🧠:<think>Alright, so I have this problem about analyzing Vestri's football performance. It's divided into two parts. Let me tackle them one by one.Starting with part 1: I need to find the sum ( S_k ) of goals scored in the first ( k ) matches, given that each match's goals ( G_k ) follow the function ( G_k = a cdot sin(bk + c) + d ). Hmm, okay. So ( S_k ) is the sum from ( m = 1 ) to ( m = k ) of ( G_m ). That is,[S_k = sum_{m=1}^{k} G_m = sum_{m=1}^{k} left( a cdot sin(bm + c) + d right)]I can split this sum into two parts: the sum of the sine terms and the sum of the constants ( d ). So,[S_k = a cdot sum_{m=1}^{k} sin(bm + c) + d cdot sum_{m=1}^{k} 1]The second sum is straightforward. The sum of 1 from 1 to ( k ) is just ( k ). So that part becomes ( d cdot k ).Now, the first sum is a bit trickier. It's the sum of sine functions with arguments in an arithmetic sequence. I remember there's a formula for the sum of sines with arguments in arithmetic progression. Let me recall it.The formula for the sum ( sum_{m=1}^{k} sin(bm + c) ) is:[frac{sinleft( frac{bk}{2} right) cdot sinleft( b cdot frac{k + 1}{2} + c right)}{sinleft( frac{b}{2} right)}]Wait, let me verify that. I think it's similar to the formula for the sum of a sine series. The general formula for the sum ( sum_{m=0}^{n-1} sin(a + md) ) is:[frac{sinleft( frac{nd}{2} right) cdot sinleft( a + frac{(n - 1)d}{2} right)}{sinleft( frac{d}{2} right)}]In our case, the sum starts at ( m = 1 ) instead of ( m = 0 ), so we can adjust accordingly. Let me set ( a = b + c ) and ( d = b ). Then, the sum becomes:[sum_{m=1}^{k} sin(bm + c) = sum_{m=1}^{k} sin(c + bm)]Let me denote ( theta = c + b ) and ( phi = b ). Then, the sum is:[sum_{m=1}^{k} sin(theta + (m - 1)phi)]Wait, no. If I set ( m' = m - 1 ), then when ( m = 1 ), ( m' = 0 ), and when ( m = k ), ( m' = k - 1 ). So,[sum_{m=1}^{k} sin(c + bm) = sum_{m'=0}^{k - 1} sin(c + b + bm') = sum_{m'=0}^{k - 1} sin(theta + m'phi)]where ( theta = c + b ) and ( phi = b ). Now, applying the formula:[sum_{m'=0}^{k - 1} sin(theta + m'phi) = frac{sinleft( frac{kphi}{2} right) cdot sinleft( theta + frac{(k - 1)phi}{2} right)}{sinleft( frac{phi}{2} right)}]Substituting back ( theta = c + b ) and ( phi = b ):[sum_{m=1}^{k} sin(c + bm) = frac{sinleft( frac{kb}{2} right) cdot sinleft( c + b + frac{(k - 1)b}{2} right)}{sinleft( frac{b}{2} right)}]Simplify the argument of the second sine:[c + b + frac{(k - 1)b}{2} = c + frac{2b + (k - 1)b}{2} = c + frac{(k + 1)b}{2}]So, putting it all together:[sum_{m=1}^{k} sin(bm + c) = frac{sinleft( frac{kb}{2} right) cdot sinleft( c + frac{(k + 1)b}{2} right)}{sinleft( frac{b}{2} right)}]Therefore, the sum ( S_k ) is:[S_k = a cdot frac{sinleft( frac{kb}{2} right) cdot sinleft( c + frac{(k + 1)b}{2} right)}{sinleft( frac{b}{2} right)} + d cdot k]I think that's the expression for ( S_k ). Let me just double-check if the formula makes sense. When ( b = 0 ), the sine terms should simplify, but ( b = 0 ) would make the denominator zero, which is undefined. However, in that case, ( G_k = a cdot sin(c) + d ), a constant, so the sum would be ( k(a sin c + d) ), which is consistent with the formula if we take the limit as ( b to 0 ). Using L’Hospital’s Rule, the limit of the sine terms would be ( k sin(c + frac{b}{2}) ), which as ( b to 0 ) becomes ( k sin c ). So, yes, that seems consistent.Moving on to part 2: We need to calculate the probability that Vestri will score between ( M - 5 ) and ( M + 5 ) goals in the upcoming season, assuming the total goals follow a normal distribution with mean ( mu = M ) and standard deviation ( sigma ).First, since the total goals are normally distributed, we can model this as ( X sim N(mu, sigma^2) ), where ( X ) is the total goals next season. We need to find ( P(M - 5 leq X leq M + 5) ).To compute this probability, we can use the Z-score formula. The Z-score for a value ( x ) is given by:[Z = frac{x - mu}{sigma}]So, for ( x = M - 5 ):[Z_1 = frac{(M - 5) - M}{sigma} = frac{-5}{sigma}]And for ( x = M + 5 ):[Z_2 = frac{(M + 5) - M}{sigma} = frac{5}{sigma}]The probability ( P(M - 5 leq X leq M + 5) ) is then equal to the probability that ( Z ) is between ( Z_1 ) and ( Z_2 ). That is,[P(Z_1 leq Z leq Z_2) = Phi(Z_2) - Phi(Z_1)]where ( Phi ) is the cumulative distribution function (CDF) of the standard normal distribution.Therefore, the probability is:[Phileft( frac{5}{sigma} right) - Phileft( frac{-5}{sigma} right)]Alternatively, since the normal distribution is symmetric, this can also be written as:[2Phileft( frac{5}{sigma} right) - 1]But I think the first expression is more straightforward.Let me recap: We converted the problem into Z-scores, used the standard normal distribution's CDF to find the probability between those two Z-scores, and that gives us the desired probability.So, summarizing both parts:1. The sum ( S_k ) is expressed in terms of sine functions and a linear term.2. The probability is expressed using the standard normal CDF evaluated at ( pm 5/sigma ).I think that covers both parts. I don't see any mistakes in my reasoning, but let me just quickly go through each step again.For part 1, I split the sum into two parts, recognized the sine summation formula, adjusted for the starting index, applied the formula, and simplified. It seems correct.For part 2, I identified the distribution, converted to Z-scores, and used the CDF. That also seems solid.Yeah, I think I'm confident with these answers.Final Answer1. The sum ( S_k ) is given by (boxed{S_k = frac{a sinleft( frac{kb}{2} right) sinleft( c + frac{(k + 1)b}{2} right)}{sinleft( frac{b}{2} right)} + dk}).2. The probability is (boxed{2Phileft( frac{5}{sigma} right) - 1}), where (Phi) is the standard normal CDF.</think>

🧠:<think>Okay, so I have this problem about the U-BT Cluj-Napoca basketball team. It has two parts. Let me tackle them one by one.Starting with part 1: They played 30 games, and the top player's scoring pattern is given by a quadratic function S(n) = an² + bn + c. We need to find a, b, and c. We know the scores for three specific games: n=1, n=15, and n=30. The scores are 20, 50, and 80 respectively.So, I can set up three equations based on these points.First, for n=1:S(1) = a(1)² + b(1) + c = a + b + c = 20.Second, for n=15:S(15) = a(15)² + b(15) + c = 225a + 15b + c = 50.Third, for n=30:S(30) = a(30)² + b(30) + c = 900a + 30b + c = 80.So now I have a system of three equations:1. a + b + c = 202. 225a + 15b + c = 503. 900a + 30b + c = 80I need to solve this system for a, b, and c.Let me write them down again:1. a + b + c = 202. 225a + 15b + c = 503. 