e) 1+2+3+...+11
=[tex]\frac{n(n+1)}{2}=\frac{11(11+1)}{2}=\frac{11*12}{2}=11*6=66[/tex]
(am simplificat 12 cu 2)
f) 1+2+3+...+50
=[tex]\frac{n(n+1)}{2} =\frac{50(50+1)}{2}=\frac{50*51}{2}=25*51=1275[/tex]
(am simplificat 50 cu 2)
g) 2+4+6+...+100
Nr. sunt din 2 in 2, deci dam factor comun 2
=2(1+2+3+...+50)
=[tex]2(\frac{n(n+1)}{2})=2(\frac{50(50+1)}{2})=2(\frac{50*51}{2})= 2*(25*51)=2*1275=2550[/tex]
(am simplificat 50 cu 2)
h) 5+10+15+...+500
Numerele sunt din 5 in 5, deci dam factor comun 5
=5(1+2+3+...+100)
=[tex]5(\frac{n(n+1)}{2})=5(\frac{100(100+1)}{2})=5(\frac{100*101}{2})=5(50*101)=5*5050=25250[/tex]
(am simplificat 100 cu 2)
