Question

Va rog ajutor dau coroana ​

Answer 1

Explicație pas cu pas:

$y = \sqrt{4 {n}^{2} + n} =$

$= \sqrt{ {(2n)}^{2} + 2 \cdot 2n \cdot \dfrac{1}{4} + {\Big( \dfrac{1}{4} \Big)}^{2} - {\Big( \dfrac{1}{4} \Big)}^{2}}$

$= \sqrt{{\Big(2n + \dfrac{1}{4} \Big)}^{2} - {\Big( \dfrac{1}{4} \Big)}^{2}} < \sqrt{{\Big(2n + \dfrac{1}{4} \Big)}^{2}}$

$2n \leqslant y < 2n + \dfrac{1}{4}$

$\implies \Big[y \Big] = 2n$

$y < 2n + \dfrac{1}{4} \iff y - \Big[y \Big] < 2n + \dfrac{1}{4} - 2n$

$\implies \Big\{y \Big\} = y - \Big[y \Big] < \dfrac{1}{4}$

q.e.d.