10. Snを含む漸化式
ex) S_n = 4a_n - 6n \\
\begin{array}{l} \\
n = 1のとき\\
S_1 = a_1 = 4a_1 - 6 \\
a_1 = 2\\
a_{n+1} = S_{n+1} - S_n \\
= (4a_{n+1} -6(n+1)) - (4a_n - 6n) \\
= 4a_{n+1} -4a_n -6 \\
\therefore a_{n+1} = \frac{4}{3} a_n + 2 （特殊型）\\
\\
特殊解\alphaは、\alpha = \frac{4}{3}\alpha + 2 \rightarrow \alpha = -6 \\
a_{n+1} - 6 = \frac{4}{3}(a_n - 6) \\
a_n - 6 = (a_1 - 6) (\frac{4}{3})^{n - 1} \\
a_n = -4 (\frac{4}{3})^{n - 1} + 6 \\
\end{array} \\
11. 隣接3項型
ex) a_{n+2} - 5 a_{n+1} + 6 a_n = 0, a_1 = 4, a_2 = 1 \\
\begin{array}{l} \\
\alpha^2 -5 \alpha + 6 = 0 を解くと、\\
(\alpha - 2)(\alpha -3) = 0 \Rightarrow \alpha = 2, 3 \\
\\
\begin{cases}
a_{n+2} - 2 a_{n+1} = 3(a_{n+1} - 2 a_n) \cdots ① \\
a_{n+2} - 3 a_{n+1} = 2(a_{n+1} - 3 a_n) \cdots ② \\
\end{cases}\\
①より\\
\begin{array}{l}
a_{n+1} -2 a_n &= (a_2 -2 a_1) 3^{n-1} \\
&= -7 \cdot 3^{n-1} \cdots ①' \\
\end{array}
\\
②より\\
\begin{array}{l}
a_{n+1} -3 a_n &= (a_2 -3 a_1) 2^{n-1} \\
&= -11 \cdot 3^{n-1} \cdots ②' \\
\end{array} \\
①' - ②' より\\
a_n = 11 \cdot 3^{n-1} -7 \cdot 3^{n-1}
\end{array} \\
-
（重解の場合）
ex) a_{n+2} - 6 a_{n+1} + 9 a_n = 0, a_1 = 1, a_2 = 5 \\
\\
\begin{array}{l} \\
\alpha^2 -6 \alpha + 9 = 0 を解くと、\\
(\alpha - 3)^2 = 0 \Rightarrow \alpha = 3 \\
\\
a_{n+2} - 3 a_{n+1} = 3(a_{n+1} - 3 a_n) \\
\begin{array}{l}
a_{n+1} -3 a_n &= (a_2 -3 a_1) 3^{n-1} \\
a_{n+1} -3 a_n &= 2 \cdot 3^{n-1} \\
a_{n+1} &= 3 a_n + \frac{2}{3} \cdot 3^n （指数型）3^{n+1}で両辺を割る\\
\frac{a_{n+1}}{3^{n+1}} &= \frac{a_n}{3^n} + \frac{2}{9} \\
\frac{a_n}{3^n} &= \frac{a_1}{3^1} + \frac{2}{9}(n - 1) \\
\frac{a_n}{3^n} &= \frac{1}{3} + \frac{2}{9}(n - 1) \\
\frac{a_n}{3^n} &= \frac{2}{9}n + \frac{1}{9} \\
a_n &= 3^n (\frac{2}{9}n + \frac{1}{9}) \\
a_n &= 3^{n-2} (2n + 1) \\
\end{array} \\
\end{array}
12.連立漸化式
ex)
\begin{cases}
a_{n+1} = 4a_n - b_n \\
b_{n+1} = 2a_n + b_n a_1 = 4, b_1 = 1\\
\end{cases}
\begin{array}{l} \\
\end{array} \\