The expression given is a summation from $n = 4$ to $n = 12$ of the form:

$\sum_{n=4}^{12} e^{-2n} \cdot (1 - e^{-2})$

To evaluate this summation, let’s break it down:

1. Factor Out Constants: Since $(1 - e^{-2})$ is independent of $n$, we can factor it out: $(1 - e^{-2}) \sum_{n=4}^{12} e^{-2n}$

2. Recognize the Remaining Summation as a Geometric Series: The term $e^{-2n}$ represents a geometric sequence with a common ratio $r = e^{-2}$ and the first term (for $n=4$) as $e^{-2 \cdot 4} = e^{-8}$.

3. Geometric Series Sum Formula: For a geometric series $a + ar + ar^2 + \dots + ar^{k-1}$, the sum is given by: $S = \frac{a(1 - r^k)}{1 - r}$ where $a = e^{-8}$, $r = e^{-2}$, and the number of terms $k = 12 - 4 + 1 = 9$.

4. Apply the Formula: $\sum_{n=4}^{12} e^{-2n} = e^{-8} \frac{1 - (e^{-2})^9}{1 - e^{-2}}$

5. Combine with the Constant Factor: Now, multiply by $(1 - e^{-2})$: $(1 - e^{-2}) \cdot e^{-8} \frac{1 - e^{-18}}{1 - e^{-2}}$

   Simplifying, the $(1 - e^{-2})$ terms cancel out, leaving: $e^{-8} (1 - e^{-18})$

6. Final Answer: $e^{-8} (1 - e^{-18})$

This is the exact form of the sum. If needed, you can further simplify by evaluating the exponents numerically.

Would you like further details or a numerical approximation?

Related Questions:

1. How can we evaluate an infinite geometric series?
2. What other types of sequences allow for summation techniques?
3. Can we approximate $e^{-8} (1 - e^{-18})$ for a numerical answer?
4. How does the summation change if the limits are different?
5. Why does factoring constants out simplify summation calculations?

Tip: Recognizing patterns, like geometric sequences, can simplify complex summations effectively.