900a + 30b + c = 80Maybe I can subtract equation 1 from equation 2 and equation 1 from equation 3 to eliminate c.Subtracting equation 1 from equation 2:(225a + 15b + c) - (a + b + c) = 50 - 20224a + 14b = 30Similarly, subtracting equation 1 from equation 3:(900a + 30b + c) - (a + b + c) = 80 - 20899a + 29b = 60So now I have two new equations:4. 224a + 14b = 305. 899a + 29b = 60Hmm, these are two equations with two variables, a and b. Let me try to solve them.First, equation 4: 224a + 14b = 30. Maybe I can simplify this equation by dividing by 14.224a /14 = 16a, 14b /14 = b, 30 /14 = 15/7 ≈ 2.1429So equation 4 becomes:16a + b = 15/7 ≈ 2.1429Similarly, equation 5: 899a + 29b = 60.I can solve equation 4 for b:b = (15/7) - 16aThen substitute this into equation 5:899a + 29[(15/7) - 16a] = 60Let me compute this step by step.First, expand the equation:899a + 29*(15/7) - 29*16a = 60Calculate 29*(15/7):29*15 = 435, so 435/7 ≈ 62.1429And 29*16 = 464, so 464a.So the equation becomes:899a + 435/7 - 464a = 60Combine like terms:(899a - 464a) + 435/7 = 60435a + 435/7 = 60Factor out 435:435(a + 1/7) = 60Divide both sides by 435:a + 1/7 = 60 / 435Simplify 60/435: both divisible by 15, so 4/29.So:a + 1/7 = 4/29Subtract 1/7 from both sides:a = 4/29 - 1/7Find a common denominator, which is 203.4/29 = 28/203, 1/7 = 29/203.So:a = 28/203 - 29/203 = (-1)/203So a = -1/203 ≈ -0.004926Now, substitute a back into equation 4 to find b.From equation 4:16a + b = 15/7So:b = 15/7 - 16aPlug in a = -1/203:b = 15/7 - 16*(-1/203) = 15/7 + 16/203Convert 15/7 to 435/203 (since 15*29=435, 7*29=203)So:b = 435/203 + 16/203 = 451/203 ≈ 2.22167Now, with a and b known, we can find c from equation 1:a + b + c = 20So:c = 20 - a - bSubstitute a and b:c = 20 - (-1/203) - 451/203Convert 20 to 4060/203 (since 20*203=4060)So:c = 4060/203 + 1/203 - 451/203 = (4060 + 1 - 451)/203 = (4061 - 451)/203 = 3610/203Simplify 3610 divided by 203:203*17 = 3451, 3610 - 3451 = 159, so 17 + 159/203.159 and 203 have a common factor? 203 is 7*29, 159 is 3*53. No common factors, so c = 3610/203 ≈ 17.783So, summarizing:a = -1/203 ≈ -0.004926b = 451/203 ≈ 2.22167c = 3610/203 ≈ 17.783Let me check these values in the original equations to make sure.First, equation 1: a + b + c ≈ -0.004926 + 2.22167 + 17.783 ≈ 20, which is correct.Equation 2: 225a + 15b + c ≈ 225*(-0.004926) + 15*(2.22167) + 17.783Calculate each term:225*(-0.004926) ≈ -1.1098515*(2.22167) ≈ 33.325So total: -1.10985 + 33.325 + 17.783 ≈ (-1.10985 + 33.325) = 32.215 + 17.783 ≈ 50, which is correct.Equation 3: 900a + 30b + c ≈ 900*(-0.004926) + 30*(2.22167) + 17.783Calculate each term:900*(-0.004926) ≈ -4.433430*(2.22167) ≈ 66.65Total: -4.4334 + 66.65 + 17.783 ≈ (66.65 - 4.4334) = 62.2166 + 17.783 ≈ 80, which is correct.So, the values seem correct.Therefore, the constants are:a = -1/203, b = 451/203, c = 3610/203.Alternatively, as decimals:a ≈ -0.004926, b ≈ 2.22167, c ≈ 17.783.Moving on to part 2: The team's winning probability function is given by P(S) = 1 / (1 + e^{-k(S - m)}). We need to find constants k and m.Given:When S = 60, P = 0.75When S = 30, P = 0.25So, plug these into the equation.First, for S=60, P=0.75:0.75 = 1 / (1 + e^{-k(60 - m)})Similarly, for S=30, P=0.25:0.25 = 1 / (1 + e^{-k(30 - m)})Let me write these equations:1. 0.75 = 1 / (1 + e^{-k(60 - m)})2. 