Given a sequence \\{a_{n}\}\ with the sum of its first \n\ terms being \S_{n}\, \a_{1}=1\, and \a_{n+1}=2S_{n}+1\ for \n\in\mathbb{N}^{*}\.
\(1)\ Find the general term formula for the sequence \\{a_{n}\}\;
\(2)\ Let \c_{n}=\log_{3}a_{2n}\ and \b_{n}= \frac{1}{c_{n}\cdot c_{n+1}}\, denote the sum of the first \n\ terms of the sequence \\{b_{n}\}\ as \T_{n}\, and find \T_{n}\.
Solution
(1) Since \a_{n+1}=2S_{n}+1\, for \n\in\mathbb{N}^{*}\ and \n\geq 2\, we have \a_{n}=2S_{n-1}+1\.
Thus, \a_{n+1}-a_n=2a_n\ which simplifies to \a_{n+1}=3a_n\.
For \n=1\, we have \a_{2}=2a_{1}+1=3=3\times a_{1}\, which satisfies the above equation.
Therefore, the sequence \\{a_{n}\}\ is a geometric sequence.
Hence, \a_{n}=3^{n-1}\.
(2) We have \c=\log_{3}a_{2n}=\log_{3}3^{2n-1}=2n-1\.
Therefore, \b_{n}=\frac{1}{c_{n}\cdot c_{n+1}}=\frac{1}{(2n-1)(2n+1)}=\frac{1}{4}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\.
To find the sum of the first \n\ terms of the sequence \\{b_{n}\}\, we have:
\ T_{n}=\frac{1}{4}\left[1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\ldots+\frac{1}{2n-3}-\frac{1}{2n+1}+\frac{1}{2n-1}-\frac{1}{2n+3}\right] \
Notice that all terms cancel except for the first and the last two, thus the sum becomes:
\ T_{n}=\frac{1}{4}\left(1+\frac{1}{3}-\frac{1}{2n+1}-\frac{1}{2n+3}\right) \
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