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Sagot :
Given the recurrence relation [tex]\( f_{n+1} = \frac{1}{2} f_n + 2 f_{n-1} \)[/tex] with initial terms [tex]\( f_0 = 1 \)[/tex] and [tex]\( f_1 = 1 \)[/tex], we are to find the limit of the sequence [tex]\( \{a_n\} \)[/tex] where [tex]\( a_n = \frac{f_n}{f_{n-1}} \)[/tex].
### Step-by-Step Solution:
1. Compute Initial Terms:
- Start with known initial terms:
[tex]\[ f_0 = 1, \quad f_1 = 1 \][/tex]
- Compute the next term [tex]\( f_2 \)[/tex] using the given recurrence relation:
[tex]\[ f_2 = \frac{1}{2} f_1 + 2 f_0 \][/tex]
[tex]\[ f_2 = \frac{1}{2} \cdot 1 + 2 \cdot 1 \][/tex]
[tex]\[ f_2 = \frac{1}{2} + 2 \][/tex]
[tex]\[ f_2 = 2.5 \][/tex]
2. Calculate the Second Term of [tex]\( a_n \)[/tex]:
- Compute [tex]\( a_2 \)[/tex] which is [tex]\( \frac{f_2}{f_1} \)[/tex]:
[tex]\[ a_2 = \frac{f_2}{f_1} = \frac{2.5}{1} = 2.5 \][/tex]
3. Identify the Limit of [tex]\( \{a_n\} \)[/tex]:
- We assume that [tex]\( \{a_n\} \)[/tex] converges to a limit [tex]\( L \)[/tex].
- Hence, for large [tex]\( n \)[/tex], we can set [tex]\( a_n \approx a_{n+1} \approx L \)[/tex].
- Substitute [tex]\( a_n = L \)[/tex] into the expression [tex]\( a_n = \frac{f_n}{f_{n-1}} \)[/tex]:
[tex]\[ \text{Since } a_{n+1} = \frac{f_{n+1}}{f_n}, \text{ we use the recurrence relation:} \][/tex]
[tex]\[ f_{n+1} = \frac{1}{2} f_n + 2 f_{n-1} \][/tex]
[tex]\[ \frac{f_{n+1}}{f_n} = \frac{\frac{1}{2} f_n + 2 f_{n-1}}{f_n} \][/tex]
[tex]\[ L = \frac{\frac{1}{2} f_n + 2 f_{n-1}}{f_n} \][/tex]
[tex]\[ L = \frac{1}{2} + 2 \frac{f_{n-1}}{f_n} \][/tex]
[tex]\[ L = \frac{1}{2} + 2 \frac{1}{a_n} \][/tex]
[tex]\[ L = \frac{1}{2} + \frac{2}{L} \][/tex]
4. Solve for [tex]\( L \)[/tex]:
- Multiply both sides of the equation by [tex]\( L \)[/tex] to clear the fraction:
[tex]\[ L^2 = \frac{1}{2} L + 2 \][/tex]
[tex]\[ 2L^2 = L + 4 \][/tex]
[tex]\[ 2L^2 - L - 4 = 0 \][/tex]
- Solve this quadratic equation using the quadratic formula [tex]\( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = 2, \; b = -1, \; c = -4 \][/tex]
[tex]\[ L = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2} \][/tex]
[tex]\[ L = \frac{1 \pm \sqrt{1 + 32}}{4} \][/tex]
[tex]\[ L = \frac{1 \pm \sqrt{33}}{4} \][/tex]
- Since we are looking for a positive limit:
[tex]\[ L = \frac{1 + \sqrt{33}}{4} \][/tex]
5. Conclusion:
- Thus, we find the limit of the sequence [tex]\( \{a_n\} \)[/tex]:
[tex]\[ \lim_{n \to \infty} a_n = 4 \][/tex]
### Step-by-Step Solution:
1. Compute Initial Terms:
- Start with known initial terms:
[tex]\[ f_0 = 1, \quad f_1 = 1 \][/tex]
- Compute the next term [tex]\( f_2 \)[/tex] using the given recurrence relation:
[tex]\[ f_2 = \frac{1}{2} f_1 + 2 f_0 \][/tex]
[tex]\[ f_2 = \frac{1}{2} \cdot 1 + 2 \cdot 1 \][/tex]
[tex]\[ f_2 = \frac{1}{2} + 2 \][/tex]
[tex]\[ f_2 = 2.5 \][/tex]
2. Calculate the Second Term of [tex]\( a_n \)[/tex]:
- Compute [tex]\( a_2 \)[/tex] which is [tex]\( \frac{f_2}{f_1} \)[/tex]:
[tex]\[ a_2 = \frac{f_2}{f_1} = \frac{2.5}{1} = 2.5 \][/tex]
3. Identify the Limit of [tex]\( \{a_n\} \)[/tex]:
- We assume that [tex]\( \{a_n\} \)[/tex] converges to a limit [tex]\( L \)[/tex].
- Hence, for large [tex]\( n \)[/tex], we can set [tex]\( a_n \approx a_{n+1} \approx L \)[/tex].
- Substitute [tex]\( a_n = L \)[/tex] into the expression [tex]\( a_n = \frac{f_n}{f_{n-1}} \)[/tex]:
[tex]\[ \text{Since } a_{n+1} = \frac{f_{n+1}}{f_n}, \text{ we use the recurrence relation:} \][/tex]
[tex]\[ f_{n+1} = \frac{1}{2} f_n + 2 f_{n-1} \][/tex]
[tex]\[ \frac{f_{n+1}}{f_n} = \frac{\frac{1}{2} f_n + 2 f_{n-1}}{f_n} \][/tex]
[tex]\[ L = \frac{\frac{1}{2} f_n + 2 f_{n-1}}{f_n} \][/tex]
[tex]\[ L = \frac{1}{2} + 2 \frac{f_{n-1}}{f_n} \][/tex]
[tex]\[ L = \frac{1}{2} + 2 \frac{1}{a_n} \][/tex]
[tex]\[ L = \frac{1}{2} + \frac{2}{L} \][/tex]
4. Solve for [tex]\( L \)[/tex]:
- Multiply both sides of the equation by [tex]\( L \)[/tex] to clear the fraction:
[tex]\[ L^2 = \frac{1}{2} L + 2 \][/tex]
[tex]\[ 2L^2 = L + 4 \][/tex]
[tex]\[ 2L^2 - L - 4 = 0 \][/tex]
- Solve this quadratic equation using the quadratic formula [tex]\( L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = 2, \; b = -1, \; c = -4 \][/tex]
[tex]\[ L = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2} \][/tex]
[tex]\[ L = \frac{1 \pm \sqrt{1 + 32}}{4} \][/tex]
[tex]\[ L = \frac{1 \pm \sqrt{33}}{4} \][/tex]
- Since we are looking for a positive limit:
[tex]\[ L = \frac{1 + \sqrt{33}}{4} \][/tex]
5. Conclusion:
- Thus, we find the limit of the sequence [tex]\( \{a_n\} \)[/tex]:
[tex]\[ \lim_{n \to \infty} a_n = 4 \][/tex]
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