Răspuns :

Miky93id
1+3+5+7+.....+2005 < n < 2(1+2+3+...+1003)

S1 < n < S2

n=(2005-1):2+1
n=2004 :2+1
n=1002 +1
n=1003  -nr de termeni 

(1+2005)*1003 /2 < n < 2 (1+1003)*1003/2

2006 *1003/2 < n < 1004 *1003

1003 * 1003 < n < 1004 *1003 

1006009 < n < 1007012  =>n={1,2,3,4...,1003}
Uvgguvgugguvguid
1+3+5+7+....+2005<n<2+4+6+...+2006

S=1+3+5+7+....+2005                            (2005-1):2+1=2004:2+1=1002+1=1003 (de nr.)
S=2005+2003+2001+1998+....+1  (le-am pus pe dos)
2S=2006+2006+2006+2006+....+2006  (le-am adunat pe fiecare coloana)
2S=2006*1003 | :2
S=1003*1003 
S=1006009

D=2+4+6+...+2006                                 (2006-2):2+1=2004:2+1=1002+1=1003 (de nr.)
D=2006+2004+2002+...+2
D=2008+2008+2008+....+2008
2D=2008*1003 | :2
D=1004*1003
D=1007012
(astea doua au fost sume Gauss)

1006009<n<1007012
n∈{1006010,1006011,....,1007011}

Sper ca te-am ajutat

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