https://file.aiquickdraw.com/mathbot/file/1729034065538ayqtgyal.jpg

Let's tackle the questions one by one based on the information given.

### Question 8: Find \( F_7, F_8, F_9, F_{10} \), and \( F_{11} \).

We know that the Fibonacci sequence is defined as:
\[
F_n = F_{n-1} + F_{n-2}
\]
Starting from the given values:
- \( F_1 = 1 \)
- \( F_2 = 1 \)
- \( F_3 = 2 \)
- \( F_4 = 3 \)
- \( F_5 = 5 \)
- \( F_6 = 8 \)

Let's continue the sequence:

1. **For \( F_7 \):**
   \[
   F_7 = F_6 + F_5 = 8 + 5 = 13
   \]

2. **For \( F_8 \):**
   \[
   F_8 = F_7 + F_6 = 13 + 8 = 21
   \]

3. **For \( F_9 \):**
   \[
   F_9 = F_8 + F_7 = 21 + 13 = 34
   \]

4. **For \( F_{10} \):**
   \[
   F_{10} = F_9 + F_8 = 34 + 21 = 55
   \]

5. **For \( F_{11} \):**
   \[
   F_{11} = F_{10} + F_9 = 55 + 34 = 89
   \]

Therefore, the terms are:
\[
F_7 = 13, \quad F_8 = 21, \quad F_9 = 34, \quad F_{10} = 55, \quad F_{11} = 89
\]

### Question 9: For all \( n \in \mathbb{Z}^+ \), prove that \( F_1 + F_3 + F_5 + \dots + F_{2n-1} = F_{2n} \).

This statement implies that the sum of the odd-indexed Fibonacci numbers up to \( F_{2n-1} \) is equal to \( F_{2n} \).

#### Proof by Induction:
1. **Base Case: \( n = 1 \)**  
   \[
   F_1 = F_2 = 1
   \]  
   So, the base case holds.

2. **Inductive Step:**  
   Assume \( F_1 + F_3 + F_5 + \dots + F_{2k-1} = F_{2k} \) holds.  
   We need to prove that \( F_1 + F_3 + F_5 + \dots + F_{2k-1} + F_{2k+1} = F_{2k+2} \).
   
   Since \( F_{2k+2} = F_{2k+1} + F_{2k} \) and by our inductive hypothesis \( F_1 + F_3 + \dots + F_{2k-1} = F_{2k} \), we have:
   \[
   F_1 + F_3 + \dots + F_{2k+1} = F_{2k} + F_{2k+1} = F_{2k+2}
   \]
   Thus, the statement holds by induction.

### Question 10: For all \( n \in \mathbb{Z}^+ \), prove that \( F_1^2 + F_2^2 + F_3^2 + \dots + F_n^2 = F_n F_{n+1} \).

This states that the sum of the squares of the first \( n \) Fibonacci numbers is equal to the product of \( F_n \) and \( F_{n+1} \).

#### Proof by Induction:
1. **Base Case: \( n = 1 \)**  
   \[
   F_1^2 = 1 \quad \text{and} \quad F_1 F_2 = 1 \times 1 = 1
   \]  
   So, the base case holds.

2. **Inductive Step:**  
   Assume \( F_1^2 + F_2^2 + \dots + F_k^2 = F_k F_{k+1} \) holds for \( n = k \).  
   We need to prove \( F_1^2 + F_2^2 + \dots + F_k^2 + F_{k+1}^2 = F_{k+1} F_{k+2} \).
   
   From the inductive hypothesis, \( F_1^2 + F_2^2 + \dots + F_k^2 = F_k F_{k+1} \).  
   Adding \( F_{k+1}^2 \) to both sides, we get:
   \[
   F_1^2 + F_2^2 + \dots + F_k^2 + F_{k+1}^2 = F_k F_{k+1} + F_{k+1}^2
   \]
   \[
   = F_{k+1}(F_k + F_{k+1}) = F_{k+1} F_{k+2}
   \]
   Therefore, the statement holds by induction.

Would you like further details on any of these proofs or have any questions?

---

### Related Questions:
1. What is the general formula for the Fibonacci sequence?
2. How can we express Fibonacci numbers using matrix exponentiation?
3. Can the sum of even-indexed Fibonacci numbers be represented similarly?
4. What is the closed-form formula for Fibonacci numbers known as Binet's formula?
5. How can we prove that every nth Fibonacci number is divisible by \( F_n \)?

### Tip:
To explore Fibonacci numbers further, try deriving their properties using Binet's formula. It provides an exact form and can be helpful in proofs involving Fibonacci identities.

Find F_7, F_8, F_9, F_10, and F_11. Prove that the sum of odd-indexed Fibonacci numbers up to F_{2n-1} equals F_{2n}, and that the sum of the squares of the first n Fibonacci numbers equals F_n * F_{n+1}.

Learn how to calculate Fibonacci numbers F_7 to F_11 and explore the properties of Fibonacci sums. Discover how the sum of odd-indexed terms relates to even-indexed terms, and how the sum of squares of Fibonacci numbers connects to their product. Includes step-by-step solutions and induction proofs.

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