Answer
$a_{1}=\frac{(1)^2-1}{(1)^2+1}=\frac{0}{2}=0$
$a_{2}=\frac{(2)^2-1}{(2)^2+1}=\frac{3}{5}$
$a_{3}=\frac{(3)^2-1}{(3)^2+1}=\frac{8}{10}$
$a_{4}=\frac{(4)^2-1}{(4)^2+1}=\frac{15}{17}$
$a_{10}=\frac{(10)^2-1}{(10)^2+1}=\frac{99}{101}$
$a_{15}=\frac{(15)^2-1}{(15)^2+1}=\frac{224}{226}$
Work Step by Step
If we want to find a term, we have to substitute $n$ by its index:
$a_{1}=\frac{(1)^2-1}{(1)^2+1}=\frac{0}{2}=0$
$a_{2}=\frac{(2)^2-1}{(2)^2+1}=\frac{3}{5}$
$a_{3}=\frac{(3)^2-1}{(3)^2+1}=\frac{8}{10}$
$a_{4}=\frac{(4)^2-1}{(4)^2+1}=\frac{15}{17}$
$a_{10}=\frac{(10)^2-1}{(10)^2+1}=\frac{99}{101}$
$a_{15}=\frac{(15)^2-1}{(15)^2+1}=\frac{224}{226}$
