Question

Va rog !!! Dau coroana !!!
Calculați :​

Answer 1

$\sqrt{n^{2} + n + 1} < \sqrt{(n^{2} + n + 1) + n} = \sqrt{n^{2} + 2n + 1} = \sqrt{(n + 1)^{2} } = n + 1$

$\sqrt{n^{2} + n + 1} > \sqrt{n^{2}} = n$

deci:

$n < \sqrt{n^{2} + n + 1} < n + 1 \ \ \Big| - 1$

$n - 1 < \sqrt{n^{2} + n + 1} - 1 < n$

$S_{n} = \sqrt{n^{2} + n + 1} - 1$

$\iff n - 1 < S_{n} < n$

$\implies \Big[S_{n}\Big] = n - 1$

q.e.d.