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!

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.