🧠:<think>Okay, so I have this problem about a light novel series with n volumes. Each volume has a certain number of pages, and the number of pages follows a quadratic relationship given by p_i = ai² + bi + c. There are two sub-problems here.Starting with Sub-problem 1: I need to find the constants a, b, and c given that the first three volumes have 150, 230, and 330 pages respectively. Hmm, okay. Since each volume corresponds to i=1, 2, 3, I can set up equations based on the given page numbers.So, for i=1: p₁ = a(1)² + b(1) + c = a + b + c = 150 For i=2: p₂ = a(2)² + b(2) + c = 4a + 2b + c = 230 For i=3: p₃ = a(3)² + b(3) + c = 9a + 3b + c = 330Now, I have a system of three equations:1) a + b + c = 150 2) 4a + 2b + c = 230 3) 9a + 3b + c = 330I need to solve for a, b, and c. Let me subtract equation 1 from equation 2 to eliminate c:(4a + 2b + c) - (a + b + c) = 230 - 150 3a + b = 80 --> Let's call this equation 4.Similarly, subtract equation 2 from equation 3:(9a + 3b + c) - (4a + 2b + c) = 330 - 230 5a + b = 100 --> Let's call this equation 5.Now, subtract equation 4 from equation 5:(5a + b) - (3a + b) = 100 - 80 2a = 20 So, a = 10.Now, plug a = 10 into equation 4:3(10) + b = 80 30 + b = 80 b = 50.Now, plug a = 10 and b = 50 into equation 1:10 + 50 + c = 150 60 + c = 150 c = 90.So, the constants are a=10, b=50, c=90.Let me double-check these values with the given page numbers:For i=1: 10(1) + 50(1) + 90 = 10 + 50 + 90 = 150 ✔️ For i=2: 10(4) + 50(2) + 90 = 40 + 100 + 90 = 230 ✔️ For i=3: 10(9) + 50(3) + 90 = 90 + 150 + 90 = 330 ✔️Looks good!Moving on to Sub-problem 2: I need to express the total number of volumes n in terms of a, b, c, k, and T, where k is pages per day and T is the total days. Then, determine n when k=50 and T=120.First, the total number of pages in the series is the sum of pages from volume 1 to volume n. Since each volume i has p_i = ai² + bi + c pages, the total pages S is:S = Σ (from i=1 to n) (ai² + bi + c) = a Σi² + b Σi + c Σ1I remember the formulas for these sums:Σi² from 1 to n is n(n+1)(2n+1)/6 Σi from 1 to n is n(n+1)/2 Σ1 from 1 to n is nSo, substituting these in:S = a [n(n+1)(2n+1)/6] + b [n(n+1)/2] + c [n]Simplify each term:First term: a * n(n+1)(2n+1)/6 Second term: b * n(n+1)/2 Third term: c * nNow, the total reading time is T days, and the reader reads k pages per day. So, total pages S = k * T.Therefore:a [n(n+1)(2n+1)/6] + b [n(n+1)/2] + c n = k TWe need to solve for n in terms of a, b, c, k, T. Hmm, this seems like a cubic equation in n, which might be complicated to solve symbolically. Maybe we can express it as a cubic equation and then, for the specific values, solve numerically.But let's see if we can write it more neatly.Let me factor out n:n [ a (n+1)(2n+1)/6 + b (n+1)/2 + c ] = k TBut even so, it's still a cubic equation. Perhaps we can write it as:(2a/6) n³ + (a/6 + b/2) n² + (a/6 + b/2 + c) n - k T = 0 Simplify coefficients:2a/6 = a/3 a/6 + b/2 = (a + 3b)/6 a/6 + b/2 + c = (a + 3b + 6c)/6So, the equation becomes:(a/3) n³ + ( (a + 3b)/6 ) n² + ( (a + 3b + 6c)/6 ) n - k T = 0Multiply both sides by 6 to eliminate denominators:2a n³ + (a + 3b) n² + (a + 3b + 6c) n - 6k T = 0So, the cubic equation is:2a n³ + (a + 3b) n² + (a + 3b + 6c) n - 6k T = 0Now, for the specific case where a=10, b=50, c=90, k=50, T=120.Plugging these into the equation:2*10 n³ + (10 + 3*50) n² + (10 + 3*50 + 6*90) n - 6*50*120 = 0 Compute each term:2*10 = 20 10 + 150 = 160 10 + 150 + 540 = 690 6*50*120 = 36000So, the equation becomes:20n³ + 160n² + 690n - 36000 = 0Simplify by dividing all terms by 10:2n³ + 16n² + 69n - 3600 = 0Now, we have to solve 2n³ + 16n² + 69n - 3600 = 0This is a cubic equation, and solving it might require some trial and error or using the rational root theorem.Possible rational roots are factors of 3600 divided by factors of 2. So, possible roots could be ±1, ±2, ±3, ..., up to ±3600, but that's a lot. Maybe we can test some integer values.Let me try n=10:2*(1000) + 16*(100) + 69*(10) - 3600 = 2000 + 1600 + 690 - 3600 = (2000 + 1600) = 3600; 3600 + 690 = 4290; 4290 - 3600 = 690 ≠ 0n=10 gives 690, which is positive.n=15:2*(3375) + 16*(225) + 69*(15) - 3600 = 6750 + 3600 + 1035 - 3600 = (6750 + 3600) = 10350; 10350 + 1035 = 11385; 11385 - 3600 = 7785 ≠ 0Still positive.n=20:2*(8000) + 16*(400) + 69*(20) - 3600 = 16000 + 6400 + 1380 - 3600 = (16000 + 6400) = 22400; 22400 + 1380 = 23780; 23780 - 3600 = 20180 ≠ 0Still positive.Wait, maybe n is less than 10? Let's try n=5:2*(125) + 16*(25) + 69*(5) - 3600 = 250 + 400 + 345 - 3600 = (250 + 400) = 650; 650 + 345 = 995; 995 - 3600 = -2605 ≠ 0Negative. So between n=5 and n=10, the function crosses from negative to positive. So, the root is between 5 and 10.Let me try n=8:2*(512) + 16*(64) + 69*(8) - 3600 = 1024 + 1024 + 552 - 3600 = (1024 + 1024) = 2048; 2048 + 552 = 2600; 2600 - 3600 = -1000 ≠ 0Still negative.n=9:2*(729) + 16*(81) + 69*(9) - 3600 = 1458 + 1296 + 621 - 3600 = (1458 + 1296) = 2754; 2754 + 621 = 3375; 3375 - 3600 = -225 ≠ 0Still negative.n=9.5:Let me compute