8. Calculate:
$$
\sum_{k=1}^{2014}\left[\frac{-3+\sqrt{8 k+1}}{4}\right]=
$$
$\qquad$
where, $[x]$ denotes the greatest integer not exceeding the real number $x$.
8. 40115 .

Let $t=\frac{-3+\sqrt{8 k+1}}{4}$. Then
$$
k=2 t^{2}+3 t+1 \text {. }
$$

Therefore, when and only when
$$
\begin{array}{l}
2 n^{2}+3 n+1 \leqslant k \\
<2(n+1)^{2}+3(n+1)+1(n \in \mathbf{N})
\end{array}
$$

$\left[\frac{-3+\sqrt{8 k+1}}{4}\right]=n$.
By $2 \times 30^{2}+3 \times 30+1=1891$, $2 \times 31^{2}+3 \times 31+1=2016$,

we know
$$
\begin{array}{l}
\sum_{k=1}^{2014}\left[\frac{-3+\sqrt{8 k+1}}{4}\right] \\
=\sum_{k=1}^{30} n\left[2(n+1)^{2}+3(n+1)+1-\right. \\
\left.\left(2 n^{2}+3 n+1\right)\right]-30 \\
=\sum_{k=1}^{30}\left(4 n^{2}+5 n\right)-30=40115 . \\
\end{array}
$$