Question

Ma ajuta cineva pls
A) 1+2+3+......+100
B)1+2+3+....+1001
C)1+2+3+........+560
D)3+6+9+.....303
E)5+10+15+.....305

Answer 1

Folosim suma lui gasu
n(n+1):2
n=ultimul numar

a)
1+2+3+..+100
n(n+1):2
100(100+1):2
100*101:2
5,050

b)
1+2+3+..+1001
n(n+1):2
1001*(1001+1):2
1001*1002:2
501,501

c)
1+2+3+...+560
n(n+1):2
560(560+1):
560*561:2
157,080
d)
3+6+9+..+303
n(n+1):2
303(303+1):2
303*304:2
46,056

e)
5+10+15+,....+305
n(n+1):2
305(305+1):2
305*306:2
46,665

Answer 2

A) 1+2+3+...+100=101*100:2=101*50=5050
B) 1+2+3+...+1001=1001*1000:2=1001*500=500500
C) 1+2+3+...+560=561*560:2=561*280=157080
D) 3+6+9+...+303=3(1+2+3+...+101)=3(101*102:2)=3(101*51)=3*5151=15453
E) 5+10+15+...305=5(1+2+3+...+61)=5(61*62:2)=5(61*31)=5*1891=9455
Folosesti suma lui Gauss:n*(n+1):2 
n este numarul si n+1 succesorul lui n