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Question

$S=\dfrac{2mu2}{1.2}+\dfrac{2mu2}{2.3}+\dfrac{2mu2}{3.4}+...+\dfrac{2mu2}{2022.2023}$(mu = mũ)

Nguyễn Ngọc Huy Toàn · Accepted Answer

$S=\dfrac{2^2}{1.2}+\dfrac{2^2}{2.3}+\dfrac{2^2}{3.4}+...+\dfrac{2^2}{2022.2023}$$S=2^2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)$$S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)$$S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)$$S=2^2.\dfrac{2022}{2023}$$S=\dfrac{2^2.2022}{2023}=\dfrac{8088}{2023}$Câu trả lời: $S=\dfrac{2^2}{1.2}+\dfrac{2^2}{2.3}+\dfrac{2^2}{3.4}+...+\dfrac{2^2}{2022.2023}$$S=2^2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)$$S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)$$S=2^2.\left(\dfrac{1}{1}-\dfrac{1}{2023}\right)$$S=2^2.\dfrac{2022}{2023}$$S=\dfrac{2^2.2022}{2023}=\dfrac{8088}{2023}$

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