The problem you've shared is related to sums, products, and basic combinatorics. Let me break down the mathematical expressions from the image:

1. The task is to compute two sums.

First Sum:

$\sum_{k=0}^{E(n/2)} \binom{n}{2k}$

Here, $E(n/2)$ represents the floor function of $n/2$, meaning the greatest integer less than or equal to $n/2$, and $\binom{n}{2k}$ is a binomial coefficient. This sum likely involves summing over even binomial coefficients.

Second Sum:

$\sum_{k=0}^{E((n-1)/2)} \binom{n}{2k+1}$

This sum is over the odd binomial coefficients, where $E((n-1)/2)$ is the floor of $(n-1)/2$, representing the largest integer less than or equal to $(n-1)/2$.

Identity Given:

Both sums have the result: $= 2^{n-1}$ and the second sum specifically equals $2^n - 1$ for $n \in \mathbb{N^*}$.

These sums are linked to binomial expansions and symmetry properties of binomial coefficients, specifically the sum of even and odd-indexed binomial coefficients.

Would you like a detailed step-by-step derivation of these sums?

Here are five related questions for better understanding:

1. What is the relationship between binomial coefficients and Pascal's Triangle?
2. How are even and odd binomial coefficients distributed in the binomial expansion?
3. Can these sums be related to combinatorial interpretations, like counting subsets?
4. Why do the sums of even and odd binomial coefficients add up to powers of two?
5. How does the symmetry in binomial coefficients simplify such summations?

Tip: The binomial theorem can often simplify the evaluation of sums involving binomial coefficients.