Welcome to Westonci.ca, where your questions are met with accurate answers from a community of experts and enthusiasts. Get immediate and reliable solutions to your questions from a community of experienced experts on our Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
Sure, let's tackle each problem step-by-step.
### Problem 1: Convergence of the Series
Given the series:
[tex]\[ 1 - \frac{1}{\sqrt{3}} + \frac{1}{3} - \frac{1}{3\sqrt{3}} + \frac{1}{9} - \frac{1}{9\sqrt{3}} + \cdots \][/tex]
To determine whether this series converges or diverges, and if it converges, find the sum, we need to identify whether it represents a geometric series.
#### Step 1: Identify the Pattern
Observe the terms of the series and how they change. We can group the series in pairs of positive and negative terms as follows:
[tex]\[ \left(1 - \frac{1}{\sqrt{3}}\right) + \left(\frac{1}{3} - \frac{1}{3\sqrt{3}}\right) + \left(\frac{1}{9} - \frac{1}{9\sqrt{3}}\right) + \cdots \][/tex]
#### Step 2: Recognize the Series as Geometric
Each pair of terms can be expressed as:
[tex]\[ \left(\frac{1}{3^n} - \frac{1}{3^n \cdot \sqrt{3}}\right) \][/tex]
Factor out [tex]\( \frac{1}{3^n} \)[/tex] from each term:
[tex]\[ \left(\frac{1}{3^n} \times 1 - \frac{1}{3^n} \times \frac{1}{\sqrt{3}}\right) = \frac{1}{3^n} \left(1 - \frac{1}{\sqrt{3}}\right) \][/tex]
#### Step 3: Simplify the Series
Let:
[tex]\[ a = 1 - \frac{1}{\sqrt{3}} \][/tex]
Then our series becomes:
[tex]\[ a \left(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots\right) \][/tex]
#### Step 4: Sum the Geometric Series
This is a geometric series with the first term [tex]\( a \)[/tex] and common ratio [tex]\( r = \frac{1}{3} \)[/tex]. The sum [tex]\( S \)[/tex] of an infinite geometric series [tex]\( a + ar + ar^2 + ar^3 + \cdots \)[/tex] can be found using the formula:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
Here, [tex]\( a = 1 - \frac{1}{\sqrt{3}} \)[/tex] and [tex]\( r = \frac{1}{3} \)[/tex]. Plugging these in:
[tex]\[ S = \frac{1 - \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{1 - \frac{1}{\sqrt{3}}}{\frac{2}{3}} = \frac{3}{2} \left(1 - \frac{1}{\sqrt{3}}\right) \][/tex]
Thus, the series converges and the sum is:
[tex]\[ \boxed{\frac{3}{2} \left(1 - \frac{1}{\sqrt{3}}\right)} \][/tex]
### Problem 2: Fractional Expansion of Repeating Decimal
Given the repeating decimal:
[tex]\[ 0.424242 \overline{42} \][/tex]
Let's convert this repeating decimal into a fraction.
#### Step 1: Express the Repeating Decimal as a Geometric Series
Notice that:
[tex]\[ 0.42424242\ldots = 0.42 + 0.0042 + 0.000042 + \cdots \][/tex]
We can write this as:
[tex]\[ 0.42 (1 + 0.01 + 0.0001 + \cdots) \][/tex]
Notice this forms a geometric series where each term is [tex]\( 0.01 \)[/tex] times the previous term.
#### Step 2: Write it as a Geometric Series
Let:
[tex]\[ a = 0.42 = \frac{42}{100} \][/tex]
[tex]\[ r = 0.01 = \frac{1}{100} \][/tex]
Thus:
[tex]\[ 0.424242\ldots = \frac{42}{100} \left(1 + \left(\frac{1}{100}\right) + \left(\frac{1}{100}\right)^2 + \cdots\right) \][/tex]
#### Step 3: Sum the Geometric Series
The sum [tex]\( S \)[/tex] of an infinite geometric series [tex]\( a + ar + ar^2 + ar^3 + \cdots \)[/tex] is:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
Here, [tex]\( a = \frac{42}{100} \)[/tex] and [tex]\( r = \frac{1}{100} \)[/tex]:
[tex]\[ S = \frac{\frac{42}{100}}{1 - \frac{1}{100}} = \frac{\frac{42}{100}}{\frac{99}{100}} = \frac{42}{99} \][/tex]
#### Step 4: Simplify the Fraction
Simplify [tex]\( \frac{42}{99} \)[/tex] by dividing the numerator and the denominator by their greatest common divisor, which is 3:
[tex]\[ \frac{42 \div 3}{99 \div 3} = \frac{14}{33} \][/tex]
Thus, the fractional expansion of the repeating decimal [tex]\( 0.424242 \overline{42} \)[/tex] is:
[tex]\[ \boxed{\frac{14}{33}} \][/tex]
### Problem 1: Convergence of the Series
Given the series:
[tex]\[ 1 - \frac{1}{\sqrt{3}} + \frac{1}{3} - \frac{1}{3\sqrt{3}} + \frac{1}{9} - \frac{1}{9\sqrt{3}} + \cdots \][/tex]
To determine whether this series converges or diverges, and if it converges, find the sum, we need to identify whether it represents a geometric series.
