Question

[tex]\text{Aflati cate numere n}\in \mathbb{N}^*\ \text{verifica inegalitatea:}\\
\frac{1}{\sqrt[3]{4}+\sqrt[3]{2}+1}+\frac{1}{\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4}}+_\dots}+\frac{1}{\sqrt[3]{(n+1)^2}+\sqrt[3]{n(n+1)}+\sqrt[3]{n^2}}\ \textless \ \sqrt{3}[/tex]

Answer 1

$\displaystyle Se~tine~cont~de~identitatea~a^3-b^3=(a-b)(a^2+ab+b^2). \\ \\ Astfel,~prin~amplificarea~fractiei~\frac{1}{ \sqrt[3]{(k+1)^2}+ \sqrt[3]{k(k+1)}+ \sqrt[3]{k^2}} \\ \\ cu~ \sqrt[3]{k+1}- \sqrt[3]{k},~obtinem: \\ \\ \frac{1}{ \sqrt[3]{(k+1)^2}+ \sqrt[3]{k(k+1)}+ \sqrt[3]{k^2}}= \frac{ \sqrt[3]{k+1} - \sqrt[3]{k} }{(k+1)-k}=\sqrt[3]{k+1} - \sqrt[3]{k}.$

$\displaystyle Inecuatia~devine: \\ \\ \sqrt[3]{2}- 1+\sqrt[3]{3}- \sqrt[3]{2} + \sqrt[3]{4}- \sqrt[3]{3}+...+ \sqrt[3]{n+1}- \sqrt[3]{n}\ \textless \ \sqrt{3} \Leftrightarrow \\ \\ \Leftrightarrow \sqrt[3]{n+1}-1\ \textless \ \sqrt{3} \Leftrightarrow \sqrt[3]{n+1}\ \textless \ \sqrt{3}+1. \\ \\ Deci~n+1\ \textless \ ( \sqrt{3}+1)^3 \\ \\ n+1\ \textless \ 10+6 \sqrt{3} \\ \\ n\ \textless \ 9+6 \sqrt 3.$

$\displaystyle Insa~n \in \mathbb{N^*},~deci~n \leq [9+6 \sqrt{3}]. \\ \\ 1,7\ \textless \ \sqrt{3}\ \textless \ 1,75 \Rightarrow 10,2\ \textless \ 6 \sqrt{3}\ \textless \ 10,5 \Rightarrow 19,2\ \textless \ 9+ 6 \sqrt{3}\ \textless \ 19,5 \Rightarrow \\ \\ \Rightarrow [9+6 \sqrt{3}]=19. \\ \\ Deci~n \in \{1,2,3,...,19\}.$

Answer 2

a³  - b³ = (a - b)( a² + ab + b²)


[tex]\it a = \sqrt[3]{n+1}, \ \ b = \sqrt[3]n \\\;\\ (\sqrt[3]{n+1})^3 - (\sqrt[3]n )^3 = (\sqrt[3]{n+1} - \sqrt[3]n )(\sqrt[3]{(n+1)^2}+\sqrt[3]{n(n+1)}+\sqrt[3]{n^2}) \\\;\\ \Rightarrow n+1-n = (\sqrt[3]{n+1} - \sqrt[3]n )(\sqrt[3]{(n+1)^2}+\sqrt[3]{n(n+1)}+\sqrt[3]{n^2}) \Rightarrow \\\;\\ \Rightarrow 1 = (\sqrt[3]{n+1} - \sqrt[3]n )(\sqrt[3]{(n+1)^2}+\sqrt[3]{n(n+1)}+\sqrt[3]{n^2}) \Rightarrow [/tex]

$\it \Rightarrow\dfrac{1} {\sqrt[3]{(n+1)^2}+\sqrt[3]{n(n+1)}+\sqrt[3]{n^2}} = \sqrt[3]{(n+1)} - \sqrt[3]n$

Particularizând, pentru fiecare termen și reducând termenii opuși,

inegalitatea dată devine:

$\it \sqrt[3]{n+1} -1 \ \textless \ \sqrt3$

[tex]\it (\sqrt[3]{n+1})^3 \ \textless \ (\sqrt3+1)^3\Leftrightarrow n+1 \ \textless \ 3\sqrt3+9+3\sqrt3+1\Leftrightarrow \\\;\\ \Leftrightarrow n -9\ \textless \ 6\sqrt3 \Leftrightarrow (n -9)^2\ \textless \ (6\sqrt3)^2 \Leftrightarrow (n -9)^2\ \textless \ 108 \Leftrightarrow \\\;\\ \Leftrightarrow n - 9 \leq [\sqrt{108}] \Leftrightarrow n-9 \leq 10 \Leftrightarrow n\leq19[/tex]

Deoarece n este natural nenul, rezultă că există 19 numere naturale care verifică relația dată.