Chapter 12: Q 50. (page 809)

In Problems 37–50, 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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In Problems 37–50, 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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In Problems 37–50, 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

Mechanics Maths

Calculus

Applied Mathematics

Decision Maths

Theoretical and Mathematical Physics

Statistics

Study anywhere. Anytime. Across all devices.

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