<s>[INST] If $a_n=\left[\frac{n^2+8n+10}{n+9}\right]$, find $ \sum_{n=1}^{30} a_n$ I am working on a problem, assuming high school math knowledge.

Let ${a_n}$ be the sequence defined by $$a_n=\left[\frac{n^2+8n+10}{n+9}\right]\,,$$ where $[x]$ denotes the largest integer which does not exceed $x$. Find the value of $ \sum\limits_{n=1}^{30} a_n$.

I honestly do not understand the text in bold. The answer provided is $445$. Could you please explain what I should be looking at here?
 [/INST] Hint
\begin{align*}
\frac{n^2+8n+10}{n+9}&=\frac{n^2+9n-n-9+19}{n+9}\\
&=n-1+\frac{19}{n+9}\\
\therefore \left\lfloor\frac{n^2+8n+10}{n+9}\right\rfloor&=n-1+\left\lfloor \frac{19}{n+9}\right\rfloor
\end{align*}
As $n$ varies from $1$ to $30$, we will have
\begin{align*}
\left\lfloor\frac{19}{\color{red}{1}+9}\right\rfloor&=\lfloor 1.9\rfloor=1\\
\left\lfloor\frac{19}{\color{red}{2}+9}\right\rfloor&=\lfloor 1.7\rfloor=1\\
\vdots & = \vdots\\
\left\lfloor\frac{19}{\color{red}{11}+9}\right\rfloor&=\lfloor 0.95\rfloor=0\\
\vdots & = \vdots\\
\end{align*}
Hopefully you can take it from here.
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