💡Alright, so I've come across this interesting problem that says: "Show that ( n ) is prime if and only if [lim_{r rightarrowinfty},lim_{s rightarrowinfty},lim_{t rightarrow infty},sum_{u=0}^{s}left(1-left(cos,frac{(u!)^{r} pi}{n} right)^{2t} right)=n."]At first glance, this seems pretty abstract, especially with all those limits and the cosine function involved. But I think if I break it down step by step, I can understand what's going on here.Okay, let's start by understanding what this expression is doing. We have a triple limit: first as ( r ) approaches infinity, then ( s ) approaches infinity, and finally ( t ) approaches infinity. Inside the limits, there's a sum from ( u = 0 ) to ( s ) of terms that involve cosine raised to the power of ( 2t ), and then subtracted from 1.So, the expression inside the sum is ( 1 - left(cosleft(frac{(u!)^{r} pi}{n}right)right)^{2t} ). Let's dissect this.First, ( u! ) is the factorial of ( u ), which grows very rapidly. Then, raising that to the power ( r ) makes it even larger as ( r ) increases. Dividing by ( n ) gives us some multiple of ( pi ), which is then fed into the cosine function.Now, cosine of an angle is a periodic function with period ( 2pi ), so the value of ( cosleft(frac{(u!)^{r} pi}{n}right) ) depends on how ( frac{(u!)^{r}}{n} ) relates to multiples of ( 2pi ). But since ( pi ) is involved, it's not immediately clear how this will behave.However, when we raise cosine to the power of ( 2t ), and then take ( t ) to infinity, this term will behave differently depending on whether the cosine is equal to 1 or not. If ( cosleft(frac{(u!)^{r} pi}{n}right) = 1 ), then ( 1 - 1^{2t} = 0 ). If the cosine is less than 1 in absolute value, then ( left(cosleft(frac{(u!)^{r} pi}{n}right)right)^{2t} ) will approach 0 as ( t ) goes to infinity, making ( 1 - 0 = 1 ).So, essentially, each term in the sum will be 1 if ( cosleft(frac{(u!)^{r} pi}{n}right) ) is not equal to 1, and 0 otherwise. Therefore, the entire sum will count the number of ( u ) values for which ( cosleft(frac{(u!)^{r} pi}{n}right) ) is not equal to 1.Now, when does ( cosleft(frac{(u!)^{r} pi}{n}right) ) equal 1? That happens when ( frac{(u!)^{r} pi}{n} ) is an integer multiple of ( 2pi ), which simplifies to ( frac{(u!)^{r}}{n} ) being an integer. So, ( n ) must divide ( (u!)^{r} ).If ( n ) is prime, then ( n ) divides ( u! ) only when ( u geq n ). This is because a prime number ( n ) will only appear in the factorial ( u! ) when ( u ) is at least ( n ). Therefore, for ( u < n ), ( n ) does not divide ( u! ), and hence ( n ) does not divide ( (u!)^{r} ) either. This means that for ( u < n ), ( frac{(u!)^{r}}{n} ) is not an integer, so ( cosleft(frac{(u!)^{r} pi}{n}right) ) is not equal to 1, and thus each term in the sum for ( u < n ) contributes 1.On the other hand, for ( u geq n ), ( n ) divides ( u! ), so ( n ) divides ( (u!)^{r} ), making ( frac{(u!)^{r}}{n} ) an integer. Therefore, ( cosleft(frac{(u!)