Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Let's study each sequence [tex]\((u_n)\)[/tex] as defined in the problem, step-by-step.
### (a) Sequence Defined by [tex]\( u_1 = 1, u_n = \frac{u_n + 2}{u_n + 1} \)[/tex]
The sequence starts with [tex]\( u_1 = 1 \)[/tex].
- Base case: [tex]\( u_1 = 1 \)[/tex].
- Compute the next terms recursively using:
[tex]\[ u_{n+1} = \frac{u_n + 2}{u_n + 1} \][/tex]
For [tex]\( n = 5 \)[/tex]:
- [tex]\( u_2 = \frac{u_1 + 2}{u_1 + 1} = \frac{1 + 2}{1 + 1} = \frac{3}{2} = 1.5 \)[/tex]
- [tex]\( u_3 = \frac{u_2 + 2}{u_2 + 1} = \frac{1.5 + 2}{1.5 + 1} = \frac{3.5}{2.5} = 1.4 \)[/tex]
- [tex]\( u_4 = \frac{u_3 + 2}{u_3 + 1} = \frac{1.4 + 2}{1.4 + 1} \approx 1.3846 \)[/tex]
- [tex]\( u_5 = \frac{u_4 + 2}{u_4 + 1} \approx \frac{1.3846 + 2}{1.3846 + 1} = 1.4137931034482758 \)[/tex]
Therefore, after 5 terms, the fifth term is approximately:
[tex]\[ u_5 \approx 1.4138 \][/tex]
### (b) Sequence Defined by [tex]\( u_0 = a, u_{n+1} = u_n^3 \)[/tex]
Given [tex]\( a=2 \)[/tex], let's find the fifth term [tex]\( u_5 \)[/tex]:
- [tex]\( u_0 = 2 \)[/tex]
- [tex]\( u_1 = u_0^3 = 2^3 = 8 \)[/tex]
- [tex]\( u_2 = u_1^3 = 8^3 = 512 \)[/tex]
- [tex]\( u_3 = u_2^3 = 512^3 = 134217728 \)[/tex]
- [tex]\( u_4 = u_3^3 = 134217728^3 = 2417851639229258349412352 \)[/tex]
- [tex]\( u_5 = u_4^3 = (2417851639229258349412352)^3 \)[/tex]
The value of [tex]\( u_5 \)[/tex] is a very large number, approximately:
[tex]\[ u_5 \approx 14134776518227074636666380005943348126619871175004951664972849610340958208 \][/tex]
### (c) Sequence Defined by [tex]\( u_0 > 0, u_{n+1} = \sqrt{u_n + \sqrt{u_{n-1} + \cdots + \sqrt{u_0}}} \)[/tex]
Given [tex]\( u_0 = 1 \)[/tex], let's calculate [tex]\( u_5 \)[/tex]:
- [tex]\( u_0 = 1 \)[/tex]
- [tex]\( u_1 = \sqrt{u_0} = \sqrt{1} = 1 \)[/tex]
- [tex]\( u_2 = \sqrt{u_1 + \sqrt{u_0}} = \sqrt{1 + \sqrt{1}} = \sqrt{1 + 1} = \sqrt{2} \approx 1.4142 \)[/tex]
- [tex]\( u_3 = \sqrt{u_2 + \sqrt{u_1 + \sqrt{u_0}}} = \sqrt{1.4142 + \sqrt{1 + 1}} \approx \sqrt{1.4142 + 1.4142} = \sqrt{2.8284} \approx 1.6818 \)[/tex]
- Continue this nested square root process for further terms.
