The problem presented in the image is asking to prove the following equation for $n \geq 1$:

$\binom{n}{0} + 2 \binom{n}{1} + 2^2 \binom{n}{2} + \cdots + 2^n \binom{n}{n} = 3^n$

This involves a binomial sum where the coefficients of the binomial terms are multiplied by powers of 2. The goal is to demonstrate that this sum equals $3^n$.

Proof Outline (Using Binomial Theorem):

The equation resembles a variant of the binomial expansion. Let's consider the binomial theorem for expanding powers of binomials:

$(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k$

If we set $x = 2$, then:

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

This expands as:

$3^n = \sum_{k=0}^{n} \binom{n}{k} 2^k$

This is exactly the equation we are asked to prove:

$\binom{n}{0} + 2 \binom{n}{1} + 2^2 \binom{n}{2} + \cdots + 2^n \binom{n}{n} = 3^n$

Thus, the expression is a direct result of the binomial expansion of $(1 + 2)^n$.

Conclusion:

The equation is a direct application of the binomial theorem, and the proof is complete by recognizing that the given sum corresponds to the expansion of $(1 + 2)^n$.

Do you want more details or have any questions?

Here are 5 related questions to further your understanding:

1. What is the binomial theorem, and how does it apply to problems like this?
2. How would the sum change if we used a different constant instead of 2 in the powers?
3. Can you derive other binomial identities from variations of this formula?
4. What happens if $n$ is very large—how does the left-hand side of the equation grow compared to $3^n$?
5. How can this type of binomial expansion be applied to real-world problems in probability or combinatorics?

Tip: When encountering sums involving binomial coefficients, always consider if the binomial theorem or a generating function might be helpful in simplifying or proving the identity.