^{r} pi}{n}right) = 1 ), and the corresponding term in the sum becomes 0.So, when ( n ) is prime, the sum from ( u = 0 ) to ( s ) will have ( n ) terms contributing 1 (for ( u = 0 ) to ( u = n - 1 )) and the rest contributing 0. As ( s ) approaches infinity, the sum will approach ( n ), and taking the limits as ( r ) and ( t ) approach infinity doesn't change this result because we've already established the behavior of each term.Now, let's consider the converse: if ( n ) is not prime, then ( n ) can be factored into smaller integers greater than 1. Let's say ( n = ab ) where ( a > 1 ) and ( b > 1 ). In this case, ( n ) divides ( u! ) for ( u geq a ), since ( a ) is a factor of ( u! ) when ( u geq a ). Therefore, for ( u geq a ), ( frac{(u!)^{r}}{n} ) is an integer, making ( cosleft(frac{(u!)^{r} pi}{n}right) = 1 ), and the corresponding term in the sum is 0.For ( u < a ), ( n ) does not divide ( u! ), so ( frac{(u!)^{r}}{n} ) is not an integer, and ( cosleft(frac{(u!)^{r} pi}{n}right) ) is not equal to 1, contributing 1 to the sum. Therefore, the sum will have ( a ) terms contributing 1 (for ( u = 0 ) to ( u = a - 1 )) and the rest contributing 0. Since ( a < n ), the sum will approach ( a ) as ( s ) approaches infinity, which is less than ( n ).Hence, if ( n ) is not prime, the limit will be less than ( n ), and if ( n ) is prime, the limit will be exactly ( n ). This establishes the equivalence.I think I've got a handle on this now. The key insight is understanding how the primality of ( n ) affects whether ( n ) divides ( (u!)^{r} ) and consequently the behavior of the cosine function raised to the power of ( 2t ). By analyzing these behaviors under the given limits, we can determine whether ( n ) is prime based on the value of the limit.

💡Alright, so I have this matrix here, and I need to compute its fourth power. The matrix is:[begin{pmatrix} 2 & -2 2 & -1 end{pmatrix}]Okay, so I need to calculate this matrix multiplied by itself four times. That sounds like a lot of work, but maybe there's a pattern or a shortcut I can use. Let me think about how matrix multiplication works. Each element in the resulting matrix is obtained by taking the dot product of the corresponding row from the first matrix and column from the second matrix.First, maybe I should compute the square of the matrix, then the cube, and then the fourth power. That seems like a step-by-step approach. Let me start by computing the square.So, let's compute:[begin{pmatrix} 2 & -2 2 & -1 end{pmatrix} times begin{pmatrix} 2 & -2 2 & -1 end{pmatrix}]Let me denote the first matrix as A and the second as A as well since it's the same matrix. So, A squared is A times A.Calculating the element in the first row, first column:(2)(2) + (-2)(2) = 4 - 4 = 0First row, second column:(2)(-2) + (-2)(-1) = -4 + 2 = -2Second row, first column:(2)(2) + (-1)(2) = 4 - 2 = 2Second row, second column:(2)(-2) + (-1)(-1) = -4 + 1 = -3So, A squared is:[begin{pmatrix} 0 & -2 2 & -3 end{pmatrix}]Alright, that wasn't too bad. Now, let's compute A cubed, which is A squared times