This question asks to prove the following identity using the binomial theorem:

$\sum_{k=0}^{n} \binom{n}{k} (-2)^k = (-1)^n$ where $n = 0, 1, 2, \dots$.

Step-by-Step Explanation:

1. The Binomial Theorem:

The binomial theorem states that for any real numbers $x$ and $y$, and any non-negative integer $n$, we have: $(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}.$

In this case, we will apply the binomial theorem with $x = -2$ and $y = 1$, which gives: $(1 + (-2))^n = \sum_{k=0}^{n} \binom{n}{k} (-2)^k 1^{n-k}.$

2. Simplify the Left-Hand Side:

On the left-hand side, $1 + (-2) = -1$, so the equation becomes: $(-1)^n = \sum_{k=0}^{n} \binom{n}{k} (-2)^k.$

3. Conclusion:

Thus, we've shown that: $\sum_{k=0}^{n} \binom{n}{k} (-2)^k = (-1)^n,$ which is exactly what the problem asks to prove.

Would you like more details on any step, or any additional explanations?

Related Questions:

1. How is the binomial coefficient $\binom{n}{k}$ defined?
2. Can you prove this identity using mathematical induction instead of the binomial theorem?
3. How does the binomial theorem apply to complex numbers?
4. What happens if we change the coefficient $-2$ to another number, like $-3$?
5. What are some applications of the binomial theorem in combinatorics?

Tip:

Whenever you encounter sums involving binomial coefficients, always check if the binomial theorem applies—it simplifies many such expressions!