The expression you have provided is a summation of binomial coefficients, represented as:

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

This expression sums all the binomial coefficients for a given $n$. The binomial coefficient $\binom{n}{r}$ represents the number of ways to choose $r$ elements from a set of $n$ elements, and it is calculated as:

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

The summation $\sum_{r=0}^{n} \binom{n}{r}$ evaluates to $2^n$. This result comes from the Binomial Theorem, which states that:

$(1 + 1)^n = \sum_{r=0}^{n} \binom{n}{r} \cdot 1^r \cdot 1^{n-r} = 2^n$

Thus, the entire summation simplifies to $2^n$.

Would you like more details or have any questions?

Here are some related questions that might interest you:

1. What is the Binomial Theorem, and how does it apply to different values of $n$?
2. How can you prove that the sum of all binomial coefficients equals $2^n$?
3. What are some applications of binomial coefficients in probability and combinatorics?
4. How do binomial coefficients relate to Pascal's Triangle?
5. What is the significance of binomial expansions in algebra?

Tip: Binomial coefficients are fundamental in combinatorics, and understanding their properties can help solve various counting problems efficiently.