After iterating sufficiently for [tex]\( n = 5 \)[/tex], we obtain:
[tex]\[ u_5 \approx 1.6343 \][/tex]
### (d) Sequence Defined by [tex]\( u_0 = 1, u_{n+1} = e^{u_n} - 2 \)[/tex]
Starting with [tex]\( u_0 = 1 \)[/tex], let's find [tex]\( u_5 \)[/tex]:
- [tex]\( u_0 = 1 \)[/tex]
- [tex]\( u_1 = e^{u_0} - 2 = e^1 - 2 = e - 2 \approx 0.7183 \)[/tex]
- [tex]\( u_2 = e^{u_1} - 2 \approx e^{0.7183} - 2 \approx 2.0511 - 2 = 0.0511 \)[/tex]
- [tex]\( u_3 = e^{u_2} - 2 \approx e^{0.0511} - 2 \approx 1.0525 - 2 = -0.9475 \)[/tex]
- Each term depends on the exponential of the previous term, which results in an increasingly negative sequence.
After iterating to [tex]\( n=5 \)[/tex]:
[tex]\[ u_5 \approx -1.8006 \][/tex]
### (e) Sequence Defined by [tex]\( u_0 > 0, u_{n+1} = \frac{u_n^2 + 3}{2(u_n + 1)} \)[/tex]
Given [tex]\( u_0 = 1 \)[/tex], let's compute [tex]\( u_5 \)[/tex]:
- [tex]\( u_0 = 1 \)[/tex]
- [tex]\( u_1 = \frac{u_0^2 + 3}{2(u_0 + 1)} = \frac{1^2 + 3}{2(1 + 1)} = \frac{4}{4} = 1 \)[/tex]
- Each subsequent term, due to the structure of the recurrence relation, will similarly evaluate to:
[tex]\[ u_n = 1 \][/tex]
Thus:
[tex]\[ u_5 = 1 \][/tex]
In conclusion, the sequences are:
(a) [tex]\( u_5 \approx 1.4138 \)[/tex]
(b) [tex]\( u_5 \approx 14134776518227074636666380005943348126619871175004951664972849610340958208 \)[/tex]
(c) [tex]\( u_5 \approx 1.6343 \)[/tex]
(d) [tex]\( u_5 \approx -1.8006 \)[/tex]
(e) [tex]\( u_5 = 1 \)[/tex]
### (a) Sequence Defined by [tex]\( u_1 = 1, u_n = \frac{u_n + 2}{u_n + 1} \)[/tex]
The sequence starts with [tex]\( u_1 = 1 \)[/tex].
- Base case: [tex]\( u_1 = 1 \)[/tex].
- Compute the next terms recursively using:
[tex]\[ u_{n+1} = \frac{u_n + 2}{u_n + 1} \][/tex]
For [tex]\( n = 5 \)[/tex]:
- [tex]\( u_2 = \frac{u_1 + 2}{u_1 + 1} = \frac{1 + 2}{1 + 1} = \frac{3}{2} = 1.5 \)[/tex]
- [tex]\( u_3 = \frac{u_2 + 2}{u_2 + 1} = \frac{1.5 + 2}{1.5 + 1} = \frac{3.5}{2.5} = 1.4 \)[/tex]
- [tex]\( u_4 = \frac{u_3 + 2}{u_3 + 1} = \frac{1.4 + 2}{1.4 + 1} \approx 1.3846 \)[/tex]
- [tex]\( u_5 = \frac{u_4 + 2}{u_4 + 1} \approx \frac{1.3846 + 2}{1.3846 + 1} = 1.4137931034482758 \)[/tex]
Therefore, after 5 terms, the fifth term is approximately:
[tex]\[ u_5 \approx 1.4138 \][/tex]
### (b) Sequence Defined by [tex]\( u_0 = a, u_{n+1} = u_n^3 \)[/tex]
Given [tex]\( a=2 \)[/tex], let's find the fifth term [tex]\( u_5 \)[/tex]:
- [tex]\( u_0 = 2 \)[/tex]
- [tex]\( u_1 = u_0^3 = 2^3 = 8 \)[/tex]
- [tex]\( u_2 = u_1^3 = 8^3 = 512 \)[/tex]
- [tex]\( u_3 = u_2^3 = 512^3 = 134217728 \)[/tex]
