Answer
$\dfrac{y^{2n}-1}{2}$
Work Step by Step
Factor the numerators and the denominators, when needed to have:
$\\=\dfrac{(y^n-2)(y^n+1)}{2(y^n-2)} \cdot \dfrac{(y^n+1)(y^n-1)}{y^n+1}$
Cancel the common factors to have:
$\\\require{cancel}=\dfrac{\cancel{(y^n-2)}\cancel{(y^n+1)}}{2\cancel{(y^n-2)}} \cdot \dfrac{(y^n+1)(y^n-1)}{\cancel{y^n+1}}
\\=\dfrac{(y^n+1)(y^n-1)}{2}
\\=\dfrac{y^{2n}-1}{2}$
