$$
\text { 10・53 Let } S_{n}=1-2+3-4+\cdots+(-1)^{n-1} n, \quad n=1,2, \cdots \text { , }
$$
then $S_{17}+S_{33}+S_{50}$ equals
(A) 0 .
(B) 1 .
(C) 6 .
(D) -1 .
(E) 2 .
(19th American High School Mathematics Examination, 1968)
[Solution] (1) When $n$ is even,
$$
\begin{aligned}
S_{n} & =1-2+3-4+\cdots+(-1)^{n-1} n \\
& =(1-2)+(3-4)+\cdots+[(n-1)-n] \\
& =-1-1-\cdots-1=-\frac{n}{2} ;
\end{aligned}
$$
(2) When $n$ is odd,
$$
\begin{aligned}
S_{n} & =1+\{(-2+3)+(-4+5)+\cdots+[-(n-1)+n]\} \\
& =1+\{1+1+\cdots+1\}=1+\frac{n-1}{2}=\frac{n+1}{2} .
\end{aligned}
$$

From the above, $S_{17}+S_{33}+S_{50}=\frac{18}{2}+\frac{34}{2}-\frac{50}{2}=1$.
Therefore, the answer is $(B)$.