与えられた漸化式で定義される数列 $\{a_n\}$ の第2項から第5項までを求める問題です。4つの異なる漸化式が与えられています。代数学数列漸化式2025/4/301. 問題の内容与えられた漸化式で定義される数列 {an}\{a_n\}{an} の第2項から第5項までを求める問題です。4つの異なる漸化式が与えられています。2. 解き方の手順(1) a1=100a_1 = 100a1=100, an+1=an−5a_{n+1} = a_n - 5an+1=an−5* a2=a1−5=100−5=95a_2 = a_1 - 5 = 100 - 5 = 95a2=a1−5=100−5=95* a3=a2−5=95−5=90a_3 = a_2 - 5 = 95 - 5 = 90a3=a2−5=95−5=90* a4=a3−5=90−5=85a_4 = a_3 - 5 = 90 - 5 = 85a4=a3−5=90−5=85* a5=a4−5=85−5=80a_5 = a_4 - 5 = 85 - 5 = 80a5=a4−5=85−5=80(2) a1=2a_1 = 2a1=2, an+1=3ana_{n+1} = 3a_nan+1=3an* a2=3a1=3⋅2=6a_2 = 3a_1 = 3 \cdot 2 = 6a2=3a1=3⋅2=6* a3=3a2=3⋅6=18a_3 = 3a_2 = 3 \cdot 6 = 18a3=3a2=3⋅6=18* a4=3a3=3⋅18=54a_4 = 3a_3 = 3 \cdot 18 = 54a4=3a3=3⋅18=54* a5=3a4=3⋅54=162a_5 = 3a_4 = 3 \cdot 54 = 162a5=3a4=3⋅54=162(3) a1=2a_1 = 2a1=2, an+1=3an+2a_{n+1} = 3a_n + 2an+1=3an+2* a2=3a1+2=3⋅2+2=8a_2 = 3a_1 + 2 = 3 \cdot 2 + 2 = 8a2=3a1+2=3⋅2+2=8* a3=3a2+2=3⋅8+2=26a_3 = 3a_2 + 2 = 3 \cdot 8 + 2 = 26a3=3a2+2=3⋅8+2=26* a4=3a3+2=3⋅26+2=80a_4 = 3a_3 + 2 = 3 \cdot 26 + 2 = 80a4=3a3+2=3⋅26+2=80* a5=3a4+2=3⋅80+2=242a_5 = 3a_4 + 2 = 3 \cdot 80 + 2 = 242a5=3a4+2=3⋅80+2=242(4) a1=1a_1 = 1a1=1, an+1=an+na_{n+1} = a_n + nan+1=an+n* a2=a1+1=1+1=2a_2 = a_1 + 1 = 1 + 1 = 2a2=a1+1=1+1=2* a3=a2+2=2+2=4a_3 = a_2 + 2 = 2 + 2 = 4a3=a2+2=2+2=4* a4=a3+3=4+3=7a_4 = a_3 + 3 = 4 + 3 = 7a4=a3+3=4+3=7* a5=a4+4=7+4=11a_5 = a_4 + 4 = 7 + 4 = 11a5=a4+4=7+4=113. 最終的な答え(1) a2=95a_2 = 95a2=95, a3=90a_3 = 90a3=90, a4=85a_4 = 85a4=85, a5=80a_5 = 80a5=80(2) a2=6a_2 = 6a2=6, a3=18a_3 = 18a3=18, a4=54a_4 = 54a4=54, a5=162a_5 = 162a5=162(3) a2=8a_2 = 8a2=8, a3=26a_3 = 26a3=26, a4=80a_4 = 80a4=80, a5=242a_5 = 242a5=242(4) a2=2a_2 = 2a2=2, a3=4a_3 = 4a3=4, a4=7a_4 = 7a4=7, a5=11a_5 = 11a5=11