A.So,[begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} times begin{pmatrix} 2 & -2 2 & -1 end{pmatrix}]First row, first column:(0)(2) + (-2)(2) = 0 - 4 = -4First row, second column:(0)(-2) + (-2)(-1) = 0 + 2 = 2Second row, first column:(2)(2) + (-3)(2) = 4 - 6 = -2Second row, second column:(2)(-2) + (-3)(-1) = -4 + 3 = -1So, A cubed is:[begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix}]Hmm, interesting. Now, moving on to A to the fourth power, which is A cubed times A.So,[begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix} times begin{pmatrix} 2 & -2 2 & -1 end{pmatrix}]First row, first column:(-4)(2) + (2)(2) = -8 + 4 = -4Wait, that doesn't seem right. Let me double-check that.First row, first column:(-4)(2) + (2)(2) = -8 + 4 = -4First row, second column:(-4)(-2) + (2)(-1) = 8 - 2 = 6Second row, first column:(-2)(2) + (-1)(2) = -4 - 2 = -6Second row, second column:(-2)(-2) + (-1)(-1) = 4 + 1 = 5So, A to the fourth power is:[begin{pmatrix} -4 & 6 -6 & 5 end{pmatrix}]Wait, that doesn't match what I thought earlier. Maybe I made a mistake in my calculations. Let me go through each step again.Calculating A squared:First row, first column: 2*2 + (-2)*2 = 4 - 4 = 0First row, second column: 2*(-2) + (-2)*(-1) = -4 + 2 = -2Second row, first column: 2*2 + (-1)*2 = 4 - 2 = 2Second row, second column: 2*(-2) + (-1)*(-1) = -4 + 1 = -3So, A squared is correct.Now, A cubed:First row, first column: 0*2 + (-2)*2 = 0 - 4 = -4First row, second column: 0*(-2) + (-2)*(-1) = 0 + 2 = 2Second row, first column: 2*2 + (-3)*2 = 4 - 6 = -2Second row, second column: 2*(-2) + (-3)*(-1) = -4 + 3 = -1So, A cubed is correct.Now, A to the fourth power:First row, first column: (-4)*2 + 2*2 = -8 + 4 = -4First row, second column: (-4)*(-2) + 2*(-1) = 8 - 2 = 6Second row, first column: (-2)*2 + (-1)*2 = -4 - 2 = -6Second row, second column: (-2)*(-2) + (-1)*(-1) = 4 + 1 = 5So, A^4 is:[begin{pmatrix} -4 & 6 -6 & 5 end{pmatrix}]Wait, but I thought earlier that the fourth power was:[begin{pmatrix} -8 & 8 0 & 3 end{pmatrix}]That's different. Did I make a mistake in my previous calculation? Let me check again.Wait, in my initial step-by-step, I computed A squared, then multiplied by A to get A cubed, and then multiplied by A again to get A^4. But in my second attempt, I got a different result. This inconsistency is confusing.Let me try another approach. Maybe diagonalizing the matrix would help, but that might be too advanced for me right now. Alternatively, perhaps I can look for a pattern or a recurrence relation.Alternatively, maybe I can compute A^4 directly by multiplying A squared by A squared.So, A squared is:[begin{pmatrix} 0 & -2 2 & -3 end{pmatrix}]So, A^4 is (A squared)^2, which is:[begin{pmatrix} 0 & -2 2 & -3 end{pmatrix} times begin{pmatrix} 0 & -2 2 & -3 end{pmatrix}]Calculating this:First row, first column: 0*0 + (-2)*2 = 0 - 4 = -4First row, second column: 0*(-2) + (-2)*(-3) = 0 + 6 = 6Second row, first column: 2*0 + (-3)*2 = 0 - 6 = -6Second row, second column: 2*(-2) + (-3)*(-3) = -4 + 9 = 5So, A^4 