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To solve for the left-hand side of \( P(1) \), we need to use the recurrence relation provided for the sequence \( f_1, f_2, f_3, \dots \):

\[
f_k = f_{k-1} + 2^k \quad \text{for each integer} \, k \geq 2 \quad \text{with} \, f_1 = 1.
\]

The left-hand side of \( P(1) \) corresponds to \( f_1 \). From the initial condition given, \( f_1 = 1 \).

Thus, the left-hand side of \( P(1) \) is \( f_1 = 1 \). Therefore, \( P(1) \) holds true, as the left-hand side equals the right-hand side, which is also 1.

Would you like further clarification or have any questions about this proof?

Here are five related questions that build upon this:

1. What is the general form of the explicit formula for \( f_n \)?
2. How do we use induction to prove a recursive formula?
3. What is the significance of the base case \( P(1) \) in mathematical induction?
4. Can we derive \( f_2 \) using the recurrence relation provided?
5. How can the inductive hypothesis help verify that \( P(k+1) \) is true?

**Tip:** In proofs by induction, it is crucial to ensure both the base case and the inductive step are correctly handled to complete the proof.

Explore the mathematical proof using induction for the recurrence relation f_k = f_{k-1} + 2^k. Understand how to solve for the left-hand side of P(1), and see how the inductive hypothesis helps verify the correctness of P(k+1). Suitable for students in Grades 10-12.

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