Cho `A=1/1.2+1/3.4+…+1/2005.2006` và `B=1/1004.2006+1/1005.2006+…+1/2006.1004`.
Chứng minh `A/B inZ`.
Cho `A=1/1.2+1/3.4+…+1/2005.2006` và `B=1/1004.2006+1/1005.2006+…+1/2006.1004`. Chứng minh `A/B inZ`.
By Josephine
By Josephine
Cho `A=1/1.2+1/3.4+…+1/2005.2006` và `B=1/1004.2006+1/1005.2006+…+1/2006.1004`.
Chứng minh `A/B inZ`.
Giải thích các bước giải:
A=$\frac{1}{1.2}$ +$\frac{1}{3.4}$ +….+$\frac{1}{2005.2006}$
A=1-$\frac{1}{2}$ +$\frac{1}{3}$ -$\frac{1}{4}$ +….+$\frac{1}{2005}$ -$\frac{1}{2006}$
A=(1+$\frac{1}{3}$ +$\frac{1}{5}$ +…+$\frac{1}{2005}$ )-($\frac{1}{2}$ +$\frac{1}{4}$ +$\frac{1}{6}$ +…..+$\frac{1}{2006}$ )
A=(1+$\frac{1}{2}$ +$\frac{1}{3}$ +…+$\frac{1}{2006}$ )-2.($\frac{1}{2}$ +$\frac{1}{4}$ +$\frac{1}{6}$ +…..+$\frac{1}{2006}$ )
A=(1+$\frac{1}{2}$ +$\frac{1}{3}$ +…+$\frac{1}{2006}$ )-(1+$\frac{1}{2}$ +$\frac{1}{3}$ …..+$\frac{1}{1003}$ )
A=$\frac{1}{1004}$ +$\frac{1}{1005}$ +….+$\frac{1}{2006}$
B=$\frac{1}{1004}$ .$\frac{1}{2006}$ +$\frac{1}{1005}$ .2006+…..+$\frac{1}{1006}$ .1004 3010
B=1004+$\frac{2006}{1004}$ .2006+1005+$\frac{2006}{1005}$ .2006+…..+2006+$\frac{1004}{2006}$ .1004 3010
B=$\frac{1}{2006}$ +$\frac{1}{1004}$ +$\frac{1}{1005}$ +$\frac{1}{1005}$ +….+$\frac{1}{1004}$ +$\frac{1}{2006}$ 3010
B=2.(1/1004+1/1005+1/1006+….+1/2006)
B=2.(1/1004+1/1005+1/1006+….+1/2006)/3010
B=(1/1004+1/1005+1/1006+….+1/2006)/1505 ⇒A/
B=(1/1004+1/1005+….+1/2006)/(1/1004+1/1005+1/1006+….+1/2006)/1505 ⇒A/B=1505
Hay A/B ∈ Z
( Mk vt P/s thế thôi nhá còn đâu bn tự viết vào giúp mk. Bh mk bận nếu bn k hiểu thì tý mk vt tiếp cho)
$ Xin Ctlhn$
$@ nhuquynh1110$
Ta có :
$A$= $\dfrac{1}{1.2}$ + $\dfrac{1}{3.4}$ + … + $\dfrac{1}{2005.2006}$
$A$ = $\dfrac{1}{1}$ – $\dfrac{1}{2}$ + $\dfrac{1}{3}$ – $\dfrac{1}{4}$ + … + $\dfrac{1}{2005}$ – $\dfrac{1}{2006}$
$A$ = ( $\dfrac{1}{1}$ + $\dfrac{1}{3}$ + $\dfrac{1}{5}$ + …. + $\dfrac{1}{2005}$ ) – ( $\dfrac{1}{2}$ + $\dfrac{1}{4}$ + … + $\dfrac{1}{2006}$ )
$A$ = ( $\dfrac{1}{1}$ + $\dfrac{1}{2}$ + $\dfrac{1}{3}$ + $\dfrac{1}{4}$ + $\dfrac{1}{5}$ + $\dfrac{1}{6}$ + …. + $\dfrac{1}{2005}$ + $\dfrac{1}{2006}$ ) – $2$ . ( $\dfrac{1}{2}$ + $\dfrac{1}{4}$ + … + $\dfrac{1}{2006}$ )
$A$ = ( $\dfrac{1}{1}$ + $\dfrac{1}{2}$ + $\dfrac{1}{3}$ + $\dfrac{1}{4}$ + $\dfrac{1}{5}$ + $\dfrac{1}{6}$ + …. + $\dfrac{1}{2005}$ + $\dfrac{1}{2006}$ ) – ( $1$ + $\dfrac{1}{2}$ + … + $\dfrac{1}{1003}$ )
$A$ = $\dfrac{1}{1004}$ + $\dfrac{1}{1005}$ + … + $\dfrac{1}{2006}$ ($1$)
Ta lại có :
$B$ = $\dfrac{1}{1004 . 2006}$ + $\dfrac{1}{1005 . 2005 }$( cái này bị sai đề bạn nha ) + $…$ + $\dfrac{1}{2006 . 1004}$
$3010B$ = $\dfrac{3010}{1004 . 2006}$ + $\dfrac{3010}{1005 . 2005 }$ + …. + $\dfrac{3010}{2006 . 1004}$
$3010B$ = $\dfrac{1}{1004}$ + $\dfrac{1}{2006}$ + $\dfrac{1}{1005}$ + $\dfrac{1}{2005}$ + … + $\dfrac{1}{2006}$ + $\dfrac{1}{1004}$ = $2$ . ( $\dfrac{1}{1004}$ + $\dfrac{1}{1005}$ + … + $\dfrac{1}{2006}$ )
⇒ $B$ = $\dfrac{2 . ( \dfrac{1}{1004} + \dfrac{1}{1005} + … + \dfrac{1}{2006} ) }{3010}$ = $\dfrac{\dfrac{1}{1004} + \dfrac{1}{1005} + … + \dfrac{1}{2006} )}{1505}$ ($2$)
Từ ($1$) , ($2$) ⇒
$\dfrac{A}{B}$ = $\dfrac{\dfrac{1}{1004} + \dfrac{1}{1005} + … + \dfrac{1}{2006}}{\dfrac{\dfrac{1}{1004} + \dfrac{1}{1005} + … + \dfrac{1}{2006} )}{1505}}$ = $1505$ ∈ $Z$
⇒ $\dfrac{A}{B}$ ∈ $Z$ ( đpcm )