https://file.aiquickdraw.com/mathbot/file/1727354414736rmvyeqc2.jpg

The problem given is to prove the following identity:

\[
\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \ldots + (-1)^n \binom{n}{n} = 2^n
\]

for a given natural number \( n \).

### **Solution and Proof:**

The left-hand side of the equation is an alternating sum of binomial coefficients. It can be expressed as:

\[
\sum_{k=0}^{n} (-1)^k \binom{n}{k}
\]

This sum is a well-known identity in combinatorics, and it equals \( (1-1)^n \). Here's why:

1. **Binomial Expansion:**
   According to the binomial theorem, for any real numbers \( a \) and \( b \):

   \[
   (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
   \]

   If we set \( a = 1 \) and \( b = -1 \), then:

   \[
   (1 + (-1))^n = \sum_{k=0}^{n} \binom{n}{k} \cdot 1^{n-k} \cdot (-1)^k
   \]

   This simplifies to:

   \[
   0^n = \sum_{k=0}^{n} (-1)^k \binom{n}{k}
   \]

   for \( n > 0 \). Therefore, when \( n \) is a positive integer, the sum of the alternating binomial coefficients is \( 0 \).

2. **Modified Problem Interpretation:**
   It appears that there may have been a typo or misunderstanding in the problem statement, as the identity shown in the image should sum to zero instead of \( 2^n \). This identity holds true for natural numbers greater than zero, where:

   \[
   \sum_{k=0}^{n} (-1)^k \binom{n}{k} = 0
   \]

Let me know if you have any specific questions or want further clarification!

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### **Relative Questions:**

1. Can you explain how the binomial theorem is applied in this proof?
2. What does \( \binom{n}{k} \) represent in combinatorics?
3. Why does \( (1 + (-1))^n = 0 \) when \( n > 0 \)?
4. Are there any other similar identities involving binomial coefficients?
5. How does the alternating sign affect the sum of the binomial coefficients?

### **Tip:**

For combinatorics problems, always consider expanding the terms using the binomial theorem to simplify and identify patterns.

Buktikan \(\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \ldots + (-1)^n \binom{n}{n} = 2^n\) untuk suatu n bilangan Asli.

Prove the identity involving binomial coefficients \(\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \ldots + (-1)^n \binom{n}{n} = 2^n\). Explore the binomial theorem and how alternating sums apply to combinatorics for natural numbers.

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