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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 \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \).

Prove, by using Binomial Theorem, that the sum of binomial coefficients from r=0 to n is equal to 2^n.

Learn how to prove that the sum of binomial coefficients from r=0 to n is equal to 2^n using the Binomial Theorem. This problem involves key algebraic and combinatoric concepts, including the Binomial Theorem and binomial coefficients. Suitable for students in Grades 10-12.

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