0.25 = 1 / (1 + e^{-k(30 - m)})Let me solve equation 1 first.Take reciprocals:1 / 0.75 = 1 + e^{-k(60 - m)}1 / 0.75 = 4/3 ≈ 1.3333So:4/3 = 1 + e^{-k(60 - m)}Subtract 1:4/3 - 1 = e^{-k(60 - m)}1/3 = e^{-k(60 - m)}Take natural logarithm:ln(1/3) = -k(60 - m)ln(1/3) = -ln(3) ≈ -1.0986So:-1.0986 = -k(60 - m)Multiply both sides by -1:1.0986 = k(60 - m) --> Equation ASimilarly, solve equation 2:0.25 = 1 / (1 + e^{-k(30 - m)})Take reciprocals:1 / 0.25 = 1 + e^{-k(30 - m)}4 = 1 + e^{-k(30 - m)}Subtract 1:3 = e^{-k(30 - m)}Take natural logarithm:ln(3) ≈ 1.0986 = -k(30 - m)Multiply both sides by -1:-1.0986 = k(30 - m) --> Equation BNow, we have two equations:Equation A: 1.0986 = k(60 - m)Equation B: -1.0986 = k(30 - m)Let me write them as:A: k(60 - m) = 1.0986B: k(30 - m) = -1.0986Let me subtract equation B from equation A:k(60 - m) - k(30 - m) = 1.0986 - (-1.0986)Simplify left side:k(60 - m - 30 + m) = k(30) = 2.1972So:30k = 2.1972Therefore, k = 2.1972 / 30 ≈ 0.07324So k ≈ 0.07324Now, plug k into equation A to find m.From equation A:k(60 - m) = 1.0986So:60 - m = 1.0986 / k ≈ 1.0986 / 0.07324 ≈ 15So:60 - m ≈ 15Therefore, m ≈ 60 - 15 = 45Let me verify with equation B:k(30 - m) ≈ 0.07324*(30 - 45) = 0.07324*(-15) ≈ -1.0986, which matches equation B.So, k ≈ 0.07324 and m ≈ 45.But let me express k more accurately.From equation A:k = 1.0986 / (60 - m)But from equation B:k = -1.0986 / (30 - m)Set them equal:1.0986 / (60 - m) = -1.0986 / (30 - m)Multiply both sides by (60 - m)(30 - m):1.0986*(30 - m) = -1.0986*(60 - m)Divide both sides by 1.0986:(30 - m) = -(60 - m)Simplify:30 - m = -60 + mBring variables to one side:30 + 60 = m + m90 = 2mSo m = 45Then, plug m=45 into equation A:k(60 - 45) = 1.0986k*15 = 1.0986k = 1.0986 / 15 ≈ 0.07324So, exact value of k is 1.0986 / 15. Since 1.0986 is approximately ln(3), which is about 1.098612289.So, k = ln(3)/15 ≈ 0.07324Therefore, exact expressions:k = ln(3)/15m = 45So, summarizing:k = ln(3)/15 ≈ 0.07324m = 45Let me double-check.If m=45 and k=ln(3)/15, then:For S=60:P(S) = 1 / (1 + e^{-k(60 - 45)}) = 1 / (1 + e^{-k*15})k*15 = ln(3), so e^{-ln(3)} = 1/3Thus, P(S) = 1 / (1 + 1/3) = 1 / (4/3) = 3/4 = 0.75, which is correct.For S=30:P(S) = 1 / (1 + e^{-k(30 - 45)}) = 1 / (1 + e^{-k*(-15)}) = 1 / (1 + e^{k*15})Again, k*15 = ln(3), so e^{ln(3)} = 3Thus, P(S) = 1 / (1 + 3) = 1/4 = 0.25, which is correct.Perfect, so the exact values are k = ln(3)/15 and m = 45.Final Answer1. The constants are ( a = boxed{-dfrac{1}{203}} ), ( b = boxed{dfrac{451}{203}} ), and ( c = boxed{dfrac{3610}{203}} ).2. The constants are ( k = boxed{dfrac{ln 3}{15}} ) and ( m = boxed{45} ).</think>

🧠:<think>Alright, so I have this problem about quadratic sequences and Paleolithic carvings. Let me try to figure it out step by step. First, the problem says that the position of the nth carving is given by a quadratic equation: P_n = an² + bn + c. We're given the first three positions: 3, 10, and 21. So, for n=1, P₁=3; n=2, P₂=10; n=3, P₃=21. I need to find the constants a, b, and c.Okay, let's write down the equations based on the given positions. For n=1:a(1)² + b(1) + c = 3Which simplifies to:a + b + c = 3 ...(1)For n=2:a(2)² + b(2) + c = 10Which is:4a + 2b + c = 10 ...(2)For n=3:a(3)² + b(3) + c = 21Which becomes:9a + 3b + c = 21 ...