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Chapter 11: Problem 52

Write the terms for each series. Evaluate the sum, given that \(x_{1}=-2, x_{2}=-1, x_{3}=0\) \(x_{4}=1,\) and \(x_{5}=2 .\) $$\sum_{i=1}^{5}-x_{i}$$

Short Answer

Expert verified
The sum is 0.

Step by step solution

01

Identify the Given Terms

List out the given terms for the series. We have the terms: \(x_{1} = -2\), \(x_{2} = -1\), \(x_{3} = 0\), \(x_{4} = 1\), and \(x_{5} = 2\).
02

Substitute and Apply the Summation

We need to find the sum \(\sum_{i=1}^{5} -x_{i}\). Substitute each \(x_i\) into the equation and apply the negative sign: \(\sum_{i=1}^{5} -x_{i} = -x_{1} - x_{2} - x_{3} - x_{4} - x_{5}\)
03

Insert the Values

Replace each \(x_i\) with its respective value: \(-(-2) - (-1) - (0) - (1) - (2) \)
04

Simplify the Expression

Evaluate the expression step-by-step: \(2 + 1 + 0 - 1 - 2 = 2 + 1 -1 -2 = 0\)
05

Calculate the Sum

Sum the final simplified values: \(2 + 1 - 1 - 2 = 0\)

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Key Concepts

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

Summation Notation
Summation notation, represented by the Greek letter sigma \(\sum\), is a compact way to express the sum of a sequence of terms. In the format \(\sum_{i=1}^{n} a_i\), \(\sum\) signifies summation, \(\i=1\) represents the starting index (usually 1), and \(\) is the last value of the index. The expression \(\sum_{i=1}^{5} -x_{i}\) tells us to sum the terms from \(i=1\) to \(i=5\) of \(-x_i\). This notation helps to succinctly represent long sums and clarify operations on sequences.

To break it down:
  • Sigma (Summation Symbol) 鈥� The \(\sum\) sign represents the sum of a series of terms.
  • Index of Summation 鈥� \(\i\) starts at 1 and ends at 5 in our exercise.
  • Expression 鈥� \(-x_i\), meaning we take the negative of each term in the series.
Understanding summation notation is fundamental for efficiently summing series in mathematics.
Arithmetic Series
An arithmetic series is the sum of the terms of an arithmetic sequence. An arithmetic sequence has a constant difference between consecutive terms. For instance, in our exercise we had values \(x_1 = -2 \, x_2 = -1 \, x_3 = 0 \, x_4 = 1 \, x_5 = 2\). These terms do not form an arithmetic sequence because their differences are not constant, but they can still be summed using summation notation.

Key takeaways about arithmetic series:
  • Consistent Difference 鈥� The difference between consecutive terms in an arithmetic sequence is always the same.
  • Sum Formula 鈥� For a series, if we have \(n\) terms, the sum is given by \[S = \frac{n}{2} (a_1 + a_n)\], but only if it's truly an arithmetic series.
Though our sequence here isn't arithmetic, understanding this helps in recognizing structured sequences and their summations.
Series Evaluation
Series evaluation is the process of finding the sum of all the terms in a series. For \(\sum_{i=1}^{5} -x_{i}\), we followed these steps to evaluate:

  • Identify the Terms: We list out the terms given: \(x_{1}=-2 \), \(x_{2}=-1 \), \(x_{3}=0 \), \(x_{4}=1 \), \(x_{5}=2\).
  • Apply the Negative: Getting \(-x_{1}, -x_{2}, -x_{3}, -x_{4}, -x_{5}\) simplifies to \(2, 1, 0, -1, -2\).
  • Simplify and Sum: Adding these values gives us \(\sum_{i=1}^{5} -x_{i} = 2 + 1 + 0 - 1 - 2 = 0\).
Evaluating a series accurately requires small step-by-step simplifications and careful application of operations like taking the negative in this instance.
Precalculus Problem-Solving
Precalculus problem-solving often involves understanding and applying algebraic concepts to solve complex problems. This problem demonstrated how to:
  • Recognize Patterns: Identify the terms in the series and understand the operations involved.
  • Apply Mathematical Rules: Use summation notation and negative values correctly.
  • Simplify Step-by-Step: Break down complex expressions into simpler parts to find the final result.
These skills are crucial for tackling more advanced topics in calculus and other higher-level mathematics courses. The ability to evaluate series, identify sequences, and logically progress through a problem are foundational skills developed in precalculus.

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Most popular questions from this chapter

Assume that \(n\) is a positive integer. Use mathematical induction to prove each statement S by following these steps. See Example \(I\). (a) Verify the statement for \(n=1\) (b) Write the statement for \(n=k\) (c) Write the statement for \(n=k+1\) (d) Assume the statement is true for \(n=k\). Use algebra to change the statement in part (b) to the statement in part (c). (e) Write a conclusion based on Steps (a)-(d). $$3+9+27+\dots+3^{n}=\frac{1}{2}\left(3^{n+1}-3\right)$$

Use the summation feature of a graphing calculator to evaluate each sum. Round to the nearest thousandth. $$\sum_{j=3}^{8} 2(0.4)^{j}$$

Use the summation properties and rules to evaluate each series. $$\sum_{i=1}^{6}\left(2+i-i^{2}\right)$$

The series $$x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots$$ can be used to approximate the value of \(\ln (1+x)\) for values of \(x\) in \((-1,1] .\) Use the first six terms of this series to approximate each expression. Compare this approximation with the value obtained on a calculator. (a) \(\ln 1.02(x=0.02)\) (b) \(\ln 0.97(x=-0.03)\)

It can be shown that $$(1+x)^{n}=1+n x+\frac{n(n-1)}{2 !} x^{2}+\frac{n(n-1)(n-2)}{3 !} x^{3}+\cdots$$ is true for any real number \(n\) (not just positive integer values) and any real number \(x\), where \(|x| < 1 .\) Use this series to approximate the given number to the nearest thousandth. $$(1.03)^{2}$$
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