Cho A= 1/2 + 1/3 + 1/4 + …. + 1/100 + 1/101
B= 1/100.1 + 1/99.2 + 1/98.3 + …. + 1/52.49 + 1/51.50
Hỏi A – 101 .B
Cho A= 1/2 + 1/3 + 1/4 + …. + 1/100 + 1/101
B= 1/100.1 + 1/99.2 + 1/98.3 + …. + 1/52.49 + 1/51.50
Hỏi A – 101 .B
Nhìn hình chứ latex lỗi
Ta có :
`A = 1/2 + 1/3 + 1/4 + ….. + 1/100 + 1/101`
`= 1/100 + 1/101 + ( 1/2 + 1/99) + ( 1/3 + 1/98 )+ …. +( 1/52 + 1/49) + ( 1/51 + 1/50)`
`= 1/100 + 1/101 + 101/( 99.2) + 101/( 98.3 )+ …. + 101/( 52.49 ) + 101/( 51.50 )`
`⇒ 101B = 101/100 + 101/( 99.2 ) + 101/( 98.3 )+ …. + 101/( 52.49 ) + 101/( 51.50 )`
`B = 1/( 100.1 ) + 1/( 99.2 ) + 1/( 98.3 )+ …. + 1/( 52.49 ) + 1/( 51.50 )`
`⇒ 101B =1 + 1/100+ 101/( 99.2 ) + 101/( 98.3 )+ …. + 101/( 52.49 ) + 101/( 51.50 )`
Ta có :
`A – 101B = (1/100 + 1/101 + 101/( 99.2 ) + 101/( 98.3 )+ …. + 101/( 52.49 ) + 101/( 51.50) – ( 1 + 1/100+ 101/( 99.2 ) + 101/( 98.3 )+ …. + 101/( 52.49 ) + 101/( 51.50)`
`= ( 1/101 – 1) + ( 1/100 – 1/100) + [(101/( 99.2 ) + 101/( 98.3 )+ …. + 101/( 52.49 ) + 101/( 51.50) – (101/( 99.2 ) + 101/( 98.3 )+ …. + 101/( 52.49) + 101/( 51.50)]`
`= −100/101`
Đáp án:
Ta có :
A = $\dfrac{1}{2}$ + $\dfrac{1}{3}$ + $\dfrac{1}{4}$ + ….. + $\dfrac{1}{100}$ + $\dfrac{1}{101}$
= $\dfrac{1}{100}$ + $\dfrac{1}{101}$ + ( $\dfrac{1}{2}$ + $\dfrac{1}{99}$) + ( $\dfrac{1}{3}$ + $\dfrac{1}{98}$ )+ …. +( $\dfrac{1}{52}$ + $\dfrac{1}{49}$) + ( $\dfrac{1}{51}$ + $\dfrac{1}{50}$)
= $\dfrac{1}{100}$ + $\dfrac{1}{101}$ + $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$
=> 101B = $\dfrac{101}{100}$ + $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$
B = $\dfrac{1}{100.1}$ + $\dfrac{1}{99.2}$ + $\dfrac{1}{98.3}$+ …. + $\dfrac{1}{52.49}$ + $\dfrac{1}{51.50}$
=> 101B =1 + $\dfrac{1}{100}$+ $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$
Ta có :
A – 101B = ($\dfrac{1}{100}$ + $\dfrac{1}{101}$ + $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$) – ( 1 + $\dfrac{1}{100}$+ $\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$)
= ($\dfrac{1}{101}$ – 1) + ( $\dfrac{1}{100}$ – $\dfrac{1}{100}$) + [($\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$) – ($\dfrac{101}{99.2}$ + $\dfrac{101}{98.3}$+ …. + $\dfrac{101}{52.49}$ + $\dfrac{101}{51.50}$)]
= $\dfrac{-100}{101}$
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