6. From the infinite sequence $\left\{\frac{1}{2^{n-1}}\right\}$, sequentially extract an infinite geometric sequence such that the sum of all its terms $S$ satisfies $\frac{4}{61}<S<\frac{1}{13}$. The number of such geometric sequences is ( ) .
(A)1
(B) 2
(C) 3
(D) infinitely many
6.( B ).

Let the first term of the required sequence be $\frac{1}{2^{m}}$, and the common ratio be $\frac{1}{2^{n}}$, then it must be that $m \geqslant 0, n \geqslant 1$ and $m, n \in \mathbf{Z}$.
Then $S=\frac{\frac{1}{2^{m}}}{1-\frac{1}{2^{n}}}=\frac{1}{2^{m}-2^{m-n}}$.
$$
\begin{array}{c}
\because \frac{4}{61}13 \text { and } 2^{m-1}<15.25 .
\end{array}
$$

Therefore, $m=4$, substituting we get $n=3$ or 4.
Hence, there are two sequences that satisfy the conditions.