To complete the table for the first 10 terms of the sequence defined by [tex]\( a_1 = 9 \)[/tex] and [tex]\( a_{n+1} = \sqrt{2 + \sqrt{a_n}} \)[/tex] with each term rounded to four decimal places, let's continue from where we left off:
Given:
[tex]\[
\begin{array}{|c|c|}
\hline
n & u(n) = a_n \\
\hline
1 & 9 \\
2 & 2.2361 \\
3 & 2.0586 \\
4 & 1.8533 \\
5 & 1.8322 \\
\hline
\end{array}
\][/tex]
From the results, the next terms are:
6. For [tex]\( a_6 \)[/tex]:
[tex]\[ a_6 = (2 + \sqrt{a_5})^{0.5} \][/tex]
[tex]\[ a_5 \approx 1.8322 \][/tex]
[tex]\[ a_6 \approx 1.8312 \][/tex]
7. For [tex]\( a_7 \)[/tex]:
[tex]\[ a_7 = (2 + \sqrt{a_6})^{0.5} \][/tex]
[tex]\[ a_6 \approx 1.8312 \][/tex]
[tex]\[ a_7 \approx 1.8312 \][/tex]
8. For [tex]\( a_8 \)[/tex]:
[tex]\[ a_8 = (2 + \sqrt{a_7})^{0.5} \][/tex]
[tex]\[ a_7 \approx 1.8312 \][/tex]
[tex]\[ a_8 \approx 1.8312 \][/tex]
9. For [tex]\( a_9 \)[/tex]:
[tex]\[ a_9 = (2 + \sqrt{a_8})^{0.5} \][/tex]
[tex]\[ a_8 \approx 1.8312 \][/tex]
[tex]\[ a_9 \approx 1.8312 \][/tex]
10. For [tex]\( a_{10} \)[/tex]:
[tex]\[ a_{10} = (2 + \sqrt{a_9})^{0.5} \][/tex]
[tex]\[ a_9 \approx 1.8312 \][/tex]
[tex]\[ a_{10} \approx 1.8312 \][/tex]
So, the completed table will look like this:
[tex]\[
\begin{array}{|c|c|}
\hline
n & u(n) = a_n \\
\hline
1 & 9 \\
2 & 2.2361 \\
3 & 2.0586 \\
4 & 1.8533 \\
5 & 1.8322 \\
6 & 1.8312 \\
7 & 1.8312 \\
8 & 1.8312 \\
9 & 1.8312 \\
10 & 1.8312 \\
\hline
\end{array}
\][/tex]
