Sure! Let's solve the given task step-by-step based on the sequence rule:
The [tex]\( n \)[/tex]th term of the sequence is given by:
[tex]\[ a_n = \frac{n^2 - 4}{n^2 + 4} \][/tex]
We need to find the first 4 terms ([tex]\(a_1, a_2, a_3, a_4\)[/tex]), as well as [tex]\(a_{10}\)[/tex] and [tex]\(a_{15}\)[/tex].
#### Step-by-Step Calculation:
1. First term ([tex]\(a_1\)[/tex]):
[tex]\[
a_1 = \frac{1^2 - 4}{1^2 + 4} = \frac{1 - 4}{1 + 4} = \frac{-3}{5} = -\frac{3}{5}
\][/tex]
So, [tex]\( a_1 = -\frac{3}{5} \)[/tex]
2. Second term ([tex]\(a_2\)[/tex]):
[tex]\[
a_2 = \frac{2^2 - 4}{2^2 + 4} = \frac{4 - 4}{4 + 4} = \frac{0}{8} = 0
\][/tex]
So, [tex]\( a_2 = 0 \)[/tex]
3. Third term ([tex]\(a_3\)[/tex]):
[tex]\[
a_3 = \frac{3^2 - 4}{3^2 + 4} = \frac{9 - 4}{9 + 4} = \frac{5}{13}
\][/tex]
So, [tex]\( a_3 = \frac{5}{13} \)[/tex]
4. Fourth term ([tex]\(a_4\)[/tex]):
[tex]\[
a_4 = \frac{4^2 - 4}{4^2 + 4} = \frac{16 - 4}{16 + 4} = \frac{12}{20} = \frac{3}{5}
\][/tex]
So, [tex]\( a_4 = \frac{3}{5} \)[/tex]
5. Tenth term ([tex]\(a_{10}\)[/tex]):
[tex]\[
a_{10} = \frac{10^2 - 4}{10^2 + 4} = \frac{100 - 4}{100 + 4} = \frac{96}{104} = \frac{24}{26} = \frac{12}{13}
\][/tex]
So, [tex]\( a_{10} = \frac{12}{13} \)[/tex]
6. Fifteenth term ([tex]\(a_{15}\)[/tex]):
[tex]\[
a_{15} = \frac{15^2 - 4}{15^2 + 4} = \frac{225 - 4}{225 + 4} = \frac{221}{229}
\][/tex]
Here, 221 and 229 do not share any common factors other than 1, so the fraction is already in its simplest form.
So, [tex]\( a_{15} = \frac{221}{229} \)[/tex]
Combining all the results, we have:
- The first term is [tex]\( a_1 = -\frac{3}{5} \)[/tex]
- The second term is [tex]\( a_2 = 0 \)[/tex]
- The third term is [tex]\( a_3 = \frac{5}{13} \)[/tex]
- The fourth term is [tex]\( a_4 = \frac{3}{5} \)[/tex]
- The tenth term is [tex]\( a_{10} = \frac{12}{13} \)[/tex]
- The fifteenth term is [tex]\( a_{15} = \frac{221}{229} \)[/tex]
