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find the simplest formula in this problem possible

The recurrence relation provided is:

\[
T(0) = 1
\]
\[
T(n) = 2 \cdot T(n-1) + 2 \quad \text{for } n \geq 1.
\]

To find the simplest closed-form solution, let's solve this step-by-step:

---

### Step 1: Expand the recurrence relation
Expand \( T(n) \) in terms of previous terms:

\[
T(n) = 2 \cdot T(n-1) + 2
\]
\[
T(n-1) = 2 \cdot T(n-2) + 2
\]

Substituting \( T(n-1) \) into \( T(n) \):

\[
T(n) = 2 \cdot \big(2 \cdot T(n-2) + 2\big) + 2
\]
\[
T(n) = 2^2 \cdot T(n-2) + 2^2 + 2.
\]

Expanding further for \( T(n-2) \):

\[
T(n-2) = 2 \cdot T(n-3) + 2
\]
\[
T(n) = 2^3 \cdot T(n-3) + 2^3 + 2^2 + 2.
\]

---

### Step 2: Generalize the pattern
By continuing this expansion, we observe the pattern:
\[
T(n) = 2^k \cdot T(n-k) + 2^k + 2^{k-1} + \cdots + 2^1 + 2,
\]
where \( k = n \) when \( T(n-k) = T(0) \).

---

### Step 3: Substitute \( T(0) = 1 \)
Substitute \( T(0) = 1 \) when \( k = n \):
\[
T(n) = 2^n \cdot T(0) + \big(2^n + 2^{n-1} + \cdots + 2^1 + 2\big).
\]
\[
T(n) = 2^n \cdot 1 + \big(2^n - 1\big),
\]
where the sum of the geometric series \( 2^n + 2^{n-1} + \cdots + 2 \) simplifies to \( 2^{n+1} - 2 \).

---

### Step 4: Simplify the closed form
\[
T(n) = 2^n + (2^{n+1} - 2).
\]
\[
T(n) = 2^{n+1} + 2^n - 2.
\]

---

### Final Closed-Form Formula:
\[
T(n) = 3 \cdot 2^n - 2.
\]

---

Would you like a detailed explanation of each step or help applying this formula to specific values?  

Here are related questions you can explore:
1. How does the closed-form formula verify \( T(0) = 1 \)?
2. What happens to \( T(n) \) as \( n \to \infty \)?
3. How do geometric series help solve recurrence relations?
4. Can you derive \( T(n) \) using iterative techniques instead?
5. How does this formula change if \( T(0) \) is a different value?

**Tip:** When solving recurrence relations, always look for patterns in expansions to identify sums or sequences like geometric progressions.

Explore the solution to the recurrence relation T(n) = 2 * T(n-1) + 2 with a detailed step-by-step explanation. Learn how to solve recurrence relations, simplify geometric sums, and derive closed-form expressions. Suitable for students in Grades 10-12.

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