If $S_{n}=1-2+3-4+5-6+\cdots+(-1)^{n+1} n$ for each positive integer $n$, then $S_{1992}+S_{1993}$ is equal to
(a) -2 ;
(b) -1 ;
(c) 0 ;
(d) 1 ;
(e) 2 .
The correct option is (d).

The expression $(-1)^{n+1}$ in the definition of $S_{n}$ has a value of 1 if $n$ is even and a value of -1 if $n$ is odd.

Solution 1: By associating consecutive terms in pairs, we obtain a sum of several terms equal to -1: $(1-2)+(3-4)+(5-6)+\cdots$. Therefore,

$$
S_{1992}=\underbrace{(1-2)+(3-4)+\cdots+(1991-1992)}_{1992 \div 2=996 \text{ terms}}=(-1) \times 996=-996
$$

and

$$
S_{1993}=(1-2)+(3-4)+\cdots+(1991-1992)+1993=-996+1993=997
$$

Thus, $S_{1992}+S_{1993}=-996+997=1$.

Solution 2: Since

$$
S_{2 n}=\underbrace{(1-2)+(3-4)+(5-6)+\cdots+[2 n-(2 n+1)]}_{n \text{ terms equal to }-1}
$$

we get $S_{2 n}=-n$ and $S_{2 n+1}=S_{2 n}+(2 n+1)=-n+2 n+1=n+1$. Therefore, $S_{2 n}+S_{2 n+1}=$ 1.