f(9.5):2*(9.5)^3 + 16*(9.5)^2 + 69*(9.5) - 3600First, 9.5³ = 857.375 9.5² = 90.25So,2*857.375 = 1714.75 16*90.25 = 1444 69*9.5 = 655.5Adding these up: 1714.75 + 1444 = 3158.75; 3158.75 + 655.5 = 3814.25 3814.25 - 3600 = 214.25 > 0So, f(9.5)=214.25, which is positive.Since f(9)=-225 and f(9.5)=214.25, the root is between 9 and 9.5.Let me try n=9.25:Compute f(9.25):9.25³ = approx 9.25*9.25=85.5625; 85.5625*9.25 ≈ 85.5625*9 + 85.5625*0.25 ≈ 770.0625 + 21.3906 ≈ 791.4531 9.25² = 85.5625So,2*791.4531 ≈ 1582.9062 16*85.5625 ≈ 1369 69*9.25 ≈ 636.75Adding these: 1582.9062 + 1369 ≈ 2951.9062; 2951.9062 + 636.75 ≈ 3588.6562 3588.6562 - 3600 ≈ -11.3438So, f(9.25)≈-11.34Close to zero. Let's try n=9.3:9.3³ = 9.3*9.3=86.49; 86.49*9.3 ≈ 804.357 9.3²=86.49So,2*804.357 ≈ 1608.714 16*86.49 ≈ 1383.84 69*9.3 ≈ 641.7Adding: 1608.714 + 1383.84 ≈ 2992.554; 2992.554 + 641.7 ≈ 3634.254 3634.254 - 3600 ≈ 34.254 >0So, f(9.3)=34.254So, between 9.25 and 9.3, f(n) crosses zero.Using linear approximation:At n=9.25, f=-11.34 At n=9.3, f=34.254The difference in n is 0.05, and the difference in f is 34.254 - (-11.34)=45.594We need to find delta such that f=0:delta = 11.34 / 45.594 ≈ 0.2486 of the interval 0.05So, delta ≈ 0.05 * 0.2486 ≈ 0.01243Thus, approximate root at n=9.25 + 0.01243 ≈ 9.2624So, n≈9.26But since n must be an integer (number of volumes), we need to check n=9 and n=10.Wait, but when n=9, the total pages would be S=2*10*9³ +16*10*9² +69*10*9 -3600? Wait, no, actually, wait.Wait, I think I confused the equation earlier.Wait, no, actually, the equation was 2n³ +16n² +69n -3600=0Wait, no, actually, the equation is 2n³ +16n² +69n -3600=0Wait, but in the specific case, I plugged in a=10, b=50, c=90, k=50, T=120.Wait, but when I plugged into the equation, I think I made a mistake. Let me double-check.Wait, the total pages S is equal to k*T, which is 50*120=6000 pages.But earlier, I set up the equation as 2a n³ + (a + 3b) n² + (a + 3b + 6c) n - 6k T = 0Plugging a=10, b=50, c=90, k=50, T=120:2*10=20 a + 3b=10 + 150=160 a + 3b + 6c=10 + 150 + 540=690 6k T=6*50*120=36000So, equation is 20n³ +160n² +690n -36000=0Then, dividing by 10: 2n³ +16n² +69n -3600=0Wait, but S is 6000 pages, so 2n³ +16n² +69n = 3600Wait, 2n³ +16n² +69n = 3600But when I plug n=9:2*(729) +16*(81) +69*(9) = 1458 + 1296 + 621 = 1458+1296=2754; 2754+621=3375 < 3600n=10: 2*1000 +16*100 +69*10=2000+1600+690=4290 >3600So, n is between 9 and 10. But n must be integer, so n=10 would give total pages 4290, which is more than 6000? Wait, wait, no.Wait, hold on, S is 6000 pages, but according to the equation, 2n³ +16n² +69n = 3600Wait, that can't be. Wait, no, hold on, I think I messed up the substitution.Wait, S = k*T = 50*120=6000But from earlier, S = a [n(n+1)(2n+1)/6] + b [n(n+1)/2] + c nWith a=10, b=50, c=90:S =10*[n(n+1)(2n+1)/6] +50*[n(n+1)/2] +90nLet me compute S for n=9:First term:10*[9*10*19/6] =10*(1710/6)=10*285=2850 Second term:50*[9*10/2]=50*45=2250 Third term:90*9=810 Total S=2850+2250+810=5910Which is less than 6000.For n=10:First term:10*[10*11*21/6]=10*(2310/6)=10*385=3850 Second term:50*[10*11/2]=50*55=2750 Third term:90*10=900 Total S=3850+2750+900=7500Which is more than 6000.So, the total pages needed is 6000, which is between n=9 and n=10.But n must be integer, so n=10 would give 7500 pages, which is more than 6000, but the reader finishes exactly in T=120 days. So, maybe n=10 is the answer because the reader can read the extra pages within the same time? Wait, but the total pages would be 7500, and reading 50 pages per day for 120 days is 6000 pages. So, the reader would finish 6000 pages, but the series has 7500 pages. That doesn't make sense.Wait, perhaps I made a mistake in the setup.Wait, the total pages S must equal k*T, which is 6000. So, S=6000.But when n=9, S=5910, which is less than 6000. When n=10, S=7500, which is more than 6000. So, there is no integer n where S=6000. Therefore, the series cannot be completed in exactly 120 days if n is an integer. But the problem says "finishes the entire series in exactly T days", so n must be such that S=6000.But since S is a cubic function in n, and for integer n, S jumps from 5910 to 7500, which skips over 6000. Therefore, there is no integer n that satisfies S=6000. Hmm, that seems problematic.Wait, maybe I made a mistake in the equation earlier.Wait, let me recast the problem.Total pages S = sum_{i=1}^n (10i² +50i +90) =10*sum i² +50*sum i +90*sum 1Which is:10*(n(n+1)(2n+1)/6) +50*(n(n+1)/2) +90nSimplify:10*(n(n+1)(2n+1)/6) = (10/6)*n(n+1)(2n+1) = (5/3)*n(n+1)(2n+1)50*(n(n+1)/2) =25*n(n+1)90n remains.So, S = (5/3)n(n+1)(2n+1) +25n(n+1) +90nLet me compute this for n=9:First term: (5/3)*9*10*19 = (5/3)*1710=5*570=2850 Second term:25*9*10=2250 Third term:90*9=810 Total:2850+2250+810=5910n=10:First term: (5/3)*10*11*21=(5/3)*2310=5*770=3850 Second term:25*10*11=2750 Third term:90*10=900 Total:3850+2750+900=7500So, same as before.Thus, S(n=9)=5910, S(n=10)=7500. So, 6000 is between these. Therefore, there is no integer n where S=6000. So, the problem might be expecting a non-integer n, but since n must be integer, perhaps the answer is n=10, but the reader would have to read only 6000 pages, skipping some pages of the 10th volume. But the problem says "finishes the entire series", so n must be such that S=6000. But since it's not possible with integer n, maybe the answer is n=10, but the reader only reads part of the 10th volume? But the problem says "finishes the entire series", so perhaps n=10 is the answer, even though it exceeds the total pages.Alternatively, maybe I made a mistake in the equation setup.Wait, let me check the equation again.From the beginning:Total pages S = sum_{i=1}^n (10i² +50i +90) =10*sum i² +50*sum i +90*sum 1Which is:10*(n(n+1)(2n+1)/6) +50*(n(n+1)/2) +90nSo, S = (10/6)n(n+1)(2n+1) + (50/2)n(n+1) +90n = (5/3)n(n+1)(2n+1) +25n(n+1) +90nLet me factor n:n [ (5/3)(n+1)(2n+1) +25(n+1) +90 ]Let me compute the expression inside the brackets:(5/3)(n+1)(2n+1) +25(n+1) +90Factor out (n+1):(n+1)[ (5/3)(2n+1) +25 ] +90Compute (5/3)(2n+1) +25:= (10n/3 +5/3) +25 = (10n/3 +5/3 +75/3) = (10n +80)/3So, the expression becomes:(n+1)(10n +80)/3 +90Multiply out:(10n² +80n +10n +80)/3 +90 = (10n² +90n +80)/3 +90 = (10n² +90n +80)/3 +270/3 = (10n² +90n +80 +270)/3 = (10n² +90n +350)/3So, S = n*(10n² +90n +350)/3Set this equal to 6000:n*(10n² +90n +350)/3 =6000 Multiply both sides by 3:n*(10n² +90n +350)=18000 Expand:10n³ +90n² +350n -18000=0 Divide by 10:n³ +9n² +35n -1800=0Now, we have a cubic equation: n³ +9n² +35n -1800=0Let me try to find integer roots. Possible roots are factors of 1800: ±1, ±2, ±3, ..., up to ±1800.Let me try n=10:1000 +900 +350 -1800=1000+900=1900; 1900+350=2250; 2250-1800=450≠0n=12:1728 + 1296 +420 -1800=1728+1296=3024; 3024+420=3444; 3444-1800=1644≠0n=15:3375 +2025 +525 -1800=3375+2025=5400; 5400+525=5925; 5925-1800=4125≠0n=9:729 +729 +315 -1800=729+729=1458; 1458+315=1773; 1773-1800=-27≠0n=8:512 +576 +280 -1800=512+576=1088; 1088+280=1368; 1368-1800=-432≠0n=11:1331 +1089 +385 -1800=1331+1089=2420; 2420+385=2805; 2805-1800=1005≠0n=13:2197 +1521 +455 -1800=2197+1521=3718; 3718+455=4173; 4173-1800=2373≠0n=7:343 +441 +245 -1800=343+441=784; 784+245=1029; 1029-1800=-771≠0n=6:216 +324 +210 -1800=216+324=540; 540+210=750; 750-1800=-1050≠0n=5:125 +225 +175 -1800=125+225=350; 350+175=525; 525-1800=-1275≠0n=14:2744 +1764 +490 -1800=2744+1764=4508; 4508+490=4998; 4998-1800=3198≠0n=16:4096 +2304 +560 -1800=4096+2304=6400; 6400+560=6960; 6960-1800=5160≠0n=17:4913 +2601 +595 -1800=4913+2601=7514; 7514+595=8109; 8109-1800=6309≠0n=18:5832 +2916 +630 -1800=5832+2916=8748; 8748+630=9378; 9378-1800=7578≠0n=19:6859 +3249 +665 -1800=6859+3249=10108; 10108+665=10773; 10773-1800=8973≠0n=20:8000 +3600 +700 -1800=8000+3600=11600; 11600+700=12300; 12300-1800=10500≠0Hmm, none of these are working. Maybe I made a mistake in the equation.Wait, let me go back.Total pages S = sum_{i=1}^n (10i² +50i +90) =10*sum i² +50*sum i +90*sum 1Which is:10*(n(n+1)(2n+1)/6) +50*(n(n+1)/2) +90nSimplify:10*(n(n+1)(2n+1)/6) = (10/6)n(n+1)(2n+1) = (5/3)n(n+1)(2n+1)50*(n(n+1)/2) =25n(n+1)90n remains.So, S = (5/3)n(n+1)(2n+1) +25n(n+1) +90nLet me compute this for n=9:(5/3)*9*10*19 = (5/3)*1710=2850 25*9*10=2250 90*9=810 Total=2850+2250+810=5910n=10:(5/3)*10*11*21=(5/3)*2310=3850 25*10*11=2750 90*10=900 Total=3850+2750+900=7500So, S(n=9)=5910, S(n=10)=7500Thus, to get S=6000, n must be between 9 and 10. But since n must be integer, there is no solution. Therefore, the answer is n=10, but the reader would have to read only part of the 10th volume. However, the problem states "finishes the entire series", which implies that n must be such that S=6000. Since it's not possible with integer n, perhaps the answer is n=10, but the reader reads only 6000 pages, which is less than the total 7500 pages. But that contradicts "finishes the entire series".Alternatively, maybe I made a mistake in the equation setup. Let me check again.Wait, the equation was:2a n³ + (a + 3b) n² + (a + 3b + 6c) n - 6k T = 0Plugging a=10, b=50, c=90, k=50, T=120:2*10=20 a +3b=10+150=160 a +3b +6c=10+150+540=690 6kT=6*50*120=36000So, equation:20n³ +160n² +690n -36000=0Divide by 10:2n³ +16n² +69n -3600=0Wait, but earlier, when I tried to compute S for n=9, I got 5910, which is less than 6000. For n=10, 7500>6000. So, the equation 2n³ +16n² +69n=3600 must hold.Wait, but 2n³ +16n² +69n=3600Let me compute for n=9:2*729 +16*81 +69*9=1458 +1296 +621=3375 <3600n=10:2*1000 +16*100 +69*10=2000+1600+690=4290>3600So, n is between 9 and 10.But since n must be integer, the answer is n=10, but the total pages would be 4290, which is more than 3600. Wait, but 3600 is the right-hand side of the equation, which was derived from S=6000.Wait, hold on, I think I messed up the equation.Wait, the equation was:2a n³ + (a + 3b) n² + (a + 3b + 6c) n - 6k T = 0But S = k*T =6000But S is also equal to (5/3)n(n+1)(2n+1) +25n(n+1) +90nWhich we set equal to 6000, leading to the equation 2n³ +16n² +69n -3600=0Wait, but 2n³ +16n² +69n=3600But when n=9, 2*729 +16*81 +69*9=1458+1296+621=33753375=3600? No, 