Q 50. In Problems 37鈥�50, a sequence ... [FREE SOLUTION]

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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 12: Q 50. (page 809)

In Problems 37鈥�50, a sequence is defined recursively. Write down the first five terms.
$a_{1} = \sqrt{2}; a_{n} = \sqrt{\frac{a_{n - 1}}{2}}$

Short Answer

Expert verified

The first five terms of the recursively defined sequence are $a_{1} = \sqrt{2}, a_{2} = \sqrt{\frac{\sqrt{2}}{2}}, a_{3} = \frac{\sqrt{\sqrt{\sqrt{2}}}}{\sqrt{2 \sqrt{2}}}, a_{4} = \frac{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}{\sqrt{2 \sqrt{2 \sqrt{2}}}}, a_{5} = \frac{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}}{\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2}}}}}$ .

Step by step solution

Step 1. Write the given information.

The given recursively defined sequence is:

$a_{1} = \sqrt{2} a_{n} = \sqrt{\frac{a_{n - 1}}{2}}$

Step 2. Determine the first term and find out the second term.

The first term is $(a_{1}) = \sqrt{2}$ .

Now substitute 2 for nin the given formula $a_{n} = \sqrt{\frac{a_{n - 1}}{2}}$ to get the second term,

$a_{2} = \sqrt{\frac{a_{2 - 1}}{2}} = \sqrt{\frac{a_{1}}{2}} a_{2} = \sqrt{\frac{\sqrt{2}}{2}}$

Step 3. Find the 3rd and 4th terms.

Similarly substitute 3, 4 for nin the given formula to get the third and fourth term,

$a_{3} = \sqrt{\frac{a_{3 - 1}}{2}} = \sqrt{\frac{a_{2}}{2}} = \sqrt{\frac{\sqrt{\frac{\sqrt{2}}{2}}}{2}} = \sqrt{\frac{\sqrt{\sqrt{2}}}{2 \sqrt{2}}} a_{3} = \frac{\sqrt{\sqrt{\sqrt{2}}}}{\sqrt{2 \sqrt{2}}} a_{4} = \sqrt{\frac{a_{4 - 1}}{2}} = \sqrt{\frac{a_{3}}{2}} = \sqrt{\frac{\frac{\sqrt{\sqrt{\sqrt{2}}}}{\sqrt{2 \sqrt{2}}}}{2}} a_{4} = \frac{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}{\sqrt{2 \sqrt{2 \sqrt{2}}}}$

Step 4. Find the 5th term.

Now substitute 5 for nin the given formula to get the 5th term,

$a_{5} = \sqrt{\frac{a_{5 - 1}}{2}} = \sqrt{\frac{a_{4}}{2}} = \sqrt{\frac{\frac{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}{\sqrt{2 \sqrt{2 \sqrt{2}}}}}{2}} a_{5} = \frac{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}}{\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2}}}}}$

Therefore, the first five terms arerole="math" localid="1646740081552" $a_{1} = \sqrt{2}, a_{2} = \sqrt{\frac{\sqrt{2}}{2}}, a_{3} = \frac{\sqrt{\sqrt{\sqrt{2}}}}{\sqrt{2 \sqrt{2}}}, a_{4} = \frac{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}{\sqrt{2 \sqrt{2 \sqrt{2}}}}, a_{5} = \frac{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}}{\sqrt{2 \sqrt{2 \sqrt{2 \sqrt{2}}}}}$ .

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91影视

In Problems 37鈥�50, a sequence is defined recursively. Write down the first five terms.
$a_{1} = \sqrt{2}; a_{n} = \sqrt{\frac{a_{n - 1}}{2}}$

Short Answer

Step by step solution

Step 1. Write the given information.

Step 2. Determine the first term and find out the second term.

Step 3. Find the 3rd and 4th terms.

Step 4. Find the 5th term.

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91影视

In Problems 37鈥�50, a sequence is defined recursively. Write down the first five terms.a1=2;an=an-12

Short Answer

Step by step solution

Step 1. Write the given information.

Step 2. Determine the first term and find out the second term.

Step 3. Find the 3rd and 4th terms.

Step 4. Find the 5th term.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Probability and Statistics

Mechanics Maths

Theoretical and Mathematical Physics

Applied Mathematics

Pure Maths

Logic and Functions

Study anywhere. Anytime. Across all devices.

In Problems 37鈥�50, a sequence is defined recursively. Write down the first five terms.
$a_{1} = \sqrt{2}; a_{n} = \sqrt{\frac{a_{n - 1}}{2}}$