle A₁B₁C₁ is isosceles with A₁B₁ = B₁C₁, so angle at A₁ is equal to angle at C₁. Since they are similar, their corresponding angles are equal. So angle BAC in triangle ABC corresponds to angle A₁B₁C₁ in triangle A₁B₁C₁, right? Wait, no, similarity preserves angles, so angle at A in ABC corresponds to angle at A₁ in A₁B₁C₁, and so on.Given that A₁B₁ is perpendicular to AC, that gives us a right angle somewhere. So A₁B₁ is perpendicular to AC, meaning that the line from A₁ to B₁ is perpendicular to AC. So in triangle ABC, AC is one of the sides, and A₁ is on AC, B₁ is on AB, and C₁ is on BC.Let me try to sketch this mentally. Triangle ABC with AB = BC, so it's an isosceles triangle with base AC. Then, triangle A₁B₁C₁ is inside ABC, with A₁ on AC, B₁ on AB, and C₁ on BC. A₁B₁ is perpendicular to AC, so that's a right angle at A₁B₁.Since the triangles are similar with a ratio of 2:1, all corresponding sides of ABC are twice as long as those of A₁B₁C₁. So AB is twice A₁B₁, BC is twice B₁C₁, and AC is twice A₁C₁.Let me denote some variables. Let me call angle BAC as α. Since triangle ABC is isosceles with AB = BC, angle BAC = angle BCA = α, so angle ABC is 180° - 2α.Similarly, in triangle A₁B₁C₁, since it's isosceles with A₁B₁ = B₁C₁, angle at A₁ is equal to angle at C₁. Let's denote angle A₁B₁C₁ as β. Since triangle A₁B₁C₁ is similar to ABC, their angles correspond. So angle at A₁ in A₁B₁C₁ corresponds to angle at A in ABC, which is α. So angle A₁B₁C₁ is also α.Wait, but in triangle A₁B₁C₁, angles at A₁ and C₁ are equal, so angle at A₁ = angle at C₁ = something. Wait, maybe I need to think differently.Since A₁B₁ is perpendicular to AC, and A₁ is on AC, B₁ is on AB, then triangle A₁B₁C is a right triangle with right angle at B₁. Wait, no, A₁B₁ is perpendicular to AC, so the angle at A₁ between A₁B₁ and AC is 90°, but A₁B₁ is a side of triangle A₁B₁C₁, which is isosceles.This is getting a bit confusing. Maybe I should assign coordinates to the points to make it easier.Let me place point A at (0,0), point C at (c,0), and since AB = BC, point B must be somewhere above the base AC. Let me find coordinates for B.Since AB = BC, the triangle is symmetric about the perpendicular bisector of AC. So the midpoint of AC is at (c/2, 0), and point B is at (c/2, h), where h is the height of the triangle.So coordinates:- A: (0,0)- C: (c,0)- B: (c/2, h)Now, triangle A₁B₁C₁ is similar to ABC with a ratio of 1:2, so all its sides are half the length of ABC's sides.Points:- A₁ is on AC, so somewhere between A and C.- B₁ is on AB, so somewhere between A and B.- C₁ is on BC, so somewhere between B and C.Given that A₁B₁ is perpendicular to AC. Since AC is the base from (0,0) to (c,0), it's along the x-axis. So A₁B₁ is perpendicular to AC, meaning it's a vertical line.Wait, if A₁B₁ is perpendicular to AC, which is horizontal, then A₁B₁ must be vertical. So point A₁ is on AC, which is the x-axis, and point B₁ is on AB. Since A₁B₁ is vertical, the x-coordinate of A₁ and B₁ must be the same.Let me denote A₁ as (a,0) on AC, so B₁ must be at (a, b) on AB. Since B₁ is on AB, which goes from (0,0) to (c/2, h), the