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Sagot :
Sure, let's address each part of the question step-by-step:
### Part (a)
Express [tex]\( 8.0\dot{5} \)[/tex] as a rational fraction using an infinite geometric series
First, let's rewrite the infinite decimal [tex]\( 8.0\dot{5} \)[/tex] as a geometric series.
[tex]\[ 8.0\dot{5} = 8 + 0.5555\ldots \][/tex]
We can express [tex]\( 0.5555\ldots \)[/tex] as a geometric series where the first term [tex]\( a \)[/tex] is [tex]\( 0.5 \)[/tex] and the common ratio [tex]\( r \)[/tex] is [tex]\( 0.1 \)[/tex].
So:
[tex]\[ 0.5555\ldots = 0.5 + 0.05 + 0.005 + \cdots \][/tex]
This is an infinite geometric series where:
[tex]\[ a = 0.5, \quad r = 0.1 \][/tex]
The sum [tex]\( S \)[/tex] of an infinite geometric series is given by:
[tex]\[ S = \frac{a}{1-r} \][/tex]
Thus,
[tex]\[ S = \frac{0.5}{1-0.1} = \frac{0.5}{0.9} = \frac{0.5}{\frac{9}{10}} = \frac{0.5 \times 10}{9} = \frac{5}{9} \][/tex]
Therefore,
[tex]\[ 0.5555\ldots = \frac{5}{9} \][/tex]
So,
[tex]\[ 8.0\dot{5} = 8 + \frac{5}{9} = \frac{72}{9} + \frac{5}{9} = \frac{77}{9} \][/tex]
### Part (b)
Determine the summation of the first [tex]\( n \)[/tex] terms of the series if [tex]\( a = 7 \)[/tex], [tex]\( b = 77 \)[/tex], and [tex]\( c = 777 \)[/tex]
Here we have the numbers [tex]\( a = 7 \)[/tex], [tex]\( b = 77 \)[/tex], and [tex]\( c = 777 \)[/tex].
Recognize the pattern in the series: [tex]\( a_n = 7 \times (10^n - 1)/9 \)[/tex].
For example:
- [tex]\( a_1 = 7 \)[/tex]
- [tex]\( a_2 = 77 \)[/tex]
- [tex]\( a_3 = 777 \)[/tex]
To find the sum of the first [tex]\( n \)[/tex] terms of this series:
[tex]\[ S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n \frac{7 (10^k - 1)}{9} \][/tex]
Let's simplify this:
[tex]\[ S_n = \frac{7}{9} \sum_{k=1}^n (10^k - 1) \][/tex]
[tex]\[ S_n = \frac{7}{9} \left( \sum_{k=1}^n 10^k - \sum_{k=1}^n 1 \right) \][/tex]
The sum of the first [tex]\( n \)[/tex] terms of the series [tex]\( 10^k \)[/tex] is:
[tex]\[ \sum_{k=1}^n 10^k = 10(1) + 10(2) + \cdots + 10(n) = 10 \left( \frac{10^n - 1}{9} \right) \][/tex]
The sum [tex]\( \sum_{k=1}^n 1 \)[/tex] is just [tex]\( n \)[/tex].
Therefore:
[tex]\[ S_n = \frac{7}{9} \left( 10 \times \frac{10^n - 1}{9} - n \right) \][/tex]
[tex]\[ S_n = \frac{7}{9} \left( \frac{10 (10^n - 1)}{9} - n \right) \][/tex]
[tex]\[ S_n = \frac{7}{81} (10^{n+1} - 10 - 9n) \][/tex]
So, the sum of the first [tex]\( n \)[/tex] terms of the series is:
[tex]\[ S_n = \frac{7}{81} ( 10^{n+1} - 10 - 9n) \][/tex]
### Part (c)
Impose a condition on [tex]\( x \)[/tex] for the sum of the infinite series and calculate the sum of the given series
The given series is:
[tex]\[ 1 + (7x+1)^{-1} + (7x+1)^{-2} + (7x+1)^{-3} + \cdots \][/tex]
For this to be a geometric series, the common ratio must satisfy [tex]\( |r| < 1 \)[/tex].
Here, the common ratio [tex]\( r = \frac{1}{7x+1} \)[/tex], so:
[tex]\[ \left| \frac{1}{7x+1} \right| < 1 \][/tex]
This gives:
[tex]\[ -1 < \frac{1}{7x+1} < 1 \][/tex]
Simplifying:
[tex]\[ 7x+1 > 1 \quad \text{and} \quad 7x+1 < -1 \][/tex]
The first inequality:
[tex]\[ 7x + 1 > 1 \][/tex]
[tex]\[ 7x > 0 \][/tex]
[tex]\[ x > 0 \][/tex]
The second inequality:
[tex]\[ 7x + 1 < -1 \][/tex]
[tex]\[ 7x < -2 \][/tex]
[tex]\[ x < -\frac{2}{7} \][/tex]
So, for the series to converge, [tex]\( x \)[/tex] must be:
[tex]\[ x < -\frac{2}{7} \quad \text{or} \quad x > 0 \][/tex]
To calculate the sum of the given series, we use the formula for the sum of an infinite geometric series:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
Here, [tex]\( a = 1 \)[/tex] and [tex]\( r = \frac{1}{7x+1} \)[/tex].
Thus, the sum of the series is:
[tex]\[ S = \frac{1}{1 - \frac{1}{7x+1}} \][/tex]
Simplify the denominator:
[tex]\[ S = \frac{1}{\frac{7x+1-1}{7x+1}} = \frac{1}{\frac{7x}{7x+1}} = \frac{7x+1}{7x} = 1 + \frac{1}{7x} \][/tex]
So the sum of the given series is:
[tex]\[ S = 1 + \frac{1}{7x} \][/tex]
### Part (a)
Express [tex]\( 8.0\dot{5} \)[/tex] as a rational fraction using an infinite geometric series
First, let's rewrite the infinite decimal [tex]\( 8.0\dot{5} \)[/tex] as a geometric series.
