题目内容
(本小题10分)设等比数列
的各项均为正值,首项
,前n项和为
,且
(1)求
的通项;(2)求
的前n项和




(1)求



(1)
(2)
。


本试题主要是考查了等比数列的前n项和与通项公式的运用。利用数列的前n项和以及通项公式的表示得到数列的首项和公比的值,得到第一问,然后由于
,则
的前n项和
,利用等差数列和等比数列分组求和,得到结论。
(1)由
得
即
可得
因为
>0,所以
解得
因而
(2)
故
的前n项和
两式相减得
即



(1)由


即

可得



解得


(2)

故



两式相减得


即


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