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Sagot :
Certainly! Let's solve the problem step-by-step.
### Part a: Generating the first six terms of the sequence
The sequence is defined as:
[tex]\[ a_1 = 2 \][/tex]
[tex]\[ a_{n+1} = \frac{1}{2} \left(a_n + 4 \right) \][/tex]
Let's calculate the first six terms.
1. For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 2 \][/tex]
2. For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = \frac{1}{2} \left(a_1 + 4 \right) = \frac{1}{2} \left(2 + 4 \right) = \frac{1}{2} \times 6 = 3 \][/tex]
3. For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = \frac{1}{2} \left(a_2 + 4 \right) = \frac{1}{2} \left(3 + 4 \right) = \frac{1}{2} \times 7 = 3.5 \][/tex]
4. For [tex]\( n = 4 \)[/tex]:
[tex]\[ a_4 = \frac{1}{2} \left(a_3 + 4 \right) = \frac{1}{2} \left(3.5 + 4 \right) = \frac{1}{2} \times 7.5 = 3.75 \][/tex]
5. For [tex]\( n = 5 \)[/tex]:
[tex]\[ a_5 = \frac{1}{2} \left(a_4 + 4 \right) = \frac{1}{2} \left(3.75 + 4 \right) = \frac{1}{2} \times 7.75 = 3.875 \][/tex]
6. For [tex]\( n = 6 \)[/tex]:
[tex]\[ a_6 = \frac{1}{2} \left(a_5 + 4 \right) = \frac{1}{2} \left(3.875 + 4 \right) = \frac{1}{2} \times 7.875 = 3.9375 \][/tex]
So, the first six terms of the sequence are:
[tex]\[ \{2, 3, 3.5, 3.75, 3.875, 3.9375\} \][/tex]
### Part b: Proving that [tex]\(a_n < 4\)[/tex] for all [tex]\(n\)[/tex]
We will prove this statement by induction.
Base case:
For [tex]\( n = 1 \)[/tex],
[tex]\[ a_1 = 2 \][/tex]
Clearly, [tex]\( 2 < 4 \)[/tex], so the base case holds.
Inductive step:
Assume [tex]\( a_n < 4 \)[/tex] for some [tex]\( n \)[/tex]. We need to show that [tex]\( a_{n+1} < 4 \)[/tex].
From the recursive definition, we have:
[tex]\[ a_{n+1} = \frac{1}{2} \left(a_n + 4 \right) \][/tex]
Since [tex]\( a_n < 4 \)[/tex], let [tex]\( a_n = 4 - \epsilon \)[/tex], where [tex]\( \epsilon > 0 \)[/tex]. Then,
[tex]\[ a_{n+1} = \frac{1}{2} \left((4 - \epsilon) + 4 \right) = \frac{1}{2} \left(8 - \epsilon \right) = 4 - \frac{\epsilon}{2} \][/tex]
Since [tex]\( \epsilon > 0 \)[/tex], it follows that [tex]\( 4 - \frac{\epsilon}{2} < 4 \)[/tex]. Hence,
[tex]\[ a_{n+1} < 4 \][/tex]
By induction, we have proven that [tex]\( a_n < 4 \)[/tex] for all [tex]\( n \)[/tex].
### Part c: Proving that [tex]\(\{ a_n \}\)[/tex] is a monotonically increasing sequence
We want to show that [tex]\( a_{n+1} \ge a_n \)[/tex] for all [tex]\( n \)[/tex].
Using the recursive definition:
[tex]\[ a_{n+1} = \frac{1}{2} \left(a_n + 4 \right) \][/tex]
Let's examine the difference [tex]\( a_{n+1} - a_n \)[/tex]:
[tex]\[ a_{n+1} - a_n = \frac{1}{2} \left(a_n + 4 \right) - a_n \][/tex]
[tex]\[ = \frac{1}{2} a_n + 2 - a_n \][/tex]
[tex]\[ = 2 - \frac{1}{2} a_n \][/tex]
Since [tex]\( a_n < 4 \)[/tex], let [tex]\( a_n = 4 - \epsilon \)[/tex] where [tex]\( \epsilon > 0 \)[/tex]. Then,
[tex]\[ 2 - \frac{1}{2} a_n = 2 - \frac{1}{2} (4 - \epsilon) = 2 - 2 + \frac{\epsilon}{2} = \frac{\epsilon}{2} \][/tex]
Since [tex]\( \epsilon > 0 \)[/tex], it follows that [tex]\( \frac{\epsilon}{2} > 0 \)[/tex]. Therefore,
[tex]\[ a_{n+1} - a_n = \frac{\epsilon}{2} > 0 \][/tex]
Thus, we have [tex]\( a_{n+1} > a_n \)[/tex], proving that the sequence [tex]\(\{ a_n \}\)[/tex] is monotonically increasing.
### Conclusion
a) The first six terms of the sequence are [tex]\(\{ 2, 3, 3.5, 3.75, 3.875, 3.9375 \}\)[/tex].
b) We proved that [tex]\( a_n < 4 \)[/tex] for all [tex]\( n \)[/tex].
c) We also proved that the sequence [tex]\(\{ a_n \}\)[/tex] is monotonically increasing.
