Question

exercitiu clasa a 9a

Answer 1

Răspuns:

a) $b_n\displaystyle\ge\frac{16}{5}\Rightarrow\frac{3n+1}{n}\ge\frac{16}{5}\Rightarrow n\le 5$

deci 5 termeni sunt cel puțin egali cu 16/5.

b)

$b_{n+1}-b_n=\displaystyle\frac{3n+4}{n+1}-\frac{3n+1}{n}=\frac{-1}{n(n+1} < 0$

c)

$\displaystyle\frac{b_{n+1}}{b_n}=\frac{\frac{3n+4}{n+1}}{\frac{3n+1}{n}}=\frac{3n^2+4n}{3n^2+4n+1} < 1$

Explicație pas cu pas: