Prove that $\frac{(n+1)!}{(n-1)!}=n(n+1)$

Question

Simplify $$\frac{(n+1)!}{(n-1)!}$$

My book shows the answer as $n(n+1)$. I don't know how does it come up?

I have tried:

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

I have been trying so hard but can't proceed more towards answer. Please help.

Answer 1

Rather $(n+1)!=(n+1)(n)(n-1)!$ now just cancel it with $(n-1)!$ thats all.

Answer 2

$$\color{red}{\frac{(n+1)!}{(n-1)!}}=\color{green}{\frac{(n+1)n(n-1)(n-2)...3\cdot 2\cdot 1}{(n-1)(n-2)...3\cdot 2\cdot 1}}=\color{blue}{(n+1)n}$$

Answer 3

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

Remember that $k!= k(k-1)(k-2)...(3)(2)(1)$

Answer 4

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

cancel $(n-1)!$ and it solves the problem

Answer 5

Based on your attempt:

$$\require{cancel}\frac{(n+1)!}{(n-1)!}=\frac{(n+1)n\cancel{(n-1)!}}{\cancel{(n-1)!}}=\color{red}{n(n+1)}$$