## Precalculus (6th Edition) Blitzer

$\frac{(2n+1)!}{2n!}$ Try to reduce the fraction before performing the multiplications. We can express (2n+1)! as (2n+1)! = (2n+1) 2n!. We can divide both the numerator and the denominator by the common factor, 2n!. $\frac{(2n+1)!}{2n!}$= $\frac{ (2n+1)2n!}{2n!}$ = 2n+1