第1章 1.8

(1)

$a = \dfrac{1}{12}$ $r = 3$なので

$a_n = ar^{n-1} = \frac{1}{12}3^{n-1} = \frac{1}{4}3^{n-2} \qquad S_n = \frac{\dfrac{1}{12}(3^n-1)}{3-1} = \frac{3^n-1}{24}$

(2)

$a = 1$ $r = \dfrac{1}{2}$なので

$a_n = \frac{1}{2^{n-1}} \qquad S_n = \frac{\left(\dfrac{1}{2^n}-1\right)}{\dfrac{1}{2}-1} = \frac{\dfrac{1-2^n}{2^n}}{-\dfrac{1}{2}} = \frac{2^n-1}{2^{n-1}}$

(3)

$a = 2$ $r = -2$なので

$a_n = 2(-2)^{n-1} = (-1)^{n-1}2^n \qquad S_n = \frac{2\{1-(-2)^n\}}{1-(-2)} = \frac{2}{3}\{1-(-2)^n\}$

(4)

$a = 2$ $r = \dfrac{1}{\sqrt{2}}$なので

$a_n = 2\left(\frac{1}{\sqrt{2}}\right)^{n-1} = \frac{2}{\sqrt{2}^{\,n-1}}$

$S_n = \frac{2\left(1-\dfrac{1}{\sqrt{2}^{\,n}}\right)}{1-\dfrac{1}{\sqrt{2}}} = \frac{2\dfrac{\sqrt{2}^{\,n}-1}{\sqrt{2}^{\,n}}}{\dfrac{\sqrt{2}-1}{\sqrt{2}}} = \frac{\sqrt{2}^{\,n}-1}{\sqrt{2}^{\,n-3}(\sqrt{2}-1)} = \frac{(\sqrt{2}^{\,n}-1)(\sqrt{2}+1)}{\sqrt{2}^{\,n-3}} = \frac{(2+\sqrt{2})(\sqrt{2}^{\,n}-1)}{\sqrt{2}^{\,n-2}}$