13.65 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 .
(China High School Mathematics League, 1994)
[Solution] 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 the common ratio $-\frac{1}{3}$.
Therefore, $S_{n}-n=\frac{8 \times\left[1-\left(-\frac{1}{3}\right)^{n}\right]}{1+\frac{1}{3}}=6-6 \times\left(-\frac{1}{3}\right)^{n}$.
$\therefore \quad\left|S_{n}-n-6\right|=6 \times\left(\frac{1}{3}\right)^{n}250>3^{5}$.
Thus, $n-1>5, n>6$, hence $n \geqslant 7$.
It is not difficult to verify that $n=7$ satisfies the requirement.
Therefore, the answer is $(C)$.