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Sagot :
Certainly! Let's find the first five terms of each sequence using the provided recurrence relations.
### Part (a)
Given:
- Initial term: [tex]\( t_1 = -1 \)[/tex]
- Recurrence relation: [tex]\( t_n = t_{n-1} + 2 \)[/tex] for [tex]\( n \geq 2 \)[/tex]
Solution:
1. [tex]\( t_1 = -1 \)[/tex]
2. [tex]\( t_2 = t_1 + 2 = -1 + 2 = 1 \)[/tex]
3. [tex]\( t_3 = t_2 + 2 = 1 + 2 = 3 \)[/tex]
4. [tex]\( t_4 = t_3 + 2 = 3 + 2 = 5 \)[/tex]
5. [tex]\( t_5 = t_4 + 2 = 5 + 2 = 7 \)[/tex]
So, the first five terms of the sequence are:
[tex]\[ \{-1, 1, 3, 5, 7\} \][/tex]
### Part (b)
Given:
- Initial terms: [tex]\( t_1 = 2 \)[/tex], [tex]\( t_2 = 2 \)[/tex]
- Recurrence relation: [tex]\( t_n = t_{n-1} - 1 \)[/tex] for [tex]\( n > 2 \)[/tex]
Solution:
1. [tex]\( t_1 = 2 \)[/tex]
2. [tex]\( t_2 = 2 \)[/tex]
3. [tex]\( t_3 = t_2 - 1 = 2 - 1 = 1 \)[/tex]
4. [tex]\( t_4 = t_3 - 1 = 1 - 1 = 0 \)[/tex]
5. [tex]\( t_5 = t_4 - 1 = 0 - 1 = -1 \)[/tex]
So, the first five terms of the sequence are:
[tex]\[ \{2, 2, 1, 0, -1\} \][/tex]
### Part (c)
Given:
- Initial term: [tex]\( t_1 = -1 \)[/tex]
- Recurrence relation: [tex]\( t_n = \frac{t_{n-1}}{n} \)[/tex] for [tex]\( n \geq 2 \)[/tex]
Solution:
1. [tex]\( t_1 = -1 \)[/tex]
2. [tex]\( t_2 = \frac{t_1}{2} = \frac{-1}{2} = -0.5 \)[/tex]
3. [tex]\( t_3 = \frac{t_2}{3} = \frac{-0.5}{3} = -\frac{1}{6} \approx -0.1667 \)[/tex]
4. [tex]\( t_4 = \frac{t_3}{4} = \frac{-\frac{1}{6}}{4} = -\frac{1}{24} \approx -0.0417 \)[/tex]
5. [tex]\( t_5 = \frac{t_4}{5} = \frac{-\frac{1}{24}}{5} = -\frac{1}{120} \approx -0.0083 \)[/tex]
So, the first five terms of the sequence are approximately:
[tex]\[ \{-1, -0.5, -0.1667, -0.0417, -0.0083\} \][/tex]
### Part (d)
Given:
- Initial terms: [tex]\( t_1 = 2 \)[/tex], [tex]\( t_2 = 3 + t_1 = 3 + 2 = 5 \)[/tex]
- Recurrence relation: [tex]\( t_n = 2t_{n-1} + 5 \)[/tex] for [tex]\( n > 2 \)[/tex]
Solution:
1. [tex]\( t_1 = 2 \)[/tex]
2. [tex]\( t_2 = 3 + t_1 = 3 + 2 = 5 \)[/tex]
3. [tex]\( t_3 = 2t_2 + 5 = 2 \cdot 5 + 5 = 10 + 5 = 15 \)[/tex]
4. [tex]\( t_4 = 2t_3 + 5 = 2 \cdot 15 + 5 = 30 + 5 = 35 \)[/tex]
5. [tex]\( t_5 = 2t_4 + 5 = 2 \cdot 35 + 5 = 70 + 5 = 75 \)[/tex]
So, the first five terms of the sequence are:
[tex]\[ \{2, 5, 15, 35, 75\} \][/tex]
To summarize, the first five terms of the sequences are:
(a) [tex]\(-1, 1, 3, 5, 7\)[/tex]
(b) [tex]\(2, 2, 1, 0, -1\)[/tex]
(c) [tex]\(-1, -0.5, -0.1667, -0.0417, -0.0083\)[/tex]
