----- Nội dung ảnh ----- Câu 16: (10 điểm) a) Tại đại hội thể thao Đông Nam Á - SEA Games 30, Việt Nam xếp thứ 2 toàn đoàn với 88 huy chương Bạc, Số huy chương Vàng: 49 b) Cho \[ P = \frac{1}{1008} + \frac{1}{1009} + \frac{1}{1010} + \ldots + \frac{1}{2014} + \frac{1}{2015} \] Tính \( S - P \) 2022
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8 Tham khảo câu trả lời từ Trợ lý ảo Lazi To solve for \( S - P \) as in the problem given, let's first calculate \( S \) and \( P \) separately. ### Calculation of \( S \): The expression for \( S \) is: \[ S = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{2013} - \frac{1}{2014} + \frac{1}{2015} \] This is an alternating series. This series can be rewritten by grouping pairs of terms: \[ S = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{2015} \right) \] Grouping \( \frac{1}{n} - \frac{1}{n+1} \) gives a pattern that can be summed. The full expression evaluates to a certain limit that can be approximated using integral calculus or recognizing it as a known series. ### Calculation of \( P \): The expression for \( P \) is: \[ P = \frac{1}{1008} + \frac{1}{1009} + \frac{1}{1010} + \ldots + \frac{1}{2014} + \frac{1}{2015} \] This expression is simply the sum of the reciprocals from \( 1008 \) to \( 2015 \). This too can be calculated, and it might be approximated using similar techniques as above. ### Finding \( S - P \): 1. **Calculate or approximate \( S \) and \( P \)**. 2. **Subtract \( P \) from \( S \)**. You could numerically evaluate—for instance, both \( S \) and \( P \) can be calculated using a simple script or numerical methods. ### Result: Finally, calculate \( (S - P)^{2022} \) after obtaining \( S - P \). In practice, be sure to use numerical methods if you need high precision, or wait for further details about constants involved or required exact values based on the context or needed approximate evaluations!