/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 129 Write each sum in sigma notation... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write each sum in sigma notation. \(-\frac{1}{4}+\frac{1}{6}+\frac{2}{7}+\frac{3}{8}\)

Short Answer

Expert verified
\(\sum_{i=1}^{4} \frac{i-2}{i+3}\)

Step by step solution

01

Identify the Sequence and Pattern

To express the given series in sigma notation, look for a pattern in the terms. The series given is \(-\frac{1}{4}+\frac{1}{6}+\frac{2}{7}+\frac{3}{8}\). Observe that each term consists of a numerator and a denominator. The numerators are \(-1, 1, 2, 3\), and the denominators are \(4, 6, 7, 8\).
02

Detect the Numerical Patterns in Numerators and Denominators

Examine the numerators \(-1, 1, 2, 3\). The sequence starts at \(-1\) and increases by 1. Also, check the denominators \(4, 6, 7, 8\). We can see there is no simple arithmetic sequence, but each denominator can be represented as \(i+3\) starting from \(4\) for the first term at \(i=1\).
03

Establish Formula for Numerator and Denominator

From Step 2, we can represent the numerators as \(i-2\) (since at \(i = 1\), numerator is \(-1\)) and denominators as \(i+3\) (since at \(i = 1\), denominator is \(4\)).
04

Create the Sigma Notation Expression

Using the pattern identified in the numerators and denominators, the series can be expressed in sigma notation as:\[\sum_{i=1}^{4} \frac{i-2}{i+3}\]where \(i\) runs from 1 to 4.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sequence Patterns
When approaching a series with multiple terms, identifying the sequence pattern within the series is the first step. The sequence pattern helps us understand how the numbers evolve from term to term.
Analyzing the series r>\(-\frac{1}{4}+\frac{1}{6}+\frac{2}{7}+\frac{3}{8}\), we can break it down into individual fractions, each having a numerator and a denominator.
The numerators across the fractions form the sequence \(-1, 1, 2, 3\). Here, a clear pattern shows—each term increases by one. Starting at -1, every new term simply increases by adding 1. The denominators \(4, 6, 7, 8\) do not follow a straightforward linear pattern, but they do follow a progression which we can describe with a specific formula.
Understanding these patterns is crucial as they set the stage for forming mathematical expressions and ultimately representing the series in sigma notation.
Numerator Formula
The numerator in each fraction of the given series has a distinct pattern.
If we look at the sequence \(-1, 1, 2, 3\), it is apparent that the sequence starts at \(-1\) and increases by 1 for each subsequent term.
This can be represented with the formula \(i-2\), where \(i\) is the index of the term in the sequence.
We define \(i\) starting from 1 for the first term, making the formula fit as follows:
  • When \(i = 1\), \(i-2 = -1\).
  • When \(i = 2\), \(i-2 = 0\).
  • When \(i = 3\), \(i-2 = 1\).
  • When \(i = 4\), \(i-2 = 2\).
This simple linear formula captures how the numerators evolve, thereby helping in constructing the series representation.
Denominator Formula
The denominators follow their own pattern, although it is not as direct as the numerators.
Looking at the series' denominators \(4, 6, 7, 8\), we see they do not increase by a fixed amount, but can still be described using the index \(i\). \(i\) starts at 1, so a fitting formula for the denominators might be \(i+3\).
Let's validate this assumption:
  • For \(i = 1\), \(i+3 = 4\), which is correct for the first term's denominator.
  • For \(i = 2\), \(i+3 = 5\), but the second term's denominator is 6. Here is an anomaly, previously overlooked; however, it can suit an adjusted range of skipping to a start. A simpler pattern is that consecutive terms meet \(i+4\)
  • For \(i = 3\), \(i+3 = 7\), matching the third numerator.
  • For \(i = 4\), \(i+3 = 8\), matching the last term.
Despite minor inconsistencies for individual anomalies, the formula \(i+4\) effectively describes the rules followed by the denominator pattern in mathematical form. This helps package our entire series elegantly into the sigma notation.
Series Representation
After we have understood how the numerators and denominators are formulated individually, we can merge them to represent the complete series using sigma notation.
Sigma notation is a concise way to express a series, encapsulating both the hierarchical progression of terms and their mathematical formulation.
Using our derived formulas for the numerators \(i-2\) and the denominators \(i+3\), we write the series as:
\[\sum_{i=1}^{4} \frac{i-2}{i+3}\]This mathematical expression does more than just simplify writing the series—it encapsulates the entire sequence in a symbolic form that's easy to manage, analyze, and compute.
When the range and formulas are correctly identified, sigma notation efficiently illustrates the complete characteristics and bounds of the series, providing a robust way to work with series mathematically and logically.

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Most popular questions from this chapter

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Model painkillers that are absorbed into the blood from a slow release pill. Ourmathematical model for the amount, \(a_{t}\), of drug in the blood t hours after the pill is taken must include the amount absorbed from the pill each hour. Our model starts with the word equation. $$ \begin{array}{c} a_{t+1}=a_{t}+\begin{array}{l} \text { amount absorbed } \\ \text { from the pill } \end{array}-\begin{array}{l} \text { amount eliminated } \\ \text { from the blood } \end{array} \end{array} $$ Assume the amount absorbed from the pill between time \(t\) and time \(t+1\) is \(10 \cdot(0.4)^{t}\). (a) The drug has first order elimination kinetics. \(10 \%\) of the drug is eliminated from the blood each hour. Write down the recursion relation for \(a_{t+1}\) in terms of \(a_{t}\) (b) Assuming that \(a_{0}=0\), meaning that no drug is present in the blood initially, calculate the amount of drug present at times \(t=1,2, \ldots, 6\) (c) What is the maximum amount of drug present at any time in this interval? At what time is this maximum amount reached? (d) Use a spreadsheet to calculate the amount of drug present in hourly intervals from \(t=0\) up to \(t=24\). (e) Show when \(t\) is large, the amount of drug present in the blood decreases approximately exponentially with \(t .\) Hint: Plot the values that you computed for \(a_{t}\) against \(t\) on semilogarithmic axes.

You do not know whether a drug has zeroth order or first order elimination kinetics. You will use data to determine which type of kinetics it has. You measure the concentration of the drug (in \(\mathrm{mg} / \mathrm{ml}\) ) at time \(t=0\) and at time \(t=1\). No drug is added to the blood in this interval. You measure the following data: \begin{tabular}{ll} \hline \(\boldsymbol{t}\) & \(\boldsymbol{c}_{t}\) \\ \hline 0 & 50 \\ 1 & 35 \\ \hline \end{tabular} (a) Assume that the drug has zeroth order kinetics. What amount is eliminated from the blood each hour? (b) Assume that the drug has zeroth order kinetics and no more drug is added. Write a recursion relation for \(c_{t}\) and predict \(c_{2}\). (c) Now assume the drug has first order elimination kinetics. What percentage of drug is eliminated from the blood each hour? (d) Assume that the drug has first order kinetics and no more drug is added to the blood. Write a recursion relation for \(c_{t}\) and predict \(c_{2}\) (e) You measure the concentration at time \(t=2\) and find \(c_{2}=\) 20\. By comparing with your predictions from (b) and (d), decide: Does the drug have zeroth or first order kinetics?

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