3375<3600n=10:2*1000 +16*100 +69*10=2000+1600+690=4290>3600So, n is between 9 and 10.But since n must be integer, perhaps the answer is n=10, but the reader would have to read only 6000 pages, which is less than the total 7500 pages. But the problem says "finishes the entire series", so n must be such that S=6000. Since it's not possible with integer n, maybe the answer is n=10, but the series is not entirely finished. But the problem states "finishes the entire series", so perhaps n=10 is the answer, even though it exceeds the total pages.Alternatively, maybe I made a mistake in the equation setup.Wait, let me try solving the cubic equation numerically.Equation:2n³ +16n² +69n -3600=0Let me use the Newton-Raphson method.Let f(n)=2n³ +16n² +69n -3600f'(n)=6n² +32n +69We need to find n such that f(n)=0.We know f(9)=2*729 +16*81 +69*9 -3600=1458+1296+621-3600=3375-3600=-225f(10)=2*1000 +16*100 +69*10 -3600=2000+1600+690-3600=4290-3600=690So, f(9)=-225, f(10)=690Let's take n0=9.5f(9.5)=2*(857.375) +16*(90.25) +69*(9.5) -3600=1714.75 +1444 +655.5 -3600=1714.75+1444=3158.75+655.5=3814.25-3600=214.25f(9.5)=214.25f'(9.5)=6*(90.25) +32*(9.5) +69=541.5 +304 +69=914.5Next approximation: n1=9.5 - f(9.5)/f'(9.5)=9.5 -214.25/914.5≈9.5 -0.234≈9.266Compute f(9.266):n=9.266n³≈9.266^3≈approx 9^3=729, 9.266-9=0.266, so using binomial approx:(9 +0.266)^3≈9³ +3*9²*0.266 +3*9*(0.266)^2 + (0.266)^3≈729 +3*81*0.266 +3*9*0.0708 +0.0189≈729 +63.858 +1.891 +0.0189≈729+63.858=792.858+1.891=794.749+0.0189≈794.768So, 2n³≈2*794.768≈1589.536n²≈9.266²≈85.8616n²≈16*85.86≈1373.7669n≈69*9.266≈638.154Total f(n)=1589.536 +1373.76 +638.154 -3600≈1589.536+1373.76≈2963.296+638.154≈3601.45-3600≈1.45So, f(9.266)≈1.45f'(9.266)=6*(85.86) +32*(9.266) +69≈515.16 +296.512 +69≈515.16+296.512≈811.672+69≈880.672Next approximation: n2=9.266 -1.45/880.672≈9.266 -0.0016≈9.2644Compute f(9.2644):n=9.2644n³≈approx 9.2644^3≈using previous approx, n=9.266 gave n³≈794.768, so n=9.2644 is slightly less, say≈794.52n³≈1589n²≈9.2644²≈85.8316n²≈1373.2869n≈69*9.2644≈638Total f(n)=1589 +1373.28 +638 -3600≈1589+1373.28≈2962.28+638≈3600.28-3600≈0.28f'(9.2644)=6*(85.83) +32*(9.2644) +69≈514.98 +296.46 +69≈514.98+296.46≈811.44+69≈880.44Next approximation: n3=9.2644 -0.28/880.44≈9.2644 -0.000318≈9.26408Compute f(9.26408):n=9.26408n³≈approx 9.26408^3≈794.52n³≈1589n²≈9.26408²≈85.8316n²≈1373.2869n≈69*9.26408≈638Total f(n)=1589 +1373.28 +638 -3600≈3600.28-3600≈0.28Wait, seems like it's converging to around n≈9.264So, n≈9.264But since n must be integer, the closest integer is n=9, but S=5910<6000, or n=10, S=7500>6000.But the problem says "finishes the entire series in exactly T days", so n must be such that S=6000. Since it's not possible with integer n, perhaps the answer is n=10, but the reader reads only 6000 pages, which is less than the total 7500. But that contradicts "finishes the entire series".Alternatively, maybe the problem expects a non-integer n, but n must be integer. So, perhaps the answer is n=10, even though it's not exact.Alternatively, maybe I made a mistake in the equation setup.Wait, let me check the initial equation.Total pages S = sum_{i=1}^n (10i² +50i +90) =10*sum i² +50*sum i +90*sum 1Which is:10*(n(n+1)(2n+1)/6) +50*(n(n+1)/2) +90nSet equal to 6000:10*(n(n+1)(2n+1)/6) +50*(n(n+1)/2) +90n =6000Multiply both sides by 6 to eliminate denominators:10*n(n+1)(2n+1) +150*n(n+1) +540n =36000Expand:10n(n+1)(2n+1)=10*(2n³ +3n² +n)=20n³ +30n² +10n150n(n+1)=150n² +150n540n remains.So, total equation:20n³ +30n² +10n +150n² +150n +540n =36000Combine like terms:20n³ + (30n² +150n²) + (10n +150n +540n) =36000 20n³ +180n² +700n =36000Divide by 10:2n³ +18n² +70n =3600Wait, earlier I had 2n³ +16n² +69n -3600=0, but now it's 2n³ +18n² +70n -3600=0Wait, which is correct?Wait, let me recompute:Original equation:10*(n(n+1)(2n+1)/6) +50*(n(n+1)/2) +90n =6000Multiply by 6:10*n(n+1)(2n+1) +150*n(n+1) +540n =36000Expand:10*(2n³ +3n² +n) +150*(n² +n) +540n =20n³ +30n² +10n +150n² +150n +540n =20n³ + (30n² +150n²) + (10n +150n +540n) =20n³ +180n² +700nSet equal to 36000:20n³ +180n² +700n =36000 Divide by 10:2n³ +18n² +70n =3600 Thus, 2n³ +18n² +70n -3600=0Earlier, I had 2n³ +16n² +69n -3600=0, which was incorrect. I must have made a mistake in the coefficients earlier.So, correct equation is 2n³ +18n² +70n -3600=0Let me try to solve this.f(n)=2n³ +18n² +70n -3600f(9)=2*729 +18*81 +70*9 -3600=1458 +1458 +630 -3600=1458+1458=2916+630=3546-3600=-54f(10)=2*1000 +18*100 +70*10 -3600=2000+1800+700-3600=4500-3600=900So, f(9)=-54, f(10)=900Thus, root between 9 and10.Let me try n=9.5:f(9.5)=2*(857.375) +18*(90.25) +70*(9.5) -3600=1714.75 +1624.5 +665 -3600=1714.75+1624.5=3339.25+665=4004.25-3600=404.25>0f(9.25):n=9.25n³=9.25³≈814.53125 n²=85.5625f(n)=2*814.53125 +18*85.5625 +70*9.25 -3600 =1629.0625 +1540.125 +647.5 -3600 =1629.0625+1540.125=3169.1875+647.5=3816.6875-3600=216.6875>0f(9.1):n=9.1n³≈9.1³≈753.571 n²≈82.81f(n)=2*753.571 +18*82.81 +70*9.1 -3600 =1507.142 +1490.58 +637 -3600 =1507.142+1490.58=2997.722+637=3634.722-3600=34.722>0f(9.05):n=9.05n³≈9.05³≈741.48 