chứng minh rằng
$\frac{1}{1×2}$+ $\frac{1}{3×4}$+…+ $\frac{1}{99 ×100}$
=$\frac{1}{51}$+ $\frac{1}{52}$+…+ $\frac{1}{100}$
chứng minh rằng
$\frac{1}{1×2}$+ $\frac{1}{3×4}$+…+ $\frac{1}{99 ×100}$
=$\frac{1}{51}$+ $\frac{1}{52}$+…+ $\frac{1}{100}$
`=`$\frac{1}{1.2}$ + $\frac{1}{3.4}$ + …+ $\frac{1}{99.100}$
`=` `1` `-` $\frac{1}{2}$ + $\frac{1}{3}$ – $\frac{1}{4}$ +…..+ $\frac{1}{99}$ – $\frac{1}{100}$
`=` (`1` + $\frac{1}{3}$ + $\frac{1}{5}$ +…+ $\frac{1}{99}$) – ($\frac{1}{2}$ + $\frac{1}{4}$ + …+ $\frac{1}{100}$ )
`=` (`1` + $\frac{1}{3}$ + $\frac{1}{5}$ +…+ $\frac{1}{99}$) – ($\frac{1}{2}$ + $\frac{1}{4}$ + …+ $\frac{1}{100}$ ) – 2. ($\frac{1}{2}$ + $\frac{1}{4}$ + … + $\frac{1}{100}$ )
`=` (`1` + $\frac{1}{2}$ + $\frac{1}{3}$ +…+ $\frac{1}{99}$ + $\frac{1}{100}$ ) – ( 1 + $\frac{1}{2}$ + $\frac{1}{3}$ + …+ $\frac{1}{50}$ )
`=` `1` + $\frac{1}{2}$ + $\frac{1}{3}$ +…+ $\frac{1}{99}$ + $\frac{1}{100}$ – 1 – $\frac{1}{2}$ – $\frac{1}{3}$ – …- $\frac{1}{50}$
`=` $\frac{1}{51}$ + $\frac{1}{52}$ + $\frac{1}{53}$ +….+ $\frac{1}{100}$
`=>` $\frac{1}{1.2}$ + $\frac{1}{3.4}$ + …+ $\frac{1}{99.100}$ = = $\frac{1}{51}$ + $\frac{1}{52}$ + $\frac{1}{53}$ +….+ $\frac{1}{100}$
1/1*2 + 1/3*4 +…+ 1/99*100
=1 – 1/2 + 1/3 – 1/4 + … + 1/99 – 1/100
=( 1 + 1/3 + 1/5 + … + 1/99 ) – ( 1/2 + 1/4 + 1/6 + … + 1/100 )
=( 1 + 1/3 + 1/5 + … + 1/99 )+ ( 1/2 + 1/4 + 1/6 + … + 1/100 )-2*( 1/2 + 1/4 + 1/6+…+1/100)
=( 1 + 1/2 + 1/3 + 1/4 + … + 1/99 + 1/100 ) – ( 1 + 1/2 + 1/3 + 1/4 + … + 1/50 )
=1+ 1/2 + 1/3 + 1/4+ … + 1/99 + 1/100 – 1- 1/2 – 1/3 – 1/4 – … – 1/50
= 1/51 + 1/52 + 1/53 + … + 1/100
=>1/1.2+1/3.4+…+1/99.100= 1/51 + 1/52 + 1/53 + … + 1/100
Mik ko bt gõ latex , bạn thông cảm
chúc học tốt!!