3. Given the sequence $\left\{a_{n}\right\}$ satisfies $3 a_{n+1}+a_{n}=4(n \geqslant 1)$, and $a_{1}=9$, the sum of the first $n$ terms is $S_{n}$, then the smallest integer $n$ that satisfies the inequality $\left|S_{n}-n-6\right|<\frac{1}{125}$ is
(A) 5 ;
(B) 6 ;
(C) 7 ;
(D) 8 .
3. (C)

From $3\left(a_{n+1}-1\right)=-\left(a_{n}-1\right), a_{1}-1=8$, we know that the sequence $\left\{a_{n}-1\right\}$ is a geometric sequence with the first term 8 and common ratio $-\frac{1}{3}$. Therefore,
$$
\begin{array}{l}
S_{n}-n=\frac{8 \times\left[1-\left(1-\frac{1}{3}\right)^{n}\right]}{1+\frac{1}{3}}=6-6 \times\left(-\frac{1}{3}\right)^{n} . \\
\therefore \quad\left|S_{n}-n-6\right|=6 \times\left(\frac{1}{3}\right)^{n}250>3^{5}, \quad \therefore n-1>5, n>6$, thus $n \geqslant 7$. It is not difficult to verify that $n=7$ meets the requirement, hence the answer is (C).
$$