is:[begin{pmatrix} -4 & 6 -6 & 5 end{pmatrix}]Hmm, so this matches my second calculation but not the first one. I must have made a mistake in my initial step-by-step. Let me go back to my first approach.In my first attempt, I computed A squared, then A cubed, then A^4. But when I computed A cubed, I got:[begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix}]Then, multiplying by A, I got:First row, first column: (-4)*2 + 2*2 = -8 + 4 = -4First row, second column: (-4)*(-2) + 2*(-1) = 8 - 2 = 6Second row, first column: (-2)*2 + (-1)*2 = -4 - 2 = -6Second row, second column: (-2)*(-2) + (-1)*(-1) = 4 + 1 = 5So, A^4 is:[begin{pmatrix} -4 & 6 -6 & 5 end{pmatrix}]But in my initial thought process, I thought the answer was:[begin{pmatrix} -8 & 8 0 & 3 end{pmatrix}]Clearly, there's a discrepancy here. I must have made a mistake in my initial calculation. Let me check my first step again.Wait, in my initial step, when I computed A squared, I got:[begin{pmatrix} 0 & -2 2 & -3 end{pmatrix}]Then, when computing A cubed, I multiplied A squared by A:First row, first column: 0*2 + (-2)*2 = 0 - 4 = -4First row, second column: 0*(-2) + (-2)*(-1) = 0 + 2 = 2Second row, first column: 2*2 + (-3)*2 = 4 - 6 = -2Second row, second column: 2*(-2) + (-3)*(-1) = -4 + 3 = -1So, A cubed is:[begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix}]Then, multiplying by A again:First row, first column: (-4)*2 + 2*2 = -8 + 4 = -4First row, second column: (-4)*(-2) + 2*(-1) = 8 - 2 = 6Second row, first column: (-2)*2 + (-1)*2 = -4 - 2 = -6Second row, second column: (-2)*(-2) + (-1)*(-1) = 4 + 1 = 5So, A^4 is:[begin{pmatrix} -4 & 6 -6 & 5 end{pmatrix}]Therefore, my initial thought process was incorrect. The correct answer should be:[begin{pmatrix} -4 & 6 -6 & 5 end{pmatrix}]I must have made a mistake in my initial calculation, perhaps in the multiplication steps. It's important to double-check each step carefully to avoid errors.Alternatively, maybe I can use another method, such as finding the eigenvalues and eigenvectors to diagonalize the matrix, which would make computing higher powers easier. But since I'm just starting out, I think sticking to step-by-step multiplication is safer for now.So, to summarize:A = [begin{pmatrix} 2 & -2 2 & -1 end{pmatrix}]A squared = [begin{pmatrix} 0 & -2 2 & -3 end{pmatrix}]A cubed = [begin{pmatrix} -4 & 2 -2 & -1 end{pmatrix}]A^4 = [begin{pmatrix} -4 & 6 -6 & 5 end{pmatrix}]I think this is correct now. I'll stick with this answer.

💡Okay, so I have this function f defined on positive integers, and I need to find all x such that f(x) = 19, with x not exceeding 2019. Then, I have to sum all those x's. Hmm, let me try to understand the function first.The function f is defined recursively. It starts with f(1) = 1. Then, for any n, f(2n) = 2f(n) and f(2n + 1) = 2f(n) - 1. So, it's kind of building up the function based on whether the number is even or odd.Maybe I can compute some values of f(n) to see if I can find a pattern. Let's start with small n:- f(1) = 1 (given)- f(2) = f(2*1) = 