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Complete the table below for the first 10 terms of the sequence whose terms are given by [tex]a_1=9[/tex] and [tex]a_{n+1}=\sqrt{2+\sqrt{a_n}}[/tex].

(Round to four decimal places as needed.)

\begin{tabular}{|c|c|}
\hline
[tex]$n$[/tex] & [tex]$a_n$[/tex] \\
\hline
1 & 9 \\
2 & [tex]$\square$[/tex] \\
3 & [tex]$\square$[/tex] \\
4 & [tex]$\square$[/tex] \\
5 & [tex]$\square$[/tex] \\
6 & [tex]$\square$[/tex] \\
7 & [tex]$\square$[/tex] \\
8 & [tex]$\square$[/tex] \\
9 & [tex]$\square$[/tex] \\
10 & [tex]$\square$[/tex] \\
\hline
\end{tabular}


Sagot :

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]