Question

Doar de punctul b am nevoie​

Answer 1

Răspuns:

Explicație pas cu pas:

Ne folosim de punctul a

Astfel vom avea

$\frac{1}{1\cdot 6} =\frac{1}{5} (\frac{1}{1}-\frac{1}{6})$

$\frac{1}{6\cdot 11} =\frac{1}{5} (\frac{1}{6}-\frac{1}{11})$

...

$\frac{1}{(5n+1)\cdot (5n+6)} =\frac{1}{5} (\frac{1}{5n+1}-\frac{1}{5n+6})$

Inlocuim in suma noastra si vom avea

$\frac{1}{5} (\frac{1}{1} -\frac{1}{6} )+\frac{1}{5} (\frac{1}{6} -\frac{1}{11} )+...+\frac{1}{5} (\frac{1}{5n+1}-\frac{1}{5n+6}) =\\\frac{1}{5} (\frac{1}{1} -\frac{1}{6}+\frac{1}{6} -\frac{1}{11} +...+\frac{1}{5n+1}-\frac{1}{5n+6})=\frac{1}{5}( \frac{1}{1} -\frac{1}{5n+6})$

$\frac{1}{5}( \frac{1}{1} -\frac{1}{5n+6})=\frac{1}{5} \cdot \frac{5n+5}{5n+6}$

$\frac{5n+5}{5n+6}<1$ este fractie subunitara⇒$\frac{1}{5}\cdot \frac{5n+5}{5n+6}<\frac{1}{5}$