Question

(4+8+12+16+......8000):(2+4+6+8+....4000)+(2 la puterea 53+ 2 la puterea 52+2 la puterea51):(2 la puterea 17) totul la a-3-a ?

Answer 1

4(1+2+3+4+...+2000):2(1+2+3+4+...+2000)+(2⁵³+2⁵²+2⁵¹):(2¹⁷)³=
Termenii in paranteza se simplifica si ramane banala impartire 4:2
Din a doua parte a exercitiului dam factor comun, care se va simplifica cu 2⁵¹
2+2⁵¹(2²+2+1):2⁵¹=
2+7=
9

Answer 2

(4+8+12+16+......8000)=4(1+2+3+4+....+2000)=4(2000+1)(2000)/2=
2x2000x2001

(2+4+6+8+....4000)=2(1+2+3+4+....+2000)=2(2000+1)(2000)/2=
2000x2001

(2 la puterea 53+ 2 la puterea 52+2 la puterea51)= 2^51(2^2+2+1)=
(2^51)x7

(2^17)^3=2^51

E=(2x2000x2001):(2000x2001) +7x(2^51):(2^51)= 
E=2+7=9