15. Let $[x]$ denote the greatest integer less than or equal to $x$, e.g. $[\pi]=3,[5.31]=5$ and $[2009]=2009$. Evaluate $\left[\sqrt{2009^{2}+1}+\sqrt{2009^{2}+2}+\cdots+\sqrt{2009^{2}+4018}\right]$.
(2 marks)
設 $[x]$ 表示小於或等於 $x$ 的最大整數，例如： $[\pi]=3 、[5.31]=5$ 、
$[2009]=2009$ 。求 $\left[\sqrt{2009^{2}+1}+\sqrt{2009^{2}+2}+\cdots+\sqrt{2009^{2}+4018}\right]$ 的値。
15. 8074171
15. Note that $\sqrt{2009^{2}+m}-2009=\frac{m}{\sqrt{2009^{2}+m}+2009}$. For $1 \leq m \leq 4018$, we have
$$
\frac{m}{4019}<\frac{m}{\sqrt{2009^{2}+m}+2009}<\frac{m}{4018} .
$$

Hence each summand is equal to 2009 plus a 'decimal part'. Summing up all the 'decimal parts', we have
$$
2009=\sum_{m=1}^{4018} \frac{m}{4019}<\left(\sum_{m=1}^{4018} \sqrt{2009^{2}+m}\right)-2009 \times 4018<\sum_{m=1}^{4018} \frac{m}{4018}=2009 \frac{1}{2}
$$
and so the answer is $2009 \times 4018+2009=8074171$.