Rainbowdashask
a fost răspuns

Cum se rezolva? 1+2+3+...+2012=?
101+102+103+...+200=?

Răspuns :

Miky93ask
S=1+2+3+4+.....+2012

S=(1+2012)*2012/2

S=2013 *1006

S=2025078

S=101+102+103+......+200

S=(101+200)*100/2

S=301*50

S=15050
Sunt sume Gauss:
1+2+.....+n=n(n+1)/2
1+2+.....+2012= 2012(2012+1)/2=1006·2013
101+102+...+200=100·100+1+2+...+100=10000+100(100+1)/2=10000+50·101=10000+5050=15050
Spor la lucru!