# 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?

• E.g. $[\pi] = 3$ while $[-\pi ] = -4$.
– xbh
Commented Jun 19, 2019 at 6:57

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.

$$[x]$$ is the largest integer which does not exceed $$x$$.

For example, $$[12.4]=12$$, $$[10.995]=10$$, $$[7]=7$$, $$[-2.3]=-3$$.

For this problem $$\displaystyle a_n=\left[\frac{(n-1)(n+9)+19}{n+9}\right]=n-1+\left[\frac{19}{n+9}\right]$$.

For $$n=1,2,\dots, 10$$, $$a_n=n$$.

For $$n=11,12,\dots, 30$$, $$a_n=n-1$$.

$$\sum\limits_{n=1}^{30} a_n =\sum\limits_{n=1}^{30}(n-1)+ \sum\limits_{n=1}^{30} [\frac{19}{n+9}]=$$

$$(1+2+3+...+29) + \sum _1^{10} (1) =435+10=445$$