1. The problem is to find the limit as $n$ approaches infinity of the sequence $$1 + 2 + 2^2 + \dots + 2^{n-1} - 3 \cdot 2^{n+1}.$$\n\n2. This is a sum of a geometric series minus a term. The geometric series sum formula for the first $m$ terms is $$S_m = \frac{a(r^m - 1)}{r - 1},$$ where $a$ is the first term and $r$ is the common ratio.\n\n3. Here, the geometric series is $$1 + 2 + 2^2 + \dots + 2^{n-1}$$ with $a=1$ and $r=2$, so the sum is $$S_n = \frac{1(2^n - 1)}{2 - 1} = 2^n - 1.$$\n\n4. Substitute this back into the original expression: $$S_n - 3 \cdot 2^{n+1} = (2^n - 1) - 3 \cdot 2^{n+1}.$$\n\n5. Simplify the expression: $$2^n - 1 - 3 \cdot 2^{n+1} = 2^n - 1 - 3 \cdot 2 \cdot 2^n = 2^n - 1 - 6 \cdot 2^n = 2^n - 6 \cdot 2^n - 1 = -5 \cdot 2^n - 1.$$\n\n6. Now, find the limit as $n \to \infty$: $$\lim_{n \to \infty} (-5 \cdot 2^n - 1).$$\n\n7. Since $2^n$ grows without bound, $-5 \cdot 2^n$ tends to $-\infty$, so the limit is $-\infty$.\n\n8. However, the problem likely expects a finite value, so let's re-examine the original expression carefully. The original sequence is $$1 + 2 + 2^2 + \dots + 2^{n-1} - 3 \cdot 2^{n+1}.$$\n\n9. The sum of the geometric series is $$2^n - 1,$$ so the expression is $$2^n - 1 - 3 \cdot 2^{n+1} = 2^n - 1 - 3 \cdot 2 \cdot 2^n = 2^n - 1 - 6 \cdot 2^n = -5 \cdot 2^n - 1.$$\n\n10. As $n$ grows large, $-5 \cdot 2^n$ dominates and the expression tends to $-\infty$, which is not among the options.\n\n11. Possibly the problem meant the sum $$1 + 2 + 2^2 + \dots + 2^{n-1} - 3 \cdot 2^{n-1}$$ instead of $3 \cdot 2^{n+1}$. Let's check that: $$S_n = 2^n - 1,$$ and subtract $$3 \cdot 2^{n-1}.$$\n\n12. Then the expression is $$2^n - 1 - 3 \cdot 2^{n-1} = 2^{n-1} \cdot 2 - 1 - 3 \cdot 2^{n-1} = 2^{n-1}(2 - 3) - 1 = -2^{n-1} - 1.$$\n\n13. As $n \to \infty$, $-2^{n-1} - 1 \to -\infty$ again.\n\n14. Since the options are finite numbers, let's consider the sum $$1 + 2 + 2^2 + \dots + 2^{n-1} - 3 \cdot 2^n + 1$$ (maybe a typo in the problem).\n\n15. Sum is $$2^n - 1,$$ subtract $$3 \cdot 2^n + 1,$$ so expression is $$2^n - 1 - 3 \cdot 2^n - 1 = -2 \cdot 2^n - 2 = -2^{n+1} - 2,$$ which tends to $-\infty$.\n\n16. Since none of these match the options, the problem likely wants the sum $$1 + 2 + 2^2 + \dots + 2^{n-1} - 3 \cdot 2^{n+1}$$ divided by $2^n$ or similar.\n\n17. Let's try dividing the entire expression by $2^n$: $$\frac{1 + 2 + 2^2 + \dots + 2^{n-1} - 3 \cdot 2^{n+1}}{2^n} = \frac{2^n - 1}{2^n} - 3 \cdot \frac{2^{n+1}}{2^n} = 1 - \frac{1}{2^n} - 3 \cdot 2 = 1 - 0 - 6 = -5.$$\n\n18. This is $-5$, still not among options.\n\n19. Given the options, the closest is 12, 14, 2, or 13. The sum of the geometric series $$1 + 2 + 4 + \dots + 2^{n-1} = 2^n - 1,$$ which tends to infinity as $n \to \infty$.\n\n20. Therefore, the limit of the sequence as given diverges to $-\infty$. None of the options match this.\n\n21. If the problem is to find the sum $$1 + 2 + 2^2 + \dots + 2^{n-1}$$ only, the limit as $n \to \infty$ is infinite.\n\n22. If the problem is to find the sum $$1 + 2 + 2^2 + \dots + 2^{n-1} - 3 \cdot 2^{n+1}$$ the limit is $-\infty$.\n\n23. Since the options are finite, the problem might have a typo or missing parentheses.\n\n24. Without further clarification, the limit does not exist finitely.\n\nFinal answer: The limit diverges to $-\infty$, which is not among the options.