次の条件によって定められる数列 $\{a_n\}$ の第2項から第5項を求めよ。 (1) $a_1 = 100, a_{n+1} = a_n - 5$ (2) $a_1 = 2, a_{n+1} = 3a_n$ (3) $a_1 = 2, a_{n+1} = 3a_n + 2$ (4) $a_1 = 1, a_{n+1} = a_n + n$

1. 問題の内容

次の条件によって定められる数列 $\{a_n\}$ の第2項から第5項を求めよ。

(1) $a_1 = 100, a_{n+1} = a_n - 5$

(2) $a_1 = 2, a_{n+1} = 3a_n$

(3) $a_1 = 2, a_{n+1} = 3a_n + 2$

(4) $a_1 = 1, a_{n+1} = a_n + n$

2. 解き方の手順

(1) $a_1 = 100, a_{n+1} = a_n - 5$

$a_2 = a_1 - 5 = 100 - 5 = 95$

$a_3 = a_2 - 5 = 95 - 5 = 90$

$a_4 = a_3 - 5 = 90 - 5 = 85$

$a_5 = a_4 - 5 = 85 - 5 = 80$

(2) $a_1 = 2, a_{n+1} = 3a_n$

$a_2 = 3a_1 = 3 \times 2 = 6$

$a_3 = 3a_2 = 3 \times 6 = 18$

$a_4 = 3a_3 = 3 \times 18 = 54$

$a_5 = 3a_4 = 3 \times 54 = 162$

(3) $a_1 = 2, a_{n+1} = 3a_n + 2$

$a_2 = 3a_1 + 2 = 3 \times 2 + 2 = 6 + 2 = 8$

$a_3 = 3a_2 + 2 = 3 \times 8 + 2 = 24 + 2 = 26$

$a_4 = 3a_3 + 2 = 3 \times 26 + 2 = 78 + 2 = 80$

$a_5 = 3a_4 + 2 = 3 \times 80 + 2 = 240 + 2 = 242$

(4) $a_1 = 1, a_{n+1} = a_n + n$

$a_2 = a_1 + 1 = 1 + 1 = 2$

$a_3 = a_2 + 2 = 2 + 2 = 4$

$a_4 = a_3 + 3 = 4 + 3 = 7$

$a_5 = a_4 + 4 = 7 + 4 = 11$

3. 最終的な答え

(1) $a_2 = 95, a_3 = 90, a_4 = 85, a_5 = 80$

(2) $a_2 = 6, a_3 = 18, a_4 = 54, a_5 = 162$

(3) $a_2 = 8, a_3 = 26, a_4 = 80, a_5 = 242$

(4) $a_2 = 2, a_3 = 4, a_4 = 7, a_5 = 11$