parametric equation of AB is x = (c/2)t, y = ht, where t ranges from 0 to 1.So point B₁ is at (a, b) = ((c/2)t, ht). But since A₁B₁ is vertical, a = (c/2)t, so t = 2a/c. Therefore, b = h*(2a/c). So B₁ is at (a, 2ha/c).Now, triangle A₁B₁C₁ is similar to ABC with ratio 1:2, so the sides of A₁B₁C₁ are half the length of ABC's sides.First, let's compute the length of A₁B₁. Since A₁ is at (a,0) and B₁ is at (a, 2ha/c), the distance between them is sqrt[(a - a)^2 + (2ha/c - 0)^2] = 2ha/c.Since A₁B₁ corresponds to AB in triangle ABC, which has length AB. Let's compute AB.Point A is at (0,0), point B is at (c/2, h). So AB = sqrt[(c/2)^2 + h^2].Given that AB : A₁B₁ = 2:1, so AB = 2*A₁B₁.So sqrt[(c/2)^2 + h^2] = 2*(2ha/c) = 4ha/c.So sqrt[(c^2)/4 + h^2] = 4ha/c.Let me square both sides:(c²)/4 + h² = (16h²a²)/c².Hmm, that's one equation.Now, since triangle A₁B₁C₁ is similar to ABC, the orientation should be the same. So the sides should correspond accordingly.Point C₁ is on BC. Let's find coordinates for C₁.Point C is at (c,0), point B is at (c/2, h). The parametric equation of BC is x = c - (c/2)s, y = 0 + h*s, where s ranges from 0 to 1.So point C₁ is at (c - (c/2)s, hs).Since triangle A₁B₁C₁ is similar to ABC, the sides should be in proportion.In triangle ABC, sides are AB, BC, AC. In triangle A₁B₁C₁, sides are A₁B₁, B₁C₁, A₁C₁.Given the ratio 1:2, so A₁B₁ = (1/2)AB, B₁C₁ = (1/2)BC, and A₁C₁ = (1/2)AC.We already have A₁B₁ = 2ha/c, and AB = sqrt[(c²)/4 + h²], so 2ha/c = (1/2)*sqrt[(c²)/4 + h²].Wait, but earlier we had sqrt[(c²)/4 + h²] = 4ha/c, so 2ha/c = (1/2)*sqrt[(c²)/4 + h²] would imply 2ha/c = (1/2)*(4ha/c) = 2ha/c, which is consistent.So that's okay.Now, let's consider side B₁C₁. It should be equal to A₁B₁ because triangle A₁B₁C₁ is isosceles with A₁B₁ = B₁C₁.So B₁C₁ = A₁B₁ = 2ha/c.Let me compute the distance between B₁ and C₁.Point B₁ is at (a, 2ha/c), point C₁ is at (c - (c/2)s, hs).So distance B₁C₁ is sqrt[(c - (c/2)s - a)^2 + (hs - 2ha/c)^2] = 2ha/c.So:sqrt[(c - (c/2)s - a)^2 + (hs - 2ha/c)^2] = 2ha/c.Let me square both sides:(c - (c/2)s - a)^2 + (hs - 2ha/c)^2 = (4h²a²)/c².That's equation two.Also, since triangle A₁B₁C₁ is similar to ABC, the angles correspond. So angle at A₁ is equal to angle at A, which is α.Wait, maybe I can use vectors or slopes to find the angles.Alternatively, since A₁B₁ is vertical, and A₁ is at (a,0), B₁ is at (a, 2ha/c). Then, the line from B₁ to C₁ should make the same angle as the line from B to C.Wait, maybe I can find the slope of BC and set it equal to the slope of B₁C₁.Slope of BC: from (c/2, h) to (c,0). So slope is (0 - h)/(c - c/2) = (-h)/(c/2) = -2h/c.Similarly, slope of B₁C₁: from (a, 2ha/c) to (c - (c/2)s, hs). So slope is (hs - 2ha/c)/(c - (c/2)s - a).Since triangle A₁B₁C₁ is similar to ABC, the slope of B₁C₁ should be equal to the slope of BC, which is -2h/c.So:(hs - 2ha/c)/(c - (c/2)s - a) = -2h/c.Let me write that:(hs - (2ha)/c) = (-2h/c)(c - (c/2)s - a).Simplify the right side:(-2h/c)(c - (c/2)s - a) = (-2h/c)*(c) + (-2h/c)*(-c/2 s) + (-2h/c)*(-a) = -2h + h s + (2ha)/c.So left side: hs - (2ha)/c.Set equal:hs - (2ha)/c = -2h + hs + (2ha)/c.Wait, subtract