Răspunsuri

2014-11-16T21:01:16+02:00
1+3+5+7+9+11+13+15+17+19+21+23+25+27+29+31+33+35+37+39+41
2014-11-16T21:02:41+02:00
( 1+2+3+4+...+41) - ( 2+4+6+..+40)=
 \frac{41 ori 42}{2} - 2· ( 1+2+3+...+20)=
 = \frac{41 ori 42}{2} - 2·  \frac{20 ori 21}{2}  

de aici cred ca te descurci :)
Msss
nu mai calculez dar iti spun formula n=ultimul nr... in cazul tau 41... n(n+1) supra 2