/*! 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 2 Find a formula for \(s_{n}, n \g... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find a formula for \(s_{n}, n \geq 1\) for the sequence \(-3,5,-7,9,-11 \ldots\)

Short Answer

Expert verified
The formula for the Sequence is s_{n}= (-1)^{n+1}(2n+1).

Step by step solution

01

- Identify the Pattern

Examine the sequence: -3, 5, -7, 9, -11, Notice the alternating signs and determine how the absolute values progress: 3, 5, 7, 9, 11. The absolute values increment by 2 each time.
02

- Establish the Rule for the Absolute Value

Since the absolute values increase by 2, the terms can be written as 2n + 1 or similar. Test this to ensure it matches the absolute values. For instance, the first term is |2(1)-1| = 3, the second term is |2(2)+1| = 5, and so forth.
03

- Determine the Sign Pattern

The signs alternate: The 1st term is negative, the 2nd term positive: Observe that (-1)^{n+1} provides the appropriate pattern.
04

- Combine the Formulas

Combine the steps to obtain the formula factoring in signs: s_{n} = (-1)^{n+1}(2n + 1). This formula ensures the correct alternation in signs while accommodating the sequence structure.

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Key Concepts

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

Pattern Recognition
Recognizing patterns is a fundamental skill in solving sequences. Patterns help us see the structure of a sequence and predict future terms. In the given sequence \[-3, 5, -7, 9, -11 \dots\], the first observation is the **alternation of signs**.
The sequence alternates between negative and positive values.
For example, terms go from -3 (negative) to 5 (positive), then back to -7 (negative).
Additionally, the sequence reveals a pattern in the absolute values of each term. These are \[3, 5, 7, 9, 11 \dots\], incrementing by 2.
Recognizing these patterns is the first step to building a formula that fits all terms of the sequence.
Sign Alternation
Sign alternation is a key feature in many sequences, including our current example. To capture this alternation:
  • Notice how the sequence alternates from negative to positive.
  • The mathematical expression \[(-1)^{n+1}\]fits this pattern perfectly.
    When n is 1, it evaluates to \[(-1)^{2} = 1\]making the term positive.
    When n is 2, it evaluates to \[(-1)^{3} = -1\], making the term negative.
This consistent switch between terms ensures the formula encapsulates the alternating nature of the sequence.
By incorporating this sign pattern into the formula, all terms will correctly alternate.
Formula Derivation
Deriving a formula involves combining the detected patterns and expressions into a cohesive whole. Here's how:
  • Start by realizing the absolute values follow \[|2n - 1|\], which was found through **Pattern Recognition**.
  • Add the **Sign Alternation** component: \[(-1)^{n+1}\].
Combining these aspects, we derive the general term's formula:
\[s_n = (-1)^{n+1}(2n - 1)\].
This formula aligns with each term in the sequence, correctly interchanging signs while matching the absolute values.
Sequence Analysis
Analyzing a sequence goes beyond just finding the next term. It involves:
  • Deeply inspecting the sequence's growth rate or pattern.
  • Deriving relationships between terms.
    For instance, in our sequence: \[-3, 5, -7, 9, -11\dots\] we identify a common difference of 2 in the absolute values.
Effective analysis helps in understanding the sequence structure, further verifying if our derived formula \[s_n = (-1)^{n+1}(2n - 1)\] holds true for different values of n.
Plugging various n values (e.g., 1, 2, 3) ensures the formula generates terms that match the initial sequence, offering complete verification.

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Most popular questions from this chapter

Compute the value of the following improper integral. If it converges, enter its value. Enter infinity if it diverges to \(\infty\), and -infinity if it diverges to \(-\infty\). Otherwise, enter diverges. $$ \int_{1}^{\infty} \frac{3 d x}{x^{2}+1}= $$ Does the series \(\sum_{n=1}^{\infty} \frac{3}{n^{2}+1}\) converge or diverge? [Choose: converges | diverges to +infinity | diverges to -infinity | diverges]

Determine the sum of the following series. $$ \sum_{n=1}^{\infty}\left(\frac{3^{n}+5^{n}}{9^{n}}\right) $$

Find the first four terms of the Taylor series for the function \(\cos (x)\) about the point \(a=\) \(-\pi / 4\). (Your answers should include the variable \(\mathrm{x}\) when appropriate.) \(\cos (x)=\) \(+\ldots\)

