Question

tính

a; A = 3/4 * 8/9 * 15/16 * ..............* 9999/10000

b; B = {1 - 1/21 } * { 1 - 1/28 } * {1 - 1/36 } * .......................*{ 1 - 1326}

c; C = { 1 + 1/1*3 } * { 1 + 1 / 2*4 } * { 1 + 1/3*5} * ...........................* { 1+ 1/99*101}

Answer 1

\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{9999}{10000}\\ =\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{99\cdot101}{100\cdot100}\\ =\dfrac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot...\cdot100\cdot100}\\ =\dfrac{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}\\ =\dfrac{1\cdot101}{100\cdot2}\\ =\dfrac{101}{200}\)

Answer 2

\(C=\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\left(1+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{1\cdot3}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{2\cdot4}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{3\cdot5}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{99\cdot101}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\left(\dfrac{2^2-1}{1\cdot3}+\dfrac{1}{1\cdot3}\right)\cdot\left(\dfrac{3^2-1}{2\cdot4}+\dfrac{1}{2\cdot4}\right)\cdot\left(\dfrac{4^2-1}{3\cdot5}+\dfrac{1}{3\cdot5}\right)\cdot...\cdot\left(\dfrac{100^2-1}{99\cdot101}+\dfrac{1}{99\cdot101}\right)\\ =\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot...\cdot100^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot99\cdot101}\\ =\dfrac{\left(2\cdot3\cdot4\cdot...\cdot100\right)\cdot\left(2\cdot3\cdot4\cdot...\cdot100\right)}{\left(1\cdot2\cdot3\cdot...\cdot99\right)\cdot\left(3\cdot4\cdot5\cdot...\cdot101\right)}\\ =\dfrac{100\cdot2}{1\cdot101}=\dfrac{200}{101}\)

Answer 3

\(B=\left(1-\dfrac{1}{21}\right)\cdot\left(1-\dfrac{1}{28}\right)\cdot\left(1-\dfrac{1}{36}\right)\cdot...\cdot\left(1-\dfrac{1}{1326}\right)\\ =\dfrac{20}{21}\cdot\dfrac{27}{28}\cdot\dfrac{35}{36}\cdot...\cdot\dfrac{1325}{1326}\\=\dfrac{40}{42}\cdot\dfrac{54}{56}\cdot\dfrac{70}{72}\cdot...\cdot\dfrac{2650}{2652}\\ =\dfrac{5\cdot8}{6\cdot7}\cdot\dfrac{6\cdot9}{7\cdot8}\cdot\dfrac{7\cdot10}{8\cdot9}\cdot...\cdot\dfrac{50\cdot53}{51\cdot52}\\ =\dfrac{5\cdot8\cdot6\cdot9\cdot7\cdot10\cdot...\cdot50\cdot53}{6\cdot7\cdot7\cdot8\cdot8\cdot9\cdot...\cdot51\cdot52}\\ =\dfrac{\left(5\cdot6\cdot7\cdot...\cdot50\right)\cdot\left(8\cdot9\cdot10\cdot...\cdot53\right)}{\left(6\cdot7\cdot8\cdot...\cdot51\right)\cdot\left(7\cdot8\cdot9\cdot...\cdot52\right)}=\dfrac{5\cdot53}{51\cdot7}=\dfrac{265}{357} \)

Answer 4

\(A=\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}.....\dfrac{9999}{10000}\)

\(A=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}.....\dfrac{99.101}{100.100}\)

\(A=\dfrac{1.2.3...99}{2.3.4...100}.\dfrac{3.4.5...101}{2.3.4...100}\)

\(A=\dfrac{1}{100}.\dfrac{2}{101}\)

\(A=\dfrac{1}{5050}\)

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