NHDTA-477 JAV Judul yang sesuai dan terpercaya dalam bahasa Indonesia adalah sangat penting. Mohon berikan konten yang lebih sesuai untuk sayaFullPath:UsersusernameDocuments<|endoftext|>Find the value of the expression [sum_{n = 1}^infty frac{1}{n(n + 2)}.] To find the value of the infinite series (sum_{n=1}^infty frac{1}{n(n+2)}), we start by using partial fraction decomposition to simplify the general term (frac{1}{n(n+2)}). We can write: [ frac{1}{n(n+2)} = frac{A}{n} + frac{B}{n+2} ] Multiplying both sides by (n(n+2)), we get: [ 1 = A(n+2) + Bn ] Expanding and combining like terms, we have: [ 1 = An + 2A + Bn = (A + B)n + 2A ] For the equation to hold for all (n), the coefficients of (n) and the constant term must be equal on both sides. This gives us the system of equations: [ A + B = 0 quad ext{and} quad 2A = 1 ] Solving these equations, we get: [ A = frac{1}{2} quad ext{and} quad B = -frac{1}{2} ] So, we can rewrite the term as: [ frac{1}{n(n+2)} = frac{1/2}{n} - frac{1/2}{n+2} = frac{1}{2} left( frac{1}{n} - frac{1}{n+2} ight) ] Now, we can write the series as: [ sum_{n=1}^infty frac{1}{n(n+2)} = frac{1}{2} sum_{n=1}^infty left( frac{1}{n} - frac{1}{n+2} ight) ] This is a telescoping series. Let's write out the first few terms to see the cancellation: [ frac{1}{2} left( left( frac{1}{1} - frac{1}{3} ight) + left( frac{1}{2} - frac{1}{4} ight) + left( frac{1}{3} - frac{1}{5} ight) + left( frac{1}{4} - frac{1}{6} ight) + cdots ight) ] Notice that most terms cancel out, leaving us with: [ frac{1}{2} left( 1 + frac{1}{2} ight) = frac{1}{2} left( frac{3}{2} ight) = frac{3}{4} ] Therefore, the value of the series is: [ oxed{frac{3}{4}} ]Fullpath: C:UsersusernameDocumentsmath problemsseries problem.txt On the island of Castor, there are 60 chess players. A quarter of the island's chess players have never lost to an AI. How many people on the island have lost to a computer, at least once? To determine how many people on the island of Castor have lost to a computer at least once, we start by finding out how many players have never lost to an AI. Given that there are 60 chess players on the island and a quarter of them have never lost to an AI, we calculate the number of players who have never lost to an AI as follows: [ frac{1}{4} imes 60 = 15 ] This means that 15 players have never lost to an AI. Therefore, the number of players who have lost to a computer at least once is the total number of players minus the number of players who have never lost to an AI: [ 60 - 15 = 45 ] Thus, the number of people on the island who have lost to a computer at least once is (oxed{45}). - Cuplikan Gratis dan Subtitle Bahasa Indonesia srt.

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