18 n=1 4+2" 3n REPETIDAS
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
find the sum of the sereis or show that it divereges
![The image depicts a mathematical series expressed using sigma notation. The series is defined as follows:
\[ \sum_{n=1}^{\infty} \frac{4 + 2^n}{3^n} \]
Here is a detailed explanation:
- The summation symbol \( \Sigma \) indicates that we are summing a series.
- The index of summation is \( n \), which starts at 1 and goes to infinity.
- The expression to be summed is \( \frac{4 + 2^n}{3^n} \).
This series can be interpreted as the sum of the terms of the form \( \frac{4 + 2^n}{3^n} \), where \( n \) takes on successive integer values starting from 1 and increasing without bound.
To clarify, for the first few terms:
- When \( n = 1 \), the term is \( \frac{4 + 2^1}{3^1} = \frac{4 + 2}{3} = 2 \)
- When \( n = 2 \), the term is \( \frac{4 + 2^2}{3^2} = \frac{4 + 4}{9} = \frac{8}{9} \)
- When \( n = 3 \), the term is \( \frac{4 + 2^3}{3^3} = \frac{4 + 8}{27} = \frac{12}{27} = \frac{4}{9} \)
This pattern continues indefinitely as \( n \) increases.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff24651e9-0a43-4fcf-8170-c5e34dc6dcab%2F937f1067-f532-4aa0-960c-124999d3aac6%2Fddfm2ul_processed.png&w=3840&q=75)
Transcribed Image Text:The image depicts a mathematical series expressed using sigma notation. The series is defined as follows:
\[ \sum_{n=1}^{\infty} \frac{4 + 2^n}{3^n} \]
Here is a detailed explanation:
- The summation symbol \( \Sigma \) indicates that we are summing a series.
- The index of summation is \( n \), which starts at 1 and goes to infinity.
- The expression to be summed is \( \frac{4 + 2^n}{3^n} \).
This series can be interpreted as the sum of the terms of the form \( \frac{4 + 2^n}{3^n} \), where \( n \) takes on successive integer values starting from 1 and increasing without bound.
To clarify, for the first few terms:
- When \( n = 1 \), the term is \( \frac{4 + 2^1}{3^1} = \frac{4 + 2}{3} = 2 \)
- When \( n = 2 \), the term is \( \frac{4 + 2^2}{3^2} = \frac{4 + 4}{9} = \frac{8}{9} \)
- When \( n = 3 \), the term is \( \frac{4 + 2^3}{3^3} = \frac{4 + 8}{27} = \frac{12}{27} = \frac{4}{9} \)
This pattern continues indefinitely as \( n \) increases.
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