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Sagot :
Let's determine the explicit formula for the given sequence step-by-step.
The sequence is initially defined as follows:
[tex]\[ f(1) = 40 \][/tex]
[tex]\[ f(n) = -f(n - 1) \][/tex]
To understand the pattern in the sequence, let's evaluate the first few terms:
1. For [tex]\( n = 1 \)[/tex]:
[tex]\[ f(1) = 40 \][/tex]
2. For [tex]\( n = 2 \)[/tex]:
[tex]\[ f(2) = -f(1) = -40 \][/tex]
3. For [tex]\( n = 3 \)[/tex]:
[tex]\[ f(3) = -f(2) = -(-40) = 40 \][/tex]
4. For [tex]\( n = 4 \)[/tex]:
[tex]\[ f(4) = -f(3) = -40 \][/tex]
5. For [tex]\( n = 5 \)[/tex]:
[tex]\[ f(5) = -f(4) = -(-40) = 40 \][/tex]
From these calculations, it's clear that the sequence alternates between [tex]\( 40 \)[/tex] and [tex]\( -40 \)[/tex].
Let's identify the pattern:
- When [tex]\( n \)[/tex] is odd (1, 3, 5,...), [tex]\( f(n) \)[/tex] is [tex]\( 40 \)[/tex].
- When [tex]\( n \)[/tex] is even (2, 4, 6,...), [tex]\( f(n) \)[/tex] is [tex]\( -40 \)[/tex].
This alternating sign can be expressed using the term [tex]\((-1)^{n - 1}\)[/tex]:
- For [tex]\( n = 1 \)[/tex], [tex]\( (-1)^{1-1} = (-1)^0 = 1 \)[/tex]
- For [tex]\( n = 2 \)[/tex], [tex]\( (-1)^{2-1} = (-1)^1 = -1 \)[/tex]
- For [tex]\( n = 3 \)[/tex], [tex]\( (-1)^{3-1} = (-1)^2 = 1 \)[/tex]
- For [tex]\( n = 4 \)[/tex], [tex]\( (-1)^{4-1} = (-1)^3 = -1 \)[/tex]
Hence, the explicit formula is:
[tex]\[ f(n) = 40 \cdot (-1)^{n - 1} \][/tex]
Among the given options, the correct explicit formula that matches our derivation is:
[tex]\[ f(n) = 40 \cdot (-1)^{n - 1} \][/tex]
Thus, the explicit formula used to define this sequence is:
[tex]\[ f(n) = 40 \cdot (-1)^{n - 1} \][/tex]
The sequence is initially defined as follows:
[tex]\[ f(1) = 40 \][/tex]
[tex]\[ f(n) = -f(n - 1) \][/tex]
To understand the pattern in the sequence, let's evaluate the first few terms:
1. For [tex]\( n = 1 \)[/tex]:
[tex]\[ f(1) = 40 \][/tex]
2. For [tex]\( n = 2 \)[/tex]:
[tex]\[ f(2) = -f(1) = -40 \][/tex]
3. For [tex]\( n = 3 \)[/tex]:
[tex]\[ f(3) = -f(2) = -(-40) = 40 \][/tex]
4. For [tex]\( n = 4 \)[/tex]:
[tex]\[ f(4) = -f(3) = -40 \][/tex]
5. For [tex]\( n = 5 \)[/tex]:
[tex]\[ f(5) = -f(4) = -(-40) = 40 \][/tex]
From these calculations, it's clear that the sequence alternates between [tex]\( 40 \)[/tex] and [tex]\( -40 \)[/tex].
Let's identify the pattern:
- When [tex]\( n \)[/tex] is odd (1, 3, 5,...), [tex]\( f(n) \)[/tex] is [tex]\( 40 \)[/tex].
- When [tex]\( n \)[/tex] is even (2, 4, 6,...), [tex]\( f(n) \)[/tex] is [tex]\( -40 \)[/tex].
This alternating sign can be expressed using the term [tex]\((-1)^{n - 1}\)[/tex]:
- For [tex]\( n = 1 \)[/tex], [tex]\( (-1)^{1-1} = (-1)^0 = 1 \)[/tex]
- For [tex]\( n = 2 \)[/tex], [tex]\( (-1)^{2-1} = (-1)^1 = -1 \)[/tex]
- For [tex]\( n = 3 \)[/tex], [tex]\( (-1)^{3-1} = (-1)^2 = 1 \)[/tex]
- For [tex]\( n = 4 \)[/tex], [tex]\( (-1)^{4-1} = (-1)^3 = -1 \)[/tex]
Hence, the explicit formula is:
[tex]\[ f(n) = 40 \cdot (-1)^{n - 1} \][/tex]
Among the given options, the correct explicit formula that matches our derivation is:
[tex]\[ f(n) = 40 \cdot (-1)^{n - 1} \][/tex]
Thus, the explicit formula used to define this sequence is:
[tex]\[ f(n) = 40 \cdot (-1)^{n - 1} \][/tex]
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