We are asked to prove the following identity using the binomial theorem:

$2^n = \sum_{k=0}^{n} (-1)^k \binom{n}{k} 3^{n-k}$

Step-by-Step Proof:

1. Recall the Binomial Theorem:

   The binomial theorem states that for any two numbers $x$ and $y$, and any non-negative integer $n$, we have:

   $(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^k y^{n-k}$

2. Apply the Binomial Theorem to $(3 - 1)^n$:

   Set $x = 3$ and $y = -1$ in the binomial expansion. We get:

   $(3 - 1)^n = \sum_{k=0}^{n} \binom{n}{k} 3^k (-1)^{n-k}$

   Since $(3 - 1) = 2$, this simplifies to:

   $2^n = \sum_{k=0}^{n} \binom{n}{k} 3^k (-1)^{n-k}$

3. Rearrange the Terms:

   Now, we change the index of the summation to make the exponents of $(-1)$ and $3$ match the desired form. Let’s rearrange the powers of $(-1)$:

   $2^n = \sum_{k=0}^{n} \binom{n}{k} (-1)^k 3^{n-k}$

   This is exactly the identity we were asked to prove:

   $2^n = \sum_{k=0}^{n} (-1)^k \binom{n}{k} 3^{n-k}$

Thus, the identity is proven using the binomial theorem.

Would you like further clarification on any part of the proof or additional details?

Related Questions:

1. How can the binomial theorem be used for other expansions like $(x + y)^n$?
2. How does the binomial coefficient $\binom{n}{k}$ relate to combinations?
3. What is the general form of the binomial theorem for any power $n$?
4. Can this proof be extended to negative values of $n$?
5. What happens if we apply the binomial theorem to complex numbers?

Tip:

The binomial theorem is a powerful tool in combinatorics and algebra. It can simplify many proofs by expanding terms in summation form.