To solve this problem, we need to complete the table by finding the first 10 terms of the sequence where [tex]\(a_1 = 11\)[/tex] and [tex]\(a_{n+1} = \sqrt{10 + \sqrt{a_n}}\)[/tex]. We will round each value to four decimal places.
Here is the step-by-step solution for each term:
1. [tex]\(a_1\)[/tex]: Given directly as 11.
2. [tex]\(a_2\)[/tex]: Use the formula [tex]\(a_{n+1} = \sqrt{10 + \sqrt{a_n}}\)[/tex] with [tex]\(a_n = 11\)[/tex].
[tex]\[
a_2 = \sqrt{10 + \sqrt{11}} \approx 3.6492
\][/tex]
3. [tex]\(a_3\)[/tex]: Use the formula with [tex]\(a_n = 3.6492\)[/tex].
[tex]\[
a_3 = \sqrt{10 + \sqrt{3.6492}} \approx 3.4511
\][/tex]
4. [tex]\(a_4\)[/tex]: Use the formula with [tex]\(a_n = 3.4511\)[/tex].
[tex]\[
a_4 = \sqrt{10 + \sqrt{3.4511}} \approx 3.4435
\][/tex]
5. [tex]\(a_5\)[/tex]: Use the formula with [tex]\(a_n = 3.4435\)[/tex].
[tex]\[
a_5 = \sqrt{10 + \sqrt{3.4435}} \approx 3.4432
\][/tex]
6. [tex]\(a_6\)[/tex]: Use the formula with [tex]\(a_n = 3.4432\)[/tex].
[tex]\[
a_6 = \sqrt{10 + \sqrt{3.4432}} \approx 3.4432
\][/tex]
7. [tex]\(a_7\)[/tex]: Use the formula with [tex]\(a_n = 3.4432\)[/tex].
[tex]\[
a_7 = \sqrt{10 + \sqrt{3.4432}} \approx 3.4432
\][/tex]
8. [tex]\(a_8\)[/tex]: Use the formula with [tex]\(a_n = 3.4432\)[/tex].
[tex]\[
a_8 = \sqrt{10 + \sqrt{3.4432}} \approx 3.4432
\][/tex]
9. [tex]\(a_9\)[/tex]: Use the formula with [tex]\(a_n = 3.4432\)[/tex].
[tex]\[
a_9 = \sqrt{10 + \sqrt{3.4432}} \approx 3.4432
\][/tex]
10. [tex]\(a_{10}\)[/tex]: Use the formula with [tex]\(a_n = 3.4432\)[/tex].
[tex]\[
a_{10} = \sqrt{10 + \sqrt{3.4432}} \approx 3.4432
\][/tex]
Now we can complete the table:
[tex]\[
\begin{tabular}{|c|c|}
\hline
\(n\) & \(u(n) = a_n\) \\
\hline
1 & 11 \\
2 & 3.6492 \\
3 & 3.4511 \\
4 & 3.4435 \\
5 & 3.4432 \\
6 & 3.4432 \\
7 & 3.4432 \\
8 & 3.4432 \\
9 & 3.4432 \\
10 & 3.4432 \\
\hline
\end{tabular}
\][/tex]
