1. Given the sequence $\left\{a_{n}\right\}$, the sum of the first $n$ terms $S_{n}=n^{2}-1$ $\left(n \in \mathbf{N}_{+}\right)$. Then $a_{1}+a_{3}+a_{5}+a_{7}+a_{9}=(\quad)$.
(A) 40
(B) 44
(C) 45
(D) 49
- 1. B.

When $n=1$, $a_{1}=S_{1}=0$;
When $n \geqslant 2$, $a_{n}=S_{n}-S_{n-1}=2 n-1$.
Therefore, $a_{1}+a_{3}+a_{5}+a_{7}+a_{9}=44$.