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 $\sum_{x=1}^{30}f(x).$
Summation of a sequence
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sequences-and-series
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2A rather straightforward method would be to compute it with the aid of, say, a computer. – 2012-06-19
1 Answers
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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 $a>2$.