2f(1) = 2*1 = 2- f(3) = f(2*1 + 1) = 2f(1) - 1 = 2*1 - 1 = 1- f(4) = f(2*2) = 2f(2) = 2*2 = 4- f(5) = f(2*2 + 1) = 2f(2) - 1 = 2*2 - 1 = 3- f(6) = f(2*3) = 2f(3) = 2*1 = 2- f(7) = f(2*3 + 1) = 2f(3) - 1 = 2*1 - 1 = 1- f(8) = f(2*4) = 2f(4) = 2*4 = 8- f(9) = f(2*4 + 1) = 2f(4) - 1 = 2*4 - 1 = 7- f(10) = f(2*5) = 2f(5) = 2*3 = 6- f(11) = f(2*5 + 1) = 2f(5) - 1 = 2*3 - 1 = 5- f(12) = f(2*6) = 2f(6) = 2*2 = 4- f(13) = f(2*6 + 1) = 2f(6) - 1 = 2*2 - 1 = 3- f(14) = f(2*7) = 2f(7) = 2*1 = 2- f(15) = f(2*7 + 1) = 2f(7) - 1 = 2*1 - 1 = 1- f(16) = f(2*8) = 2f(8) = 2*8 = 16Hmm, interesting. It seems like f(n) is related to the binary representation of n. For example, f(1) = 1, which is 1 in binary. f(2) = 2, which is 10 in binary. f(3) = 1, which is 11 in binary. Wait, maybe f(n) is the difference between the highest power of 2 less than or equal to n and n itself?Wait, let's see:- For n=1: highest power is 1, so 1 - 1 = 0, but f(1)=1. Hmm, not quite.- For n=2: highest power is 2, 2 - 2=0, but f(2)=2.- For n=3: highest power is 2, 2 - 3 = -1, but f(3)=1.- For n=4: highest power is 4, 4 - 4=0, but f(4)=4.- For n=5: highest power is 4, 4 - 5 = -1, but f(5)=3.Wait, maybe it's the other way around. Maybe f(n) is the highest power of 2 less than or equal to n minus n? But that doesn't fit either.Wait, let's think differently. Maybe f(n) is the number of times you can divide n by 2 until it becomes odd, but that doesn't seem to fit either.Wait, looking back at the values:n: 1, f(n):1n:2, f(n):2n:3, f(n):1n:4, f(n):4n:5, f(n):3n:6, f(n):2n:7, f(n):1n:8, f(n):8n:9, f(n):7n:10, f(n):6n:11, f(n):5n:12, f(n):4n:13, f(n):3n:14, f(n):2n:15, f(n):1n:16, f(n):16Hmm, it's like f(n) counts down from the highest power of 2. For example, starting at n=8 (which is 2^3), f(n) goes 8,7,6,5,4,3,2,1 as n goes from 8 to 15. Then at n=16, it resets to 16.So, in general, for numbers from 2^k to 2^{k+1}-1, f(n) = 2^k - (n - 2^k). So f(n) = 2^{k+1} - n - 1? Wait, let me check.Wait, for n=8 to 15:f(8)=8, which is 2^3f(9)=7, which is 2^3 -1f(10)=6, which is 2^3 -2...f(15)=1, which is 2^3 -7So, f(n) = 2^{k} - (n - 2^{k}) where k is such that 2^{k} ≤ n < 2^{k+1}Wait, that would be f(n) = 2^{k} - (n - 2^{k}) = 2^{k+1} - nYes, that seems to fit.So, f(n) = 2^{k+1} - n, where k is the integer such that 2^{k} ≤ n < 2^{k+1}So, for example, for n=5, which is between 4 and 8, k=2, so f(5)=2^{3} -5=8-5=3, which matches.Similarly, for n=10, which is between 8 and 16, k=3, so f(10)=2^{4} -10=16-10=6, which matches.Okay, so general formula: f(n) = 2^{k+1} - n, where k = floor(log2(n))So, if f(n) =19, then 2^{k+1} - n =19, so n=2^{k+1} -19But n must be a positive integer, so 2^{k+1} -19 ≥1, so 2^{k+1} ≥20So, 2^{k+1} ≥20, so k+1 ≥5, since 2^4=16 <20 and 2^5=32 ≥20Thus, k+1 ≥5, so k ≥4Also, n must be less than 2^{k+1}, so 2^{k+1} -19 <2^{k+1}, which is always true.But also, n must be ≥2^{k}, so 2^{k+1} -19 ≥2^{k}So, 2^{k+1} -19 ≥2^{k}Which simplifies to 2^{k} ≥19So, 2^{k} ≥19, so k ≥5, since 2^4=16 <19 and 2^5=32 ≥19So, k must be ≥5Thus, k can be 5,6,7,8,9,...But n must be ≤2019, so 2^{k+1} -19 ≤2019So, 2^{k+1} ≤2019 +19=2038So, 2^{k+1} ≤2038What's the largest k such that 2^{k+1} ≤2038?Compute 2^10=1024, 2^11=20482048>2038, so k+1 ≤10, so k ≤9Thus, k can be 5,6,7,8,9Therefore, the possible k's are 5,6,7,8,9Thus, for each k from 5 to9, compute n=2^{k+1} -19Compute:For k=5: n=2^6 -19=64-19=45k=6: n=2^7 -19=128-19=109k=7: n=2^8 -19=256-19=237k=8: n=2^9 -19=512-19=493k=9: n=2^{10} -19=1024-19=1005Wait, but 2^{10}=1024, so n=1024-19=1005Wait, 1005 is less than 2019, so that's fine.Wait, but what about k=10? 2^{11}=2048, so n=2048-19=2029, which is greater than 2019, so we can't include that.So, the solutions are n=45,109,237,493,1005Now, sum them up:45 +109=154154 +237=391391 +493=884884 +1005=1889So, the sum is 1889Wait, let me double-check the calculations:45 +109=154154 +237: 154+200=354, 354+37=391391 +493: 391+400=791, 791+93=884884 +1005: 884+1000=1884, 1884+5=1889Yes, that seems correct.So, the sum of all positive integer solutions to f(x)=19 that do not exceed 2019 is 1889.**Final Answer**boxed{1889}

💡Alright, so I've got this problem about a geometric sequence and some inequality to prove. Let me try to break it down step by step.**Problem (1):** Find the general term formula of the sequence {a_n}.Okay, so we know that the sum of the first n terms, S_n, lies on the curve y = b^x + r. That means for each n, S_n = b^n + r. Since it's a geometric sequence, the sum S_n should follow the formula for the sum of a geometric series. Wait, the sum of a geometric series is S_n = a_1*(1 - r^n)/(1 - r) if r ≠ 1. But here, S_n is given as b^n + r. Hmm, that seems different. Maybe I need to relate these two expressions.Let me write down what I know:- S_n = b^n + r- For a geometric sequence, S_n = a_1*(1 - q^n)/(1 - q), where q is the common ratio.So, setting these equal: b^n + r = a_1*(1 - q^n)/(1 - q). Hmm, that seems tricky. Maybe I can find a relationship between b and q, and also find r in terms of a_1 and q.Alternatively, maybe I can find a_n directly. Since S_n = b^n + r, then a_n = S_n - S_{n-1}. Let's compute that.a_n = S_n - S_{n-1} = (b^n + r) - (b^{n-1} + r) = b^n - b^{n-1} = b^{n-1}(b - 1).Oh, that seems straightforward! So the general term a_n is b^{n-1}(b - 1). That makes sense because it's a geometric sequence with first term (b - 1) and common ratio b.Wait, let me check for n=1. If n=1, a_1 = b^{0}(b - 1) = 1*(b - 1) = b - 1. Then S_1 should be a_1 = b - 1, but according to the given S_n = b^1 + r = b + r. So, b - 1 = b + r => r = -1.Oh, so r is actually -1. That's interesting. So the sum S_n is b^n - 1. Let me verify for n=2.a_2 = b^{1}(b - 1) = b(b - 1). Then S_2 = a_1 + a_2 = (b - 1) + b(b - 1) = (b - 1)(1 + b) = b^2 - 1. Which matches S_2 = b^2 - 1. Perfect, that works.So, in conclusion, the general term is a_n = b^{n-1}(b - 1), and r = -1.