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[SPAM] Re: [obm-l] Soma !!!
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This is a multi-part message in MIME format.
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charset="iso-8859-1"
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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 =
rescrever 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=E9trica. 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
=20
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=20
=20
----- Original Message -----=20
From: saulo nilson=20
To: obm-l@xxxxxxxxxxxxxx=20
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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charset="iso-8859-1"
Content-Transfer-Encoding: quoted-printable
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META http-equiv=3DContent-Type content=3D"text/html; =
charset=3Diso-8859-1">
<META content=3D"MSHTML 6.00.2900.2180" name=3DGENERATOR>
<STYLE></STYLE>
</HEAD>
<BODY bgColor=3D#ffffff>
<DIV><FONT face=3DArial size=3D2>Essa quest=E3o deu muito trabalho =E0 =
tres semana, mais=20
no fim deu certo.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2> 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=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2> S_n =3D =
1.(10^1 - 1)/9=20
+2.(10^2 - 1)/9 +............+n.(10^n - 1)/9</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2> =
S_n =3D 1/9.[=20
<STRONG>(1.10^1 +2.10^2+.......+n.10^n)</STRONG> -=20
(1+2+3+.......+n)]</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2> Esta parte que eat=E1 em negrito =
=E9 : S=E9rie=20
aritm=E9tico - geom=E9trica. Voc=EA aplica a sguinte =
f=F3rmula:</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial=20
size=3D2> 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=3DArial size=3D2> </FONT></DIV>
<DIV><FONT face=3DArial=20
size=3D2> &nbs=
p; =20
obs:a_0=3D0 , a_1=3D1 e q=3D10</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2> Portanto,</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2> <STRONG>S_n=3D 1/9 {10/81( =
1+9n.10^n -=20
10^n) - [n(n+1)]/2}</STRONG></FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2> Testei =
com n=3D1,2,3 e=20
deu certo </FONT></DIV>
<DIV><FONT face=3DArial =
size=3D2> =20
</FONT></DIV>
<BLOCKQUOTE=20
style=3D"PADDING-RIGHT: 0px; PADDING-LEFT: 5px; MARGIN-LEFT: 5px; =
BORDER-LEFT: #000000 2px solid; MARGIN-RIGHT: 0px">
<DIV style=3D"FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV=20
style=3D"BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: =
black"><B>From:</B>=20
<A title=3Dsaulo.nilson@xxxxxxxxx =
href=3D"mailto:saulo.nilson@xxxxxxxxx";>saulo=20
nilson</A> </DIV>
<DIV style=3D"FONT: 10pt arial"><B>To:</B> <A =
title=3Dobm-l@xxxxxxxxxxxxxx=20
href=3D"mailto:obm-l@xxxxxxxxxxxxxx";>obm-l@xxxxxxxxxxxxxx</A> </DIV>
<DIV style=3D"FONT: 10pt arial"><B>Sent:</B> Tuesday, April 08, 2008 =
11:26=20
PM</DIV>
<DIV style=3D"FONT: 10pt arial"><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=3Dgmail_quote>2008/4/8 Pedro J=FAnior <<A=20
=
href=3D"mailto:pedromatematico06@xxxxxxxxx";>pedromatematico06@xxxxxxxxx</=
A>>:<BR>
<BLOCKQUOTE class=3Dgmail_quote=20
style=3D"PADDING-LEFT: 1ex; MARGIN: 0px 0px 0px 0.8ex; BORDER-LEFT: =
#ccc 1px solid">Engalhei=20
na seguinte soma:<BR><BR>J=E1 usei aquele exerc=EDcio do livro do =
Lidisk, 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></BODY></HTML>
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