## Precalculus (6th Edition) Blitzer

$\left( n+3 \right)$.
We know that $n!=n\left( n-1 \right)\left( n-2 \right)\left( n-3 \right)\cdots 3\times 2\times 1$. Now to evaluate fractions with factorials in the numerator and the denominator, we have to reduce the fraction before performing the multiplications. Here, for $\frac{\left( n+3 \right)!}{\left( n+2 \right)!}$ write $\left( n+3 \right)!$as$\left( n+3 \right)\left( n+2 \right)!$. After that, we can divide both the numerator and the denominator by the common factor and therefore we can reduce the multiplication. We are given that, $\frac{\left( n+3 \right)!}{\left( n+2 \right)!}$ Now we can write it as $\frac{(n+3)(n+2)!}{(n+2)!}$. Now, divide the numerator and denominator by $(n+2)!$ to get $\frac{\left( n+3 \right)!}{\left( n+2 \right)!}=\left( n+3 \right)$.