Câu hỏi của Nguyễn Thị Phương Linh

Question

$A=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+.....+\dfrac{1}{10000}$

$B=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+....+\dfrac{2}{99\cdot101}$

阮草~๖ۣۜDαɾƙ · Accepted Answer

$B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+....+\dfrac{2}{99.101}$

$B=2.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)$

$B=\dfrac{2}{2}.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{99.101}\right)$

$B=1.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)$

$B=1.\left(1-\dfrac{1}{101}\right)$

$B=1.\dfrac{100}{101}$

$B=\dfrac{100}{101}$Câu trả lời: $B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+....+\dfrac{2}{99.101}$

$B=2.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}\right)$

$B=\dfrac{2}{2}.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{99.101}\right)$

$B=1.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)$

$B=1.\left(1-\dfrac{1}{101}\right)$

$B=1.\dfrac{100}{101}$

$B=\dfrac{100}{101}$

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