6. Let $a_{1}=6, a_{n+1}=\left[\frac{5}{4} a_{n}+\frac{3}{4} \sqrt{a_{n}^{2}-2}\right]$ $\left(n \in \mathbf{N}_{+}\right)$, where $[x]$ denotes the greatest integer not exceeding $x$. Then the units digit of $a_{1}+a_{2}+\cdots+a_{2000}$ is ( ).
(A) 1
(B) 2
(C) 3
(D) 4
6.B.

Given $a_{1}=6=5 \times 2^{1-1}+1, a_{2}=11=5 \times 2^{2-1}+1, \cdots \cdots$, we conjecture: $a_{n}=5 \times 2^{n-1}+1$.

From the known recurrence relation, it is easy to prove by mathematical induction (omitted).
Thus, when $n>1$, $a_{n} \equiv 1(\bmod 10)$.
Therefore, $a_{1}+a_{2}+\cdots+a_{2007} \equiv 6+2006 \equiv 2(\bmod 10)$.