5. Let the sequence $\left\{a_{n}\right\}$ satisfy: $a_{1}=6, a_{n+1}=\left[\frac{5}{4} a_{n}+\frac{3}{4} \sqrt{a_{n}^{2}-2}\right], n \in \mathbf{N}^{*}$, where $[x]$ denotes the greatest integer not exceeding the real number $x$, and $S_{n}$ is the sum of the first $n$ terms of $\left\{a_{n}\right\}$. Then the unit digit of $S_{2016}$ is
A. 1
B. 2
C. 5
D. 6
For $a_{n} \geqslant 6$, let $0<\delta=\frac{3}{4} a_{n}-\frac{3}{4} \sqrt{a_{n}^{2}-2}=\frac{3}{2} \cdot \frac{1}{a_{n}+\sqrt{a_{n}^{2}-2}}<0.13$ $\Rightarrow a_{n+1}=\left[\frac{5}{4} a_{n}+\frac{3}{4} a_{n}-\delta\right]=2 a_{n}-1$.
Thus, $a_{n+1}-1=2\left(a_{n}-1\right) \Rightarrow a_{n}=5 \cdot 2^{n-1}+1$, hence $S_{2016}=5 \cdot 2^{2016}+2011$. Therefore, the unit digit of $S_{2016}$ is 1, so the answer is $A$.