Answer :
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]
