Question

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

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:

$$\begin{gathered} a_1=\sqrt{2} \\ {a}_n=\sqrt{\frac{a_{n-1}}{2}} \end{gathered}$$

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

The first term is $\left(a_1\right)=\sqrt{2}$.

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

$$\begin{gathered} a_2=\sqrt{\frac{a_{2-1}}{2}} \\ =\sqrt{\frac{a_1}{2}} \\ {a}_2=\sqrt{\frac{\sqrt{2}}{2}} \end{gathered}$$

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

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

$$\begin{gathered} a_3=\sqrt{\frac{a_{3-1}}{2}} \\ =\sqrt{\frac{a_2}{2}} \\ =\sqrt{\frac{\sqrt{\frac{\sqrt{2}}{2}}}{2}} \\ =\sqrt{\frac{\sqrt{\sqrt{2}}}{2\sqrt{2}}} \\ {a}_3=\frac{\sqrt{\sqrt{\sqrt{2}}}}{\sqrt{2\sqrt{2}}} \\ {a}_4=\sqrt{\frac{a_{4-1}}{2}} \\ =\sqrt{\frac{a_3}{2}} \\ =\sqrt{\frac{\frac{\sqrt{\sqrt{\sqrt{2}}}}{\sqrt{2\sqrt{2}}}}{2}} \\ {a}_4=\frac{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}{\sqrt{2\sqrt{2\sqrt{2}}}} \end{gathered}$$

Step 4. Find the 5th term.

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

$$\begin{gathered} a_5=\sqrt{\frac{a_{5-1}}{2}} \\ =\sqrt{\frac{a_4}{2}} \\ =\sqrt{\frac{\frac{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}{\sqrt{2\sqrt{2\sqrt{2}}}}}{2}} \\ {a}_5=\frac{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}}{\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}} \end{gathered}$$

Therefore, the first five terms arerole="math" localid="1646740081552" $a_1=\sqrt{2},a_2=\sqrt{\frac{\sqrt{2}}{2}},a_3=\frac{\sqrt{\sqrt{\sqrt{2}}}}{\sqrt{2\sqrt{2}}},a_4=\frac{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}{\sqrt{2\sqrt{2\sqrt{2}}}},a_5=\frac{\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}}{\sqrt{2\sqrt{2\sqrt{2\sqrt{2}}}}}$.