**Problem (2):** When b=2, define b_n = 2(log_2 a_n + 1). Prove that the product (b_1 +1)/b_1 * (b_2 +1)/b_2 * ... * (b_n +1)/b_n > sqrt(n +1).Alright, let's substitute b=2 into a_n. From part (1), a_n = 2^{n-1}(2 - 1) = 2^{n-1}.So, log_2 a_n = log_2 (2^{n-1}) = n - 1. Therefore, b_n = 2(log_2 a_n + 1) = 2(n - 1 + 1) = 2n.So, b_n = 2n. Therefore, each term in the product is (2n + 1)/(2n). So the product becomes:Product from k=1 to n of (2k + 1)/(2k) = (3/2)*(5/4)*(7/6)*...*((2n + 1)/(2n)).I need to show this product is greater than sqrt(n + 1).Hmm, how can I approach this? Maybe take the logarithm of the product and compare it to (1/2) ln(n + 1). Or perhaps use induction.Alternatively, notice that each term (2k + 1)/(2k) = 1 + 1/(2k). So, the product is the product of (1 + 1/(2k)) from k=1 to n.I recall that the product (1 + x_k) can be bounded using inequalities. Maybe use the inequality that for positive x_k, the product is greater than 1 + sum x_k. But that might not be tight enough.Alternatively, consider that (1 + 1/(2k)) = sqrt(1 + 1/k) * sqrt(1 + 1/(2k)^2). Hmm, not sure.Wait, another idea: pair terms. For example, (3/2)*(5/4) = (3*5)/(2*4) = 15/8. Hmm, not sure if that helps.Alternatively, think about telescoping products. Maybe relate it to factorials or double factorials.Wait, the product (2n + 1)!! / (2n)!!, where !! denotes double factorial. Because (2n + 1)!! = 1*3*5*...*(2n +1) and (2n)!! = 2*4*6*...*2n.So, the product is (2n +1)!! / (2n)!!.I remember that (2n +1)!! = (2n +1)! / (2^n n!) and (2n)!! = 2^n n!.So, (2n +1)!! / (2n)!! = (2n +1)! / (2^n n!) / (2^n n!) ) = (2n +1)! / (4^n (n!)^2).Hmm, so our product is (2n +1)! / (4^n (n!)^2).I also recall that (2n +1)! / (n! n! ) = (2n +1) * (2n)! / (n! n! ) = (2n +1) * C(2n, n), where C is combination.So, (2n +1)!! / (2n)!! = (2n +1) * C(2n, n) / 4^n.But I'm not sure if that helps. Maybe use Stirling's approximation for factorials?Stirling's formula: n! ≈ sqrt(2πn) (n/e)^n.So, let's approximate (2n +1)! ≈ sqrt(2π(2n +1)) ( (2n +1)/e )^{2n +1}.Similarly, (n!)^2 ≈ (2πn) (n^{2n} / e^{2n}).So, (2n +1)! / (n! n! ) ≈ sqrt(2π(2n +1)) ( (2n +1)/e )^{2n +1} / (2πn) (n^{2n} / e^{2n} )).Simplify:= sqrt( (2n +1)/(2πn) ) * ( (2n +1)^{2n +1} / e^{2n +1} ) / (n^{2n} / e^{2n} )= sqrt( (2n +1)/(2πn) ) * ( (2n +1)^{2n +1} / e ) / n^{2n}= sqrt( (2n +1)/(2πn) ) * (2n +1) * ( (2n +1)^{2n} / n^{2n} ) / e= sqrt( (2n +1)/(2πn) ) * (2n +1) * ( ( (2n +1)/n )^{2n} ) / e= sqrt( (2n +1)/(2πn) ) * (2n +1) * ( (2 + 1/n)^{2n} ) / eAs n becomes large, (2 + 1/n)^{2n} approaches e^{4}, since (1 + 1/n)^{2n} approaches e^{2}, so (2 + 1/n)^{2n} = (2(1 + 1/(2n)))^{2n} = 2^{2n} * (1 + 1/(2n))^{2n} ≈ 4^n * e.So, putting it all together:≈ sqrt( (2n +1)/(2πn) ) * (2n +1) * (4^n * e) / e= sqrt( (2n +1)/(2πn) ) * (2n +1) * 4^nBut our product is (2n +1)! / (4^n (n!)