## Subject IV. (15 points)

Let $a_{k}=\sqrt{\left(k^{2}-k-2\right)\left(k^{2}+7 k+10\right)+36}, k \in \mathbb{N}^{*}$. Determine $n \in \mathbb{N}^{*}$ for which:

$\left[\sqrt{a_{1}}\right]+\left[\sqrt{a_{2}}\right]+\left[\sqrt{a_{3}}\right]+\cdots+\left[\sqrt{a_{n}}\right]=490$, where $[a]$ represents the integer part of the real number $a$.

Prof. Violin Gorcea, Avram Iancu Theoretical High School, Cluj-Napoca

## Grading Scale for 9th Grade (OLM 2015 - Local Stage)

## Official $10 p$

## Translation of the text into English, preserving the original line breaks and formatting, as requested.
## Subject IV. (15 points)

$a_{k}=\sqrt{\left(k^{2}-k-2\right)\left(k^{2}+7 k+10\right)+36}=\sqrt{(k-2)(k+1)(k+2)(k+5)+36}=$

$=\sqrt{\left(k^{2}+3 k-10\right)\left(k^{2}+3 k+2\right)+36}$

Let $k^{2}+3 k-10=t \Rightarrow a_{t}=\sqrt{t(t+12)+36}=t+6$

Thus, $a_{k}=k^{2}+3 k-4$

For $k=1$ we have $\left[\sqrt{a_{1}}\right]=0$, for $k \in\{2,3,4\}$ we have $\left[\sqrt{a_{k}}\right]=k$.

For $k \geq 5 \Rightarrow k+1 \leq \sqrt{k^{2}+3 k-4}<k+2 \Rightarrow\left[\sqrt{a_{k}}\right]=k+1$.

We obtain $2+3+4+6+7+\cdots+n+1=490 \Leftrightarrow(n+1)(n+2)=992 \Rightarrow n=30$