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[SPAM] Re: [obm-l] Soma !!!
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------=_Part_18739_6325195.1207766505825
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Eita mund=E3o da matem=E1tica...
Rapaz 1=AA vez que vi esta f=F3rmula, nossa, mas faz sentido claro...
vou verificar valeu mesmo, s=F3 uma perguntinha, onde vc encontrou essa
quest=E3o mesmo?
pois encontrei numa lista de exerc=EDcio por a=ED, e coloquei na minha por=
=E9m n=E3o
havia resolvido antes.
resultado nome da quest=E3o: UM PROFESSOR EM APUROS!!!
KKKKKKKKKKK
Bom, agrade=E7o bastante a colabora=E7=E3o e vou apicar indu=E7=E3o afim de=
verificar
se vale para todo n.
abra=E7os
E a caminhada continua!
2001/11/1 Pedro <npc1972@xxxxxxxxx>:
> Essa quest=E3o deu muito trabalho =E0 tres semana, mais no fim deu certo=
.
>
> Seja S_n =3D 1.11^0 + 2.11^1 +3.11^2 +...........+n.11111111111 rescrev=
er
> de uma maneira para facilitar a solu=E7=E3o:
>
> S_n =3D 1.(10^1 - 1)/9 +2.(10^2 - 1)/9 +............+n.(10^n - 1)/9
>
> S_n =3D 1/9.[ *(1.10^1 +2.10^2+.......+n.10^n)* - (1+2+3+.......+n=
)]
>
> Esta parte que eat=E1 em negrito =E9 : S=E9rie aritm=E9tico - geom=E9tr=
ica. Voc=EA
> aplica a sguinte f=F3rmula:
>
> S_n=3D[ a_1(1 - q^n)/1- q] + rq[1 - nq^(n - 1) +(n - 1).q^n]/=
(1
> - q)^2
>
> obs:a_0=3D0 , a_1=3D1 e q=3D10
>
> Portanto,
>
> *S_n=3D 1/9 {10/81( 1+9n.10^n - 10^n) - [n(n+1)]/2}*
>
> Testei com n=3D1,2,3 e deu certo
>
>
> ----- Original Message -----
> *From:* saulo nilson <saulo.nilson@xxxxxxxxx>
> *To:* obm-l@xxxxxxxxxxxxxx
> *Sent:* Tuesday, April 08, 2008 11:26 PM
> *Subject:* Re: [obm-l] Soma !!!
>
> (1+n)n/2+(2+n)(n-1)/2+(3+n)(n-3)/2,,,
> soma(k+n)(n-(k-1))/2=3D1/2soma(n^2-k^2)+n+k=3D
> =3D1/2(n^3+n^2+(1+n)n/2-n(n+1)(2n+1)/6=3D
> =3D3n(n+1)(6n+3-(2n+1))=3D12n(n+1)^2
>
> 2008/4/8 Pedro J=FAnior <pedromatematico06@xxxxxxxxx>:
>
> > Engalhei na seguinte soma:
> >
> > J=E1 usei aquele exerc=EDcio do livro do Lidisk, mas aquela soma =E9 de=
1 + 11
> > + 111 + ... + (111...1), onde (111...1) tem exatamente n d=EDgitos, mas=
mesmo
> > assim ainda n=E3o saiu!
> >
> >
> > S_n =3D 1 + 22 + 333 + 4444 + ... + n ( 111...1)
> >
> >
> > onde (111...1) tem exatamente n d=EDgitos.
> >
> > Desde J=E1 agrade=E7o!!!
> >
>
>
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Eita mund=E3o da matem=E1tica...<br><br>Rapaz 1=AA vez que vi esta f=F3rmul=
a, nossa, mas faz sentido claro...<br><br>vou verificar valeu mesmo, s=F3 u=
ma perguntinha, onde vc encontrou essa quest=E3o mesmo?<br><br>pois encontr=
ei numa lista de exerc=EDcio por a=ED, e coloquei na minha por=E9m n=E3o ha=
via resolvido antes.<br>
<br>resultado nome da quest=E3o: <font color=3D"#cc0000"><font size=3D"4">U=
M PROFESSOR EM APUROS!!!<br><br>KKKKKKKKKKK<br><br><font color=3D"#000000">=
<font size=3D"2">Bom, agrade=E7o bastante a colabora=E7=E3o e vou apicar in=
du=E7=E3o afim de verificar se vale para todo n.<br>
<br>abra=E7os<br><br>E a caminhada continua!<br></font></font></font></font=
><br><div class=3D"gmail_quote">2001/11/1 Pedro <<a href=3D"mailto:npc19=
72@xxxxxxxxx">npc1972@xxxxxxxxx</a>>:<br><blockquote class=3D"gmail_quot=
e" style=3D"border-left: 1px solid rgb(204, 204, 204); margin: 0pt 0pt 0pt =
0.8ex; padding-left: 1ex;">
<div bgcolor=3D"#ffffff">
<div><font face=3D"Arial" size=3D"2">Essa quest=E3o deu muito trabalho =E0 =
tres semana, mais=20
no fim deu certo.</font></div>
<div><font face=3D"Arial" size=3D"2"></font> </div>
<div><font face=3D"Arial" size=3D"2"> Seja S_n =3D 1.11^0 + 2.11^1 +3=
.11^2=20
+...........+n.11111111111 rescrever de uma maneira para facilitar a=20
solu=E7=E3o:</font></div>
<div><font face=3D"Arial" size=3D"2"></font> </div>
<div><font face=3D"Arial" size=3D"2"> S_n =3D 1.(10=
^1 - 1)/9=20
+2.(10^2 - 1)/9 +............+n.(10^n - 1)/9</font></div>
<div><font face=3D"Arial" size=3D"2"></font> </div>
