Question

aratati ca numarul A= log2(2/1)+log2(3/2)+log2(4/3)+...+log2(8/7) este natural

Answer 1

Răspuns:

Explicație pas cu pas:

=$log_{2}{(\frac{2}{1}*\frac{3}{2}*\frac{4}{3}*...*\frac{8}{7})}=log_{2}{8}=3$ ∈N

Answer 2

Metoda I:

log₂(2/1) + log₂(3/2) + log₂(4/3) + ... + log₂(8/7) =

= (log₂2 - log₂1) + (log₂3 - log₂2) + (log₂4 - log₂3) + ... + (log₂8 - log₂7) =

= log₂2 + log₂3 + log₂4 ... + log₂8 -  log₂1 - log₂2 - log₂3 - ... - log₂7 =

= log₂8 -  log₂1 =

= 3 - 0 =

= 3 ∈ ℕ

Metoda II:

log₂(2/1) + log₂(3/2) + log₂(4/3) + ... + log₂(8/7) =

= log₂[(2/1)·(3/2)·(4/3)·...·(8/7)] =

= log₂(8/1) =

= log₂8 =

= 3 ∈ ℕ