Problem 41

To solve the given problem, we will use the binomial theorem, which states that for any positive integer $n$ and any real numbers $a$ and $b$, we have:

$$(a + b)^n = \sum_{k=0}^{n} C_n^k a^{n-k}b^k.$$

Let's apply the binomial theorem with $a=1$ and $b=2$:

$$\begin{align*} (1 + 2)^n &= \sum_{k=0}^{n} C_n^k 1^{n-k}2^k \\ &= \sum_{k=0}^{n} C_n^k 2^k \\ &= C_n^0 + 2C_n^1 + 2^2C_n^2 + \ldots + 2^nC_n^n. \end{align*}$$

Therefore, we can simplify the left-hand side to obtain:

$$\boxed{3^n}.$$