6. 记 S_{n} 为等比数列 \left\{a_{n}\right\} 的前 n 项和. 若 a_{5}-a_{3}=12, a_{6}-a_{4}=24, 则 \frac{S_{n}}{a_{n}}=(\quad)

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Answer: B. 2-2^{1-n} 【详解】设等比数列的公比为 q, 由 a_{5}-a_{3}=12, a_{6}-a_{4}=24 可得: \left\{\begin{array}{l}a_{1} q^{4}-a_{1} q^{2}=12 \ a_{1} q^{5}-a_{1} q^{3}=24\end{array} \Rightarrow\left\{\begin{array}{l}q=2 \ a_{1}=1\end{array}ight.ight., 所以 a_{n}=a_{1} q^{n-1}=2^{n-1}, S_{n}=\frac{a_{1}\left(1-q^{n}ight)}{1-q}=\frac{1-2^{n}}{1-2}=2^{n}-1, 因此 \frac{S_{n}}{a_{n}}=\frac{2^{n}-1}{2^{n-1}}=2-2^{1-n}. 故选: B.

6. 记 $S_{n}$ 为等比数列 $\left\{a_{n}\right\}$ 的前 $n$ 项和. 若 $a_{5}-a_{3}=12, a_{6}-a_{4}=24$, 则 $\frac{S_{n}}{a_{n}}=(\quad)$

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