14. (2019 - 四川高考模拟 (文) ) 已知数列 $\left\{a_{n}\right\}_{\text {的前 }} n_{\text {项和为 }} S_{n}$, 且满足 $2 S_{n}=-a_{n}+n\left(n \in N^{*}\right)$.

(I) 求证: 数列 $\left\{a_{n}-\frac{1}{2}\right\}$ 为等比数列;

(II) 求数列 $\left\{a_{n}-1\right\}$ 的前 $n$ 项和 $T_{n}$.
( I ) $2 S_{n}=-a_{n}+n$,

当 $n \geq 2$ 时, $2 S_{n-1}=-a_{n-1}+n-1$,

两式相减, 得 $2 a_{n}=-a_{n}+a_{n-1}+1$, 即 $a_{n}=\frac{1}{3} a_{n-1}+\frac{1}{3}$.

$\therefore a_{n}-\frac{1}{2}=\frac{1}{3}\left(a_{n-1}-\frac{1}{2}\right)$, 所以数列 $\left\{a_{n}-\frac{1}{2}\right\}$ 为等比数列.

(II) 由 $2 S_{1}=-a_{1}+1$, 得 $a_{1}=\frac{1}{3}$. 由 (I) 知, 数列 $\left\{a_{n}-\frac{1}{2}\right\}$ 是以 $-\frac{1}{6}$ 为首项, $\frac{1}{3}$ 为公比的等比数列.

所以 $a_{n}-\frac{1}{2}=-\frac{1}{6}\left(\frac{1}{3}\right)^{n-1}=-\frac{1}{2}\left(\frac{1}{3}\right)^{n}$,

$\therefore a_{n}=-\frac{1}{2}\left(\frac{1}{3}\right)^{n}+\frac{1}{2}$,
$\therefore a_{n}-1=-\frac{1}{2}\left(\frac{1}{3}\right)^{n}-\frac{1}{2}$,

$\therefore T_{n}=\frac{-\frac{1}{6}\left[1-\left(\frac{1}{3}\right)^{n}\right]}{1-\frac{1}{3}}-\frac{n}{2}=\frac{1}{4}\left[\left(\frac{1}{3}\right)^{n}-1\right]-\frac{n}{2}$.