指数型の漸化式なので両辺4^(n +1)で割ると
an +1/4^(n +1)＝(3/4)an/4^n +(1/4)より
bn＝an/4^nとおくと
bn +1＝(3/4)bn +1/4となって特性型の漸化式に帰着できます
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![基本例題の[2]についてです。ここの条件ですが、
a/3<=1<=4a/3とありますが4a... - 1](https://d1e9oo257tadp1.cloudfront.net/uploads/qa_question_image/file/1810404/thumb_l_webp_70236300-B387-4A8D-A246-76C58E626EEE.webp?Expires=1787684624&Signature=KyaSQMZeAEM2-1WRvHSQsSvqIhvHGlnRDxHVCHNJjjlX8hnZB7m1UXLcrO9cWDUAULkOl0r3Ss05sSeQeLVoPYmv~M56X5y52NOViwT0CRlDb58TO4cCONAHs1~s8p7-YAvOGq5AVdfg7Ekm922Z2AdprgyZAmsp8A8mi5GaINVnGh99L91wwmLCLVqCX5CJEwpwROXnOLNKqCfGRv9m7~xjzLS8Gal3YaD3zOJuTZtqdJEgT99tQdFD5-WTRLOMBdUlgr1ckt4KdC4e4OifXSxzRKDUJfW1iJkUmmETUsCoN8rR8KquNrkACOU5PnI438IuEW7rw6a4ZNB3m8S48Q__&Key-Pair-Id=K2W722D70GJS8W&Expires=1787684624&Signature=MWrwOhNtMfQh3m0rvAshR4OwfRpOhOOsvXfT1YcNym4m24-M9qQMmEG2Qv20NhLUJTiXhZ9X0DstuDq0xkEQtE6AYKbQ5iPcwrABsMvIoRe4gn5XzCQn4I7q7h~b5U980XwgTjjE9vdLpGdLrDnxSAnJgAJUAsp63vQf5js9U8Yf7E-wEsecjxC7EdQlBEHpAF5lFmiWdu3zU6r85rlusFVczoJ41Amxrydu222garBIBP1J-XW57CQNT0dHtCSIWO7GJ6ZiWQh5B-bxddNEWV1whk6o7B-04K3JICU0mpbk~QkrROSjYm74C~3RmxWfljiF9E464~~441Gb-A0K3A__&Key-Pair-Id=K2W722D70GJS8W)
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