#### Step 1: Identify the Pattern
Observe the terms of the series and how they change. We can group the series in pairs of positive and negative terms as follows:
[tex]\[ \left(1 - \frac{1}{\sqrt{3}}\right) + \left(\frac{1}{3} - \frac{1}{3\sqrt{3}}\right) + \left(\frac{1}{9} - \frac{1}{9\sqrt{3}}\right) + \cdots \][/tex]
#### Step 2: Recognize the Series as Geometric
Each pair of terms can be expressed as:
[tex]\[ \left(\frac{1}{3^n} - \frac{1}{3^n \cdot \sqrt{3}}\right) \][/tex]
Factor out [tex]\( \frac{1}{3^n} \)[/tex] from each term:
[tex]\[ \left(\frac{1}{3^n} \times 1 - \frac{1}{3^n} \times \frac{1}{\sqrt{3}}\right) = \frac{1}{3^n} \left(1 - \frac{1}{\sqrt{3}}\right) \][/tex]
#### Step 3: Simplify the Series
Let:
[tex]\[ a = 1 - \frac{1}{\sqrt{3}} \][/tex]
Then our series becomes:
[tex]\[ a \left(1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots\right) \][/tex]
#### Step 4: Sum the Geometric Series
This is a geometric series with the first term [tex]\( a \)[/tex] and common ratio [tex]\( r = \frac{1}{3} \)[/tex]. The sum [tex]\( S \)[/tex] of an infinite geometric series [tex]\( a + ar + ar^2 + ar^3 + \cdots \)[/tex] can be found using the formula:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
Here, [tex]\( a = 1 - \frac{1}{\sqrt{3}} \)[/tex] and [tex]\( r = \frac{1}{3} \)[/tex]. Plugging these in:
[tex]\[ S = \frac{1 - \frac{1}{\sqrt{3}}}{1 - \frac{1}{3}} = \frac{1 - \frac{1}{\sqrt{3}}}{\frac{2}{3}} = \frac{3}{2} \left(1 - \frac{1}{\sqrt{3}}\right) \][/tex]
Thus, the series converges and the sum is:
[tex]\[ \boxed{\frac{3}{2} \left(1 - \frac{1}{\sqrt{3}}\right)} \][/tex]
### Problem 2: Fractional Expansion of Repeating Decimal
Given the repeating decimal:
[tex]\[ 0.424242 \overline{42} \][/tex]
Let's convert this repeating decimal into a fraction.
#### Step 1: Express the Repeating Decimal as a Geometric Series
Notice that:
[tex]\[ 0.42424242\ldots = 0.42 + 0.0042 + 0.000042 + \cdots \][/tex]
We can write this as:
[tex]\[ 0.42 (1 + 0.01 + 0.0001 + \cdots) \][/tex]
Notice this forms a geometric series where each term is [tex]\( 0.01 \)[/tex] times the previous term.
#### Step 2: Write it as a Geometric Series
Let:
[tex]\[ a = 0.42 = \frac{42}{100} \][/tex]
[tex]\[ r = 0.01 = \frac{1}{100} \][/tex]
Thus:
[tex]\[ 0.424242\ldots = \frac{42}{100} \left(1 + \left(\frac{1}{100}\right) + \left(\frac{1}{100}\right)^2 + \cdots\right) \][/tex]
#### Step 3: Sum the Geometric Series
The sum [tex]\( S \)[/tex] of an infinite geometric series [tex]\( a + ar + ar^2 + ar^3 + \cdots \)[/tex] is:
[tex]\[ S = \frac{a}{1 - r} \][/tex]
Here, [tex]\( a = \frac{42}{100} \)[/tex] and [tex]\( r = \frac{1}{100} \)[/tex]:
[tex]\[ S = \frac{\frac{42}{100}}{1 - \frac{1}{100}} = \frac{\frac{42}{100}}{\frac{99}{100}} = \frac{42}{99} \][/tex]
#### Step 4: Simplify the Fraction
Simplify [tex]\( \frac{42}{99} \)[/tex] by dividing the numerator and the denominator by their greatest common divisor, which is 3:
[tex]\[ \frac{42 \div 3}{99 \div 3} = \frac{14}{33} \][/tex]
Thus, the fractional expansion of the repeating decimal [tex]\( 0.424242 \overline{42} \)[/tex] is:
[tex]\[ \boxed{\frac{14}{33}} \][/tex]
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