hs from both sides:- (2ha)/c = -2h + (2ha)/c.Bring terms with (2ha)/c to one side:- (2ha)/c - (2ha)/c = -2h.So - (4ha)/c = -2h.Multiply both sides by -1:(4ha)/c = 2h.Divide both sides by h (assuming h ≠ 0):4a/c = 2.So 4a = 2c => a = c/2.Wait, so a is c/2. So point A₁ is at (c/2, 0), which is the midpoint of AC.Hmm, that's interesting. So A₁ is the midpoint of AC.Then, point B₁ is at (a, 2ha/c) = (c/2, 2h*(c/2)/c) = (c/2, h).Wait, that's the same as point B. But point B₁ is supposed to be on AB, not coinciding with B.Wait, that can't be. Did I make a mistake?Wait, let's go back.We had:Slope of B₁C₁ = slope of BC.Which gave us:(hs - 2ha/c)/(c - (c/2)s - a) = -2h/c.Then we simplified:hs - 2ha/c = (-2h/c)(c - (c/2)s - a).Which became:hs - 2ha/c = -2h + h s + 2ha/c.Subtracting hs from both sides:-2ha/c = -2h + 2ha/c.Bringing terms with ha/c to the left:-2ha/c - 2ha/c = -2h.So -4ha/c = -2h.Divide both sides by -2h (assuming h ≠ 0):(4a)/c = 2 => a = (2c)/4 = c/2.So a = c/2.So point A₁ is at (c/2, 0), which is the midpoint of AC. Then, point B₁ is at (c/2, 2h*(c/2)/c) = (c/2, h). But that's the same as point B, which is at (c/2, h). So B₁ coincides with B, but B₁ is supposed to be on AB, which is from A(0,0) to B(c/2, h). So if B₁ is at (c/2, h), that's point B itself. But in the problem, B₁ is on AB, so it can be anywhere from A to B, including B. But if B₁ is at B, then triangle A₁B₁C₁ would have points A₁ at midpoint of AC, B₁ at B, and C₁ somewhere on BC.But then, triangle A₁B₁C₁ would have sides A₁B₁, B₁C₁, and A₁C₁. A₁B₁ is from midpoint of AC to B, which is length sqrt[(c/2 - c/2)^2 + (h - 0)^2] = h. But earlier, we had A₁B₁ = 2ha/c, which with a = c/2 is 2h*(c/2)/c = h. So that's consistent.But then, B₁C₁ is from B to C₁ on BC. Since triangle A₁B₁C₁ is isosceles with A₁B₁ = B₁C₁, so B₁C₁ must also be equal to h.But point C₁ is on BC. Let's compute the distance from B to C₁.Point B is at (c/2, h), point C is at (c, 0). So BC has length sqrt[(c - c/2)^2 + (0 - h)^2] = sqrt[(c/2)^2 + h²].We need B₁C₁ = h. So the distance from B to C₁ is h.Let me parametrize point C₁ on BC. Let me use parameter s, where s=0 is at B and s=1 is at C.So point C₁ is at (c/2 + (c/2)s, h - h s).So distance from B to C₁ is sqrt[(c/2 s)^2 + (-h s)^2] = s*sqrt[(c/2)^2 + h²].We need this equal to h:s*sqrt[(c/2)^2 + h²] = h.So s = h / sqrt[(c/2)^2 + h²].Therefore, point C₁ is at:x = c/2 + (c/2)*(h / sqrt[(c/2)^2 + h²]) = c/2 + (c h)/(2 sqrt[(c²)/4 + h²])y = h - h*(h / sqrt[(c²)/4 + h²]) = h - h² / sqrt[(c²)/4 + h²]Now, since triangle A₁B₁C₁ is similar to ABC, the ratio should be 1:2.We have A₁B₁ = h, and AB = sqrt[(c/2)^2 + h²]. So the ratio is h / sqrt[(c²)/4 + h²] = 1/2.So:h / sqrt[(c²)/4 + h²] = 1/2.Multiply both sides by sqrt[(c²)/4 + h²]:h = (1/2) sqrt[(c²)/4 + h²].Square both sides:h² = (1/4)[(c²)/4 + h²].Multiply both sides by 4:4h² = (c²)/4 + h².Subtract h²:3h² = (c²)/4.So c² = 12 h².Thus, c = 2 h sqrt(3).So the base AC is 2 h sqrt(3), and the height is h.Now, in triangle ABC, the sides AB and BC are equal, each is sqrt[(c/2)^2 + h²] = sqrt[(h sqrt(3))² + h²] = sqrt[3 h² + h²] = sqrt[4 h²] = 