Key Concepts and Formulas

• Greatest Integer Function (Floor Function): For any real number $x$, $[x]$ (or $\lfloor x \rfloor$) is the greatest integer less than or equal to $x$.
• Properties of Floor Function:
  • If $n$ is an integer, $[n] = n$.
  • If $x = n + f$, where $n$ is an integer and $0 \le f < 1$, then $[x] = n$.
  • For any real number $x$, $[x+n] = [x] + n$ if n is an integer.
  • If $x$ is not an integer, $[-x] = -[x] - 1$.
• Summation Notation: $\sum\limits_{n=a}^{b} f(n) = f(a) + f(a+1) + ... + f(b)$.

Step-by-Step Solution

Step 1: Define the term $a_n$

Let $a_n = \left[ \frac{(-1)^n n}{2} \right]$. We need to analyze this expression for even and odd values of $n$.

Step 2: Analyze the case when $n$ is even

If $n$ is even, let $n = 2k$ where $k$ is an integer. Then, substitute into the expression for $a_n$: $a_{2k} = \left[ \frac{(-1)^{2k} (2k)}{2} \right] = \left[ \frac{(1)(2k)}{2} \right] = [k]$ Since $k$ is an integer, $[k] = k$. So, $a_{2k} = k$.

Step 3: Analyze the case when $n$ is odd

If $n$ is odd, let $n = 2k+1$ where $k$ is an integer. Substitute into the expression for $a_n$: $a_{2k+1} = \left[ \frac{(-1)^{2k+1} (2k+1)}{2} \right] = \left[ \frac{(-1)(2k+1)}{2} \right] = \left[ -\frac{2k+1}{2} \right]$ Rewrite the fraction: $a_{2k+1} = \left[ -k - \frac{1}{2} \right]$ Since $k$ is an integer, we can use the property of the floor function: $a_{2k+1} = \left[ -k - \frac{1}{2} \right] = -k + \left[ -\frac{1}{2} \right] = -k -1$

Step 4: Express the summation in terms of $k$

The given summation is $\sum\limits_{n = 8}^{100} a_n = \sum\limits_{n = 8}^{100} \left[ \frac{(-1)^n n}{2} \right]$. We need to determine the values of $k$ for the given range of $n$.

When $n=8$, $n$ is even, so $2k = 8 \implies k = 4$. When $n=100$, $n$ is even, so $2k = 100 \implies k = 50$. Therefore, for even terms, $k$ ranges from 4 to 50.

When $n=9$, $n$ is odd, so $2k+1 = 9 \implies 2k = 8 \implies k = 4$. When $n=99$, $n$ is odd, so $2k+1 = 99 \implies 2k = 98 \implies k = 49$. Therefore, for odd terms, $k$ ranges from 4 to 49.

Step 5: Rewrite the sum as a sum of even and odd terms

We can rewrite the sum as: $S = \sum_{k=4}^{50} a_{2k} + \sum_{k=4}^{49} a_{2k+1} = \sum_{k=4}^{50} k + \sum_{k=4}^{49} (-k-1) = \sum_{k=4}^{50} k - \sum_{k=4}^{49} (k+1)$ $S = \sum_{k=4}^{50} k - \sum_{k=4}^{49} k - \sum_{k=4}^{49} 1$

Step 6: Simplify the summation

$S = \left(\sum_{k=4}^{49} k + 50\right) - \sum_{k=4}^{49} k - \sum_{k=4}^{49} 1 = 50 - \sum_{k=4}^{49} 1$ The number of terms in $\sum_{k=4}^{49} 1$ is $49 - 4 + 1 = 46$. Therefore, $\sum_{k=4}^{49} 1 = 46$.

Step 7: Calculate the final sum

$S = 50 - 46 = 4$

Common Mistakes & Tips

• Carefully handle the negative signs when dealing with the floor function. Remember that $[-x] = -[x] - 1$ if $x$ is not an integer.
• Pay attention to the range of the summation. Make sure to adjust the limits of summation when substituting $n = 2k$ or $n = 2k+1$.
• Grouping terms strategically can simplify the calculation.

Summary

We analyzed the given summation by considering even and odd values of $n$ separately. We expressed the sum in terms of the greatest integer function and then simplified it using properties of the floor function. By strategically grouping the terms, we were able to evaluate the sum and found it to be equal to 4.

Final Answer

The final answer is $\boxed{4}$, which corresponds to option (B).