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Chapter 9: Problem 43

Find a recursive definition for the sequence. $$3,5,9,17,33, \ldots$$

Short Answer

Expert verified
The recursive formula is \(a_n = a_{n-1} + 2^{n-1}\), starting with \(a_1 = 3\).

Step by step solution

01

Identify the Pattern

First, look at the differences between successive terms in the sequence: \(5 - 3 = 2\), \(9 - 5 = 4\), \(17 - 9 = 8\), \(33 - 17 = 16\). The differences are 2, 4, 8, 16, which are powers of 2.
02

Formulate the Recursive Formula

Use the pattern in differences to write a recursive formula. If the first term \(a_1 = 3\) and each subsequent term is obtained by adding a power of 2 to the previous term, then \(a_n = a_{n-1} + 2^{n-1}\).
03

Verify the Recursive Formula

Check the recursive formula \(a_n = a_{n-1} + 2^{n-1}\) against the sequence: \(a_1 = 3\), \(a_2 = 3 + 2^1 = 5\), \(a_3 = 5 + 2^2 = 9\), \(a_4 = 9 + 2^3 = 17\), \(a_5 = 17 + 2^4 = 33\). The formula correctly generates the sequence.

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Key Concepts

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

Difference Patterns
When trying to understand a sequence, observing the difference between successive terms is often a great starting point. This sequence begins with the numbers 3, 5, 9, 17, 33... Let's take a closer look at how these numbers change from one to the next. This is known as examining 'difference patterns.'

By calculating the differences between consecutive terms, we get:
  • 5 - 3 = 2
  • 9 - 5 = 4
  • 17 - 9 = 8
  • 33 - 17 = 16
Notice that these differences themselves form a sequence: 2, 4, 8, 16... which are all powers of 2! Recognizing such patterns is key and helps us predict how each term is constructed based on its predecessor. This ability to detect patterns to find formulas is crucial in solving sequences logically.
Recursive Formula
Once the pattern in a sequence has been identified, it's time to formulate the sequence mathematically. This is done through a 'recursive formula' which defines each term based on the previous term. For our sequence, we observed that each term is 2 raised to some power added to the previous term.

The recursive formula can be established as follows:
  • Start by recognizing that the very first term, denoted as \(a_1 = 3\).
  • Next, understand that each subsequent term \(a_n\) is produced by taking the previous term \(a_{n-1}\) and adding the next power of 2: \(a_n = a_{n-1} + 2^{n-1}\).
This simple equation lets us calculate each new term in the sequence by knowing only the preceding term. It's a concise and efficient way of expressing any sequence that follows a regular pattern.
Powers of Two
The term 'powers of two' refers to numbers that can be expressed as \(2^n\), where \(n\) is a non-negative integer. They follow a straightforward pattern: 1, 2, 4, 8, 16, 32, and so on. In our sequence, powers of two played a crucial role in determining the differences between consecutive terms.

As we noticed, the differences from one term to the next were these very powers of two:
  • The first difference is \(2 = 2^1\)
  • The second difference is \(4 = 2^2\)
  • The third difference is \(8 = 2^3\)
  • The fourth difference is \(16 = 2^4\)
Recognizing powers of two within differences of a sequence often hints at exponential growth or systematic addition. This is a common theme in sequences and plays into both understanding them better and effectively constructing recursive formulas.

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Most popular questions from this chapter

Converge by the alternating series test. Use Theorem 9.9 to find how many terms give a partial sum, \(S_{n},\) within 0.01 of the sum, \(S,\) of the series. $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2 n) !}$$

Determine whether the series converges. $$\sum_{n=1}^{\infty} \frac{(2 n) !}{(n !)^{2}}$$

Decide if the statements are true or false. Give an explanation for your answer.If \(0 \leq a_{n} \leq b_{n}\) for all \(n\) and \(\sum a_{n}\) diverges, then \(\sum b_{n}\) diverges.

Determine whether the series converges. $$\sum_{n=1}^{\infty} \frac{1}{n^{2}} \tan \left(\frac{1}{n}\right)$$

Are true or false. Give an explanation for your answer. If the power series \(\sum C_{n} x^{n}\) converges for \(x=2,\) then it converges for \(x=1.\)
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