If $$[x]$$ be the greatest integer less than or equal to $$x$$, then $$\sum_{n=8}^{100} \left[\frac{(-1)^n n}{2}\right]$$ is equal to:

We have to evaluate the finite sum

$$\sum_{n=8}^{100}\left[\frac{(-1)^n\,n}{2}\right]$$

where $$[x]$$ denotes the greatest integer less than or equal to $$x$$ (the “floor” of $$x$$).

First we separate the index $$n$$ into the two possible parities because the factor $$(-1)^n$$ behaves differently for even and odd indices.

For an even integer $$n$$ we can write $$n=2k$$. Then $$(-1)^n = 1$$, so

$$\frac{(-1)^n\,n}{2}= \frac{1\cdot 2k}{2}=k,$$ and since $$k$$ is already an integer, its greatest-integer value is

$$\left[\frac{(-1)^n\,n}{2}\right]=k=\frac{n}{2}.$$

For an odd integer $$n$$ we can write $$n=2k+1$$. Then $$(-1)^n = -1$$, so

$$\frac{(-1)^n\,n}{2}= \frac{-1\cdot(2k+1)}{2}= -\left(k+\tfrac12\right).$$

The number $$-\!\left(k+\tfrac12\right)$$ lies strictly between the two consecutive integers $$-(k+1)$$ and $$-k$$ and is less than both of them, hence its greatest-integer value is

$$\left[\frac{(-1)^n\,n}{2}\right]=-(k+1)= -\frac{n+1}{2}.$$

So, term by term,

$$ \left[\frac{(-1)^n\,n}{2}\right]= \begin{cases} \dfrac{n}{2}, & n \text{ even},\\[6pt] -\dfrac{n+1}{2}, & n \text{ odd}. \end{cases} $$

Now we split the required sum into its even and odd parts:

$$ \sum_{n=8}^{100}\Bigl[\tfrac{(-1)^n n}{2}\Bigr] =\sum_{\substack{n=8\\ n\text{ even}}}^{100}\frac{n}{2} +\sum_{\substack{n=8\\ n\text{ odd}}}^{100}\!\!\!\!\left(-\frac{n+1}{2}\right). $$

The even values of $$n$$ run from 8 to 100: 8, 10, 12, …, 100.

There are

$$\frac{100-8}{2}+1=\frac{92}{2}+1=46+1=47$$

such even integers. For each of them we take half, giving the consecutive integers

$$4,\,5,\,6,\ldots,\,50$$

(because $$8/2=4$$ and $$100/2=50$$). Their sum is obtained by the formula for an arithmetic progression:

$$ \sum_{\text{even }n}\frac{n}{2} =\frac{\text{number of terms}}{2}\,( \text{first term}+\text{last term}) =\frac{47}{2}\,(4+50) =\frac{47}{2}\times54 =47\times27 =1269. $$

The odd values of $$n$$ run from 9 to 99:

9, 11, 13, …, 99.

There are 46 of them, because the total count 93 minus the 47 evens gives 46 odds. For each odd $$n$$ we substitute $$-\dfrac{n+1}{2}$$. Now $$(n+1)/2$$ for these odds is the consecutive list

$$5,\,6,\,7,\ldots,\,50,$$

and taking the negative gives

$$-5,\,-6,\,-7,\ldots,\,-50.$$

Again using the arithmetic-progression formula, the sum of these 46 integers is

$$ \sum_{\text{odd }n}\left(-\frac{n+1}{2}\right) =-\,\frac{46}{2}\,(5+50) =-\,23\times55 =-1265. $$

Combining the even and odd contributions:

$$ \sum_{n=8}^{100}\Bigl[\tfrac{(-1)^n n}{2}\Bigr] =1269 + (-1265)=4. $$

Hence, the correct answer is Option B.