2. Simplify: a) $\frac{(n+1)!}{n!}$; b) $\frac{(n+1)!}{(n-1)!}$; c) $\frac{1}{(n+1)!}+\frac{1}{n!}$.
Solution. a) Given that $(n+1)!=1 \cdot 2 \cdot 3 \ldots n(n+1)$, and $n!=1 \cdot 2 \cdot 3 \ldots n$, we can simplify the fraction; $\frac{(n+1)!}{n!}=n+1$.

6) Since $(n+1)!=1 \cdot 2 \cdot 3 \ldots(n-1) n(n+1)$, after simplification we get $\frac{(n+1)!}{(n-1)!}=n(n+1)$.

c) We have $(n+1)!=1 \cdot 2 \cdot 3 \ldots n(n+1), n!=1 \cdot 2 \cdot 3 \ldots n$. Let's bring the fraction to a common denominator, which we will take as $(n+1)!$. Then we get

$$
\frac{1}{(n+1)!}+\frac{1}{n!}=\frac{1+(n+1)}{(n+1)!}=\frac{n+2}{(n+1)!}
$$

3-5. Calculate: