已知数列an的前n项为Sn,且a1=4/3,(4^n-1)an=3(4^(n-1))Sn,求Sn的通项式和当bn=n/3
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已知数列an的前n项为Sn,且a1=4/3,(4^n-1)an=3(4^(n-1))Sn,求Sn的通项式和当bn=n/3an时bn的前n项和Tn
![已知数列an的前n项为Sn,且a1=4/3,(4^n-1)an=3(4^(n-1))Sn,求Sn的通项式和当bn=n/3](/uploads/image/z/2812824-0-4.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E4%B8%BASn%2C%E4%B8%94a1%3D4%2F3%2C%284%5En-1%29an%3D3%284%5E%28n-1%29%29Sn%2C%E6%B1%82Sn%E7%9A%84%E9%80%9A%E9%A1%B9%E5%BC%8F%E5%92%8C%E5%BD%93bn%3Dn%2F3)
(4^n-1)an=3(4^(n-1))Sn,即(4^n-1)(Sn-S(n-1))=3(4^(n-1))Sn,计算后得Sn=(4^n-1)S(n-1)/(4^(n-1))将S(n-1)类似的展开可得Sn=(4^n-1)/3S1=((4^n-1)/3)(4/3),寝室快熄灯了,后面的题目就没时间写出来了,望谅解!
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