(2013 大纲) 已知数列 $\left\{a_{n}\right\}$ 满足 $3 a_{n+1}+a_{n}=0, a_{2}=-\frac{4}{3}$, 则 $\left\{a_{n}\right\}$ 的前 10 项和等于 ()
 $\because a_{n+1}=-\frac{1}{3} a_{n}, \therefore\left\{a_{n}\right\}$ 是等比数列

又 $a_{2}=-\frac{4}{3}, \therefore a_{1}=4, \therefore S_{10}=\frac{4\left(1-\left(-\frac{1}{3}\right)^{10}\right)}{1+\frac{1}{3}}=3\left(1-3^{-10}\right)$, 故选 C.