Concursul de matematică rdquoValea Sălăuțeirdquo

Liceul Tehnologic Telciu 16 noiembrie 2013 Clasa a V-a

1 a) Demonstraţi că

b) Scrieţi pe ca sumă de trei pătrate perfecte nenule

2 Fie şi

a) Arătaţi că şi sunt pătrate perfecte

b) Arătaţi că 2013 + lt 4

3 Se dă şirul 9 13 17 21 25 hellip

a) Care este al 10-lea termen al şirului

b) Aflaţi termenul care ocupă locul 2013 icircn şir

c) Verificaţi dacă numerele 1000 şi 2013 aparţin şirului

d) Calculaţi suma primilor 100 de termeni ai şirului

BAREM

1a) 2300

= 23100

= (23)100

= 8100

helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

3200

= 32100

= (32)

100 = 9

100helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

Deoarece 8100

lt 9100

=gt 2300

lt 3200

helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

b) Se scrie 29 ca sumă de trei pătrate perfecte 29 = 22+3

2+4

2helliphelliphelliphelliphelliphellip1p

292013

=29292012

=helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

=(22+3

2+4

2) 29

2012=2

229

2012+3

229

2012 +4

229

2012=helliphelliphelliphelliphelliphelliphelliphellip1p

= (2291006

)2 + (329

1006)

2 + (429

1006)

2helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

2 a) a= 2013 + 2( 1+2+3+4+hellip+2012) =2013 + 220122013 2=helliphelliphelliphellip 1p

= 2013 + 20122013 = 2013(1 + 2012) = 20132013 = 20132helliphelliphelliphellip 1p

b = 1+3+5+7+hellip+2013=1+2+3+4+hellip+2013-(2+4+hellip+2012)=

=201320142 - 2(1+2+3+4+hellip+1006)= helliphelliphellip helliphelliphelliphelliphelliphelliphelliphelliphellip1p

=20131007-210061007 2=20131007-10061007=

=1007(2013-1006)= 10071007=10072helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

b) 2013 + a = 2013 + 20132=2013(1 + 2013)= 20132014helliphelliphelliphelliphelliphelliphellip1p

4b=410072=2

21007

2=(21007)

2=2014

2helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

Deoarece 20132014lt20142 =gt 2013 + a lt 4bhelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

3 a) 9 13 17 21 252933374145 al 10-lea termen este 45helliphelliphelliphelliphellip1p

b) a1=9

a2=9+41

a3=9+42

a4=9+43

helliphelliphellip

a2013=9+42012=9+8048=8057helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip2p

c) Din 9+4x=1000 4x=991 imp icircn N deci 1000 nu aparţine şiruluihellip1p

Din 9+4y=2013 4x=2004x=501 deci 2013 aparţine şiruluihelliphellip1p

d) S= a1+ a2+ a3+hellip+ a100=9+9+41+9+42+9+43+hellip+9+499=helliphelliphelliphelliphellip1p

=9100+4(1+2+3+4+hellip+99)=900+4991002=20700helliphelliphelliphelliphelliphelliphellip1p

BAREM

1a) 2300

= 23100

= (23)100

= 8100

helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

3200

= 32100

= (32)

100 = 9

100helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

Deoarece 8100

lt 9100

=gt 2300

lt 3200

helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

b) Se scrie 29 ca sumă de trei pătrate perfecte 29 = 22+3

2+4

2helliphelliphelliphelliphelliphellip1p

292013

=29292012

=helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

=(22+3

2+4

2) 29

2012=2

229

2012+3

229

2012 +4

229

2012=helliphelliphelliphelliphelliphelliphelliphellip1p

= (2291006

)2 + (329

1006)

2 + (429

1006)

2helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

2 a) a= 2013 + 2( 1+2+3+4+hellip+2012) =2013 + 220122013 2=helliphelliphelliphellip 1p

= 2013 + 20122013 = 2013(1 + 2012) = 20132013 = 20132helliphelliphelliphellip 1p

b = 1+3+5+7+hellip+2013=1+2+3+4+hellip+2013-(2+4+hellip+2012)=

=201320142 - 2(1+2+3+4+hellip+1006)= helliphelliphellip helliphelliphelliphelliphelliphelliphelliphelliphellip1p

=20131007-210061007 2=20131007-10061007=

=1007(2013-1006)= 10071007=10072helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

b) 2013 + a = 2013 + 20132=2013(1 + 2013)= 20132014helliphelliphelliphelliphelliphelliphellip1p

4b=410072=2

21007

2=(21007)

2=2014

2helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

Deoarece 20132014lt20142 =gt 2013 + a lt 4bhelliphelliphelliphelliphelliphelliphelliphelliphelliphellip1p

3 a) 9 13 17 21 252933374145 al 10-lea termen este 45helliphelliphelliphelliphellip1p

b) a1=9

a2=9+41

a3=9+42

a4=9+43

helliphelliphellip

a2013=9+42012=9+8048=8057helliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphelliphellip2p

c) Din 9+4x=1000 4x=991 imp icircn N deci 1000 nu aparţine şiruluihellip1p

Din 9+4y=2013 4x=2004x=501 deci 2013 aparţine şiruluihelliphellip1p

d) S= a1+ a2+ a3+hellip+ a100=9+9+41+9+42+9+43+hellip+9+499=helliphelliphelliphelliphellip1p

=9100+4(1+2+3+4+hellip+99)=900+4991002=20700helliphelliphelliphelliphelliphelliphellip1p