Answer
$0, \frac{3}{5}, \frac{4}{5}, \frac{15}{17},$ and $\frac{12}{13}$
Work Step by Step
To find the first five terms of the sequence $a_n = \frac{n^2 - 1}{n^2 + 1}$, we must plug in $n=1, n=2, n=3, n=4,$ and $n=5.$
$a_1 = \frac{1^2 - 1}{1^2 + 1}=\frac{1-1}{1+1}=0$
$a_2 = \frac{2^2 - 1}{2^2 + 1}=\frac{4-1}{4+1}=\frac{3}{5}$
$a_3 = \frac{3^2 - 1}{3^2 + 1}=\frac{9-1}{9+1}=\frac{4}{5}$
$a_4 = \frac{4^2 - 1}{4^2 + 1}=\frac{16-1}{16+1}=\frac{15}{17}$
$a_5 = \frac{5^2 - 1}{5^2 + 1}=\frac{25-1}{251}=\frac{24}{26}=\frac{12}{13}$
Hence we see that the first five terms are $0, \frac{3}{5}, \frac{4}{5}, \frac{15}{17},$ and $\frac{12}{13}.$
