第2章 2.19

(1)

$a_1=1 \quad a_2=1 \quad a_3=2 \quad a_4=3 \quad a_5=5 \quad a_6=8 \quad a_7=13$

$a_8=21 \quad a_9=34 \quad a_{10}=55$

(2)

$n=1$のとき

$a_1 = \frac{1}{\sqrt5}\left\{ \frac12(1+\sqrt5) - \frac12(1-\sqrt5) \right\} = \frac{1}{\sqrt5}\cdot\sqrt5 = 1 \qquad \text{よって成立する}$

$n=k$のとき与式が成り立つと仮定すると

$a_k = \frac{1}{\sqrt5}(\alpha^k - \beta^k) \qquad \text{両辺に}a_{k+1}\text{を加えると}$

$a_k + a_{k+1} = \frac{1}{\sqrt5}(\alpha^k-\beta^k) + a_{k+1} = \frac{1}{\sqrt5}(\alpha^k - \beta^k + \alpha^{k+1} - \beta^{k+1})$

$= \frac{1}{\sqrt5}\left\{ \left(\frac12\right)^k (1+\sqrt5)^k - \left(\frac12\right)^k(1-\sqrt5)^k + \left(\frac12\right)^{k+1}(1+\sqrt5)^{k+1} - \left(\frac12\right)^{k+1}(1-\sqrt5)^{k+1} \right\}$

$= \frac{1}{\sqrt5}\cdot\left(\frac12\right)^k \left\{ (1+\sqrt5)^k - (1-\sqrt5)^k + \frac12\cdot(1+\sqrt5)^{k+1} - \frac12\cdot(1-\sqrt5)^{k+1} \right\}$

$= \frac{1}{\sqrt5}\cdot\left(\frac12\right)^k \left\{ (1+\sqrt5)^k\left(1+\frac12\cdot(1+\sqrt5)\right) - (1-\sqrt5)^k\left(1-\frac12\cdot(1-\sqrt5)\right) \right\}$

$= \frac{1}{\sqrt5}\cdot\left(\frac12\right)^k \left\{ (1+\sqrt5)^k\left(\frac32+\frac{\sqrt5}{2}\right) - (1-\sqrt5)^k\left(\frac32-\frac{\sqrt5}{2}\right) \right\}$

$= \frac{1}{\sqrt5}\cdot\left(\frac12\right)^{k+2} \left\{ (1+\sqrt5)^k(6+2\sqrt5) - (1-\sqrt5)^k(6-2\sqrt5) \right\}$

$= \frac{1}{\sqrt5}\cdot\left(\frac12\right)^{k+2} \left\{ (1+\sqrt5)^k(1+\sqrt5)^2 - (1-\sqrt5)^k(1-\sqrt5)^2 \right\}$

$= \frac{1}{\sqrt5}\cdot\left(\frac12\right)^{k+2} \left\{ (1+\sqrt5)^{k+2} - (1-\sqrt5)^{k+2} \right\} = \frac{1}{\sqrt5}(\alpha^{k+2} - \beta^{k+2})$

よって$n=k+1$のときにおいても成立する。

これから全ての自然数で与式は成立する。

(3)

$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{\dfrac{1}{\sqrt{5}}(\alpha^{n+1} - \beta^{n+1})}{\dfrac{1}{\sqrt{5}}(\alpha^{n} - \beta^{n})}$

$= \lim_{n \to \infty} \frac{\left(\dfrac{1}{2}\right)^{n+1}(1+\sqrt{5})^{n+1} - \left(\dfrac{1}{2}\right)^{n+1}(1-\sqrt{5})^{n+1}}{\left(\dfrac{1}{2}\right)^{n}(1+\sqrt{5})^{n} - \left(\dfrac{1}{2}\right)^{n}(1-\sqrt{5})^{n}} \qquad 1+\sqrt{5} = A、\ 1-\sqrt{5} = B \text{とおくと}$

$= \lim_{n \to \infty} \frac{\dfrac{1}{2}A^{n+1} - \dfrac{1}{2}B^{n+1}}{A^{n} - B^{n}} = \lim_{n \to \infty} \frac{\dfrac{1}{2}A - \dfrac{1}{2}\cdot B \cdot \left(\dfrac{B}{A}\right)^{n}}{1 - \left(\dfrac{B}{A}\right)^{n}}$

ここで $\displaystyle \lim_{n \to \infty}\left(\frac{B}{A}\right)^{n} = 0$ なので

$= \lim_{n \to \infty} \frac{\dfrac{1}{2}\cdot A - 0}{1 - 0} = \frac{1}{2}\cdot A = \frac{1+\sqrt{5}}{2}$