<div><font face=3D"Arial" size=3D"2"> S=
_n =3D 1/9.[=20
<b>(1.10^1 +2.10^2+.......+n.10^n)</b> -=20
(1+2+3+.......+n)]</font></div>
<div><font face=3D"Arial" size=3D"2"></font> </div>
<div><font face=3D"Arial" size=3D"2"> Esta parte que eat=E1 em negrit=
o =E9 : S=E9rie=20
aritm=E9tico - geom=E9trica. Voc=EA aplica a sguinte f=F3rmula:</font></div=
>
<div><font face=3D"Arial" size=3D"2"></font> </div>
<div><font face=3D"Arial" size=3D"2"> &n=
bsp; S_n=3D[ a_1(1 -=20
q^n)/1- q] + rq[1 - nq^(n - 1) +(n - 1).q^n]/(1 - q)^2</font></=
div>
<div><font face=3D"Arial" size=3D"2"> </font></div>
<div><font face=3D"Arial" size=3D"2"> &n=
bsp; =20
obs:a_0=3D0 , a_1=3D1 e q=3D10</font></div>
<div><font face=3D"Arial" size=3D"2"></font> </div>
<div><font face=3D"Arial" size=3D"2"> Portanto,</font></div>
<div><font face=3D"Arial" size=3D"2"></font> </div>
<div><font face=3D"Arial" size=3D"2"> <b>S_n=3D 1/9 {10/81( 1+9n.10^n=
-=20
10^n) - [n(n+1)]/2}</b></font></div>
<div><font face=3D"Arial" size=3D"2"></font> </div>
<div><font face=3D"Arial" size=3D"2"> Testei =
com n=3D1,2,3 e=20
deu certo </font></div><div><div></div><div class=3D"Wj3C7c">
<div><font face=3D"Arial" size=3D"2"> &n=
bsp; =20
</font></div>
<blockquote style=3D"border-left: 2px solid rgb(0, 0, 0); padding-right: 0p=
x; padding-left: 5px; margin-left: 5px; margin-right: 0px;">
<div style=3D"font-family: arial; font-style: normal; font-variant: norma=
l; font-weight: normal; font-size: 10pt; line-height: normal; font-size-adj=
ust: none; font-stretch: normal;">----- Original Message ----- </div>
<div style=3D"background: rgb(228, 228, 228) none repeat scroll 0% 50%; -=
moz-background-clip: -moz-initial; -moz-background-origin: -moz-initial; -m=
oz-background-inline-policy: -moz-initial; font-family: arial; font-style: =
normal; font-variant: normal; font-weight: normal; font-size: 10pt; line-he=
ight: normal; font-size-adjust: none; font-stretch: normal;">
<b>From:</b>=20
<a title=3D"saulo.nilson@xxxxxxxxx" href=3D"mailto:saulo.nilson@xxxxxxxxx=
" target=3D"_blank">saulo=20
nilson</a> </div>
<div style=3D"font-family: arial; font-style: normal; font-variant: norma=
l; font-weight: normal; font-size: 10pt; line-height: normal; font-size-adj=
ust: none; font-stretch: normal;"><b>To:</b> <a title=3D"obm-l@xxxxxxxxxxxx=
br" href=3D"mailto:obm-l@xxxxxxxxxxxxxx"; target=3D"_blank">obm-l@xxxxxxxxxx=
o.br</a> </div>
<div style=3D"font-family: arial; font-style: normal; font-variant: norma=
l; font-weight: normal; font-size: 10pt; line-height: normal; font-size-adj=
ust: none; font-stretch: normal;"><b>Sent:</b> Tuesday, April 08, 2008 11:2=
6=20
PM</div>
<div style=3D"font-family: arial; font-style: normal; font-variant: norma=
l; font-weight: normal; font-size: 10pt; line-height: normal; font-size-adj=
ust: none; font-stretch: normal;"><b>Subject:</b> Re: [obm-l] Soma !!!</div=
>
<div><br></div>
<div>(1+n)n/2+(2+n)(n-1)/2+(3+n)(n-3)/2,,,</div>
<div>soma(k+n)(n-(k-1))/2=3D1/2soma(n^2-k^2)+n+k=3D</div>
<div>=3D1/2(n^3+n^2+(1+n)n/2-n(n+1)(2n+1)/6=3D</div>
<div>=3D3n(n+1)(6n+3-(2n+1))=3D12n(n+1)^2<br><br></div>
<div class=3D"gmail_quote">2008/4/8 Pedro J=FAnior <<a href=3D"mailto:=
pedromatematico06@xxxxxxxxx" target=3D"_blank">pedromatematico06@xxxxxxxxx<=
/a>>:<br>
<blockquote class=3D"gmail_quote" style=3D"border-left: 1px solid rgb(204=
, 204, 204); margin: 0px 0px 0px 0.8ex; padding-left: 1ex;">Engalhei=20
na seguinte soma:<br><br>J=E1 usei aquele exerc=EDcio do livro do Lidis=
k, mas=20
aquela soma =E9 de 1 + 11 + 111 + ... + (111...1), onde (111...1) tem=
=20
exatamente n d=EDgitos, mas mesmo assim ainda n=E3o saiu!<br><br><br>S_=
n =20
=3D 1 + 22 + 333 + 4444 + ... + n ( 111...1)<br><br><br>onde (111=
...1)=20
tem exatamente n d=EDgitos.<br><br>Desde J=E1=20
agrade=E7o!!!<br></blockquote></div><br></blockquote></div></div></div>
</blockquote></div><br>
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