^2 ) ≈ [sqrt(2π(2n +1)) ( (2n +1)/e )^{2n +1} ] / [4^n (2πn) (n^{2n}/e^{2n}) ) ]Wait, this is getting too complicated. Maybe instead of approximating, think about known inequalities for binomial coefficients.I recall that C(2n, n) ≤ 4^n / sqrt(π n). So, (2n +1)!! / (2n)!! = (2n +1) * C(2n, n) / 4^n ≤ (2n +1) * (4^n / sqrt(π n)) / 4^n = (2n +1)/sqrt(π n).But we need a lower bound, not an upper bound. Hmm.Alternatively, maybe use induction. Let me try that.Base case: n=1.Product = (3/2) = 1.5. sqrt(2) ≈ 1.414. So 1.5 > 1.414, which holds.Assume for n=k, the product is greater than sqrt(k +1). Then for n=k+1, the product is previous product * (2(k+1) +1)/(2(k+1)) = previous product * (2k +3)/(2k +2).We need to show that previous product * (2k +3)/(2k +2) > sqrt(k +2).But since previous product > sqrt(k +1), we need sqrt(k +1) * (2k +3)/(2k +2) > sqrt(k +2).Let me square both sides to eliminate the square roots:(k +1) * ( (2k +3)/(2k +2) )^2 > k +2.Compute (2k +3)/(2k +2) = 1 + 1/(2k +2). So, (1 + 1/(2k +2))^2 = 1 + 2/(2k +2) + 1/(2k +2)^2 = 1 + 1/(k +1) + 1/(2k +2)^2.So, left side becomes (k +1)*(1 + 1/(k +1) + 1/(2k +2)^2 ) = (k +1) + 1 + (k +1)/(2k +2)^2.Simplify: (k +1) + 1 = k + 2. So, left side is k + 2 + (k +1)/(2k +2)^2.Which is k + 2 + something positive. Therefore, left side > k + 2, which is exactly the right side. Therefore, the inequality holds for n=k+1.Thus, by induction, the inequality holds for all n.Wait, that seems a bit hand-wavy. Let me check with n=2.For n=2, product = (3/2)*(5/4) = 15/8 = 1.875. sqrt(3) ≈ 1.732. So 1.875 > 1.732, which holds.n=3: product = (3/2)*(5/4)*(7/6) = 105/48 ≈ 2.1875. sqrt(4) = 2. So 2.1875 > 2, holds.n=4: product ≈ 2.1875*(9/8) = 2.1875*1.125 ≈ 2.4609. sqrt(5) ≈ 2.236. So holds.Seems to be working. So, I think induction is a valid approach here.So, in conclusion, the product is indeed greater than sqrt(n +1) for all n.**Final Answer**(1) The general term is boxed{a_n = b^{n-1}(b - 1)}.(2) The inequality holds as proven.

💡First, I recognize that the TV screen is a rectangle with a diagonal measurement of 50 inches and a width-to-height ratio of 16:9. To find the horizontal length, I can use the Pythagorean theorem, which relates the width, height, and diagonal of a rectangle.I'll denote the width as 16x and the height as 9x, where x is a scaling factor. Plugging these into the Pythagorean theorem:(16x)² + (9x)² = 50²Expanding and simplifying:256x² + 81x² = 2500337x² = 2500x² = 2500 / 337 ≈ 7.4184x ≈ √7.4184 ≈ 2.724Finally, the horizontal length is:16x ≈ 16 * 2.724 ≈ 43.584 inchesSo, the horizontal length of the TV is approximately 43.6 inches.

💡First, I need to simplify the given complex number ( z = frac{1+i}{3-4i} ) to make it easier to work with. To do this, I'll multiply both the numerator and the denominator by the complex conjugate of the denominator, which is ( 3+4i ). This step helps eliminate the imaginary part in the denominator.After simplifying, I'll have ( z ) expressed in the standard form ( a + bi ). Once I have ( z ) in this form, finding its complex conjugate ( bar{z} ) is straightforward—it involves changing the sign of the imaginary part.Finally, to find the modulus of ( bar{z} ), I'll use the formula ( |bar{z}| = sqrt{a^2 + b^2} ), where ( a ) and ( b ) are the real and imaginary parts of ( bar{z} ), respectively. This will give me the magnitude of the complex conjugate.