Question

(4+8+12+16+...+4000)÷(2+4+6+8+....+2000)+(2011+37)÷2^10 URGENT CU REZOLAVARE SI TOT

Answer 1

4(1+2+3+...+1000)÷2(1+2+3+...+1000)+2048÷1024=

4×1000×1001÷2÷2×1000×1001÷2+2=

2002000÷2÷2×1000×1001÷2+2=

501.000.500.002

Answer 2

Explicație pas cu pas:

Factor comun :

(4+8+12+16+...+4000)= 4(1+2+3+...+1000)

(2+4+6+8+....+2000)=2(1+2+3+...+1000)

4(1+2+3+...+1000):2(1+2+3+...+1000)+ (2011+37)÷2^10=2+2048:2^10=2+2048:1024=2+2=4