Find the closed form for each T(n) given as a recurrence: 1. : n=1 (Т(п - 1) + 2 : п22 2 T(n) 2. : n=1 T'п - 1) + 4n — 3 : п>1 2 T(n) 3. : n= 1 2 (п — 1) — 1 : п>2 2 T(n) = {}
Find the closed form for each T(n) given as a recurrence: 1. : n=1 (Т(п - 1) + 2 : п22 2 T(n) 2. : n=1 T'п - 1) + 4n — 3 : п>1 2 T(n) 3. : n= 1 2 (п — 1) — 1 : п>2 2 T(n) = {}
Computer Networking: A Top-Down Approach (7th Edition)
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Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
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![### Recurrence Relations: Closed Form Solutions
#### Problem Statement:
Find the closed form for each \( T(n) \) given as a recurrence:
1.
\[
T(n) =
\begin{cases}
2 & : \; n = 1 \\
T(n-1) + 2 & : \; n \ge 2
\end{cases}
\]
2.
\[
T(n) =
\begin{cases}
2 & : \; n = 1 \\
T(n-1) + 4n - 3 & : \; n > 1
\end{cases}
\]
3.
\[
T(n) =
\begin{cases}
2 & : \; n = 1 \\
2T(n-1) - 1 & : \; n \ge 2
\end{cases}
\]
4.
\[
T(m) =
\begin{cases}
0 & : \; m = 1 \\
2T(m-1) + m - 1 & : \; m > 1
\end{cases}
\]
5. Let \( n = 2^m - 1 \). Rewrite your answer of the previous problem as a function of \( n \).
### Detailed Insights:
1. **Recurrence Relation 1:**
- Initial condition: \( T(1) = 2 \)
- Recursive step: \( T(n) = T(n-1) + 2 \) for \( n \ge 2 \)
2. **Recurrence Relation 2:**
- Initial condition: \( T(1) = 2 \)
- Recursive step: \( T(n) = T(n-1) + 4n - 3 \) for \( n > 1 \)
3. **Recurrence Relation 3:**
- Initial condition: \( T(1) = 2 \)
- Recursive step: \( T(n) = 2T(n-1) - 1 \) for \( n \ge 2 \)
4. **Recurrence Relation 4:**
- Initial condition: \( T(m) = 0 \)
- Recursive step: \( T(m) = 2T(m-1) + m - 1 \) for \( m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc8241d13-b770-4dad-946c-6942db4056c7%2Fc5807206-4a34-46ba-852b-910d76ec8015%2Fu2kn92j_processed.png&w=3840&q=75)
Transcribed Image Text:### Recurrence Relations: Closed Form Solutions
#### Problem Statement:
Find the closed form for each \( T(n) \) given as a recurrence:
1.
\[
T(n) =
\begin{cases}
2 & : \; n = 1 \\
T(n-1) + 2 & : \; n \ge 2
\end{cases}
\]
2.
\[
T(n) =
\begin{cases}
2 & : \; n = 1 \\
T(n-1) + 4n - 3 & : \; n > 1
\end{cases}
\]
3.
\[
T(n) =
\begin{cases}
2 & : \; n = 1 \\
2T(n-1) - 1 & : \; n \ge 2
\end{cases}
\]
4.
\[
T(m) =
\begin{cases}
0 & : \; m = 1 \\
2T(m-1) + m - 1 & : \; m > 1
\end{cases}
\]
5. Let \( n = 2^m - 1 \). Rewrite your answer of the previous problem as a function of \( n \).
### Detailed Insights:
1. **Recurrence Relation 1:**
- Initial condition: \( T(1) = 2 \)
- Recursive step: \( T(n) = T(n-1) + 2 \) for \( n \ge 2 \)
2. **Recurrence Relation 2:**
- Initial condition: \( T(1) = 2 \)
- Recursive step: \( T(n) = T(n-1) + 4n - 3 \) for \( n > 1 \)
3. **Recurrence Relation 3:**
- Initial condition: \( T(1) = 2 \)
- Recursive step: \( T(n) = 2T(n-1) - 1 \) for \( n \ge 2 \)
4. **Recurrence Relation 4:**
- Initial condition: \( T(m) = 0 \)
- Recursive step: \( T(m) = 2T(m-1) + m - 1 \) for \( m
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