5. Let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=6, \\
a_{n+1}=\left[\frac{5}{4} a_{n}+\frac{3}{4} \sqrt{a_{n}^{2}-2}\right]\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
$$

If $S_{n}$ is the sum of the first $n$ terms of $\left\{a_{n}\right\}$, then the units digit of $S_{2016}$ is ( ).
(A) 1
(B) 2
(C) 5
(D) 6
5. A.

Given $a_{1}=6, a_{2}=11, a_{3}=21$, we conjecture:
$$
a_{n}=5 \times 2^{n-1}+1\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

This can be easily proven by mathematical induction.
Thus, for $n \geqslant 2$, $a_{n} \equiv 1(\bmod 10)$.
Therefore, $S_{2016} \equiv 6+2015 \equiv 1(\bmod 10)$.