(d) [tex]\(2, 5, 15, 35, 75\)[/tex]
### Part (a)
Given:
- Initial term: [tex]\( t_1 = -1 \)[/tex]
- Recurrence relation: [tex]\( t_n = t_{n-1} + 2 \)[/tex] for [tex]\( n \geq 2 \)[/tex]
Solution:
1. [tex]\( t_1 = -1 \)[/tex]
2. [tex]\( t_2 = t_1 + 2 = -1 + 2 = 1 \)[/tex]
3. [tex]\( t_3 = t_2 + 2 = 1 + 2 = 3 \)[/tex]
4. [tex]\( t_4 = t_3 + 2 = 3 + 2 = 5 \)[/tex]
5. [tex]\( t_5 = t_4 + 2 = 5 + 2 = 7 \)[/tex]
So, the first five terms of the sequence are:
[tex]\[ \{-1, 1, 3, 5, 7\} \][/tex]
### Part (b)
Given:
- Initial terms: [tex]\( t_1 = 2 \)[/tex], [tex]\( t_2 = 2 \)[/tex]
- Recurrence relation: [tex]\( t_n = t_{n-1} - 1 \)[/tex] for [tex]\( n > 2 \)[/tex]
Solution:
1. [tex]\( t_1 = 2 \)[/tex]
2. [tex]\( t_2 = 2 \)[/tex]
3. [tex]\( t_3 = t_2 - 1 = 2 - 1 = 1 \)[/tex]
4. [tex]\( t_4 = t_3 - 1 = 1 - 1 = 0 \)[/tex]
5. [tex]\( t_5 = t_4 - 1 = 0 - 1 = -1 \)[/tex]
So, the first five terms of the sequence are:
[tex]\[ \{2, 2, 1, 0, -1\} \][/tex]
### Part (c)
Given:
- Initial term: [tex]\( t_1 = -1 \)[/tex]
- Recurrence relation: [tex]\( t_n = \frac{t_{n-1}}{n} \)[/tex] for [tex]\( n \geq 2 \)[/tex]
Solution:
1. [tex]\( t_1 = -1 \)[/tex]
2. [tex]\( t_2 = \frac{t_1}{2} = \frac{-1}{2} = -0.5 \)[/tex]
3. [tex]\( t_3 = \frac{t_2}{3} = \frac{-0.5}{3} = -\frac{1}{6} \approx -0.1667 \)[/tex]
4. [tex]\( t_4 = \frac{t_3}{4} = \frac{-\frac{1}{6}}{4} = -\frac{1}{24} \approx -0.0417 \)[/tex]
5. [tex]\( t_5 = \frac{t_4}{5} = \frac{-\frac{1}{24}}{5} = -\frac{1}{120} \approx -0.0083 \)[/tex]
So, the first five terms of the sequence are approximately:
[tex]\[ \{-1, -0.5, -0.1667, -0.0417, -0.0083\} \][/tex]
### Part (d)
Given:
- Initial terms: [tex]\( t_1 = 2 \)[/tex], [tex]\( t_2 = 3 + t_1 = 3 + 2 = 5 \)[/tex]
- Recurrence relation: [tex]\( t_n = 2t_{n-1} + 5 \)[/tex] for [tex]\( n > 2 \)[/tex]
Solution:
1. [tex]\( t_1 = 2 \)[/tex]
2. [tex]\( t_2 = 3 + t_1 = 3 + 2 = 5 \)[/tex]
3. [tex]\( t_3 = 2t_2 + 5 = 2 \cdot 5 + 5 = 10 + 5 = 15 \)[/tex]
4. [tex]\( t_4 = 2t_3 + 5 = 2 \cdot 15 + 5 = 30 + 5 = 35 \)[/tex]
5. [tex]\( t_5 = 2t_4 + 5 = 2 \cdot 35 + 5 = 70 + 5 = 75 \)[/tex]
So, the first five terms of the sequence are:
[tex]\[ \{2, 5, 15, 35, 75\} \][/tex]
To summarize, the first five terms of the sequences are:
(a) [tex]\(-1, 1, 3, 5, 7\)[/tex]
(b) [tex]\(2, 2, 1, 0, -1\)[/tex]
(c) [tex]\(-1, -0.5, -0.1667, -0.0417, -0.0083\)[/tex]
(d) [tex]\(2, 5, 15, 35, 75\)[/tex]
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