(3)Now, I have three equations:1) a + b + c = 32) 4a + 2b + c = 103) 9a + 3b + c = 21I need to solve this system of equations for a, b, and c. Let me subtract equation (1) from equation (2) to eliminate c.Equation (2) - Equation (1):(4a + 2b + c) - (a + b + c) = 10 - 3Which is:3a + b = 7 ...(4)Similarly, subtract equation (2) from equation (3):Equation (3) - Equation (2):(9a + 3b + c) - (4a + 2b + c) = 21 - 10Which simplifies to:5a + b = 11 ...(5)Now, I have two equations:4) 3a + b = 75) 5a + b = 11Let me subtract equation (4) from equation (5) to eliminate b:(5a + b) - (3a + b) = 11 - 7Which gives:2a = 4So, a = 2Now that I have a, I can plug it back into equation (4) to find b.From equation (4):3(2) + b = 76 + b = 7So, b = 1Now, with a=2 and b=1, I can find c using equation (1):2 + 1 + c = 33 + c = 3So, c = 0Wait, so a=2, b=1, c=0? Let me check if these satisfy all three original equations.For equation (1): 2 + 1 + 0 = 3 ✔️Equation (2): 4(2) + 2(1) + 0 = 8 + 2 = 10 ✔️Equation (3): 9(2) + 3(1) + 0 = 18 + 3 = 21 ✔️Perfect, so the quadratic formula is P_n = 2n² + n.Alright, that was part 1. Now, moving on to part 2.The historian wants to find the position of the carving that represents the year 10,000 BC. The sequence started from 20,000 BC, and each carving corresponds to a decrease of 100 years. So, each carving is 100 years apart, going from 20,000 BC, 19,900 BC, 19,800 BC, and so on.So, I need to figure out which term in the sequence corresponds to 10,000 BC. Let me think about how many carvings are between 20,000 BC and 10,000 BC, decreasing by 100 years each time.The time difference between 20,000 BC and 10,000 BC is 10,000 years. Since each step is 100 years, the number of steps is 10,000 / 100 = 100 steps. But wait, does that mean the 100th carving is 10,000 BC? Or is it the 101st?Wait, let's think. If the first carving is 20,000 BC, then the second is 19,900 BC, the third is 19,800 BC, and so on. So, each nth carving corresponds to 20,000 - 100(n-1) BC.So, to find n when the year is 10,000 BC:10,000 = 20,000 - 100(n - 1)Let me solve for n:10,000 = 20,000 - 100n + 100Wait, that would be:10,000 = 20,100 - 100nWait, hold on, maybe I should write it as:Year = 20,000 - 100*(n - 1)So, 10,000 = 20,000 - 100(n - 1)Let me subtract 20,000 from both sides:10,000 - 20,000 = -100(n - 1)-10,000 = -100(n - 1)Divide both sides by -100:100 = n - 1So, n = 101Therefore, the 101st carving corresponds to 10,000 BC.Now, I need to find the position P₁₀₁ using the quadratic formula we found earlier: P_n = 2n² + n.So, P₁₀₁ = 2*(101)² + 101Let me compute that.First, compute 101 squared:101² = 10,201Multiply by 2:2*10,201 = 20,402Add 101:20,402 + 101 = 20,503So, the position is 20,503.Wait, that seems really large. Let me double-check my calculations.Wait, 101 squared is indeed 10,201. Multiply by 2: 20,402. Add 101: 20,503. Hmm, that seems correct. But is it reasonable?Given that the first three positions are 3, 10, 21, which are quadratic, so the positions increase quadratically. So, by the 101st term, it's going to be a large number. So, 20,503 is plausible.Wait, but let me think again about the term number. If n=1 corresponds to 20,000 BC, then n=101 corresponds to 20,000 - 100*(101 - 1) = 20,000 - 100*100 = 20,000 - 10,000 = 10,000 BC. So, yes, n=101 is correct.Therefore, P₁₀₁ = 2*(101)^2 + 101 = 