Calculul unor sume in gimnaziu
Calculul unor sume in gimnaziu
Exercitii in care se cere calcularea unei sume de mai multi
termeni sunt intalnite chiar in manualele de clasa a-IV-a sau
a-V-a.Am considerat necesara demonstrarea unor formule de calcul
pentru acestea ,altele decat cele ce folosesc inductia matematica
sau o pseudo-inductie matematica,in ideea de a le folosi in
rezolvarea unor probleme propuse pentru diferite concursuri.
Calculul unor sume de numere
1. S= 1 +2 +3 + +(n-2) +(n-1) +n
S=n +(n-1)+(n-2)+ + 3 + 2 + 1
2S=n+1+n+1+n+1++n+1+n+1+n+1
2S=n(n+1)
S=
2
)
1
(
+
n
n
2. S=1 + 3 + 5 +..+(2n-5)+(2n-3)+(2n-1)
S=(2n-1)+(2n-3)+(2n-5)++ 5 + 3 + 1
2S=2n + 2n +2n ++ 2n + 2n + 2n
2S=2n.n
S=
n
2
3. S=1 +
x
+
EMBED Equation.3
x
2
++
x
x
n
2
-
+
EMBED Equation.3
x
n
1
-
+
x
n
Sx=
EMBED Equation.3 EMBED Equation.3
x
x
x
x
x
n
n
+
+
+
+
-
+
1
3
.
...
2
Sx-S =
1
1
-
+
x
n
S(x-1) =
1
1
-
+
x
n
S=(
x
n
1
+
-1)/(
x
-1)
4. S=
1
2
+
2
2
+
3
2
++
n
2
Folosind suma primelor n numere naturale impare putem scrie:
1
2
=1
2
2
=1+3
3
2
=1+3+5
.
k
2
=1+3+5++(2k-1)
;..
n
2
=1+3+5++(2k-1)++(2n-1)
Adunand membru cu membru obtinem:
S=n.1+(n-1).3+(n-2).5++(n-k+1).(2k-1)++2.(2n-3)+(2n-1)
Termenul general are forma:(2k-1).(n-k+1) si poate fi scris:
(2k-1).(n-k+1)=(n+1).(2k-1)-2
k
2
+k,atunci:
S=(n+1).(1+3+5++2n-1)-2(
1
2
+
2
2
+
EMBED Equation.3
3
2
++
n
2
)+(1+2+3++n)
3S=(n+1).
n
2
+n(n+1)/2
6S=2.(n+1).
n
2
+n.(n+1)
6S=n(n+1)(2n+1)
S=
6
)
1
2
)(
1
(
+
+
n
n
n
5. S=
2
.
1
1
+
3
.
2
1
+
4
.
3
1
++
)
1
(
1
+
n
n
Se demonstreaza usor ca:
)
1
(
1
+
n
n
=
n
1
-
1
1
+
n
EMBED Equation.3
S=
1
1
-
2
1
+
2
1
-
3
1
++
n
1
-
1
1
+
n
=
1
1
-
1
1
+
n
=
1
+
n
n
Generalizare:
)
(
k
n
n
k
+
=
n
1
-
k
n
+
1
Aplicatii:
a) Calculati suma cifrelor numarului:
x=9+99+999++99..99,unde ultimul termen are 2008 cifre.
Numarul x se mai poate scrie:
EMBED Equation.3
x=10-1+
10
2
-1+
10
3
-1++
10
2008
-1=(10+
10
2
+
10
3
++
10
2008
-1=
=(10+
10
2
+
10
3
++
10
2008
)-2008=10(1+10+
10
2
++
10
2007
)-2008=
=10.
1
10
1
10
2008
-
-
-2008=10.
9
99
..
999
-2008=10.11111-2008=1111109102.In rezultat apare de 2004
ori,deci suma cifrelor va fi :2016.
Generalizare:
Pentru a calcula: S=a+
aa
+
aaa
++
aa
aa
...
se calculeaza:
9
a
(9+99+999++999)
b)Calculati: S=
4
.
1
3
+
9
.
4
5
+
16
.
9
7
++
1849
.
1764
85
Se foloseste relatia:
)
(
k
n
n
k
+
=
n
1
-
k
n
+
1
si avem:
S=
1
1
-
4
1
+
4
1
-
9
1
+
9
1
-
16
1
++
1764
1
-
1849
1
=
1849
1848
c)Sa se calculeze:
S=
)
1
.(
1
1
+
k
+
EMBED Equation.3
)
1
2
)(
1
(
1
+
+
k
k
+
)
1
3
)(
1
2
(
1
+
+
k
k
++
)
1
](
1
)
[(
1
+
+
-
nk
k
n
Se observa ca diferenta dintre factorii de la numitor este
k,deci vom inmulti cu k si obtinem:
Sk=
)
1
.(
1
+
k
k
+
)
1
2
)(
1
(
+
+
k
k
k
+
EMBED Equation.3
)
1
3
)(
1
2
(
+
+
k
k
k
++
)
1
](
1
)
1
[(
+
+
-
nk
k
n
k
=
=
1
1
-
1
1
+
k
+
1
1
+
k
-
1
2
1
+
k
+
1
2
1
+
k
-
1
3
1
+
k
++
1
)
1
(
1
+
-
k
n
-
1
1
+
nk
=
=
1
1
-
1
1
+
nk
=
1
1
1
+
-
+
nk
nk
=
1
+
nk
nk
,de unde:S=
1
+
nk
n
.
d)Aratati ca numarul :
N=1+2+
2
2
+
2
3
++
2
2006
nu este patrat perfect.
