Problem 1. Let the sequence $\left(a_{n}\right)_{n \in N^{*}}$ be defined by $a_{1}=\sqrt[4]{2015}, a_{n+1}=\sqrt[4]{2015+\sqrt[n+1]{a_{n}}}$ for any $n \geq 1$. Calculate $\left[a_{1}\right]+\left[a_{2}\right]+\ldots+\left[a_{2015}\right]$, where $[x]$ is the integer part of the real number $x$.
Solution. $6^{4}=1296<2015<2401=7^{4} \Leftrightarrow 6<\sqrt[4]{2015}<7 \Rightarrow\left[a_{1}\right]=6$

$$
\begin{aligned}
& 6<a_{1}<7 \Rightarrow 2<\sqrt{6}<\sqrt{a_{1}}<\sqrt{7}<3 \Rightarrow 6<\sqrt[4]{2017}<\sqrt[4]{2015+\sqrt{a_{1}}}<\sqrt[4]{2018}<7 \Rightarrow \\
& \quad \Rightarrow\left[a_{2}\right]=6
\end{aligned}
$$

Assume that $\left[a_{k}\right]=6$ and we prove that $\left[a_{k+1}\right]=6$ for any $k \geq 2$.

$$
\begin{aligned}
& 6<a_{k}<7 \Rightarrow 1<\sqrt[k+1]{6}<\sqrt[k+1]{a_{k}}<\sqrt[k+1]{7}<2 \Rightarrow \sqrt[4]{2016}<\sqrt[4]{2015+\sqrt[k+1]{a_{k}}}<\sqrt[4]{2017} \Rightarrow \\
& \Rightarrow\left[\sqrt[4]{2015+\sqrt[k+1]{a_{k}}}\right]=6 \Leftrightarrow\left[a_{k+1}\right]=6 \Rightarrow\left[a_{n}\right]=6,(\forall) n \geq 1 \\
& {\left[a_{1}\right]+\left[a_{2}\right]+\ldots+\left[a_{2015}\right]=6 \cdot 2015=12090}
\end{aligned}
$$