[tex]\[ 8.0\dot{5} = 8 + 0.5555\ldots \][/tex]
We can express [tex]\( 0.5555\ldots \)[/tex] as a geometric series where the first term [tex]\( a \)[/tex] is [tex]\( 0.5 \)[/tex] and the common ratio [tex]\( r \)[/tex] is [tex]\( 0.1 \)[/tex].
So:
[tex]\[ 0.5555\ldots = 0.5 + 0.05 + 0.005 + \cdots \][/tex]
This is an infinite geometric series where:
[tex]\[ a = 0.5, \quad r = 0.1 \][/tex]
The sum [tex]\( S \)[/tex] of an infinite geometric series is given by:
[tex]\[ S = \frac{a}{1-r} \][/tex]
Thus,
[tex]\[ S = \frac{0.5}{1-0.1} = \frac{0.5}{0.9} = \frac{0.5}{\frac{9}{10}} = \frac{0.5 \times 10}{9} = \frac{5}{9} \][/tex]
Therefore,
[tex]\[ 0.5555\ldots = \frac{5}{9} \][/tex]
So,
[tex]\[ 8.0\dot{5} = 8 + \frac{5}{9} = \frac{72}{9} + \frac{5}{9} = \frac{77}{9} \][/tex]
### Part (b)
Determine the summation of the first [tex]\( n \)[/tex] terms of the series if [tex]\( a = 7 \)[/tex], [tex]\( b = 77 \)[/tex], and [tex]\( c = 777 \)[/tex]
Here we have the numbers [tex]\( a = 7 \)[/tex], [tex]\( b = 77 \)[/tex], and [tex]\( c = 777 \)[/tex].
Recognize the pattern in the series: [tex]\( a_n = 7 \times (10^n - 1)/9 \)[/tex].
For example:
- [tex]\( a_1 = 7 \)[/tex]
- [tex]\( a_2 = 77 \)[/tex]
- [tex]\( a_3 = 777 \)[/tex]
To find the sum of the first [tex]\( n \)[/tex] terms of this series:
[tex]\[ S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n \frac{7 (10^k - 1)}{9} \][/tex]
Let's simplify this:
[tex]\[ S_n = \frac{7}{9} \sum_{k=1}^n (10^k - 1) \][/tex]
[tex]\[ S_n = \frac{7}{9} \left( \sum_{k=1}^n 10^k - \sum_{k=1}^n 1 \right) \][/tex]
The sum of the first [tex]\( n \)[/tex] terms of the series [tex]\( 10^k \)[/tex] is:
[tex]\[ \sum_{k=1}^n 10^k = 10(1) + 10(2) + \cdots + 10(n) = 10 \left( \frac{10^n - 1}{9} \right) \][/tex]
The sum [tex]\( \sum_{k=1}^n 1 \)[/tex] is just [tex]\( n \)[/tex].
Therefore:
[tex]\[ S_n = \frac{7}{9} \left( 10 \times \frac{10^n - 1}{9} - n \right) \][/tex]
[tex]\[ S_n = \frac{7}{9} \left( \frac{10 (10^n - 1)}{9} - n \right) \][/tex]
[tex]\[ S_n = \frac{7}{81} (10^{n+1} - 10 - 9n) \][/tex]
So, the sum of the first [tex]\( n \)[/tex] terms of the series is:
[tex]\[ S_n = \frac{7}{81} ( 10^{n+1} - 10 - 9n) \][/tex]
### Part (c)
Impose a condition on [tex]\( x \)[/tex] for the sum of the infinite series and calculate the sum of the given series
The given series is:
[tex]\[ 1 + (7x+1)^{-1} + (7x+1)^{-2} + (7x+1)^{-3} + \cdots \][/tex]
For this to be a geometric series, the common ratio must satisfy [tex]\( |r| < 1 \)[/tex].
Here, the common ratio [tex]\( r = \frac{1}{7x+1} \)[/tex], so:
[tex]\[ \left| \frac{1}{7x+1} \right| < 1 \][/tex]
This gives:
[tex]\[ -1 < \frac{1}{7x+1} < 1 \][/tex]
Simplifying:
[tex]\[ 7x+1 > 1 \quad \text{and} \quad 7x+1 < -1 \][/tex]
The first inequality:
[tex]\[ 7x + 1 > 1 \][/tex]
[tex]\[ 7x > 0 \][/tex]
[tex]\[ x > 0 \][/tex]
The second inequality:
[tex]\[ 7x + 1 < -1 \][/tex]
[tex]\[ 7x < -2 \][/tex]
[tex]\[ x < -\frac{2}{7} \][/tex]
So, for the series to converge, [tex]\( x \)[/tex] must be:
[tex]\[ x < -\frac{2}{7} \quad \text{or} \quad x > 0 \][/tex]
To calculate the sum of the given series, we use the formula for the sum of an infinite geometric series:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
Here, [tex]\( a = 1 \)[/tex] and [tex]\( r = \frac{1}{7x+1} \)[/tex].
Thus, the sum of the series is:
[tex]\[ S = \frac{1}{1 - \frac{1}{7x+1}} \][/tex]
Simplify the denominator:
[tex]\[ S = \frac{1}{\frac{7x+1-1}{7x+1}} = \frac{1}{\frac{7x}{7x+1}} = \frac{7x+1}{7x} = 1 + \frac{1}{7x} \][/tex]
So the sum of the given series is:
[tex]\[ S = 1 + \frac{1}{7x} \][/tex]
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