2h.So AB = BC = 2h, and AC = 2 h sqrt(3).Therefore, triangle ABC has sides AB = BC = 2h, AC = 2 h sqrt(3). So it's an isosceles triangle with two sides equal to 2h and base 2 h sqrt(3).Wait, but in such a triangle, the angles can be found using the Law of Cosines.Let me compute angle B, which is at point B.Using Law of Cosines:AC² = AB² + BC² - 2 AB BC cos(angle B).But AB = BC = 2h, AC = 2 h sqrt(3).So:(2 h sqrt(3))² = (2h)² + (2h)² - 2*(2h)*(2h)*cos(angle B).Compute:4 h² * 3 = 4 h² + 4 h² - 8 h² cos(angle B).12 h² = 8 h² - 8 h² cos(angle B).Subtract 8 h²:4 h² = -8 h² cos(angle B).Divide both sides by -8 h²:-4 h² / 8 h² = cos(angle B).Simplify:-1/2 = cos(angle B).So angle B = arccos(-1/2) = 120°.Wait, that's 120 degrees.But let me verify.Alternatively, since triangle ABC has sides AB = BC = 2h, AC = 2 h sqrt(3). So it's an isosceles triangle with two sides equal to 2h and base 2 h sqrt(3). The ratio of sides is 2h : 2h : 2 h sqrt(3), which simplifies to 1 : 1 : sqrt(3). That's a well-known triangle, which is a 30-30-120 triangle.Yes, because in a triangle with sides 1,1,sqrt(3), the angles opposite the equal sides are 30°, and the angle opposite the sqrt(3) side is 120°. So angle B is 120°.Therefore, the measure of angle B is 120 degrees.**Final Answer**The measure of angle ( B ) is (boxed{120^circ}).

answer：Alright, so I've got this math problem here with some equations, and I need to figure out a pattern and then answer three questions based on that pattern. Let me take it step by step.First, let me look at the equations given:1st equation: (2^2 - 1 = 1 times 3)2nd equation: (3^2 - 1 = 2 times 4)3rd equation: (4^2 - 1 = 3 times 5)4th equation: (5^2 - 1 = 4 times 6)5th equation: (6^2 - 1 = 5 times 7)And it says there are more like this. So, I need to see the pattern here.Looking at the left side of each equation, it's the square of a number minus 1. The right side is the product of two numbers. Let me see if I can relate them.For the first equation: (2^2 - 1 = 4 - 1 = 3), and (1 times 3 = 3). So, 3 equals 3.Second equation: (3^2 - 1 = 9 - 1 = 8), and (2 times 4 = 8). So, 8 equals 8.Third equation: (4^2 - 1 = 16 - 1 = 15), and (3 times 5 = 15). So, 15 equals 15.Fourth equation: (5^2 - 1 = 25 - 1 = 24), and (4 times 6 = 24). So, 24 equals 24.Fifth equation: (6^2 - 1 = 36 - 1 = 35), and (5 times 7 = 35). So, 35 equals 35.Okay, so the pattern seems to be that for the nth equation, the left side is ((n+1)^2 - 1), and the right side is (n times (n+2)). Let me check that.For n=1: ((1+1)^2 - 1 = 4 - 1 = 3), and (1 times (1+2) = 1 times 3 = 3). Correct.For n=2: ((2+1)^2 - 1 = 9 - 1 = 8), and (2 times (2+2) = 2 times 4 = 8). Correct.For n=3: ((3+1)^2 - 1 = 16 - 1 = 15), and (3 times (3+2) = 3 times 5 = 15). Correct.Okay, so that seems to be the general rule. So, for any natural number n, the nth equation is ((n+1)^2 - 1 = n times (n+2)).Now, moving on to the questions.1) Write down the 6th equation.Well, using the pattern, for n=6, the equation would be:((6+1)^2 - 1 = 6 times (6+2))Calculating that:(7^2 - 1 = 49 - 1 = 48)And (6 times 8 = 48)So, the 6th equation is (7^2 - 1 = 6 times 8).2) Let