Question

Đã trả lời bởi chuyên gia

Answer 1

$\frac{1}{100 \times 99} - \frac{1}{99 \times 98} - \frac{1}{98 \times 97} - . . . - \frac{1}{3 \times 2} - \frac{1}{2 \times 1} = (\frac{1}{99} - \frac{1}{100}) - (\frac{1}{98} - \frac{1}{99}) - (\frac{1}{97} - \frac{1}{98}) - . . . - (\frac{1}{2} - \frac{1}{3}) - (1 - \frac{1}{2}) = \frac{1}{99} - \frac{1}{100} - \frac{1}{98} + \frac{1}{99} - \frac{1}{97} + \frac{1}{98} - \frac{1}{96} + \frac{1}{97} - . . . - \frac{1}{2} + \frac{1}{3} - 1 + \frac{1}{2} = \frac{1}{99} - \frac{1}{100} - \frac{1}{98} + \frac{1}{99} - 1 = \frac{- 485101}{485100}$ ..

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Answer 2

$\frac{1}{100 \times 99} - \frac{1}{99 \times 98} - \frac{1}{98 \times 97} - . . . - \frac{1}{3 \times 2} - \frac{1}{2 \times 1} = (\frac{1}{99} - \frac{1}{100}) - (\frac{1}{98} - \frac{1}{99}) - (\frac{1}{97} - \frac{1}{98}) - . . . - (\frac{1}{2} - \frac{1}{3}) - (1 - \frac{1}{2}) = \frac{1}{99} - \frac{1}{100} - \frac{1}{98} + \frac{1}{99} - \frac{1}{97} + \frac{1}{98} - \frac{1}{96} + \frac{1}{97} - . . . - \frac{1}{2} + \frac{1}{3} - 1 + \frac{1}{2} = \frac{1}{99} - \frac{1}{100} - \frac{1}{98} + \frac{1}{99} - 1 = \frac{- 485101}{485100}$ ..

Answer 3

Ta có:

$\frac{1}{100 \times 99} - \frac{1}{99 \times 98} - \frac{1}{98 \times 97} - . . . - \frac{1}{3 \times 2} - \frac{1}{2 \times 1} = - (\frac{1}{100 \times 99} + \frac{1}{99 \times 98} + \frac{1}{98 \times 97} + . . . + \frac{1}{3 \times 2} + \frac{1}{2 \times 1}) + \frac{2}{99 \times 100} = - [(\frac{1}{99} - \frac{1}{100}) + (\frac{1}{98} - \frac{1}{99}) + (\frac{1}{97} - \frac{1}{98}) + . . . + (\frac{1}{2} - \frac{1}{3}) + (1 - \frac{1}{2})] + (\frac{1}{99} - \frac{1}{100}) = - (\frac{1}{99} - \frac{1}{100} + \frac{1}{98} - \frac{1}{99} + \frac{1}{97} - \frac{1}{98} + . . . + \frac{1}{2} - \frac{1}{3} + 1 - \frac{1}{2}) + \frac{2}{99} - \frac{1}{50} = - (1 - \frac{1}{100}) + \frac{1}{4950} = - \frac{99}{100} + \frac{1}{4950} = \frac{- 9799}{9900}$

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Answer 4

Ta có:

$\frac{1}{100 \times 99} - \frac{1}{99 \times 98} - \frac{1}{98 \times 97} - . . . - \frac{1}{3 \times 2} - \frac{1}{2 \times 1} = - (\frac{1}{100 \times 99} + \frac{1}{99 \times 98} + \frac{1}{98 \times 97} + . . . + \frac{1}{3 \times 2} + \frac{1}{2 \times 1}) + \frac{2}{99 \times 100} = - [(\frac{1}{99} - \frac{1}{100}) + (\frac{1}{98} - \frac{1}{99}) + (\frac{1}{97} - \frac{1}{98}) + . . . + (\frac{1}{2} - \frac{1}{3}) + (1 - \frac{1}{2})] + (\frac{1}{99} - \frac{1}{100}) = - (\frac{1}{99} - \frac{1}{100} + \frac{1}{98} - \frac{1}{99} + \frac{1}{97} - \frac{1}{98} + . . . + \frac{1}{2} - \frac{1}{3} + 1 - \frac{1}{2}) + \frac{2}{99} - \frac{1}{50} = - (1 - \frac{1}{100}) + \frac{1}{4950} = - \frac{99}{100} + \frac{1}{4950} = \frac{- 9799}{9900}$

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