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Sagot :
To determine the values of [tex]\( k \)[/tex] for which the given geometric series converges, we need to analyze the series step-by-step.
The given series is:
[tex]\[ 4 \left( \frac{1-k}{5} \right) + 8 \left( \frac{1-k}{5} \right)^2 + 16 \left( \frac{1-k}{5} \right)^3 + \ldots \][/tex]
First, we need to identify the first term ([tex]\(a\)[/tex]) and the common ratio ([tex]\(r\)[/tex]) of the geometric series.
1. Identify the first term [tex]\(a\)[/tex]:
The first term [tex]\(a\)[/tex] in the series is:
[tex]\[ a = 4 \left( \frac{1-k}{5} \right) \][/tex]
2. Identify the common ratio [tex]\(r\)[/tex]:
To find the common ratio [tex]\(r\)[/tex], observe the ratio between successive terms:
[tex]\[ r = \frac{\text{second term}}{\text{first term}} = \frac{8 \left( \frac{1-k}{5} \right)^2}{4 \left( \frac{1-k}{5} \right)} = \frac{8 \left( \frac{1-k}{5} \right)^2}{4 \left( \frac{1-k}{5} \right)} = 2 \left( \frac{1-k}{5} \right) \][/tex]
Since in a geometric series, all terms are scaled by the common ratio [tex]\(r\)[/tex]:
[tex]\[ r = \frac{1-k}{5} \][/tex]
3. Convergence of the Geometric Series:
A geometric series converges if the absolute value of the common ratio [tex]\(r\)[/tex] is less than 1:
[tex]\[ |r| < 1 \][/tex]
Hence, we need to solve:
[tex]\[ \left| \frac{1-k}{5} \right| < 1 \][/tex]
4. Solving the inequality:
[tex]\[ \left| \frac{1-k}{5} \right| < 1 \][/tex]
This absolute value inequality can be broken down into two separate inequalities:
[tex]\[ -1 < \frac{1-k}{5} < 1 \][/tex]
Multiply all parts of the inequality by 5:
[tex]\[ -5 < 1 - k < 5 \][/tex]
Now, solve for [tex]\(k\)[/tex]:
[tex]\[ -5 < 1 - k \quad \text{and} \quad 1 - k < 5 \][/tex]
For the left part, add [tex]\(k\)[/tex] and 5 to both sides:
[tex]\[ -5 + k + 5 < 1 - k + k + 5 \][/tex]
Simplifies to:
[tex]\[ k < 6 \][/tex]
For the right part, subtract 1 from both sides:
[tex]\[ 1 - k - 1 < 5 - 1 \][/tex]
Simplifies to:
[tex]\[ -k < 4 \][/tex]
Multiply by -1 (which reverses the inequality):
[tex]\[ k > -4 \][/tex]
Thus, combining both parts, the values of [tex]\(k\)[/tex] for which the series converges are:
[tex]\[ -4 < k < 6 \][/tex]
Therefore, [tex]\(k\)[/tex] must satisfy the inequality:
[tex]\[ -4 < k < 6 \][/tex]
This is the range of values for [tex]\(k\)[/tex] that ensure the convergence of the given geometric series.
The given series is:
[tex]\[ 4 \left( \frac{1-k}{5} \right) + 8 \left( \frac{1-k}{5} \right)^2 + 16 \left( \frac{1-k}{5} \right)^3 + \ldots \][/tex]
First, we need to identify the first term ([tex]\(a\)[/tex]) and the common ratio ([tex]\(r\)[/tex]) of the geometric series.
1. Identify the first term [tex]\(a\)[/tex]:
The first term [tex]\(a\)[/tex] in the series is:
[tex]\[ a = 4 \left( \frac{1-k}{5} \right) \][/tex]
2. Identify the common ratio [tex]\(r\)[/tex]:
To find the common ratio [tex]\(r\)[/tex], observe the ratio between successive terms:
[tex]\[ r = \frac{\text{second term}}{\text{first term}} = \frac{8 \left( \frac{1-k}{5} \right)^2}{4 \left( \frac{1-k}{5} \right)} = \frac{8 \left( \frac{1-k}{5} \right)^2}{4 \left( \frac{1-k}{5} \right)} = 2 \left( \frac{1-k}{5} \right) \][/tex]
Since in a geometric series, all terms are scaled by the common ratio [tex]\(r\)[/tex]:
[tex]\[ r = \frac{1-k}{5} \][/tex]
3. Convergence of the Geometric Series:
A geometric series converges if the absolute value of the common ratio [tex]\(r\)[/tex] is less than 1:
[tex]\[ |r| < 1 \][/tex]
Hence, we need to solve:
[tex]\[ \left| \frac{1-k}{5} \right| < 1 \][/tex]
4. Solving the inequality:
[tex]\[ \left| \frac{1-k}{5} \right| < 1 \][/tex]
This absolute value inequality can be broken down into two separate inequalities:
[tex]\[ -1 < \frac{1-k}{5} < 1 \][/tex]
Multiply all parts of the inequality by 5:
[tex]\[ -5 < 1 - k < 5 \][/tex]
Now, solve for [tex]\(k\)[/tex]:
[tex]\[ -5 < 1 - k \quad \text{and} \quad 1 - k < 5 \][/tex]
For the left part, add [tex]\(k\)[/tex] and 5 to both sides:
[tex]\[ -5 + k + 5 < 1 - k + k + 5 \][/tex]
Simplifies to:
[tex]\[ k < 6 \][/tex]
For the right part, subtract 1 from both sides:
[tex]\[ 1 - k - 1 < 5 - 1 \][/tex]
Simplifies to:
[tex]\[ -k < 4 \][/tex]
Multiply by -1 (which reverses the inequality):
[tex]\[ k > -4 \][/tex]
Thus, combining both parts, the values of [tex]\(k\)[/tex] for which the series converges are:
[tex]\[ -4 < k < 6 \][/tex]
Therefore, [tex]\(k\)[/tex] must satisfy the inequality:
[tex]\[ -4 < k < 6 \][/tex]
This is the range of values for [tex]\(k\)[/tex] that ensure the convergence of the given geometric series.
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