To complete the table for the first 10 terms of the sequence where [tex]\( a_1 = 9 \)[/tex] and [tex]\( a_{n+1} = \sqrt{2 + \sqrt{a_n}} \)[/tex], we need to follow the recursive formula. Let's go through each step:
1. [tex]\( a_1 = 9 \)[/tex]
2. To find [tex]\( a_2 \)[/tex]:
[tex]\[
a_2 = \sqrt{2 + \sqrt{a_1}} = \sqrt{2 + \sqrt{9}} = \sqrt{2 + 3} = \sqrt{5} \approx 2.2361
\][/tex]
3. To find [tex]\( a_3 \)[/tex]:
[tex]\[
a_3 = \sqrt{2 + \sqrt{a_2}} = \sqrt{2 + \sqrt{2.2361}} \approx \sqrt{2 + 1.4954} = \sqrt{3.4954} \approx 1.8696
\][/tex]
4. To find [tex]\( a_4 \)[/tex]:
[tex]\[
a_4 = \sqrt{2 + \sqrt{a_3}} = \sqrt{2 + \sqrt{1.8696}} \approx \sqrt{2 + 1.3670} = \sqrt{3.3670} \approx 1.8350
\][/tex]
5. To find [tex]\( a_5 \)[/tex]:
[tex]\[
a_5 = \sqrt{2 + \sqrt{a_4}} = \sqrt{2 + \sqrt{1.8350}} \approx \sqrt{2 + 1.3546} = \sqrt{3.3546} \approx 1.8316
\][/tex]
6. To find [tex]\( a_6 \)[/tex]:
[tex]\[
a_6 = \sqrt{2 + \sqrt{a_5}} = \sqrt{2 + \sqrt{1.8316}} \approx \sqrt{2 + 1.3534} = \sqrt{3.3534} \approx 1.8312
\][/tex]
7. To find [tex]\( a_7 \)[/tex]:
[tex]\[
a_7 = \sqrt{2 + \sqrt{a_6}} = \sqrt{2 + \sqrt{1.8312}} \approx \sqrt{2 + 1.3532} = \sqrt{3.3532} \approx 1.8312
\][/tex]
8. To find [tex]\( a_8 \)[/tex]:
[tex]\[
a_8 = \sqrt{2 + \sqrt{a_7}} = \sqrt{2 + \sqrt{1.8312}} \approx \sqrt{2 + 1.3532} = \sqrt{3.3532} \approx 1.8312
\][/tex]
9. To find [tex]\( a_9 \)[/tex]:
[tex]\[
a_9 = \sqrt{2 + \sqrt{a_8}} = \sqrt{2 + \sqrt{1.8312}} \approx \sqrt{2 + 1.3532} = \sqrt{3.3532} \approx 1.8312
\][/tex]
10. To find [tex]\( a_{10} \)[/tex]:
[tex]\[
a_{10} = \sqrt{2 + \sqrt{a_9}} = \sqrt{2 + \sqrt{1.8312}} \approx \sqrt{2 + 1.3532} = \sqrt{3.3532} \approx 1.8312
\][/tex]
So, the completed table is:
[tex]\[
\begin{array}{|c|c|}
\hline
n & a_n \\
\hline
1 & 9 \\
2 & 2.2361 \\
3 & 1.8696 \\
4 & 1.8350 \\
5 & 1.8316 \\
6 & 1.8312 \\
7 & 1.8312 \\
8 & 1.8312 \\
9 & 1.8312 \\
10 & 1.8312 \\
\hline
\end{array}
\][/tex]