Calculand N obtinem: N=
2
2007
-1
U(
2
2007
-1)=U(U(
2
2007
)-1)=7.Cum nici un patrat perfect nu se termina in 2,3,7,8
rezulta N nu este patrat perfect.
e)Sa se calculeze suma:
S=
1
2
+
3
2
+
5
2
++
)
1
2
(
2
-
n
EMBED Equation.3
Se porneste de la
)
1
2
(
2
-
n
=4.
n
2
-4.n+1 avem:
1
2
=4.
1
2
-4.1+1
3
2
=4.
2
2
-4.2+1
5
2
=4.
3
2
-4.3+1
.
)
1
2
(
2
-
n
=4.
n
2
-4n+1
Adunand membru cu membru obtinem:
S=4(
1
2
+
2
2
+
3
2
++
n
2
)-4(1+2+3++n)+n=
= 4.
6
)
1
2
)(
1
(
+
+
n
n
n
-4.
2
)
1
(
+
n
n
+n=
3
)
1
2
)(
1
(
2
+
+
n
n
n
-2n(n+1)+n=
=
3
3
)
1
(
6
)
1
2
)(
1
(
2
n
n
n
n
n
n
+
+
-
+
+
=
=
3
)
3
6
6
2
4
2
4
(
2
+
-
-
+
+
+
n
n
n
n
n
=
3
)
1
4
(
2
-
n
n
.
f) Calculati:
S=
2
2
+
4
2
+
6
2
++
2008
2
.Suma mai poate fi scrisa:
S=
)
1
2
(
2
+
)
2
2
(
2
+
)
3
2
(
2
++
)
1004
2
(
2
=
2
2
.
1
2
+
2
2
.
2
2
+
2
2
.
3
2
++
+
2
2
.
1004
2
=
2
2
(
1
2
+
2
2
+
3
2
++
1004
2
)=
6
2009
.
1005
.
1004
.
4
=
=1004.670.2009.
g) Calculati: S=
2
2
+
6
2
+
10
2
++
4014
2
.Suma se mai scrie:
S=
)
1
.
2
(
2
+
EMBED Equation.3
)
3
.
2
(
2
+
)
5
.
2
(
2
++
)
2007
.
2
(
2
=
2
2
.
1
2
+
2
2
.
3
2
+
+ +
2
2
.
2007
2
=4(
1
2
+
3
2
++
2007
2
)=
3
)
1
.
4
(
1004
.
4
1004
2
-
=
=
3
)
1
(
1004
.
4
2008
2
-
=
3
2009
.
2007
.
1004
.
4
=4.1004.669.2009
h) S=1+
2
1
1
+
+
3
2
1
1
+
+
++
2008
...
3
2
1
1
+
+
+
+
=
=1+
2
/
)
3
.
2
(
1
+
2
/
)
4
.
3
(
1
++
2
/
)
2009
.
2008
(
1
=
=1+
3
.
2
2
+
4
.
3
2
++
2009
.
2008
2
=1+2(
+
3
.
2
1
EMBED Equation.3
4
.
3
1
++
2009
.
2008
1
)=
=1+2(
2
1
-
3
1
+
3
1
-
4
1
++
2008
1
-
2009
1
)=1+2(
2
1
-
2009
1
)=1+
2009
2007
=
4009
4016
.
i) (S=1+
x
1
+
x
2
1
+
x
3
1
++
x
n
1
. Suma se mai poate scrie:
S=
x
x
x
n
n
n
x
1
...
1
+
+
+
+
-
=
)
1
(
1
1
-
-
+
x
x
x
n
n
(Aratati ca numarul:
x=
3
1
2
2
+
-
n
n
-
3
2
2
2
4
2
+
-
n
n
--
3
10
2
2
4
2
+
-
n
n
este patrat perfect.
Numarul poate fi scris: x=
3
)
1
(
2
-
n
-
3
)
1
(
2
2
2
-
n
--
3
)
1
(
10
2
2
-
n
=
=
)
1
(
2
-
n
(
3
1
-
3
2
2
--
3
10
2
)=
)
1
(
2
-
n
)[
3
1
-
3
2
2
(1+
3
1
+
3
2
1
++
3
8
1
)]=
=
)
1
(
2
-
n
(
3
1
-
3
2
2
.
3
3
8
9
.
2
1
-
)=
)
1
(
2
-
n
(
3
3
10
9
1
3
1
-
-
)=
)
1
(
2
-
n
.
3
10
1
=patrat perfect.
j) Calculati :S=3+7+11++8035.
Se observa ca diferenta intre factori este 4,ne gandim la
teorema impartirii cu rest si constatam:
3=4.0+3
7=4.1+3
11=4.2+3
.
8035=4.2008+3
S=4.0+3+4.1+3+4.2+3++4.2008+3=4(1+2+3+.+2008)+
+2009.3=
2
2009
.
2008
.
4
+6027=4016.2009+6027=2009.4019
Concluzionand in calculul unei sume de mai multi termeni sunt
necesare parcurgerea urmatoarelor etape:
_stabilirea numarului de termeni ai sumei;
_identificarea termenului general sau a regulii dupa care sunt
construiti termenii sumei;
_identificarea formulei sau lucru pe termenul general si
repetarea pe fiecare termen in parte
EMBED Equation.3 EMBED Equation.3
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