Question

vă rog să mă ajutați este urgent dau coroană ))​

Answer 1

Explicație pas cu pas:

a)

$\frac{1}{8} < \frac{1}{3} \leqslant \frac{1}{3} \\ \frac{1}{8} < \frac{1}{4} < \frac{1}{3} \\\frac{1}{8} < \frac{1}{5} < \frac{1}{3} \\\frac{1}{8} < \frac{1}{6} < \frac{1}{3} \\\frac{1}{8} < \frac{1}{7} < \frac{1}{3} \\\frac{1}{8} \leqslant \frac{1}{8} < \frac{1}{3}$

adunăm, pe verticală:

$\frac{6}{8} < \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} < \frac{6}{3}$

$\frac{3}{4} < \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} < 2$

b)

$\frac{1}{27} < \frac{1}{10} \leqslant \frac{1}{10} \\ \frac{1}{27} < \frac{1}{11} < \frac{1}{10} \\\frac{1}{27} < \frac{1}{12} < \frac{1}{10} \\ ... \\\frac{1}{27} < \frac{1}{26} < \frac{1}{10} \\\frac{1}{27} \leqslant \frac{1}{27} < \frac{1}{10}$

adunăm, pe verticală:

$\frac{18}{27} < \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + ... + \frac{1}{27} < \frac{18}{10}$

$\frac{2}{3} < \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + ... + \frac{1}{27} < \frac{9}{5}$