20,503.So, the position is 20,503.Wait, but just to make sure, let me compute 2*(101)^2 + 101 step by step.First, 101 squared:101 * 101:100*100 = 10,000100*1 = 1001*100 = 1001*1 = 1So, 10,000 + 100 + 100 + 1 = 10,201Yes, that's correct.Multiply by 2: 10,201 * 2 = 20,402Add 101: 20,402 + 101 = 20,503Yes, that's correct.So, the position is 20,503.Wait, just to make sure, let me check if I interpreted the problem correctly.The sequence started from 20,000 BC, each carving is 100 years apart. So, the first carving is 20,000 BC, the second is 19,900 BC, ..., the nth carving is 20,000 - 100(n-1) BC.So, solving for n when year is 10,000 BC:10,000 = 20,000 - 100(n - 1)So, 10,000 - 20,000 = -100(n - 1)-10,000 = -100(n - 1)Divide both sides by -100: 100 = n - 1 => n=101.Yes, that's correct.Therefore, the position is P₁₀₁ = 2*(101)^2 + 101 = 20,503.So, I think that's the answer.Final AnswerThe position of the carving representing the year 10,000 BC is boxed{20503}.</think>

🧠:<think>Alright, so I've got this problem about a retired skater's career, and I need to figure out their total cumulative score over 15 years. Then, I also have to compare it to a hypothetical skater who scored 85 every year. Hmm, okay, let's break this down step by step.First, the skater's average score each year is modeled by the function ( S(t) = 75 times (1.02)^t ), where ( t ) is the number of years since they started. They competed in an average of 12 competitions each year. So, to find the total cumulative score, I think I need to calculate the average score each year, multiply it by the number of competitions that year, and then sum all those up over 15 years.Wait, but the number of competitions per year is constant at 12, right? So, actually, for each year ( t ), the total score would be ( 12 times S(t) ). Then, the cumulative total would be the sum of these yearly totals from year 0 to year 14 (since it's 15 years total). Hmm, let me make sure about the time indices. If ( t = 0 ) is the first year, then ( t ) goes from 0 to 14 for 15 years.So, the total cumulative score ( C ) would be:[C = sum_{t=0}^{14} 12 times 75 times (1.02)^t]Simplifying that, it's:[C = 12 times 75 times sum_{t=0}^{14} (1.02)^t]Which is:[C = 900 times sum_{t=0}^{14} (1.02)^t]Okay, now I need to compute this sum. This looks like a geometric series. The formula for the sum of a geometric series is:[sum_{t=0}^{n} r^t = frac{r^{n+1} - 1}{r - 1}]Where ( r ) is the common ratio, which is 1.02 here, and ( n ) is 14. So plugging in the numbers:[sum_{t=0}^{14} (1.02)^t = frac{(1.02)^{15} - 1}{1.02 - 1} = frac{(1.02)^{15} - 1}{0.02}]I need to calculate ( (1.02)^{15} ). Let me see, I can use logarithms or just approximate it. Alternatively, I remember that ( (1.02)^{15} ) is approximately 1.34685. Let me verify that with a calculator:1.02^1 = 1.021.02^2 = 1.04041.02^3 ≈ 1.0612081.02^4 ≈ 1.0824321.02^5 ≈ 1.1040891.02^6 ≈ 1.1261711.02^7 ≈ 1.1486941.02^8 ≈ 1.1718681.02^9 ≈ 1.1952051.02^10 ≈ 1.2189091.02^11 ≈ 1.2430871.02^12 ≈ 1.2679491.02^13 ≈ 1.2933081.02^14 ≈ 1.3191741.02^15 ≈ 1.345893Okay, so approximately 1.345893. So plugging that back in:[frac{1.345893 - 1}{0.02} = frac{0.345893}{0.02} = 17.29465]So the sum of the geometric series is approximately 17.29465. Therefore, the total cumulative score is:[C = 900 times 17.29465 ≈ 900 times 17.29465]Calculating that:First, 900 * 17 = 15,300Then, 900 * 0.29465 ≈ 900 * 0.29465Calculating 900 * 0.2 = 180900 * 0.09465 ≈ 900 * 0.09 = 81, and 900 * 0.00465 ≈ 4.185So total for 0.29465 is approximately 180 + 81 + 4.185 ≈ 265.185So total cumulative score is approximately 15,300 + 265.185 ≈ 15,565.185So approximately 15,565.19.Wait, let me check that multiplication again because 900 * 17.29465.Alternatively, 900 * 17.29465 = 900 * 17 + 900 * 0.2946517 * 900 = 15,3000.29465 * 900: Let's compute 0.2 * 900 = 180, 0.09 * 900 = 81, 0.00465 * 900 ≈ 4.185So 180 + 81 = 261, plus 4.185 is 265.185So total is 15,300 + 265.185 = 15,565.185So approximately 15,565.19.Therefore, the total cumulative score is approximately 15,565.19.Wait, but let me make sure about the geometric series calculation. Maybe I should use a calculator for more precision.Alternatively, perhaps I can use the formula more accurately.Let me compute ( (1.02)^{15} ) more precisely.Using logarithms:ln(1.02) ≈ 0.0198026So ln(1.02^15) = 15 * 0.0198026 ≈ 0.297039Then exponentiate: e^0.297039 ≈ 1.34589So that's consistent with my earlier calculation.So the sum is (1.34589 - 1)/0.02 = 0.34589 / 0.02 = 17.2945So, 17.2945 multiplied by 900 is indeed 15,565.05So, approximately 15,565.05So, rounding to the nearest whole number, it's 15,565.Wait, but the problem says "calculate the total cumulative score", so maybe we need to present it as a whole number or keep it as a decimal. The question doesn't specify, so perhaps we can present it as 15,565.05 or 15,565.19 depending on rounding.But let me double-check my steps because sometimes when dealing with geometric series, it's easy to make an off-by-one error.Wait, the skater's career is 15 years, so t goes from 0 to 14, inclusive, which is 15 terms. So the sum from t=0 to t=14 is correct.So, the sum is (1.02^15 - 1)/0.02, which is approximately 17.2945.Multiplying by 900 gives approximately 15,565.05.So, I think that's correct.Now, moving on to part 2. The journalist wants to compare this skater to a hypothetical skater who scored 85 every year. So, the hypothetical skater's total cumulative score would be 12 competitions per year times 85 average score, over 15 years.So, that would be:Total score = 12 * 85 * 15Calculating that:12 * 85 = 1,0201,020 * 15 = 15,300So, the hypothetical skater has a total cumulative score of 15,300.The retired skater has approximately 15,565.05.So, the difference is 15,565.05 - 15,300 = 265.05So, the retired skater has a higher cumulative score by approximately 265.05 points.Wait, but let me make sure about the hypothetical skater's calculation. If the skater's average score is 85 each year, and they compete 12 times each year, then each year's total score is 12 * 85 = 1,020. Over 15 years, that's 1,020 * 15 = 15,300. That seems correct.So, the difference is 15,565.05 - 15,300 = 265.05.So, approximately 265.05 points higher for the retired skater.But let me check if the skater's cumulative score is indeed higher. Since the skater started at 75 and increased by 2% each year, their scores would have been lower in the beginning and higher towards the end. The hypothetical skater is at 85 every year, which is higher than the skater's starting point but lower than the skater's later years.So, it's possible that the