Question

8．?dāng)?shù)列$1，\frac{1}{{\sqrt{2}}}，\frac{1}{{\sqrt{3}}}，\frac{1}{2}，\frac{1}{{\sqrt{5}}}，…$的通項(xiàng)公式a_n=（　�。�

• A． a_n=$\frac{1}{{\sqrt{n+1}}}$
• B． a_n=$\frac{1}{{\sqrt{n-1}}}$
• C． ${a_n}=\frac{1}{{\sqrt{n}}}$
• D． ${a_n}=\frac{1}{{\sqrt{2n-1}}}$

Answer 1

分析 根據(jù)數(shù)列項(xiàng)的規(guī)律即可得到結(jié)論．

解答 解：數(shù)列$1，\frac{1}{{\sqrt{2}}}，\frac{1}{{\sqrt{3}}}，\frac{1}{2}，\frac{1}{{\sqrt{5}}}，…$即為$\frac{1}{1}$，$\frac{1}{\sqrt{2}}$，$\frac{1}{\sqrt{3}}$，$\frac{1}{\sqrt{4}}$，$\frac{1}{\sqrt{5}}$，…，
∴通項(xiàng)公式a_n=$\frac{1}{\sqrt{n}}$，
故選：C

點(diǎn)評(píng) 本題主要考查數(shù)列通項(xiàng)公式的求法，利用條件找到項(xiàng)的規(guī)律是解決本題的關(guān)鍵．