Question

Se considera sirul (aₙ), aₙ = ln ( 1 + $\frac{2}{n}$ ), n ≥ 1.
Sa se calculeze bₙ = a₁ + a₂ + ... + aₙ , n ≥ 1

Answer 1

Răspuns:

a1=ln(1+2/1)=ln(1+2)=ln3

a2=ln(1+2/2)=ln2

a3=ln(1+2/3)=ln5/3

a4=ln(1+2/4)=ln(1+1/2)=ln3/2

aa5=ln(1+2/5)=ln7/5

......................

an=ln(1+2/n)

bn=ln3+ln2+ln(5/3+ln3/2+ln7/5+...+ln(1+2/n)=

ln3*2*5/3*3/2*7/5*....*ln(1+2/n)=

ln6**(1+2/n)

Explicație pas cu pas: