5. In the sequence $\left\{a_{n}\right\}$, $a_{1}=5, S_{n-1}=a_{n}\left(n \geqslant 2, n \in \mathbf{N}_{+}\right)$, the general term formula for $S_{n}$ is ( ).
A. $5 \cdot 2^{n}$
B. $5 \cdot 2^{n}-5$
C. $5 \cdot 2^{n-1}$
D. $5 \cdot 2^{n+1}$
5. C.

From $S_{n-1}=a_{n}$, we can get $S_{n-1}=S_{n}-S_{n-1}$, which means $S_{n}=2 S_{n-1}(n \geqslant 2)$. Therefore, $S_{n}=S_{1} \cdot 2^{n-1}$, and since $S_{1}=a_{1}=5$, we have $S_{n}=5 \cdot 2^{n-1}$.