7.95 Let $f(n)=\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}$, where $n$ is a positive integer. If $x_{k}=(-1)^{k}, k=1,2, \cdots, n$, then the set of possible values of $f(n)$ is
(A) $\{0\}$.
(B) $\left\{\frac{1}{n}\right\}$.
(C) $\left\{0,-\frac{1}{n}\right\}$.
(D) $\left\{0, \frac{1}{n}\right\}$.
(E) $\left\{1, \frac{1}{n}\right\}$.
(19th American High School Mathematics Examination, 1968)
［Solution］Suppose we have
$$
x_{k}+x_{k+1}=(-1)^{k}+(-1)^{k+1}=(-1)^{k}(1-1)=0 \text {. }
$$

Therefore,
$$
\sum_{k=1}^{n} x_{k}=\left\{\begin{array}{ll}
0, & n \text { is even; } \\
-1, & n \text { is odd. }
\end{array}\right.
$$

Thus, $f(2 k)=0, f(2 k+1)=-\frac{1}{2 k+1}$, so the value set of $f(n)$ is contained in $\left\{0,-\frac{1}{n}\right\}$. Therefore, the answer is (C).