n²≈81.9025f(n)=2*741.48 +18*81.9025 +70*9.05 -3600 =1482.96 +1474.245 +633.5 -3600 =1482.96+1474.245=2957.205+633.5=3590.705-3600≈-9.295So, f(9.05)≈-9.295f(9.1)=34.722Thus, root between 9.05 and9.1Using linear approximation:Between n=9.05 (f=-9.295) and n=9.1 (f=34.722)Slope=34.722 - (-9.295)=44.017 per 0.05We need to find delta where f=0:delta=9.295 /44.017≈0.211 of 0.05≈0.01055Thus, root≈9.05 +0.01055≈9.06055So, n≈9.06But since n must be integer, n=9 would give S=5910, which is less than 6000, and n=10 gives S=7500>6000.Thus, the answer is n=10, but the reader would have to read only 6000 pages, which is less than the total 7500. But the problem says "finishes the entire series", so perhaps n=10 is the answer, even though it's not exact.Alternatively, maybe the problem expects a non-integer n, but n must be integer. So, the answer is n=10.Wait, but let me check S(n=9.06):Using the equation S= (5/3)n(n+1)(2n+1) +25n(n+1) +90nFor n=9.06:First term: (5/3)*9.06*10.06*(2*9.06+1)= (5/3)*9.06*10.06*19.12Compute 9.06*10.06≈91.1436 91.1436*19.12≈1743.05 (5/3)*1743.05≈2905.08Second term:25*9.06*10.06≈25*91.1436≈2278.59Third term:90*9.06≈815.4Total S≈2905.08+2278.59+815.4≈2905.08+2278.59=5183.67+815.4≈5999.07≈6000So, n≈9.06 gives S≈6000But since n must be integer, the answer is n=10, but the reader reads only part of the 10th volume. However, the problem states "finishes the entire series", so perhaps n=10 is the answer, even though it's not exact.Alternatively, maybe the problem expects the answer to be n=10, as the next integer.But given the problem's context, I think the answer is n=10.So, final answer for Sub-problem 2 is n=10.</think>

🧠:<think>Okay, so I have this problem where an AI researcher is trying to build a machine learning model to automate debugging in software development. There are two parts to this problem: modeling the distribution of bugs and predicting potential bugs using a neural network. Let me try to tackle each part step by step.Starting with the first part: Modeling the Distribution. It says that X is a random variable representing the number of bugs in a code module, and it follows a Poisson distribution with an unknown rate λ. We have a sample of n code modules with the number of bugs x₁, x₂, ..., xₙ. The task is to derive the Maximum Likelihood Estimator (MLE) for λ.Hmm, I remember that the MLE is a method to estimate the parameters of a statistical model. For a Poisson distribution, the probability mass function is given by P(X = x) = (λ^x e^{-λ}) / x!. So, the likelihood function is the product of the probabilities for each observed data point. So, the likelihood function L(λ) would be the product from i=1 to n of (λ^{x_i} e^{-λ}) / x_i!. To make it easier to work with, we usually take the natural logarithm of the likelihood function, which turns the product into a sum. That gives us the log-likelihood function, l(λ) = sum_{i=1}^n [x_i ln(λ) - λ - ln(x_i!)].To find the MLE, we need to take the derivative of the log-likelihood with respect to λ, set it equal to zero, and solve for λ. Let's compute the derivative:dl/dλ = sum_{i=1}^n [x_i / λ - 1]. Setting this equal to zero:sum_{i=1}^n (x_i / λ - 1) = 0.Let me simplify this equation. The sum can be split into two parts:sum_{i=1}^n x_i / λ - sum_{i=1}^n 1 = 0.Which is (1/λ) sum x_i - n = 0. Solving for λ, we get:(1/λ) sum x_i = n => sum x_i = nλ => λ = (sum x_i) / n.So, the MLE for λ is the sample mean of the observed bugs. That makes sense because the Poisson distribution's mean is equal to its rate parameter λ. So, the MLE estimator is just the average number of bugs across the sample modules.Alright, that seems straightforward. I think I got that part.Moving on to the second part: Predicting Potential Bugs. The researcher uses a neural network where the output is f(w, x), with w being the weights and x the features of the code module. The loss function is the mean squared error (MSE) between the predicted and actual number of bugs. We need to formulate the optimization problem and provide the gradient descent update rule for the weights w.Okay, so the MSE loss function for a neural network is typically defined as the average of the squared differences between the predictions and the actual values. If we have m training examples, the loss L can be written as:L(w) = (1/m) * sum_{j=1}^m (f(w, x_j) - y_j)^2,where y_j is the actual number of bugs for the j-th example.The optimization problem is to find the weights w that minimize this loss function. So, we can write it as:minimize_{w} L(w) = (1/m) * sum_{j=1}^m (f(w, x_j) - y_j)^2.To find the minimum, we use gradient descent, which updates the weights in the direction of the negative gradient of the loss with respect to the weights. The update rule is:w := w - η * ∇_w L(w),where η is the learning rate, and ∇_w L(w) is the gradient of the loss with respect to the weights.But wait, how do we compute the gradient ∇_w L(w)? Since L is a function of f(w, x_j), which is the output of the neural network, we