n (n ≥ 1) represent a natural number, the nth equation can be represented as...From the pattern, as I figured out earlier, it's ((n+1)^2 - 1 = n times (n+2)).So, that's the general form.3) Using the above conclusion, calculate: (frac{1}{2^2 - 1} + frac{1}{4^2 - 1} + frac{1}{6^2 - 1} + cdots + frac{1}{2024^2 - 1}).Hmm, okay. So, this is a sum of reciprocals of squares minus 1, where the bases are even numbers starting from 2 up to 2024.First, let's see if we can express each term using the pattern we found.From the nth equation, we have:((n+1)^2 - 1 = n(n+2))So, (frac{1}{(n+1)^2 - 1} = frac{1}{n(n+2)})Therefore, each term in the sum can be rewritten as (frac{1}{n(n+2)}).But in our sum, the denominators are (2^2 - 1, 4^2 - 1, 6^2 - 1, ldots, 2024^2 - 1).So, let's see:For (2^2 - 1), n+1 = 2, so n = 1.Thus, (frac{1}{2^2 - 1} = frac{1}{1 times 3})Similarly, for (4^2 - 1), n+1 = 4, so n = 3.Thus, (frac{1}{4^2 - 1} = frac{1}{3 times 5})Wait, that's interesting. So, the terms in the sum correspond to n = 1, 3, 5, ..., up to some number.Wait, let's see:The denominators are (2^2 - 1, 4^2 - 1, 6^2 - 1, ldots, 2024^2 - 1).So, the bases are 2, 4, 6, ..., 2024.So, each base is 2k where k goes from 1 to 1012, because 2024 divided by 2 is 1012.So, the sum is from k=1 to k=1012 of (frac{1}{(2k)^2 - 1}).But from the nth equation, we have:((n+1)^2 - 1 = n(n+2))So, (frac{1}{(n+1)^2 - 1} = frac{1}{n(n+2)})But in our sum, the denominators are ((2k)^2 - 1), which is ((2k)^2 - 1 = 4k^2 - 1).Wait, but in the nth equation, ((n+1)^2 - 1 = n(n+2)). So, if we set (n+1 = 2k), then n = 2k - 1.So, (frac{1}{(2k)^2 - 1} = frac{1}{(2k - 1)(2k + 1)})Therefore, each term in the sum can be expressed as (frac{1}{(2k - 1)(2k + 1)}), where k goes from 1 to 1012.So, the sum becomes:(sum_{k=1}^{1012} frac{1}{(2k - 1)(2k + 1)})Now, this looks like a telescoping series. To see that, we can use partial fractions.Let me express (frac{1}{(2k - 1)(2k + 1)}) as (frac{A}{2k - 1} + frac{B}{2k + 1}).So, let's solve for A and B:(frac{1}{(2k - 1)(2k + 1)} = frac{A}{2k - 1} + frac{B}{2k + 1})Multiplying both sides by ((2k - 1)(2k + 1)):1 = A(2k + 1) + B(2k - 1)Now, let's solve for A and B.Let me set up equations by choosing suitable values for k.First, let k = 1/2:Then, 1 = A(2*(1/2) + 1) + B(2*(1/2) - 1) = A(1 + 1) + B(1 - 1) = 2A + 0So, 1 = 2A => A = 1/2Similarly, let k = -1/2:Then, 1 = A(2*(-1/2) + 1) + B(2*(-1/2) - 1) = A(-1 + 1) + B(-1 - 1) = 0 + B(-2)So, 1 = -2B => B = -1/2Therefore, the partial fractions are:(frac{1}{(2k - 1)(2k + 1)} = frac{1/2}{2k - 1} - frac{1/2}{2k + 1})So, the sum becomes:(sum_{k=1}^{1012} left( frac{1/2}{2k - 1} - frac{1/2}{2k + 1} right))This is a telescoping series because when we write out the terms, many will cancel out.Let's write out the first few terms and the last few terms to see the pattern.For k=1:(frac{1/2}{1} - frac{1/2}{3})For k=2:(frac{1/2}{3} - frac{1/2}{5})For k=3:(frac{1/2}{5} - frac{1/2}{7})...For k=1012:(frac{1/2}{2023} - frac{1/2}{2025})Now, when we add all these up, let's see what cancels:The -1/2 * 1/3 from k=1 cancels with the +1/2 * 1/3 from k=2.Similarly, the -1/2 * 1/5 from k=2 cancels with the +1/2 * 1/5 from k=3.This pattern continues all the way to the end.So, all the intermediate terms cancel out, and we're left with the first term of the first expression and the last term of the last expression.Therefore, the sum simplifies to:(frac{1/2}{1} - frac{1/2}{2025})Simplify that:(frac{1}{2} - frac{1}{2 times 2025} = frac{1}{2} - frac{1}{4050})To combine these, find a common denominator, which is 4050.