1. Arrange positive integers as shown in Figure 1. Then the sum of the numbers in the $n$-th row $S_{n}$ is ( ).
(A) $\frac{n^{2}(n+1)}{2}$
(B) $\frac{n\left(n^{2}+1\right)}{2}$
(C) $\frac{n(n+1)^{2}}{2}$
(D) $\frac{n^{3}+1}{2}$
-、1.B.
The last natural number in the $n$th row, which is the $\frac{n(n+1)}{2}$th number, is
$$
\frac{n(n+1)}{2}=a_{1}+(n-1) \text {. }
$$

Thus, $a_{1}=\frac{n(n+1)}{2}-n+1=\frac{n(n-1)}{2}+1$ is the first term of the $n$th row. Therefore,
$$
S_{n}=\frac{n}{2}\left[\frac{n(n+1)}{2}+\frac{n(n-1)}{2}+1\right]=\frac{n\left(n^{2}+1\right)}{2} .
$$