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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
n & [tex]$a_n$[/tex] \\
\hline
1 & 9 \\
2 & 2.2361 \\
3 & 2.0586 \\
4 & 1.8533 \\
5 & 1.8322 \\
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 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]