(frac{1}{2} = frac{2025}{4050})So,(frac{2025}{4050} - frac{1}{4050} = frac{2024}{4050})Simplify this fraction by dividing numerator and denominator by 2:(frac{1012}{2025})So, the sum is equal to (frac{1012}{2025}).Let me double-check my steps to make sure I didn't make a mistake.1. Identified the pattern correctly: ((n+1)^2 - 1 = n(n+2)). That seems right.2. For the 6th equation, substituted n=6: (7^2 - 1 = 6 times 8). That checks out.3. For the general form, yes, it's ((n+1)^2 - 1 = n(n+2)).4. For the sum, recognized that each term can be expressed as (frac{1}{n(n+2)}), then used partial fractions to split it into (frac{1/2}{n} - frac{1/2}{n+2}). Wait, but in the denominator, it's (2k - 1) and (2k + 1), so actually, I think I might have made a slight miscalculation there.Wait, let's go back to the partial fractions step.I had:(frac{1}{(2k - 1)(2k + 1)} = frac{A}{2k - 1} + frac{B}{2k + 1})Solved for A and B and got A=1/2 and B=-1/2.So, the partial fractions are:(frac{1}{2(2k - 1)} - frac{1}{2(2k + 1)})Therefore, each term is (frac{1}{2(2k - 1)} - frac{1}{2(2k + 1)})So, when we sum from k=1 to k=1012, it's:(sum_{k=1}^{1012} left( frac{1}{2(2k - 1)} - frac{1}{2(2k + 1)} right))Which can be written as:(frac{1}{2} sum_{k=1}^{1012} left( frac{1}{2k - 1} - frac{1}{2k + 1} right))This is indeed a telescoping series. Let's write out the terms:For k=1: (frac{1}{2(1)} - frac{1}{2(3)} = frac{1}{2} - frac{1}{6})For k=2: (frac{1}{2(3)} - frac{1}{2(5)} = frac{1}{6} - frac{1}{10})For k=3: (frac{1}{2(5)} - frac{1}{2(7)} = frac{1}{10} - frac{1}{14})...For k=1012: (frac{1}{2(2023)} - frac{1}{2(2025)} = frac{1}{4046} - frac{1}{4050})Now, when we add all these up, the intermediate terms cancel:- The (-frac{1}{6}) from k=1 cancels with the (+frac{1}{6}) from k=2.- The (-frac{1}{10}) from k=2 cancels with the (+frac{1}{10}) from k=3.This pattern continues until the last term.So, what's left is the first term of the first expression and the last term of the last expression.Therefore, the sum simplifies to:(frac{1}{2} - frac{1}{2 times 2025} = frac{1}{2} - frac{1}{4050})As before, converting to a common denominator:(frac{1}{2} = frac{2025}{4050})So,(frac{2025}{4050} - frac{1}{4050} = frac{2024}{4050})Simplify by dividing numerator and denominator by 2:(frac{1012}{2025})Yes, that seems correct.So, the final answer for the sum is (frac{1012}{2025}).I think that's it. Let me just recap:1. Identified the pattern in the equations.2. Used the pattern to express each term in the sum as a telescoping series.3. Summed the series by canceling out intermediate terms.4. Simplified the remaining terms to get the final result.Everything seems to check out. I don't see any mistakes in my reasoning.

