$10 \cdot 30$ If $x_{k+1}=x_{k}+\frac{1}{2}$, where $k=1,2, \cdots, n-1$, and $x_{1}=1$, then $x_{1}+x_{2}+\cdots+x_{n}$ is
(A) $\frac{n+1}{2}$.
(B) $\frac{n+3}{2}$.
(C) $\frac{n^{2}-1}{2}$.
(D) $\frac{n^{2}+n}{4}$.
(E) $\frac{n^{2}+3 n}{4}$.
(13th American High School Mathematics Examination, 1962)
[Solution] Given $x_{1}=1, x_{2}=\frac{3}{2}, x_{3}=\frac{4}{2}, \cdots, x_{n}=$ $\frac{n+1}{2}$.
Then $\sum_{k=1}^{n} x_{k}=\frac{1}{2} \sum_{k=1}^{n}(k+1)=\frac{n^{2}+3 n}{4}$. Therefore, the answer is $(E)$.