skater's cumulative score is higher because in the later years, their scores surpass 85.Wait, let's check when the skater's score surpasses 85.We can solve for t when 75*(1.02)^t = 85So, (1.02)^t = 85/75 ≈ 1.1333Taking natural logs:t * ln(1.02) = ln(1.1333)t ≈ ln(1.1333)/ln(1.02) ≈ 0.1255 / 0.0198026 ≈ 6.34So, around year 6.34, the skater's score would reach 85. So, starting from year 7, the skater's average score would be above 85.Therefore, in the first 6 years, the skater's scores are below 85, and from year 7 onwards, they are above 85.So, the cumulative score difference would be negative in the first 6 years and positive from year 7 to 15.But since we're summing over all 15 years, the total difference is positive, as we calculated, 265.05.So, that seems consistent.Wait, but let me think again. The skater's cumulative score is higher, but is it by 265? Let me verify the calculations again.Retired skater: ~15,565.05Hypothetical skater: 15,300Difference: ~265.05Yes, that seems correct.Alternatively, perhaps I should compute the difference each year and sum them up, but that would be more time-consuming.Alternatively, since the hypothetical skater's total is 12*85*15=15,300, and the retired skater's total is ~15,565.05, the difference is ~265.05.So, the total difference in cumulative scores is approximately 265.05 in favor of the retired skater.Therefore, the answers are:1. Total cumulative score: approximately 15,565.052. Difference: approximately 265.05But let me check if I should present these as exact numbers or if I need to calculate them more precisely.Wait, the sum of the geometric series was approximated as 17.2945, but perhaps I can compute it more accurately.Let me use more precise value for (1.02)^15.Using a calculator:1.02^1 = 1.021.02^2 = 1.04041.02^3 = 1.0612081.02^4 = 1.082432161.02^5 = 1.104089891.02^6 = 1.126171691.02^7 = 1.148695121.02^8 = 1.171868021.02^9 = 1.195205381.02^10 = 1.218909491.02^11 = 1.243087671.02^12 = 1.267949431.02^13 = 1.293308421.02^14 = 1.319174591.02^15 = 1.34589308So, (1.02)^15 ≈ 1.34589308Therefore, the sum is (1.34589308 - 1)/0.02 = 0.34589308 / 0.02 = 17.294654So, 17.294654Then, 900 * 17.294654 = ?Calculating 900 * 17 = 15,300900 * 0.294654 = ?0.2 * 900 = 1800.09 * 900 = 810.004654 * 900 ≈ 4.1886So, 180 + 81 = 261 + 4.1886 ≈ 265.1886So, total is 15,300 + 265.1886 ≈ 15,565.1886So, approximately 15,565.19So, the total cumulative score is approximately 15,565.19Hypothetical skater: 15,300Difference: 15,565.19 - 15,300 = 265.19So, approximately 265.19Therefore, the answers are:1. Total cumulative score: 15,565.192. Difference: 265.19But since the problem might expect integer values, perhaps we can round to the nearest whole number.So, 15,565 and 265.Alternatively, maybe we can present them as exact decimals.But let me check if I can express the sum more precisely.Alternatively, perhaps I can use the formula for the sum of a geometric series without approximating (1.02)^15.But in any case, the approximate value is sufficient for the problem.So, summarizing:1. The total cumulative score is approximately 15,565.192. The difference is approximately 265.19So, I think that's the solution.Final Answer1. The total cumulative score is boxed{15565.19}.2. The total difference in cumulative scores is boxed{265.19}.</think>