Answer :
### 3.1.2 Write the series in sigma notation.
Unfortunately, the specific series in question is not provided. To illustrate how to write a series in sigma notation, let's use an example. Suppose you have the series:
[tex]\[3 + 6 + 12 + 24 + \ldots + 384\][/tex]
This is a geometric series where the first term [tex]\(a = 3\)[/tex] and the common ratio [tex]\(r = 2\)[/tex]. If the last term is [tex]\(384\)[/tex], we can set up the series in sigma notation as follows:
[tex]\[ 3 \cdot 2^{0} + 3 \cdot 2^{1} + 3 \cdot 2^{2} + \ldots + 3 \cdot 2^{k} = 384 \][/tex]
To find the value of [tex]\(k\)[/tex]:
[tex]\[3 \cdot 2^{k} = 384\][/tex]
Dividing both sides by 3:
[tex]\[2^{k} = 128\][/tex]
Thus,
[tex]\[k = \log_{2}(128) = 7\][/tex]
So, the series in sigma notation can be written as:
[tex]\[ \sum_{n=0}^{7} 3 \cdot 2^{n} \][/tex]
### 3.2 Prove that the formula for the sum of a geometric series is given by:
[tex]\[ S_n = \frac{a(1 - r^n)}{1 - r} \text{ for } r \neq 1 \][/tex]
Let's consider a geometric series defined by:
[tex]\[ S_n = a + ar + ar^2 + \cdots + ar^{n-1} \][/tex]
To find the sum, we can multiply both sides of the equation by [tex]\(r\)[/tex]:
[tex]\[ rS_n = ar + ar^2 + ar^3 + \cdots + ar^n \][/tex]
Next, subtract the second equation from the first:
[tex]\[ S_n - rS_n = a + ar + ar^2 + \cdots + ar^{n-1} - (ar + ar^2 + \cdots + ar^n) \][/tex]
This simplifies to:
[tex]\[ S_n (1 - r) = a - ar^n \][/tex]
Solving for [tex]\(S_n\)[/tex]:
[tex]\[ S_n = \frac{a(1 - r^n)}{1 - r} \][/tex]
Thus, we have proven that the sum of the first [tex]\(n\)[/tex] terms of a geometric series is:
[tex]\[ S_n = \frac{a(1 - r^n)}{1 - r} \text{ for } r \neq 1 \][/tex]
### 3.3 The first two terms of a geometric sequence are: [tex]\(\left(\tan 45^\circ\right)\)[/tex] and [tex]\(\left(\sin 45^\circ\right)\)[/tex].
### 3.3.1 Determine the sum of the first eight terms of the sequence (leave your answer in simplified surd form).
The first term [tex]\(a\)[/tex] of the geometric sequence is:
[tex]\[ a = \tan 45^\circ = 1 \][/tex]
The second term is:
[tex]\[ a_2 = \sin 45^\circ = \frac{\sqrt{2}}{2} \][/tex]
To find the common ratio [tex]\(r\)[/tex]:
[tex]\[ r = \frac{a_2}{a} = \frac{\frac{\sqrt{2}}{2}}{1} = \frac{\sqrt{2}}{2} \][/tex]
Using the formula for the sum of the first [tex]\(n\)[/tex] terms of a geometric series:
[tex]\[ S_n = \frac{a(1 - r^n)}{1 - r} \][/tex]
Given [tex]\(a = 1\)[/tex], [tex]\(r = \frac{\sqrt{2}}{2}\)[/tex], and [tex]\(n = 8\)[/tex]:
[tex]\[ S_8 = \frac{1 \left(1 - \left(\frac{\sqrt{2}}{2}\right)^8\right)}{1 - \frac{\sqrt{2}}{2}} \][/tex]
First, simplify [tex]\(\left(\frac{\sqrt{2}}{2}\right)^8\)[/tex]:
[tex]\[ \left(\frac{\sqrt{2}}{2}\right)^8 = \left(\frac{\sqrt{2}}{2}\right)^8 = \frac{(\sqrt{2})^8}{2^8} = \frac{2^4}{2^8} = \frac{16}{256} = \frac{1}{16} \][/tex]
Substituting [tex]\(\left(\frac{\sqrt{2}}{2}\right)^8\)[/tex]:
[tex]\[ S_8 = \frac{1 \left(1 - \frac{1}{16}\right)}{1 - \frac{\sqrt{2}}{2}} \][/tex]
[tex]\[ S_8 = \frac{1 \cdot \frac{15}{16}}{1 - \frac{\sqrt{2}}{2}} \][/tex]
Simplify the denominator:
[tex]\[ 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2} \][/tex]
Substituting back:
[tex]\[ S_8 = \frac{\frac{15}{16}}{\frac{2 - \sqrt{2}}{2}} \][/tex]
[tex]\[ S_8 = \frac{15}{16} \cdot \frac{2}{2 - \sqrt{2}} \][/tex]
Multiply by the conjugate of the denominator:
[tex]\[ S_8 = \frac{15}{16} \cdot \frac{2}{2 - \sqrt{2}} \cdot \frac{2 + \sqrt{2}}{2 + \sqrt{2}} \][/tex]
[tex]\[ S_8 = \frac{15 \cdot 2(2 + \sqrt{2})}{16(4 - 2)} \][/tex]
[tex]\[ S_8 = \frac{30(2 + \sqrt{2})}{32} \][/tex]
[tex]\[ S_8 = \frac{15(2 + \sqrt{2})}{16} \][/tex]
Therefore, the sum of the first eight terms of the geometric sequence is:
[tex]\[ S_8 = \frac{15(2 + \sqrt{2})}{16} \][/tex]
Unfortunately, the specific series in question is not provided. To illustrate how to write a series in sigma notation, let's use an example. Suppose you have the series:
[tex]\[3 + 6 + 12 + 24 + \ldots + 384\][/tex]
This is a geometric series where the first term [tex]\(a = 3\)[/tex] and the common ratio [tex]\(r = 2\)[/tex]. If the last term is [tex]\(384\)[/tex], we can set up the series in sigma notation as follows:
[tex]\[ 3 \cdot 2^{0} + 3 \cdot 2^{1} + 3 \cdot 2^{2} + \ldots + 3 \cdot 2^{k} = 384 \][/tex]
To find the value of [tex]\(k\)[/tex]:
[tex]\[3 \cdot 2^{k} = 384\][/tex]
Dividing both sides by 3:
[tex]\[2^{k} = 128\][/tex]
Thus,
[tex]\[k = \log_{2}(128) = 7\][/tex]
So, the series in sigma notation can be written as:
[tex]\[ \sum_{n=0}^{7} 3 \cdot 2^{n} \][/tex]
### 3.2 Prove that the formula for the sum of a geometric series is given by:
[tex]\[ S_n = \frac{a(1 - r^n)}{1 - r} \text{ for } r \neq 1 \][/tex]
Let's consider a geometric series defined by:
[tex]\[ S_n = a + ar + ar^2 + \cdots + ar^{n-1} \][/tex]
To find the sum, we can multiply both sides of the equation by [tex]\(r\)[/tex]:
[tex]\[ rS_n = ar + ar^2 + ar^3 + \cdots + ar^n \][/tex]
Next, subtract the second equation from the first:
[tex]\[ S_n - rS_n = a + ar + ar^2 + \cdots + ar^{n-1} - (ar + ar^2 + \cdots + ar^n) \][/tex]
This simplifies to:
[tex]\[ S_n (1 - r) = a - ar^n \][/tex]
Solving for [tex]\(S_n\)[/tex]:
[tex]\[ S_n = \frac{a(1 - r^n)}{1 - r} \][/tex]
Thus, we have proven that the sum of the first [tex]\(n\)[/tex] terms of a geometric series is:
[tex]\[ S_n = \frac{a(1 - r^n)}{1 - r} \text{ for } r \neq 1 \][/tex]
### 3.3 The first two terms of a geometric sequence are: [tex]\(\left(\tan 45^\circ\right)\)[/tex] and [tex]\(\left(\sin 45^\circ\right)\)[/tex].
### 3.3.1 Determine the sum of the first eight terms of the sequence (leave your answer in simplified surd form).
The first term [tex]\(a\)[/tex] of the geometric sequence is:
[tex]\[ a = \tan 45^\circ = 1 \][/tex]
The second term is:
[tex]\[ a_2 = \sin 45^\circ = \frac{\sqrt{2}}{2} \][/tex]
To find the common ratio [tex]\(r\)[/tex]:
[tex]\[ r = \frac{a_2}{a} = \frac{\frac{\sqrt{2}}{2}}{1} = \frac{\sqrt{2}}{2} \][/tex]
Using the formula for the sum of the first [tex]\(n\)[/tex] terms of a geometric series:
[tex]\[ S_n = \frac{a(1 - r^n)}{1 - r} \][/tex]
Given [tex]\(a = 1\)[/tex], [tex]\(r = \frac{\sqrt{2}}{2}\)[/tex], and [tex]\(n = 8\)[/tex]:
[tex]\[ S_8 = \frac{1 \left(1 - \left(\frac{\sqrt{2}}{2}\right)^8\right)}{1 - \frac{\sqrt{2}}{2}} \][/tex]
First, simplify [tex]\(\left(\frac{\sqrt{2}}{2}\right)^8\)[/tex]:
[tex]\[ \left(\frac{\sqrt{2}}{2}\right)^8 = \left(\frac{\sqrt{2}}{2}\right)^8 = \frac{(\sqrt{2})^8}{2^8} = \frac{2^4}{2^8} = \frac{16}{256} = \frac{1}{16} \][/tex]
Substituting [tex]\(\left(\frac{\sqrt{2}}{2}\right)^8\)[/tex]:
[tex]\[ S_8 = \frac{1 \left(1 - \frac{1}{16}\right)}{1 - \frac{\sqrt{2}}{2}} \][/tex]
[tex]\[ S_8 = \frac{1 \cdot \frac{15}{16}}{1 - \frac{\sqrt{2}}{2}} \][/tex]
Simplify the denominator:
[tex]\[ 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2} \][/tex]
Substituting back:
[tex]\[ S_8 = \frac{\frac{15}{16}}{\frac{2 - \sqrt{2}}{2}} \][/tex]
[tex]\[ S_8 = \frac{15}{16} \cdot \frac{2}{2 - \sqrt{2}} \][/tex]
Multiply by the conjugate of the denominator:
[tex]\[ S_8 = \frac{15}{16} \cdot \frac{2}{2 - \sqrt{2}} \cdot \frac{2 + \sqrt{2}}{2 + \sqrt{2}} \][/tex]
[tex]\[ S_8 = \frac{15 \cdot 2(2 + \sqrt{2})}{16(4 - 2)} \][/tex]
[tex]\[ S_8 = \frac{30(2 + \sqrt{2})}{32} \][/tex]
[tex]\[ S_8 = \frac{15(2 + \sqrt{2})}{16} \][/tex]
Therefore, the sum of the first eight terms of the geometric sequence is:
[tex]\[ S_8 = \frac{15(2 + \sqrt{2})}{16} \][/tex]