need to use backpropagation to compute the gradients efficiently. The gradient of the loss with respect to each weight is the derivative of the loss with respect to the output, multiplied by the derivative of the output with respect to the weight, summed over all training examples.In mathematical terms, for each weight w_k, the gradient component is:∂L/∂w_k = (2/m) * sum_{j=1}^m (f(w, x_j) - y_j) * ∂f/∂w_k.So, the update rule becomes:w_k := w_k - η * (2/m) * sum_{j=1}^m (f(w, x_j) - y_j) * ∂f/∂w_k.But in practice, when implementing gradient descent, especially with mini-batch or stochastic gradient descent, we often compute the gradient over a subset of the data (a mini-batch) to make the computation more efficient.Alternatively, if we consider the derivative of the loss with respect to each weight, it's the chain rule applied through the network. So, the gradient is computed by backpropagating the error from the output layer through each layer, multiplying by the derivative of the activation functions and the inputs.But since the question just asks for the gradient descent update rule, I think it's sufficient to express it in terms of the gradient of the loss with respect to the weights, without delving into the specifics of backpropagation.So, to summarize, the optimization problem is to minimize the MSE loss, and the gradient descent update rule subtracts the gradient scaled by the learning rate from the current weights.Wait, let me make sure I'm not missing anything. The loss is the MSE, so the derivative of the loss with respect to each weight is the derivative of the squared error with respect to the output, times the derivative of the output with respect to the weight. That is, for each example, the error term (f - y) multiplied by the derivative of f with respect to w. Then, averaged over all examples, multiplied by 2/m.But in the update rule, it's usually expressed as the average gradient over the batch. So, if we're doing batch gradient descent, we compute the gradient over all examples and update the weights accordingly. If it's stochastic, we do it one example at a time, and mini-batch is somewhere in between.But the question doesn't specify, so I think it's safe to present the general form of the update rule, which is:w = w - η * (1/m) * sum_{j=1}^m 2*(f(w, x_j) - y_j) * ∂f/∂w.But actually, the 2 comes from the derivative of the squared error. So, yes, the gradient is 2*(f - y)*df/dw.Wait, let me double-check. The derivative of (f - y)^2 with respect to f is 2*(f - y). Then, the derivative with respect to w is 2*(f - y)*df/dw. So, the gradient for each example is 2*(f - y)*df/dw, and then we average over all examples, so the gradient is (2/m)*sum_{j=1}^m (f_j - y_j)*df_j/dw.Therefore, the update rule is:w := w - η * (2/m) * sum_{j=1}^m (f(w, x_j) - y_j) * ∂f/∂w.Alternatively, sometimes the 2 is absorbed into the learning rate, but I think it's more precise to include it here.So, putting it all together, the optimization problem is to minimize the MSE, and the update rule is as above.I think that covers both parts. For the first part, the MLE is the sample mean, and for the second part, the optimization is via gradient descent with the specified update rule.Final Answer1. The Maximum Likelihood Estimator for (lambda) is (boxed{hat{lambda} = frac{1}{n} sum_{i=1}^{n} x_i}).2. The optimization problem is to minimize the Mean Squared Error (MSE) loss function, and the gradient descent update rule for the weights (mathbf{w}) is (boxed{mathbf{w} := mathbf{w} - eta cdot frac{2}{m} sum_{j=1}^{m} (f(mathbf{w}, mathbf{x}_j) - y_j) cdot frac{partial f}{partial mathbf{w}}}).</think>

🧠:<think>Okay, so I have this problem about the Hamister hotel project in Niagara Falls. There are two parts: first, I need to find the total projected revenue over the first 3 years using integration. Then, in the second part, I have to calculate the net profit by subtracting the operational costs from the revenue. Let me try to work through each part step by step.Starting with part 1: The revenue function is given as ( R(t) = 50 + 30 sinleft(frac{pi t}{12}right) ), where ( t ) is the number of months since the hotel opened. I need to find the total revenue over the first 3 years. Since 3 years is 36 months, I'll need to integrate this function from ( t = 0 ) to ( t = 36 ).So, the total revenue ( text{Total Revenue} = int_{0}^{36} R(t) , dt = int_{0}^{36} left(50 + 30 sinleft(frac{pi t}{12}right)right) dt ).Let me break this integral into two parts for easier computation: the integral of 50 and the integral of ( 30 sinleft(frac{pi t}{12}right) ).First, integrating 50 with respect to ( t ) is straightforward. The integral of 50 dt is ( 50t ).Next, for the integral of ( 30 sinleft(frac{pi t}{12}right) ), I'll need to use substitution. Let me set ( u = frac{pi t}{12} ). Then, ( du = frac{pi}{12} dt ), which means ( dt = frac{12}{pi} du ).Substituting into the integral, we have:( int 30 sin(u) cdot frac{12}{pi} du = 30 cdot frac{12}{pi} int sin(u) du ).The integral of ( sin(u) ) is ( -cos(u) ), so this becomes:( 30 cdot frac{12}{pi} (-cos(u)) + C = -frac{360}{pi} cosleft(frac{pi t}{12}right) + C ).Putting it all together, the integral of ( R(t) ) is:( 50t - frac{360}{pi} cosleft(frac{pi t}{12}right) + C ).Now, I need to evaluate this from 0 to 36.Calculating at ( t = 36 ):( 50 times 36 - frac{360}{pi} cosleft(frac{pi times 36}{12}right) ).Simplify ( frac{pi times 36}{12} = 3pi ). So, ( cos(3pi) ) is equal to ( -1 ).So, substituting:( 50 times 36 - frac{360}{pi} times (-1) = 1800 + frac{360}{pi} ).Now, calculating at ( t = 0 ):( 50 times 0 - frac{360}{pi} cos(0) = 0 - frac{360}{pi} times 1 = -frac{360}{pi} ).Subtracting the lower limit from the upper limit:( left(1800 + frac{360}{pi}right) - left(-frac{360}{pi}right) = 1800 + frac{360}{pi} + frac{360}{pi} = 1800 + frac{720}{pi} ).So, the total projected revenue over the first 3 years is ( 1800 + frac{720}{pi} ) million dollars. Let me compute this numerically to get an approximate value, but since the question asks for the exact value, I can leave it in terms of ( pi ).Wait, actually, let me double-check my calculations. The integral of ( sin ) function is correct, and the substitution seems right. The evaluation at 36 and 0 also looks correct. So, the exact value is ( 1800 + frac{720}{pi} ) million dollars.Moving on to part 2: I need to find the net profit over the first 3 years. The net profit is total revenue minus total costs. The revenue I have is already calculated as ( 1800 + frac{720}{pi} ) million dollars. However, the cost function is given in terms of years, ( C(t) = 0.5t^2 + 10t + 200 ), where ( t ) is the number of years since the hotel opened.So, to find the total cost over 3 years, I need to integrate ( C(t) ) from ( t = 0 ) to ( t = 3 ). Let me write that down:Total Cost ( = int_{0}^{3} C(t) , dt = int_{0}^{3} left(0.5t^2 + 10t + 200right) dt ).Again, I can break this integral into three parts:1. Integral of ( 0.5t^2 ) dt2. Integral of ( 10t ) dt3. Integral of 200 dtCalculating each:1. Integral of ( 0.5t^2 ) is ( 0.5 times frac{t^3}{3} = frac{t^3}{6} ).2. Integral of ( 10t ) is ( 10 times frac{t^2}{2} = 5t^2 ).3. Integral of 200 is ( 200t ).Putting it all together, the integral is:( frac{t^3}{6} + 5t^2 + 200t ) evaluated from 0 to 3.Calculating at ( t = 3 ):( frac{3^3}{6} + 5 times 3^2 + 200 times 3 = frac{27}{6} + 5 times 9 + 600 = 4.5 + 45 + 600 = 649.5 ).Calculating at ( t = 0 ):All terms become 0, so the total cost is 649.5 million dollars.Wait, hold on. The revenue was in millions of dollars, and the cost is given as ( C(t) = 0.5t^2 + 10t + 200 ). Is this in millions as well? The problem says "annual operational cost", so I think it's in millions of dollars per year. So, integrating over 3 years would give total cost in millions.So, total cost is 649.5 million dollars.Now, net profit is total revenue minus total cost. So:Net Profit ( = left(1800 + frac{720}{pi}right) - 649.5 ).Simplify this:First, compute 1800 - 649.5 = 1150.5.So, Net Profit ( = 1150.5 + frac{720}{pi} ) million dollars.Again, if I want to write it as an exact value, it's ( 1150.5 + frac{720}{pi} ). Alternatively, I can write 1150.5 as ( frac{2301}{2} ), so:Net Profit ( = frac{2301}{2} + frac{720}{pi} ).But maybe it's better to keep it as 1150.5 for simplicity.Wait, let me just verify the calculations again.Total revenue was ( 1800 + frac{720}{pi} ) million dollars over 36 months, which is 3 years. Total cost over 3 years is 649.5 million dollars. So, subtracting, 1800 - 649.5 is indeed 1150.5. Then, adding the ( frac{720}{pi} ), so net profit is ( 1150.5 + frac{720}{pi} ) million dollars.Just to make sure, let's compute ( frac{720}{pi} ) approximately. Since ( pi ) is approximately 3.1416, so 720 divided by 3.1416 is roughly 229.183. So, total net profit is approximately 1150.5 + 229.183 = 1379.683 million dollars. But since the question asks for the exact value, we should keep it in terms of ( pi ).So, summarizing:1. Total projected revenue over the first 3 years is ( 1800 + frac{720}{pi} ) million dollars.2. Net profit over the first 3 years is ( 1150.5 + frac{720}{pi} ) million dollars.Wait, hold on, actually, I just realized something. The revenue function is given in terms of months, while the cost function is given in terms of years. So, when calculating the total revenue, I integrated over 36 months, which is 3 years, so that's correct. The cost function is integrated over 3 years, so that's also correct. So, the units are consistent.But just to double-check, the revenue is in millions of dollars per month? Wait, no, the revenue function is ( R(t) = 50 + 30 sinleft(frac{pi t}{12}right) ), and it's in millions of dollars. So, is this per month or total? Wait, the function is given as revenue, so I think it's the revenue per month. So, integrating over t months gives total revenue in millions of dollars.Similarly, the cost function is ( C(t) = 0.5t^2 + 10t + 200 ), where t is in years, and it's in millions of dollars per year? Or is it total cost? Wait, the problem says "annual operational cost", so I think it's the cost per year, so integrating over t years gives the total cost over that period.So, integrating both over the same time period (3 years) is correct.Therefore, my calculations seem correct.Final Answer1. The total projected revenue over the first 3 years is boxed{1800 + dfrac{720}{pi}} million dollars.2. The net profit over the first 3 years is boxed{1150.5 + dfrac{720}{pi}} million dollars.</think>

