The problem is asking to prove, using the Binomial Theorem, that:

$\sum_{r=0}^{n} \binom{n}{r} = 2^n$

Proof using the Binomial Theorem:

The Binomial Theorem states that for any positive integer $n$ and any real numbers $x$ and $y$:

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

Now, let’s apply the Binomial Theorem for the special case where $x = 1$ and $y = 1$. In this case, we get:

$(1 + 1)^n = \sum_{r=0}^{n} \binom{n}{r} 1^{n-r} 1^r$

Since $1^{n-r} = 1$ and $1^r = 1$ for all $r$, the equation simplifies to:

$2^n = \sum_{r=0}^{n} \binom{n}{r}$

This is the required result.

Thus, we have proven that:

$\sum_{r=0}^{n} \binom{n}{r} = 2^n$

Would you like further details or clarification?

Here are some related questions:

1. What is the general form of the Binomial Theorem?
2. How can we apply the Binomial Theorem when $x \neq 1$ and $y \neq 1$?
3. What are some applications of the Binomial Theorem in combinatorics?
4. How does Pascal's Triangle relate to binomial coefficients?
5. Can the Binomial Theorem be extended for negative exponents?

Tip: The binomial coefficients $\binom{n}{r}$ can be found using Pascal's Triangle or directly from the formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$.