Finding limits of convergent sequences can be a challenge. However, there is a useful tool we can adapt from our study of limits of continuous functions at infinity to use to find limits of sequences. We illustrate in this exercise with the example of the sequence $$ \frac{\ln (n)}{n} $$ a. Calculate the first 10 terms of this sequence. Based on these calculations, do you think the sequence converges or diverges? Why? b. For this sequence, there is a corresponding continuous function \(f\) defined by $$ f(x)=\frac{\ln (x)}{x} $$ Draw the graph of \(f(x)\) on the interval [0,10] and then plot the entries of the sequence on the graph. What conclusion do you think we can draw about the sequence \(\left\\{\frac{\ln (n)}{n}\right\\}\) if \(\lim _{x \rightarrow \infty} f(x)=L ?\) Explain. c. Note that \(f(x)\) has the indeterminate form \(\frac{\infty}{\infty}\) as \(x\) goes to infinity. What idea from differential calculus can we use to calculate \(\lim _{x \rightarrow \infty} f(x) ?\) Use this method to find \(\lim _{x \rightarrow \infty} f(x) .\) What, then, is \(\lim _{n \rightarrow \infty} \frac{\ln (n)}{n} ?\)

Airy's equation \(^{2}\) $$ y^{\prime \prime}-x y=0 $$ can be used to model an undamped vibrating spring with spring constant \(x\) (note that \(y\) is an unknown function of \(x\) ). So the solution to this differential equation will tell us the behavior of a spring-mass system as the spring ages (like an automobile shock absorber). Assume that a solution \(y=f(x)\) has a Taylor series that can be written in the form $$ y=\sum_{k=0}^{\infty} a_{k} x^{k} $$ where the coefficients are undetermined. Our job is to find the coefficients. (a) Differentiate the series for \(y\) term by term to find the series for \(y^{\prime}\). Then repeat to find the series for \(y^{\prime \prime}\). (b) Substitute your results from part (a) into the Airy equation and show that we can write Equation \((8.6 .4)\) in the form$$ \sum_{k=2}^{\infty}(k-1) k a_{k} x^{k-2}-\sum_{k=0}^{\infty} a_{k} x^{k+1}=0 $$ (c) At this point, it would be convenient if we could combine the series on the left in \((8.6 .5)\), but one written with terms of the form \(x^{k-2}\) and the other with terms in the form \(x^{k+1}\). Explain why $$ \sum_{k=2}^{\infty}(k-1) k a_{k} x^{k-2}=\sum_{k=0}^{\infty}(k+1)(k+2) a_{k+2} x^{k} $$ (d) Now show that $$ \sum_{k=0}^{\infty} a_{k} x^{k+1}=\sum_{k=1}^{\infty} a_{k-1} x^{k} $$ (e) We can now substitute \((8.6 .6)\) and \((8.6 .7)\) into \((8.6 .5)\) to obtain $$ \sum_{n=0}^{\infty}(n+1)(n+2) a_{n+2} x^{n}-\sum_{n=1}^{\infty} a_{n-1} x^{n}=0 $$ Combine the like powers of \(x\) in the two series to show that our solution must satisfy $$ 2 a_{2}+\sum_{k=1}^{\infty}\left[(k+1)(k+2) a_{k+2}-a_{k-1}\right] x^{k}=0 $$ (f) Use equation (8.6.9) to show the following: i. \(a_{3 k+2}=0\) for every positive integer \(k\), ii. \(a_{3 k}=\frac{1}{(2)(3)(5)(6) \cdots(3 k-1)(3 k)} a_{0}\) for \(k \geq 1\), iii. \(a_{3 k+1}=\frac{1}{(3)(4)(6)(7) \cdots(3 k)(3 k+1)} a_{1}\) for \(k \geq 1\). (g) Use the previous part to conclude that the general solution to the Airy equation \((8.6 .4)\) is $$ \begin{aligned} y=& a_{0}\left(1+\sum_{k=1}^{\infty} \frac{x^{3 k}}{(2)(3)(5)(6) \cdots(3 k-1)(3 k)}\right) \\ &+a_{1}\left(x+\sum_{k=1}^{\infty} \frac{x^{3 k+1}}{(3)(4)(6)(7) \cdots(3 k)(3 k+1)}\right) \end{aligned} $$ Any values for \(a_{0}\) and \(a_{1}\) then determine a specific solution that we can approximate as closely as we like using this series solution.

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