The expression in the image is:

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

The goal is to prove that this equation holds true for $n \in \mathbb{N}$.

To prove this, we can use mathematical induction on $n$.

Step 1: Base Case ( $n = 1$ )

For $n = 1$: $\sum_{k=1}^1 2^{k+1} = 2^{1+1} = 2^2 = 4$ Now, check the right side for $n = 1$: $2^2 (2^1 - 1) = 2^2 \cdot (2 - 1) = 2^2 \cdot 1 = 4$ Both sides match, so the base case is true.

Step 2: Inductive Hypothesis

Assume the statement is true for some $n = m$, i.e., $\sum_{k=1}^m 2^{k+1} = 2^2 (2^m - 1)$

Step 3: Inductive Step ( $n = m + 1$ )

We need to show that $\sum_{k=1}^{m+1} 2^{k+1} = 2^2 (2^{m+1} - 1)$

Starting from the left side: $\sum_{k=1}^{m+1} 2^{k+1} = \left( \sum_{k=1}^m 2^{k+1} \right) + 2^{(m+1)+1}$ By the inductive hypothesis, we can substitute $\sum_{k=1}^m 2^{k+1} = 2^2 (2^m - 1)$: $= 2^2 (2^m - 1) + 2^{m+2}$ Now, factor out $2^2$: