Responder
The sum of the infinite geometric series is \( 16 + 8\sqrt{2} \). For the finite series with 20 terms, the sum \( x \) is \( 8,\!716,\!961,\!000 \).
Solución
Let's address both parts of your question step by step.
---
## **Part 1: Infinite Geometric Series**
### **Given:**
- **First term (\( a \))**: 8
- **Second term**: \( \frac{1}{\sqrt{2}} \)
### **Objective:**
Prove that the sum of the infinite geometric series is \( 16 + 8\sqrt{2} \) without using a calculator.
### **Solution:**
1. **Determine the Common Ratio (\( r \)):**
The second term of a geometric sequence is given by:
\[
a_2 = a \times r
\]
Plugging in the known values:
\[
\frac{1}{\sqrt{2}} = 8 \times r \\
\Rightarrow r = \frac{1}{8\sqrt{2}} = \frac{\sqrt{2}}{16}
\]
However, this leads to a very small ratio, which doesn't align with the expected sum. It's likely there's a typographical error in the problem statement. Assuming the second term is intended to be \( \frac{8}{\sqrt{2}} \):
\[
\frac{8}{\sqrt{2}} = 8 \times r \\
\Rightarrow r = \frac{1}{\sqrt{2}}
\]
2. **Verify the Ratio (\( |r| < 1 \)):**
\[
\left| \frac{1}{\sqrt{2}} \right| = \frac{1}{\sqrt{2}} < 1
\]
Since the absolute value of \( r \) is less than 1, the infinite geometric series converges.
3. **Calculate the Sum to Infinity (\( S \)):**
The formula for the sum of an infinite geometric series is:
\[
S = \frac{a}{1 - r}
\]
Substituting the known values:
\[
S = \frac{8}{1 - \frac{1}{\sqrt{2}}} = \frac{8}{\frac{\sqrt{2} - 1}{\sqrt{2}}} = 8 \times \frac{\sqrt{2}}{\sqrt{2} - 1}
\]
4. **Rationalize the Denominator:**
Multiply the numerator and denominator by \( \sqrt{2} + 1 \) to rationalize:
\[
S = 8 \times \frac{\sqrt{2} (\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = 8 \times \frac{2 + \sqrt{2}}{2 - 1} = 8 \times (2 + \sqrt{2}) = 16 + 8\sqrt{2}
\]
Thus, the sum of the infinite series is:
\[
S = 16 + 8\sqrt{2}
\]
---
## **Part 2: Finite Geometric Series (20 Terms)**
### **Given:**
- **Series**: \( x = 5 + 15 + 45 + \dots \) (20 terms)
### **Objectives:**
1. **Write the series in sigma notation.**
2. **Calculate the value of \( x \).**
### **Solution:**
#### **1. Sigma Notation:**
A geometric series can be expressed in sigma (summation) notation as follows:
\[
x = \sum_{k=0}^{n-1} a \times r^k
\]
where:
- \( a \) = first term
- \( r \) = common ratio
- \( n \) = number of terms
For the given series:
- **First term (\( a \))**: 5
- **Common ratio (\( r \))**: \( \frac{15}{5} = 3 \)
- **Number of terms (\( n \))**: 20
Thus, the sigma notation is:
\[
x = \sum_{k=0}^{19} 5 \times 3^k
\]
*Alternatively, it can also be written as:*
\[
x = \sum_{k=1}^{20} 5 \times 3^{k-1}
\]
#### **2. Calculating the Value of \( x \):**
The sum of the first \( n \) terms of a geometric series is given by:
\[
S_n = a \times \frac{r^n - 1}{r - 1}
\]
Plugging in the known values:
\[
S_{20} = 5 \times \frac{3^{20} - 1}{3 - 1} = 5 \times \frac{3^{20} - 1}{2}
\]
**Simplifying:**
\[
x = \frac{5}{2} \times (3^{20} - 1)
\]
Given that \( 3^{20} \) is a large number (approximately \( 3.486 \times 10^9 \)), the exact numerical value would be:
\[
3^{20} = 3^{10} \times 3^{10} = 59,049 \times 59,049 = 3,486,784,401
\]
Thus:
\[
x = \frac{5}{2} \times (3,486,784,401 - 1) = \frac{5}{2} \times 3,486,784,400 = \frac{5 \times 3,486,784,400}{2} = \frac{17,433,922,000}{2} = 8,716,961,000
\]
**Final Answer:**
\[
x = 8,\!716,\!961,\!000
\]
---
**Summary:**
1. **Infinite Geometric Series Sum:** \( 16 + 8\sqrt{2} \)
2. **Finite Geometric Series (20 terms):**
- **Sigma Notation:** \( \sum_{k=0}^{19} 5 \times 3^k \)
- **Sum (\( x \))**: \( 8,\!716,\!961,\!000 \)
Revisado y aprobado por el equipo de tutoría de UpStudy
Explicar
Simplifique esta solución
