14. 已知数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=1, a_{n+1}=2 a_{n}+1$,

(1)求证: 数列 $\left\{a_{n}+1\right\}$ 是等比数列;

(2)求 $a_{n}$ 的表达式.
(1)证明 $\because a_{n+1}=2 a_{n}+1$,

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

$\therefore \frac{a_{n+1}+1}{a_{n}+1}=2$.

$\therefore\left\{a_{n}+1\right\}$ 是等比数列, 公比为 2 , 首项为 2 .

(2)解由(1)知 $\left\{a_{n}+1\right\}$ 是等比数列.

公比为 2 , 首项 $a_{1}+1=2$.

$\therefore a_{n}+1=\left(a_{1}+1\right) \cdot 2^{n^{-1}}=2^{n}$.

$\therefore a_{n}=2^{n}-1$.哈哈