a) [(1+2011)*1006]/2 =2012*1006/2=1006*1006=1006 la a doua
Formula e primul plus ultimul , totul inmultit cu nr termenilor , totul supra 2
b)2005+2(1+2+3+...+2004)=2005+2[2004(2004+1)]/2=2005+2(2004/2005)/2=2005+2004*2005=2005(2004+1)=2005*2005=2005 la a doua
doiurile s-au simplificat
c)1996+2(1+2+3+...+1995)=1996+2[1995(1995+1)]2=1996+1995*1996=1996(1995+1)=1996*1996=1996 la a doua
d)2(11+12+13+...+249)=2[(11+249)*239]/2=260*249=64740
Ultima nu stiu sigur daca e corecta.Restul se fac tot la fel
