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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 $$\sum_{x=1}^{30}f(x).$$

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A rather straightforward method would be to compute it with the aid of, say, a computer. – akkkk Jun 19 '12 at 17:01

Since

$$\frac{a^2+8a+10}{a+9}=a-1+\frac{11}{a+9}\;,$$

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 $a>2$.

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+1 Simply beautiful.. and simple – DonAntonio Jun 20 '12 at 2:12

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