### Part a: Generating the first six terms of the sequence
The sequence is defined as:
[tex]\[ a_1 = 2 \][/tex]
[tex]\[ a_{n+1} = \frac{1}{2} \left(a_n + 4 \right) \][/tex]
Let's calculate the first six terms.
1. For [tex]\( n = 1 \)[/tex]:
[tex]\[ a_1 = 2 \][/tex]
2. For [tex]\( n = 2 \)[/tex]:
[tex]\[ a_2 = \frac{1}{2} \left(a_1 + 4 \right) = \frac{1}{2} \left(2 + 4 \right) = \frac{1}{2} \times 6 = 3 \][/tex]
3. For [tex]\( n = 3 \)[/tex]:
[tex]\[ a_3 = \frac{1}{2} \left(a_2 + 4 \right) = \frac{1}{2} \left(3 + 4 \right) = \frac{1}{2} \times 7 = 3.5 \][/tex]
4. For [tex]\( n = 4 \)[/tex]:
[tex]\[ a_4 = \frac{1}{2} \left(a_3 + 4 \right) = \frac{1}{2} \left(3.5 + 4 \right) = \frac{1}{2} \times 7.5 = 3.75 \][/tex]
5. For [tex]\( n = 5 \)[/tex]:
[tex]\[ a_5 = \frac{1}{2} \left(a_4 + 4 \right) = \frac{1}{2} \left(3.75 + 4 \right) = \frac{1}{2} \times 7.75 = 3.875 \][/tex]
6. For [tex]\( n = 6 \)[/tex]:
[tex]\[ a_6 = \frac{1}{2} \left(a_5 + 4 \right) = \frac{1}{2} \left(3.875 + 4 \right) = \frac{1}{2} \times 7.875 = 3.9375 \][/tex]
So, the first six terms of the sequence are:
[tex]\[ \{2, 3, 3.5, 3.75, 3.875, 3.9375\} \][/tex]
### Part b: Proving that [tex]\(a_n < 4\)[/tex] for all [tex]\(n\)[/tex]
We will prove this statement by induction.
Base case:
For [tex]\( n = 1 \)[/tex],
[tex]\[ a_1 = 2 \][/tex]
Clearly, [tex]\( 2 < 4 \)[/tex], so the base case holds.
Inductive step:
Assume [tex]\( a_n < 4 \)[/tex] for some [tex]\( n \)[/tex]. We need to show that [tex]\( a_{n+1} < 4 \)[/tex].
From the recursive definition, we have:
[tex]\[ a_{n+1} = \frac{1}{2} \left(a_n + 4 \right) \][/tex]
Since [tex]\( a_n < 4 \)[/tex], let [tex]\( a_n = 4 - \epsilon \)[/tex], where [tex]\( \epsilon > 0 \)[/tex]. Then,
[tex]\[ a_{n+1} = \frac{1}{2} \left((4 - \epsilon) + 4 \right) = \frac{1}{2} \left(8 - \epsilon \right) = 4 - \frac{\epsilon}{2} \][/tex]
Since [tex]\( \epsilon > 0 \)[/tex], it follows that [tex]\( 4 - \frac{\epsilon}{2} < 4 \)[/tex]. Hence,
[tex]\[ a_{n+1} < 4 \][/tex]
By induction, we have proven that [tex]\( a_n < 4 \)[/tex] for all [tex]\( n \)[/tex].
### Part c: Proving that [tex]\(\{ a_n \}\)[/tex] is a monotonically increasing sequence
We want to show that [tex]\( a_{n+1} \ge a_n \)[/tex] for all [tex]\( n \)[/tex].
Using the recursive definition:
[tex]\[ a_{n+1} = \frac{1}{2} \left(a_n + 4 \right) \][/tex]
Let's examine the difference [tex]\( a_{n+1} - a_n \)[/tex]:
[tex]\[ a_{n+1} - a_n = \frac{1}{2} \left(a_n + 4 \right) - a_n \][/tex]
[tex]\[ = \frac{1}{2} a_n + 2 - a_n \][/tex]
[tex]\[ = 2 - \frac{1}{2} a_n \][/tex]
Since [tex]\( a_n < 4 \)[/tex], let [tex]\( a_n = 4 - \epsilon \)[/tex] where [tex]\( \epsilon > 0 \)[/tex]. Then,
[tex]\[ 2 - \frac{1}{2} a_n = 2 - \frac{1}{2} (4 - \epsilon) = 2 - 2 + \frac{\epsilon}{2} = \frac{\epsilon}{2} \][/tex]
Since [tex]\( \epsilon > 0 \)[/tex], it follows that [tex]\( \frac{\epsilon}{2} > 0 \)[/tex]. Therefore,
[tex]\[ a_{n+1} - a_n = \frac{\epsilon}{2} > 0 \][/tex]
Thus, we have [tex]\( a_{n+1} > a_n \)[/tex], proving that the sequence [tex]\(\{ a_n \}\)[/tex] is monotonically increasing.
### Conclusion
a) The first six terms of the sequence are [tex]\(\{ 2, 3, 3.5, 3.75, 3.875, 3.9375 \}\)[/tex].
b) We proved that [tex]\( a_n < 4 \)[/tex] for all [tex]\( n \)[/tex].
c) We also proved that the sequence [tex]\(\{ a_n \}\)[/tex] is monotonically increasing.
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