Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $f(a)$ be the sequence defined by $$f(a)=\left[\frac{a^2+8a+10}{a+9}\right]$$ where $[x]$ is the largest integer that does not exceed $x$.

Find the value of $$\sum_{x=1}^{30}f(x).$$

share|cite|improve this question
Welcome to math.SE. Since you are new, I want to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many users find the use of the imperative ("Find", "Show", etc) to be rude when asking for help. Please consider rewriting your post. – Arturo Magidin Jun 19 '12 at 16:58
A rather straightforward method would be to compute it with the aid of, say, a computer. – akkkk Jun 19 '12 at 17:01



we have

$$\left\lfloor\frac{a^2+8a+10}{a+9}\right\rfloor=a-1+\left\lfloor\frac{11}{a+9}\right\rfloor$$ whenever $a$ is an integer. Thus,

$$\begin{align*} \sum_{a=1}^{30}\frac{a^2+8a+10}{a+9}&=\sum_{a=1}^{30}\left(a-1+\left\lfloor\frac{11}{a+9}\right\rfloor\right)\\\\ &=\sum_{a=1}^{29}a+\sum_{a=1}^{30}\left\lfloor\frac{11}{a+9}\right\rfloor\\\\ &=\frac12(29)(30)+2\\\\ &=437\;, \end{align*}$$

since $\dfrac{11}{a+9}<1$ for $a>2$.

share|cite|improve this answer
+1 Simply beautiful.. and simple – DonAntonio Jun 20 '12 at 2:12

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.