exercitiul 1.
x=7 daca
1/2+1/6+1/12+1/20+…+1/x(x+1)=2010/2011
1/2+1/6+1/12+1/20+…+ 1/7(7+1)=2010/2011
1/2+1/6+1/12+1/20+…+1/56=2010/2010
1/1*2+1/2*3+1/3*4+1/4*5+…+1/7*8=2010/2011
1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+….+1/7-1/8=2010/2011
am simplificat termenii asemenea =>
1/1-1/8=2010/2011
in partea dreapta am amplificat cu 8 =>
7/8 = 2010/2011
si mai departe ..nu mai stiu.
exercitiul 2.
n=?
2/1*3 + 2/3*5+….+2/(2n-1)(2n+1) = 2010/2011
2/1-2/3+2/3-2/5+…+2/2n-1-2/2n+1=2010/2011
am simplificat termenii asemenea
2/1-2/2n+1=2010/2011
amplific pe 2/1 cu 2n+1
4n+2-2/2n+1=2010/2011
4n/2n+1 = 2010/2011
am facut produsul mezilor cu cel al extremilor =>
4n*2011 = (2n+1)*2010 => 8044n=4020n+2010=>4020n+2010=8044n
=> 8044n-4020n=2010=>4024n=2010 =>
si de aici nu mai stiu .
Nu stiu sigur daca am rezolvat corect si daca ma puteti ajuta ,astept raspunsul dumneavoastra..
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Baiatule urmareste calculele nu doar rezultatul(adica inlocuieste k cu 1 pana la x si vezi cum se reduc termeni)…
Noroc.
Am urmat toti pasii si dupa ce am inlocuit pe k de la 1 la 7 am obtinut:
S=1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8=2010/2011
Se reduc termenii si am obt : 1-1/8=2010/2011
Indicatia dvs. S= 1-1/x+1=2010/2011 ,nu stiu de unde a reiesit 1-1/x+1
Cred ca m-am incurcat in propriile ite …acum m-am dezlegat.Uitati:
1/2+1/6+1/12+1/20+….+1/x(x+1)=2010/2011
1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+….+1/x-1/x+1=2010/2011
reduc fractiile =>
1-1/x+1 = 2010/2011
am amplificat pe 1 cu x+1 =?
x+1-1/x+1 = 2010/2011
am facut produsul intre mezi si extremi :
2011x=2010(x+1)
2011x=2010x+2010
2010x+2010=2011x
2011x-2010x=2010
x=2010
( raspunsul este exact ca in carte )
dar la exercitiul 2 am facut binee ???
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exercitiul 2.
n=?🙁
2/1*3 + 2/3*5+….+2/(2n-1)(2n+1) = 2010/2011
2/1-2/3+2/3-2/5+…+2/2n-1-2/2n+1=2010/2011
am simplificat termenii asemenea
2/1-2/2n+1=2010/2011
amplific pe 2/1 cu 2n+1
4n+2-2/2n+1=2010/2011
4n/2n+1 = 2010/2011
am facut produsul mezilor cu cel al extremilor =>
4n*2011 = (2n+1)*2010 => 8044n=4020n+2010=>4020n+2010=8044n
=> 8044n-4020n=2010=>4024n=2010 =>
si de aici nu mai stiu .
Nu stiu sigur daca am rezolvat corect si daca ma puteti ajuta ,astept raspunsul dumneavoastra..
2/1*3 + 2/3*5+….+2/(2n-1)(2n+1) = 2010/2011
Observam ca
2/1*3=1/1-1/3
2/3*5=1/3-1/5
.
2/(2n-1)(2n+1)=1/(2n-1)-1/(2n+1)
Deci S= 1/1-1/3+1/3-1/5+.+1/(2n-1)-1/(2n+1)=2010/2011
1-1/(2n+1)=2010/2011 => 2n/(2n+1)=2010/2011 => 2n*2011=2n*2010+2010
2n=2010 => n=1005
Va multumesc foarte mult pentru raspuns !
Am inteles !