🧠:<think>Alright, so I have this problem where I need to analyze the research productivity and collaboration patterns across three countries: A, B, and C. The data is given in terms of functions for each country over a 5-year period. Let me try to break this down step by step.First, the problem has two sub-problems. The first one is to find the total number of research papers published by all three countries combined at the end of 5 years and also calculate the average citation impact for each country at that time. The second sub-problem is about calculating an overall research impact score using a given formula and determining which country has the highest score.Starting with the first sub-problem. I need to compute the total number of papers published by each country at t=5 and then sum them up. Additionally, I need to find the average citation impact for each country at t=5.Let me write down the given functions for each country:For Country A:- P_A(t) = 50 + 10t- C_A(t) = 5 + t- I_A(t) = 3 + 0.5tFor Country B:- P_B(t) = 40 + 8t- C_B(t) = 6 + 0.8t- I_B(t) = 2.5 + 0.4tFor Country C:- P_C(t) = 60 + 12t- C_C(t) = 4 + 1.2t- I_C(t) = 3.5 + 0.6tSo, for each country, I can plug in t=5 into their respective P(t) functions to get the number of papers at the end of 5 years.Starting with Country A:P_A(5) = 50 + 10*5 = 50 + 50 = 100 papers.Country B:P_B(5) = 40 + 8*5 = 40 + 40 = 80 papers.Country C:P_C(5) = 60 + 12*5 = 60 + 60 = 120 papers.So, the total number of papers is 100 + 80 + 120. Let me add that up: 100 + 80 is 180, and 180 + 120 is 300. So, the total number of research papers published by all three countries combined at the end of 5 years is 300.Next, I need to calculate the average citation impact for each country at t=5. The average citation impact is given by I(t), so I just need to plug t=5 into each I(t) function.For Country A:I_A(5) = 3 + 0.5*5 = 3 + 2.5 = 5.5.Country B:I_B(5) = 2.5 + 0.4*5 = 2.5 + 2 = 4.5.Country C:I_C(5) = 3.5 + 0.6*5 = 3.5 + 3 = 6.5.So, the average citation impacts are 5.5 for A, 4.5 for B, and 6.5 for C.That completes the first sub-problem.Moving on to the second sub-problem. I need to calculate the overall research impact score R(t) for each country at t=5 using the formula R(t) = P(t) * I(t) + 2C(t). Then, determine which country has the highest R(t).First, let me note the formula again: R(t) = P(t) * I(t) + 2*C(t). So, for each country, I need to compute P(t) * I(t) and then add twice the number of collaborations C(t).I already have P(t) and I(t) at t=5 for each country, so let me use those values.Starting with Country A:P_A(5) = 100, I_A(5) = 5.5, C_A(5) = 5 + 5 = 10.So, R_A(5) = 100 * 5.5 + 2*10.Calculating that: 100 * 5.5 is 550, and 2*10 is 20. So, 550 + 20 = 570.Country A's R(t) is 570.Country B:P_B(5) = 80, I_B(5) = 4.5, C_B(5) = 6 + 0.8*5 = 6 + 4 = 10.So, R_B(5) = 80 * 4.5 + 2*10.Calculating: 80 * 4.5 is 360, and 2*10 is 20. So, 360 + 20 = 380.Country B's R(t) is 380.Country C:P_C(5) = 120, I_C(5) = 6.5, C_C(5) = 4 + 1.2*5 = 4 + 6 = 10.So, R_C(5) = 120 * 6.5 + 2*10.Calculating: 120 * 6.5. Hmm, 120*6 is 720, and 120*0.5 is 60, so total is 720 + 60 = 780. Then, 2*10 is 20. So, 780 + 20 = 800.Country C's R(t) is 800.Now, comparing the R(t) scores: Country A has 570, Country B has 380, and Country C has 800. So, clearly, Country C has the highest impact score.Wait a second, let me double-check my calculations to make sure I didn't make any errors.For Country A:P_A(5) = 100, I_A(5) = 5.5. 100 * 5.5 is indeed 550. C_A(5) is 10, so 2*10 is 20. Total R is 570. That seems correct.Country B:P_B(5) = 80, I_B(5) = 4.5. 80 * 4.5 is 360. C_B(5) is 10, so 2*10 is 20. Total R is 380. Correct.Country C:P_C(5) = 120, I_C(5) = 6.5. 120 * 6.5: Let me compute this again. 100*6.5 is 650, and 20*6.5 is 130, so 650 + 130 is 780. Then, 2*10 is 20, so total R is 800. Correct.Yes, so Country C has the highest overall research impact score.Just to recap:1. Total papers: 300.2. Average citation impacts: A=5.5, B=4.5, C=6.5.3. Impact scores: A=570, B=380, C=800. So, C is the highest.I think that's all. I don't see any mistakes in my calculations.Final Answer1. The total number of research papers published by all three countries combined at the end of the 5-year period is boxed{300}. The average citation impacts for each country are Country A: boxed{5.5}, Country B: boxed{4.5}, and Country C: boxed{6.5}.2. The overall research impact scores for each country at the end of the 5-year period are Country A: boxed{570}, Country B: boxed{380}, and Country C: boxed{800}. Therefore, Country C has the highest impact score.</think>