1. For the geometric sequence $a_{n}=\frac{1}{3^{n-1}}$, the sum of the first $n$ terms $S_{n}=\sum_{k=1}^{n} a_{k}, S_{k+1}$ and $S_{k}$ do not satisfy the recursive relationship ( ).
A. $S_{k+1}=S_{k}+\frac{1}{3^{k+1}}$
B. $S_{k+1}=1+\frac{1}{3} S_{k}$
C. $S_{k+1}=S_{k}+a_{k+1}$
D. $S_{k+1}=3 S_{k}-3+a_{k}+a_{k+1}$
1. A. Since $a_{k+1}=\frac{1}{3^{k}}$, thus $S_{k+1}=a_{k+1}+S_{k}=S_{k}+\frac{1}{3^{k}} \neq S_{k}+\frac{1}{3^{k+1}}$.