Question

17 va rog

Mulțumesc!

Answer 1

a) (1+2+3+...+100)/2(1+2+3+...+100)=1/2

am simplificat parantezele

b) (1+2+3+...+100)/[100×100+(1+2+3+...+100)=

5050/15050=20100

1+2+3+...-100=100×101/2=5050

c) (1+3+5+.. 99)/2(1+2+3+...50)=49²/(2×50×51/2)=

2401/2550

99=2n+1

2n=98

n=49

1+3+5+...+99=49²