IDNLearn.com is your trusted platform for finding reliable answers. Discover thorough and trustworthy answers from our community of knowledgeable professionals, tailored to meet your specific needs.
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]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Thank you for trusting IDNLearn.com. We’re dedicated to providing accurate answers, so visit us again for more solutions.
