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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.

Use the binomial theorem to prove that 2^n = ∑_{k=0}^{n} (-1)^k (n choose k) 3^{n-k}.

Learn how to apply the binomial theorem to prove the identity 2^n = ∑_{k=0}^{n} (-1)^k (n choose k) 3^{n-k}. This detailed step-by-step solution demonstrates the use of combinatorics and binomial expansion. Suitable for students in Grades 11-12.

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