第1章 1.12

(1)

初項$a=1$、公差$d=3$なので

$a_n = a+(n+-1)d = 1+(n-1)\cdot 3 = 3n-2$

(2)

初項$a=1$、公比$r=2$なので

$a_n = 1\cdot 2^{n-1} = 2^{n-1}$

(3)

初項$a=1$、公差$d=-1$の等差数列と$2n$の和と考えると

$a_2 = (a_1-1)+2(2-1) = a_1-1+2\cdot 1$

$a_3 = (a_2-1)+2(3-1) = \{(a_1-1+2\cdot 1)-1\}+2\cdot 2 = (a_1-1-1)+2\cdot 1+2\cdot 2$

となるので

$a_n = a_1+(n-1)d+\sum_{k=1}^{n-1}2k = 1+(n-1)\cdot(-1)+2\cdot\frac{n(n+1)}{2}-2n = -n+2+n(n+1)-2n = n^2-2n+2$

(4)

$a_2 = 1+2^1 \qquad a_3 = 1+2^1+2^2 \qquad a_4 = 1+2^1+2^2+2^3$

$a_n = 1+(2^1+2^2+2^3+\cdots 2^{n-1}) = 1+\frac{2(1-2^{n-1})}{1-2} = 1-2+2^n = 2^n-1$