（本小题满分 14 分） 设数列 a_ n 的前 n 项和…——2012 高考数学第 19 题答案解析

【解答】
（1）在 $2 S_{n}=a_{n+1}-2^{n+1}+1$ 中

令 $n=1$ 得： $2 S_{1}=a_{2}-2^{2}+1$

令 $n=2$ 得： $2 S_{2}=a_{3}-2^{3}+1$

解得：$a_{2}=2 a_{1}+3, a_{3}=6 a_{1}+13$

又 $2\left(a_{2}+5\right)=a_{1}+a_{3}$

解得 $a_{1}=1$
②由 $2 S_{n}=a_{n+1}-2^{n+1}+1$
$2 S_{n+1}=a_{n+2}-2^{n+2}+1$ 得

$$ a_{n+2}=3 a_{n+1}+2^{n+1} $$

又 $a_{1}=1, a_{2}=5$ 也满足 $a_{2}=3 a_{1}+2^{1}$
所以 $a_{n+1}=3 a_{n}+2^{n}$ 对 $n \in N^{*}$ 成立

$$ \begin{aligned} & \therefore a_{n+1}+2^{n+1}=3\left(a_{n}+2^{n}\right) \\ & \therefore a_{n}+2^{n}=3^{n} \\ & \therefore a_{n}=3^{n}-2^{n} \end{aligned} $$

③

$$ \begin{aligned} (\text { 法一) } & \because a_{n}=3^{n}-2^{n}=(3-2)\left(3^{n-1}+3^{n-2} \times 2+3^{n-3} \times 2^{2}+\ldots+2^{n-1}\right) \geq 3^{n-1} \\ & \therefore \frac{1}{a_{n}} \leq \frac{1}{3^{n-1}} \end{aligned} $$

$\left(\right.$ 法二）$\because a_{n+1}=3^{n+1}-2^{n+1}>2 \times 3^{n}-2^{n+1}=2 a_{n}$

$$ \therefore \frac{1}{a_{n+1}}<\frac{1}{2} \cdot \frac{1}{a_{n}} $$

当 $n \geq 2$ 时，$\frac{1}{a_{3}}<\frac{1}{2} \cdot \frac{1}{a_{2}}$

$$ \begin{aligned} & \frac{1}{a_{4}}<\frac{1}{2} \cdot \frac{1}{a_{3}} \\ & \frac{1}{a_{5}}<\frac{1}{2} \cdot \frac{1}{a_{4}} \end{aligned} $$

$$ \frac{1}{a_{n}}<\frac{1}{2} \cdot \frac{1}{a_{n-1}} $$

累乘得：$\frac{1}{a_{n}}<\left(\frac{1}{2}\right)^{n-2} \cdot \frac{1}{a_{2}}$

$\therefore \frac{1}{a_{1}}+\frac{1}{a_{2}}+\frac{1}{a_{3}}+\ldots \frac{1}{a_{n}} \leq 1+\frac{1}{5}+\frac{1}{2} \times \frac{1}{5}+\ldots+\left(\frac{1}{2}\right)^{n-2} \times \frac{1}{5}<\frac{7}{5}<\frac{3}{2}$