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Chapter 13: Problem 34

Find the sum of the first five terms of the sequence. $$b_{n}=2^{-n}$$

Short Answer

Expert verified
The sum is \( \frac{31}{32} \).

Step by step solution

01

Identify the Sequence and Terms

The sequence given is \( b_n = 2^{-n} \). This means the terms are \( b_1 = 2^{-1} \), \( b_2 = 2^{-2} \), \( b_3 = 2^{-3} \), \( b_4 = 2^{-4} \), and \( b_5 = 2^{-5} \). Our goal is to find the sum of these first five terms.
02

Calculate Each Term

Let's calculate each term of the sequence:- \( b_1 = 2^{-1} = \frac{1}{2} \)- \( b_2 = 2^{-2} = \frac{1}{4} \)- \( b_3 = 2^{-3} = \frac{1}{8} \)- \( b_4 = 2^{-4} = \frac{1}{16} \)- \( b_5 = 2^{-5} = \frac{1}{32} \)
03

Add the Calculated Terms

Sum the first five terms: \[ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} \] Find a common denominator, which is 32, and convert each fraction:- \( \frac{1}{2} = \frac{16}{32} \)- \( \frac{1}{4} = \frac{8}{32} \)- \( \frac{1}{8} = \frac{4}{32} \)- \( \frac{1}{16} = \frac{2}{32} \)- \( \frac{1}{32} = \frac{1}{32} \) Now add them:\[ \frac{16}{32} + \frac{8}{32} + \frac{4}{32} + \frac{2}{32} + \frac{1}{32} = \frac{31}{32} \]
04

Conclusion

The sum of the first five terms of the sequence \( b_n = 2^{-n} \) is \( \frac{31}{32} \).

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Key Concepts

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

Understanding a Geometric Sequence
In a geometric sequence, each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. This is what makes geometric sequences distinct from other types of sequences. The sequence provided in our exercise is special because it features fractions: the terms are powers of two with negative exponents.
This means each term is found by dividing by the base number (2), rather than multiplying by a whole number.
  • The first term is typically denoted as \( a \). In our scenario, \( a = \frac{1}{2} \).
  • The common ratio \( r \) in this sequence is also \( \frac{1}{2} \), because each term is half of the previous term.
  • The formula for the nth term of a geometric sequence is \( a_n = a \cdot r^{n-1} \). In the given sequence, it simplifies to \( b_n = 2^{-n} \).
Recognizing and understanding the common ratio is key to understanding how each term in the sequence relates to the others.
Sequence Calculation Basics
Finding the sum of a geometric sequence involves calculating each term separately and then summing them up. In this exercise, we started by identifying each term using the formula \( b_n = 2^{-n} \).
For example:
  • \( b_1 = 2^{-1} = \frac{1}{2} \)
  • \( b_2 = 2^{-2} = \frac{1}{4} \)
  • \( b_3 = 2^{-3} = \frac{1}{8} \)
  • \( b_4 = 2^{-4} = \frac{1}{16} \)
  • \( b_5 = 2^{-5} = \frac{1}{32} \)
Once all terms are identified, we need to add them together. This requires precise sequence calculation to effectively manage the fractional terms, which we'll cover next. These steps are crucial in sequence calculation, especially with fractions.
Simplifying Through Fraction Addition
Adding fractions can seem daunting, but with a common denominator, it becomes a straightforward process.
In our sequence, each term is a fraction: \( \frac{1}{2} \), \( \frac{1}{4} \), \( \frac{1}{8} \), \( \frac{1}{16} \), \( \frac{1}{32} \). To add these fractions efficiently, we use the least common denominator (LCD), which in this case is 32. Here's how each fraction converts to have a denominator of 32:
  • \( \frac{1}{2} = \frac{16}{32} \)
  • \( \frac{1}{4} = \frac{8}{32} \)
  • \( \frac{1}{8} = \frac{4}{32} \)
  • \( \frac{1}{16} = \frac{2}{32} \)
  • \( \frac{1}{32} = \frac{1}{32} \)
Once the fractions are rewritten with a common denominator, add the numerators: \[ 16 + 8 + 4 + 2 + 1 = 31 \] This results in the sum \( \frac{31}{32} \). With practice, fraction addition becomes a simple matter of adjusting to a common language through common denominators.

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Most popular questions from this chapter

Find the indicated roots. Express the results in rectangular form. If \(z=r(\cos \theta+i \sin \theta)\) and \(z\) is not zero, show that \(\frac{1}{z}=\frac{1}{r}(\cos \theta-i \sin \theta)\) Hint: \(1 / z=[1(\cos 0+i \sin 0)] /[r(\cos \theta+i \sin \theta)]\)

Use DeMoivre's theorem to find the indicated roots. Express the results in rectangular form. Cube roots of \(-27 i\)

Carry out the indicated operations. Express your results in rectangular form for those cases in which the trigonometric functions are readily evaluated without tables or a calculator. $$\sqrt{2}\left(\cos \frac{1}{3} \pi+i \sin \frac{1}{3} \pi\right) \times \sqrt{2}\left(\cos \frac{4}{3} \pi+i \sin \frac{4}{3} \pi\right)$$

Carry out the indicated operations. Express your results in rectangular form for those cases in which the trigonometric functions are readily evaluated without tables or a calculator. $$\sqrt{3}\left(\cos 140^{\circ}+i \sin 140^{\circ}\right) \div 3\left(\cos 5^{\circ}+i \sin 5^{\circ}\right)$$

Find the third term in the expansion of \((a-b)^{30}\).
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