- [tex]\( u_4 = u_3^3 = 134217728^3 = 2417851639229258349412352 \)[/tex]
- [tex]\( u_5 = u_4^3 = (2417851639229258349412352)^3 \)[/tex]
The value of [tex]\( u_5 \)[/tex] is a very large number, approximately:
[tex]\[ u_5 \approx 14134776518227074636666380005943348126619871175004951664972849610340958208 \][/tex]
### (c) Sequence Defined by [tex]\( u_0 > 0, u_{n+1} = \sqrt{u_n + \sqrt{u_{n-1} + \cdots + \sqrt{u_0}}} \)[/tex]
Given [tex]\( u_0 = 1 \)[/tex], let's calculate [tex]\( u_5 \)[/tex]:
- [tex]\( u_0 = 1 \)[/tex]
- [tex]\( u_1 = \sqrt{u_0} = \sqrt{1} = 1 \)[/tex]
- [tex]\( u_2 = \sqrt{u_1 + \sqrt{u_0}} = \sqrt{1 + \sqrt{1}} = \sqrt{1 + 1} = \sqrt{2} \approx 1.4142 \)[/tex]
- [tex]\( u_3 = \sqrt{u_2 + \sqrt{u_1 + \sqrt{u_0}}} = \sqrt{1.4142 + \sqrt{1 + 1}} \approx \sqrt{1.4142 + 1.4142} = \sqrt{2.8284} \approx 1.6818 \)[/tex]
- Continue this nested square root process for further terms.
After iterating sufficiently for [tex]\( n = 5 \)[/tex], we obtain:
[tex]\[ u_5 \approx 1.6343 \][/tex]
### (d) Sequence Defined by [tex]\( u_0 = 1, u_{n+1} = e^{u_n} - 2 \)[/tex]
Starting with [tex]\( u_0 = 1 \)[/tex], let's find [tex]\( u_5 \)[/tex]:
- [tex]\( u_0 = 1 \)[/tex]
- [tex]\( u_1 = e^{u_0} - 2 = e^1 - 2 = e - 2 \approx 0.7183 \)[/tex]
- [tex]\( u_2 = e^{u_1} - 2 \approx e^{0.7183} - 2 \approx 2.0511 - 2 = 0.0511 \)[/tex]
- [tex]\( u_3 = e^{u_2} - 2 \approx e^{0.0511} - 2 \approx 1.0525 - 2 = -0.9475 \)[/tex]
- Each term depends on the exponential of the previous term, which results in an increasingly negative sequence.
After iterating to [tex]\( n=5 \)[/tex]:
[tex]\[ u_5 \approx -1.8006 \][/tex]
### (e) Sequence Defined by [tex]\( u_0 > 0, u_{n+1} = \frac{u_n^2 + 3}{2(u_n + 1)} \)[/tex]
Given [tex]\( u_0 = 1 \)[/tex], let's compute [tex]\( u_5 \)[/tex]:
- [tex]\( u_0 = 1 \)[/tex]
- [tex]\( u_1 = \frac{u_0^2 + 3}{2(u_0 + 1)} = \frac{1^2 + 3}{2(1 + 1)} = \frac{4}{4} = 1 \)[/tex]
- Each subsequent term, due to the structure of the recurrence relation, will similarly evaluate to:
[tex]\[ u_n = 1 \][/tex]
Thus:
[tex]\[ u_5 = 1 \][/tex]
In conclusion, the sequences are:
(a) [tex]\( u_5 \approx 1.4138 \)[/tex]
(b) [tex]\( u_5 \approx 14134776518227074636666380005943348126619871175004951664972849610340958208 \)[/tex]
(c) [tex]\( u_5 \approx 1.6343 \)[/tex]
(d) [tex]\( u_5 \approx -1.8006 \)[/tex]
(e) [tex]\( u_5 = 1 \)[/tex]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. We hope this was helpful. Please come back whenever you need more information or answers to your queries. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
