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Chapter 10: Problem 5

Write out each finite series. $$ \sum_{n=1}^{6} \frac{1-(-1)^{n}}{2} $$

Short Answer

Expert verified
The sum of the series is 3.

Step by step solution

01

Understand the Series

The series given is \( \sum_{n=1}^{6} \frac{1-(-1)^{n}}{2} \). We need to compute the value for each term from \( n=1 \) to \( n=6 \).
02

Calculate Individual Terms

For each \( n \):- When \( n = 1 \), \( (-1)^1 = -1 \), so \( \frac{1 - (-1)^1}{2} = \frac{1 + 1}{2} = 1 \).- When \( n = 2 \), \( (-1)^2 = 1 \), so \( \frac{1 - (-1)^2}{2} = \frac{1 - 1}{2} = 0 \).- When \( n = 3 \), \( (-1)^3 = -1 \), so \( \frac{1 - (-1)^3}{2} = \frac{1 + 1}{2} = 1 \).- When \( n = 4 \), \( (-1)^4 = 1 \), so \( \frac{1 - (-1)^4}{2} = \frac{1 - 1}{2} = 0 \).- When \( n = 5 \), \( (-1)^5 = -1 \), so \( \frac{1 - (-1)^5}{2} = \frac{1 + 1}{2} = 1 \).- When \( n = 6 \), \( (-1)^6 = 1 \), so \( \frac{1 - (-1)^6}{2} = \frac{1 - 1}{2} = 0 \).
03

Write the Series

The series can now be written as: \( 1 + 0 + 1 + 0 + 1 + 0 \).
04

Calculate the Sum

Add the terms: \( 1 + 0 + 1 + 0 + 1 + 0 = 3 \).

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Key Concepts

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

Arithmetic Sequences
An arithmetic sequence is a list of numbers where each term differs from the next by a constant value. This difference is known as the common difference. For example, in the sequence 2, 4, 6, 8, the common difference is 2.
  • To find any term in an arithmetic sequence, you use the formula: \( a_n = a_1 + (n-1) \cdot d \), where \( a_1 \) is the first term, \( d \) is the common difference, and \( n \) is the term number.
  • Arithmetic sequences are useful when you need to model situations where something increases or decreases by the same amount every time, like savings plans or sequences in geometry.
  • In the context of finite series, arithmetic sequences can be summed up using the formula for the sum of an arithmetic series: \( S_n = \frac{n}{2} \cdot (a_1 + a_n) \), where \( S_n \) is the sum of the first \( n \) terms.
This concept of arithmetic sequences is important because it sets the foundation for understanding more complex sequences and series.
Summation Notation
Summation notation is a way to denote the sum of a sequence of numbers. It is represented by the symbol "Sigma" \( \Sigma \). This notation helps simplify expressions where you'd otherwise have to write out long additions.
  • The expression \( \sum_{n=1}^{6} \frac{1-(-1)^{n}}{2} \) is an example of summation notation. It tells you to find the sum of terms generated by the expression \( \frac{1-(-1)^{n}}{2} \) as \( n \) goes from 1 to 6.
  • Summation is a key concept in mathematics because it provides a concise way to represent the sum of a potentially large number of terms. For finite series, it guarantees that the process of summing up the terms will eventually end, unlike infinite series.
  • To break down summations further, it often helps to calculate the value of the expression for each individual \( n \) and then sum up the results, just as we've done in this exercise.
Understanding summation notation is crucial, especially as you venture into more advanced areas of mathematics where handling large data sets or complicated formulas is necessary.
Calculus Basics
Calculus is a branch of mathematics focused on studying rates of change (differential calculus) and accumulation of quantities (integral calculus). It stands as a fundamental tool in mathematics with applications across science and engineering.
  • Although our exercise deals with a finite series, calculus is essential in understanding infinite series and the concept of convergence, where series approach a limit.
  • Differentiation, one of the main operations in calculus, helps find the rate at which a quantity changes relative to another. Integration, its counterpart, computes the total accumulation of a quantity over an interval.
  • In more advanced topics, calculus makes use of summation notation to express integrals and derivatives, depicting areas under curves and slopes of functions respectively.
Even though the series in the exercise does not directly require calculus, the foundations of summation and series form a bridge to it, enhancing our understanding of more complex mathematical concepts.

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Most popular questions from this chapter

Suppose that \(f(0)=1\) and \(f^{(n)}(0)=n !\) for \(n=1,2,3, \ldots\) What will be the Taylor series for fat \(x=0\) ? [Hint: See page 669.] Do you know a formula from earlier in the chapter to express this function in a different form?

Find the third Taylor polynomial at \(x=0\) of each function. $$ f(x)=7+x-3 x^{2} $$

True or False: If the partial sums \(S_{n}\) of an infinite series all satisfy \(S_{n} \leq 10,\) then the sum \(S\) also satisfies \(\quad S \leq 10\).

Show that the Taylor series for \(\ln x\) $$ (x-1)-\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{3}-\frac{(x-1)^{4}}{4}+\cdots $$ converges at \(x=2\) by carrying out the following steps. a. Evaluate the series at \(x=2\) to obtain $$ 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots $$ b. Group the terms in pairs to obtain \(\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\cdots\) and explain why this expresses the series as a sum of positive terms. c. Then group the terms as \(1-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{5}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\cdots\) and explain why this shows that the sum must be less than 1 It can be shown that if a series of positive terms remains below a (finite) number, then the series converges.

Find the third Taylor polynomial at \(x=0\) of each function. $$ f(x)=2-x+3 x^{2}-x^{3} $$
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