Question

calculati urmatoarele sume a) S= 1+2+3+4...+89 b)S=3+6+9+12...+402, c) S= 3+5+7...+1001-2-4-6-...-1000vreau urgent

Answer 1

formula lui gauss S=[n(n+1)]:2
a)S=(89*90):2=4005

b)se da factor comun 3
3(1+2+3+.....+134)=3×(134*135):2=27135

c)
S=3+5+7+........+1001-(2+4+6+......+1000)
S=S1+S2
S1=3+5+7+........+1001-progresie aritmetica
an=a1+(n-1)*r
1001=3+(n-1)*2
1001=3+2n-2
1000=2n
n=500
S1=[(a1+an)*n]:2
S1=(3+1001)*500:2
S1=251000
-2-4-6-...-1000=-2(1+2+3+.......500)=-2(501*500):2=- 501*502=-251502

S1=251000
S2=-251502\
S=S1-S2=251000-251502
S=- 502