2. Determine $\mathrm{n} \in \mathrm{N}^{*}-\{1\}$ such that the number $a=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{n-1}+\sqrt{n}}$ is an element of the set $A=\{x \in R /|x|<\sqrt{11-6 \sqrt{2}}\}$.
2. Determine $\mathrm{n} \in \mathrm{N}^{*}-\{1\}$ such that the number $a=\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\ldots+\frac{1}{\sqrt{n-1}+\sqrt{n}}$ is an element

of

the set

$A=\{x \in R /|x|<\sqrt{11-6 \sqrt{2}}\}$.

$(7 p)$

## SOLUTION

$\sqrt{11-6 \sqrt{2}}=3-\sqrt{2}$

(2p)

$-3+\sqrt{2}<x<3-\sqrt{2}$ and since $x$ is a real number $\Rightarrow A=(-3+\sqrt{2} ; 3-\sqrt{2})$

(2p)
$a=\frac{\sqrt{2}-1}{1}+\frac{\sqrt{3}-\sqrt{2}}{1}+\ldots+\frac{\sqrt{n}-\sqrt{n-1}}{1}$

(1p)

$a=\sqrt{n}-1 \in(-3+\sqrt{2} ; 3-\sqrt{2}) \Rightarrow \sqrt{n}<4-\sqrt{2}$, but $\mathrm{n} \in \mathrm{N}^{*}-\{1\} \Rightarrow \mathrm{n} \in\{2 ; 3 ; 4 ; 5\}$

$(2 p)$