【迎14高考】每日一题 数列1

castelu

设数列$\{a_n\}$的前$n$项和为$S_n$，已知$a_1=1$，$S_{n+1}=4a_n+2$
（1）设$b_n=a_{n+1}-2a_n$，证明数列$\{b_n\}$是等比数列
（2）求数列$\{a_n\}$的通项公式

castelu

$$S_{n+1}=4a_n+2，a_1=1$$
$$a_1+a_2=4a_1+2，a_2=5$$
$$a_{n+1}=S_{n+1}-S_n=4a_n+2-(4a_{n-1}+2)$$
$$a_{n+1}=4a_n-4a_{n-1}$$
$$a_{n+1}-2a_n=2(a_n-2a_{n-1})$$
$$b_n=a_{n+1}-2a_n，b_1=a_2-2a_1=3$$
$$\frac{b_n}{b_{n-1}}=2，b_n=3 \cdot 2^{n-1}$$
$$\left\{ \begin{array}{l}
a_{n+1}-2a_n=3 \cdot 2^{n-1}\\
2a_n - 2^2a_{n-1}=3 \cdot 2^{n-1}\\
\cdots \\
2^{n-1}a_2-2^na_1=3 \cdot 2^{n-1}
\end{array} \right.\\$$
$${a_{n + 1}} - {2^n}{a_1} = 3n{2^{n - 1}}$$
$${a_{n + 1}} = 3n{2^{n - 1}} + {2^n}$$
$${a_n} = 3(n - 1){2^{n - 2}} + {2^{n - 1}},n \in {N^*}$$