Question

calculati : [ radical din 1 *radical din 2 ] + [ radical din 2 * radical din 3] + [radical din 3 *radical din 4 ] + [ radical din 4 * radical din 5 ]

Answer 1

Răspuns:

Explicație pas cu pas:

$k\leq\sqrt{k(k+1)}<k+1,\forall k\in\mathbb{N}\Rightarrow {[\sqrt{k(k+1)}]}=k\\\texttt{Ecuatia devine:}\\{[\sqrt{1\cdot 2}]}+{[\sqrt{2\cdot 3}]}+{[\sqrt{3\cdot 4}]}+{[\sqrt{4\cdot 5}]}=1+2+3+4=\dfrac{4\cdot 5}{2}=\boxed{10}$

Answer 2

$E = [\sqrt 1\cdot \sqrt 2]+[\sqrt 2\cdot \sqrt 3]+[\sqrt 3\cdot \sqrt 4]+[\sqrt 4\cdot \sqrt 5] \\ \\\\ \dfrac{k+(k+1)}{2}> \sqrt{k(k+1)},\,\,\,\,k\in \mathbb{N}^*\quad -\quad \text{inegalitatea mediilor} \\ \\ \dfrac{2k+1}{2} >\sqrt{k(k+1)} \\ \\ k+\dfrac{1}{2} >(\sqrt{k(k+1)}=\sqrt{k^2+k})>k \Rightarrow k<\sqrt{k(k+1)}< k+\dfrac{1}{2}\\ \\\\\Rightarrow\Big[\sqrt{k(k+1)}\Big] = k,\quad k\in \mathbb{N}^*\\ \\\Rightarrow E = 1+2+3+4 = 13+7 = \boxed{10}$