1. **Problem Statement:** Determine which of the following statements is correct: $$\sum_{k=0}^n (-1)^k \binom{n}{k} 2^{n-k} = 1, \quad \sum_{k=0}^n (-1)^k \binom{n}{k} 2^{n-k} = (-1)^n, \quad \sum_{k=0}^n (-1)^k \binom{n}{k} 2^{n-k} = 3^n$$ 2. **Recall the Binomial Theorem:** $$(x+y)^n = \sum_{k=0}^n \binom{n}{k} x^k y^{n-k}$$ 3. **Rewrite the sum to match the binomial form:** Given the sum: $$\sum_{k=0}^n (-1)^k \binom{n}{k} 2^{n-k}$$ We can write $(-1)^k = (-1)^k$ and $2^{n-k} = 2^{n-k}$, so the sum is: $$\sum_{k=0}^n \binom{n}{k} (-1)^k 2^{n-k}$$ 4. **Identify $x$ and $y$ in the binomial expansion:** If we let $x = -1$ and $y = 2$, then: $$(x + y)^n = (-1 + 2)^n = 1^n = 1$$ Expanding this using the binomial theorem: $$\sum_{k=0}^n \binom{n}{k} (-1)^k 2^{n-k} = 1$$ 5. **Conclusion:** The sum equals 1, so the correct statement is: $$\sum_{k=0}^n (-1)^k \binom{n}{k} 2^{n-k} = 1$$ which corresponds to statement I. **Final answer:** Statement I is correct.