Question

In Problems 37–50, a sequence is defined recursively. Write down the first five terms.

$a_1=\sqrt{2};a_n=\sqrt{2+a_{n-1}}$

Short answer

The first five terms of the recursively defined sequence are$a_1=\sqrt{2},a_2=\sqrt{2+\sqrt{2}},a_3=\sqrt{2+\sqrt{2+\sqrt{2}}},a_4=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}},a_5=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}$.

Step-by-step solution

Step 1. Write the given information.

The given recursively defined sequence is:

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

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 $a_n=\sqrt{2+a_{n-1}}$to get the second term,

$$\begin{gathered} a_2=\sqrt{2+a_{2-1}} \\ =\sqrt{2+a_1} \\ =\sqrt{2+\sqrt{2}} \end{gathered}$$

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

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

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

Step 4. Find the 5th term.

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

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

Therefore, the first five terms are$a_1=\sqrt{2},a_2=\sqrt{2+\sqrt{2}},a_3=\sqrt{2+\sqrt{2+\sqrt{2}}},a_4=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}},a_5=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}$.