93. Calculate the sum:

$$
S=1-2+3-4+\cdots+(-1)^{n+1} \cdot n
$$
$\triangle$ Clearly, the answer depends on whether $n$ is even or odd. Therefore, we will consider two cases.

1) Let $n$ be even. Then:

$$
S=(1-2)+(3-4)+(5-6)+\cdots+((n-1)-n)=-1-1-\ldots-1=-\frac{n}{2}
$$

2) Let $n$ be odd. We will have:

$$
S=(1-2+3-4+\cdots-(n-1))+n=-\frac{n-1}{2}+n=\frac{n+1}{2}
$$

Answer: if $n$ is even, then $S=-n / 2$; if $n$ is odd, then $S=(n+1) / 2$.

Think: can the two obtained expressions for $S$ be combined into one formula?