Find the solution to each of these recurrence relations with the given initial conditions. Use an iterative approach. a) a n =-a n-1 , a_{0} = 5 b ) a n =a n-1 +3 , a_{0} = 1 ) a n =a n-1 -n , a_{0} = 4 d) a n =2a n-1 -3 , a_{0} = - 1 e) a n =(n+1)a n-1 , a_{0} = 2 f) an 2nan a_{0} = 3

Let's solve each of the recurrence relations using an iterative approach and the given initial conditions.

### (a) \( a_n = -a_{n-1} \), \( a_0 = 5 \)

#### Step-by-step calculation:
- \( a_0 = 5 \) (given)
- \( a_1 = -a_0 = -5 \)
- \( a_2 = -a_1 = 5 \)
- \( a_3 = -a_2 = -5 \)
- \( a_4 = -a_3 = 5 \)

Thus, the sequence alternates between 5 and -5:
\[
a_n = (-1)^n \cdot 5
\]

---

### (b) \( a_n = a_{n-1} + 3 \), \( a_0 = 1 \)

#### Step-by-step calculation:
- \( a_0 = 1 \) (given)
- \( a_1 = a_0 + 3 = 1 + 3 = 4 \)
- \( a_2 = a_1 + 3 = 4 + 3 = 7 \)
- \( a_3 = a_2 + 3 = 7 + 3 = 10 \)
- \( a_4 = a_3 + 3 = 10 + 3 = 13 \)

This is an arithmetic progression with a common difference of 3. The general formula is:
\[
a_n = 1 + 3n
\]

---

### (c) \( a_n = a_{n-1} - n \), \( a_0 = 4 \)

#### Step-by-step calculation:
- \( a_0 = 4 \) (given)
- \( a_1 = a_0 - 1 = 4 - 1 = 3 \)
- \( a_2 = a_1 - 2 = 3 - 2 = 1 \)
- \( a_3 = a_2 - 3 = 1 - 3 = -2 \)
- \( a_4 = a_3 - 4 = -2 - 4 = -6 \)

This is a decreasing sequence, and the general solution can be found by summing up the decreasing values of \(n\):
\[
a_n = 4 - \frac{n(n+1)}{2}
\]

---

### (d) \( a_n = 2a_{n-1} - 3 \), \( a_0 = -1 \)

#### Step-by-step calculation:
- \( a_0 = -1 \) (given)
- \( a_1 = 2a_0 - 3 = 2(-1) - 3 = -2 - 3 = -5 \)
- \( a_2 = 2a_1 - 3 = 2(-5) - 3 = -10 - 3 = -13 \)
- \( a_3 = 2a_2 - 3 = 2(-13) - 3 = -26 - 3 = -29 \)
- \( a_4 = 2a_3 - 3 = 2(-29) - 3 = -58 - 3 = -61 \)

This sequence follows the pattern:
\[
a_n = -2^{n+1} + 1
\]

---

### (e) \( a_n = (n+1)a_{n-1} \), \( a_0 = 2 \)

#### Step-by-step calculation:
- \( a_0 = 2 \) (given)
- \( a_1 = (1+1) \cdot a_0 = 2 \cdot 2 = 4 \)
- \( a_2 = (2+1) \cdot a_1 = 3 \cdot 4 = 12 \)
- \( a_3 = (3+1) \cdot a_2 = 4 \cdot 12 = 48 \)
- \( a_4 = (4+1) \cdot a_3 = 5 \cdot 48 = 240 \)

The sequence grows very quickly. The general formula is:
\[
a_n = 2 \cdot (n+1)!
\]

---

### (f) \( a_n = 2na_{n-1} \), \( a_0 = 3 \)

#### Step-by-step calculation:
- \( a_0 = 3 \) (given)
- \( a_1 = 2 \cdot 1 \cdot a_0 = 2 \cdot 1 \cdot 3 = 6 \)
- \( a_2 = 2 \cdot 2 \cdot a_1 = 2 \cdot 2 \cdot 6 = 24 \)
- \( a_3 = 2 \cdot 3 \cdot a_2 = 2 \cdot 3 \cdot 24 = 144 \)
- \( a_4 = 2 \cdot 4 \cdot a_3 = 2 \cdot 4 \cdot 144 = 1152 \)

The general form for this sequence is:
\[
a_n = 3 \cdot 2^n \cdot n!
\]

---

Would you like more details or have any questions?  
Here are some relative questions:

1. How do we derive the closed-form solutions for recurrence relations?
2. What is the significance of factorials in sequences like part (e)?
3. How does the iterative method help in understanding patterns in recurrence relations?
4. Can you solve non-linear recurrence relations using similar approaches?
5. How would a recurrence relation behave with a different initial condition?

**Tip:** For recurrence relations, always check if there's a pattern in the first few terms to help you guess a closed-form solution.

Find the solution to each of these recurrence relations with the given initial conditions. Use an iterative approach. a) a_n = -a_{n-1}, a_0 = 5 b) a_n = a_{n-1} + 3, a_0 = 1 c) a_n = a_{n-1} - n, a_0 = 4 d) a_n = 2a_{n-1} - 3, a_0 = -1 e) a_n = (n+1)a_{n-1}, a_0 = 2 f) a_n = 2na_{n-1}, a_0 = 3

Step-by-step solutions to recurrence relations like a_n = -a_{n-1}, a_n = a_{n-1} + 3, and others. Learn how to use an iterative method to find sequences and closed-form formulas. Suitable for high school students in Grades 10-12.

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