answer：To compute (cos 210^circ), I start by recognizing that (210^circ) is in the third quadrant of the unit circle. In the third quadrant, cosine values are negative.Next, I find the reference angle for (210^circ) by subtracting (180^circ):[210^circ - 180^circ = 30^circ]So, the reference angle is (30^circ).I know that (cos 30^circ = frac{sqrt{3}}{2}). Since (210^circ) is in the third quadrant where cosine is negative, I apply the appropriate sign:[cos 210^circ = -frac{sqrt{3}}{2}]Thus, the value of (cos 210^circ) is (-frac{sqrt{3}}{2}).

answer：First, I need to determine the breadth of the rectangular plot. Since the length is 20 meters more than the breadth and the length is given as 60 meters, I can calculate the breadth by subtracting 20 meters from the length.Next, with both the length and breadth known, I can calculate the perimeter of the plot using the formula for the perimeter of a rectangle: Perimeter = 2 × (Length + Breadth).After finding the perimeter, I will calculate the total cost of fencing by multiplying the perimeter by the cost per meter of fencing, which is 26.50.Finally, I will present the total cost as the final answer.

answer：First, I need to understand what a visible factor number is. It's a number that is divisible by each of its non-zero digits. So, for example, if a number is 24, it should be divisible by both 2 and 4.Next, I'll focus on the range from 200 to 250. This range includes all three-digit numbers where the first digit is 2, the second digit (T) can be 0 to 5, and the third digit (U) can be 0 to 9.I'll break this down into cases based on the value of T:1. **Case 1: T = 0** - The number is of the form 20U. - It must be divisible by 2 and U (if U ≠ 0). - Possible values for U are 0, 1, 2, 4, 5.2. **Case 2: T = 1** - The number is of the form 21U. - It must be divisible by 2, 1, and U (if U ≠ 0). - Possible values for U are 0, 1, 2, 5.3. **Case 3: T = 2** - The number is of the form 22U. - It must be divisible by 2, 2, and U (if U ≠ 0). - Possible values for U are 0, 2, 4, 6, 8.4. **Case 4: T = 3** - The number is of the form 23U. - It must be divisible by 2, 3, and U (if U ≠ 0). - Possible values for U are 0, 3, 6.5. **Case 5: T = 4** - The number is of the form 24U. - It must be divisible by 2, 4, and U (if U ≠ 0). - Possible values for U are 0, 4, 8.6. **Case 6: T = 5** - The only number in this case is 250. - It must be divisible by 2, 5, and 0 (but 0 is ignored since division by zero is undefined). - 250 is a valid visible factor number.Finally, I'll count all the valid numbers from each case and sum them up to find the total number of visible factor numbers between 200 and 250.

answer：First, I need to determine the total amount of carbon dioxide released by the canvas bag in ounces. Since the canvas bag releases 600 pounds of carbon dioxide and there are 16 ounces in a pound, I can calculate this by multiplying 600 by 16.Next, I need to find out how many plastic bags Marla uses in total over 300 shopping trips. If she uses 8 plastic bags per trip, then over 300 trips, she will use 8 multiplied by 300, which equals 2400 plastic bags.Finally, to find out how much carbon dioxide each plastic bag releases, I divide the total carbon dioxide released by the canvas bag (in ounces) by the total number of plastic bags used. This means dividing 9600 ounces by 2400 bags to get the carbon dioxide released per plastic bag.