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Chapter 11: Problem 14

Find all terms of each finite sequence. $$a_{n}=\left(\frac{1}{2}\right)^{3-n}, 1 \leq n \leq 7$$

Short Answer

Expert verified
The terms are \( \frac{1}{4}, \frac{1}{2}, 1, 2, 4, 8, 16 \) for \(n\) from 1 to 7.

Step by step solution

01

- Understand the Sequence Formula

The given sequence formula is \(a_n = \left( \frac{1}{2} \right)^{3-n}\). This formula will be used to find the terms of the sequence for each value of \(n\) from 1 to 7.
02

- Calculate the First Term

For \(n=1\): \[a_1 = \left( \frac{1}{2} \right)^{3-1} = \left( \frac{1}{2} \right)^2 = \frac{1}{4}\]
03

- Calculate the Second Term

For \(n=2\): \[a_2 = \left( \frac{1}{2} \right)^{3-2} = \left( \frac{1}{2} \right)^1 = \frac{1}{2}\]
04

- Calculate the Third Term

For \(n=3\): \[a_3 = \left( \frac{1}{2} \right)^{3-3} = \left( \frac{1}{2} \right)^0 = 1\]
05

- Calculate the Fourth Term

For \(n=4\): \[a_4 = \left( \frac{1}{2} \right)^{3-4} = \left( \frac{1}{2} \right)^{-1} = 2\]
06

- Calculate the Fifth Term

For \(n=5\): \[a_5 = \left( \frac{1}{2} \right)^{3-5} = \left( \frac{1}{2} \right)^{-2} = 4\]
07

- Calculate the Sixth Term

For \(n=6\): \[a_6 = \left( \frac{1}{2} \right)^{3-6} = \left( \frac{1}{2} \right)^{-3} = 8\]
08

- Calculate the Seventh Term

For \(n=7\): \[a_7 = \left( \frac{1}{2} \right)^{3-7} = \left( \frac{1}{2} \right)^{-4} = 16\]

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Key Concepts

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

sequence formulas
A sequence formula helps you compute the terms in a sequence by plugging values into the formula. In this problem, the formula provided is: \[a_n = \bigg(\frac{1}{2}\bigg)^{3-n}\]To solve a sequence, you change the value of \(n\) within the given range. Here, the range of \(n\) is from 1 to 7.
  • When \(n\) is 1, you substitute 1 into the formula to find the first term.
  • Next, changing \(n\) to 2 gives the second term, and so on until you reach the seventh term.
So the formula serves as a blueprint for calculating each term individually by simply adjusting \(n\).
exponents
Exponents represent how many times you multiply a number by itself. In the given sequence, we encounter various exponents on the fraction \(\frac{1}{2}\). Here's a brief explanation of the key points:
  • When an exponent is positive, like 2, it means you multiply the base (\(\frac{1}{2}\)) by itself multiple times.
  • When an exponent is zero, any non-zero number raised to the power of zero is 1, so \(\bigg(\frac{1}{2}\bigg)^0 = 1\).
  • When an exponent is negative, it indicates a reciprocal. For example, \(\bigg(\frac{1}{2}\bigg)^{-2}\) means we take the reciprocal of \(\bigg(\frac{1}{2}\bigg)^2\), which is 4.
In our sequence, as \(n\) increases, the exponent \(3-n\) transitions from positive to negative, which inverts the fraction.
term calculations
To find each term in the sequence, you need to substitute different values of \(n\) into the sequence formula and simplify:

  • For \(n = 1\): \[a_1 = \bigg(\frac{1}{2}\bigg)^{3-1} = \bigg(\frac{1}{2}\bigg)^2 = \frac{1}{4}\]
  • For \(n = 2\):\[a_2 = \bigg(\frac{1}{2}\bigg)^{3-2} = \bigg(\frac{1}{2}\bigg)^1 = \frac{1}{2}\]
  • For \(n = 3\):\[a_3 = \bigg(\frac{1}{2}\bigg)^{3-3} = \bigg(\frac{1}{2}\bigg)^0 = 1\]
  • For \(n = 4\):\[a_4 = \bigg(\frac{1}{2}\bigg)^{3-4} = \bigg(\frac{1}{2}\bigg)^{-1} = 2\]
  • For \(n = 5\):\[a_5 = \bigg(\frac{1}{2}\bigg)^{3-5} = \bigg(\frac{1}{2}\bigg)^{-2} = 4\]
  • For \(n = 6\):\[a_6 = \bigg(\frac{1}{2}\bigg)^{3-6} = \bigg(\frac{1}{2}\bigg)^{-3} = 8\]
  • For \(n = 7\):\[a_7 = \bigg(\frac{1}{2}\bigg)^{3-7} = \bigg(\frac{1}{2}\bigg)^{-4} = 16\]
Each calculation follows the rule of exponents and order of operations. By carefully replacing \(n\) and simplifying, you can find each term in the sequence.

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