If $S_{n}=1-2+3-4+5-6+\ldots+(-1)^{n+1} n$, where $n$ is a positive integer, then $S_{1992}+S_{1993}$ is:
(a) -2
(b) -1
(c) 0
(d) 1
(e) 2
Solution 1: Remember that

$(-1)^{n+1}= \begin{cases}1 & \text { if } n \text { is odd } \\ -1 & \text { if } n \text { is even }\end{cases}$

Observe that by grouping consecutive terms in pairs,

$$
(1-2)+(3-4)+(5-6)+\cdots
$$

we get a sum of $n$ terms all equal to -1. Therefore,
$S_{1992}=\underbrace{(1-2)+(3-4)+(5-6)+\cdots+(1991-1992)}_{1992 \div 2=996 \text { terms }}=(-1) \times 996=-996$.

$S_{1993}=(1-2)+(3-4)+(5-6)+\cdots+(1991-1992)+1993=-996+1993=997$.

Thus, $S_{1992}+S_{1993}=-996+997=1$. The correct option is (d).

Solution 2: Note that

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

We obtain that $S_{2 n}=-n$ and $S_{2 n+1}=S_{2 n}+(2 n+1)=-n+2 n+1=n+1$. Therefore, $S_{2 n}+S_{2 n+1}=1$.