https://file.aiquickdraw.com/mathbot/file/1730431203924dt9wmqyn.jpg

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 \( 2^2 \):
\[
= 2^2 \left( 2^m - 1 + 2^{m} \right)
= 2^2 (2^{m+1} - 1)
\]
This matches the right side, completing the inductive step.

Since both the base case and inductive step hold, the statement is true for all \( n \in \mathbb{N} \).

Would you like more details or have questions on any part of this proof?

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### Related Questions:

1. How does mathematical induction work, and when is it used?
2. What is the formula for the sum of a geometric series?
3. How can we verify if a summation expression is correct?
4. What other methods exist for proving summation identities?
5. How does factorization help simplify expressions in proofs?

### Tip:
When proving summation identities, start by checking simple values of \( n \) to verify the pattern before diving into an induction proof.

Prove that \( \sum_{k=1}^n 2^{k+1} = 2^2 (2^n - 1) \) for \( n \in \mathbb{N} \).

Learn how to prove the summation identity \( \sum_{k=1}^n 2^{k+1} = 2^2 (2^n - 1) \) using mathematical induction. This problem covers important concepts in algebra and exponential functions, suitable for high school students in Grades 11-12.

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