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数列

{a_{n}}

满足

a_{n + 1} + 2 a_{n} a_{n + 1} - a_{n} = 0 (n \in N^{*}), a_{1} = 1

, 则下列结论正确的是

A.

若

b_{n} = 3^{\frac{1}{a_{n}}}

, 则

{b_{n}}

为等比数列

B.

若

c_{n} = \frac{1}{n} (\frac{1}{a_{1}} + \frac{1}{a_{2}} + \dots + \frac{1}{a_{n}})

, 则

{c_{n}}

为等差数列

C.

a_{n} = 2 n - 1

D.

\frac{1}{a_{1}} + \frac{1}{a_{2}} + \dots + \frac{1}{a_{2 n - 1}} = \frac{2 n - 1}{a_{n}}

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