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Mesaj |
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Postati aici intrebari legate de problemele din aceasta varianta.
--- Pitagora,
Pro-Didactician
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[Citat] Postati aici intrebari legate de problemele din aceasta varianta. |
m-ati putea ajuta la exercitiul 2c).multumesc anticipat
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f(x)=(1+x+x^2)^1004=a0+a1x+a2x^2+a3x^3+a4x^4+...+a2008x^2008
f(1)=a0+a1+a2+a3+a4+a5...+a2008
-f(1)=-a0+a1-a2+a3-a4+a5...-a2008
-if(i)=-ia0+a1+ia2-a3-ia4+a5...+ia2008
if(-i)=ia0+a1-ia2-a3+ia4+a5...-ia2008
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f(1)-f(1)-if(i)+if(-i)=4a1+4a5+...+4a2005
(3^1004 - 1)/4=a1+a5+...+a2005
f(1)=a0+a1+a2+a3+a4+a5...+a2008
-f(1)=-a0+a1-a2+a3-a4+a5...-a2008
if(i)=ia0-a1-ia2+a3+ia4-a5...-ia2008
-if(-i)=-ia0-a1+ia2+a3-ia4-a5...+ia2008
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f(1)-f(1)-if(i)+if(-i)=4a3+4a7+...+4a2007
(3^1004 -1)/4=a3+a7+...+a2007
=>a1+a5+...+a2005=a3+a7+...+a2007
f(1)=3^1004 , f(-1)=1, f(i)=1, f(-i)=1
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[Citat] f(x)=(1+x+x^2)^1004=a0+a1x+a2x^2+a3x^3+a4x^4+...+a2008x^2008
f(1)=a0+a1+a2+a3+a4+a5...+a2008
-f(1)=-a0+a1-a2+a3-a4+a5...-a2008
-if(i)=-ia0+a1+ia2-a3-ia4+a5...+ia2008
if(-i)=ia0+a1-ia2-a3+ia4+a5...-ia2008
_____________________________________________________________________________________+
f(1)-f(1)-if(i)+if(-i)=4a1+4a5+...+4a2005
(3^1004 - 1)/4=a1+a5+...+a2005
f(1)=a0+a1+a2+a3+a4+a5...+a2008
-f(1)=-a0+a1-a2+a3-a4+a5...-a2008
if(i)=ia0-a1-ia2+a3+ia4-a5...-ia2008
-if(-i)=-ia0-a1+ia2+a3-ia4-a5...+ia2008
_____________________________________________________________________________________+
f(1)-f(1)-if(i)+if(-i)=4a3+4a7+...+4a2007
(3^1004 -1)/4=a3+a7+...+a2007
=>a1+a5+...+a2005=a3+a7+...+a2007
f(1)=3^1004 , f(-1)=1, f(i)=1, f(-i)=1 |
eu tot nu am inteles!!! nu cumva s-a modificat aceasta varianta???
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punctul 2c s-a modificat
nu mai sunt probleme cu nr.complexe
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