Find the Value of n in a Unique Factorial Problem with (n+1)!/(n-1)! = 56

[ThomsonKevin]

Ok, I had never seen a factorial problem like this, and the answer(n=7) didn't help me much in understand the solution either.

If (n+1)!/(n-1)! = 56 , what's the value of n?

 

[evinda]

Ok, I had never seen a factorial problem like this, and the answer(n=7) didn't help me much in understand the solution either.

If (n+1)!/(n-1)! = 56 , what's the value of n?

(Wave)It is known that $n!=1 \cdot 2 \cdots n$.

$$\frac{(n+1)!}{(n-1)!}=56 \Rightarrow \frac{1 \cdot 2 \cdots (n-1) \cdot n \cdot (n+1)}{1 \cdot 2 \cdot 3 \cdots (n-1)}=56 \Rightarrow \frac{n \cdot (n+1)}{1}=56 \Rightarrow n \cdot (n+1)=56 \\ \Rightarrow n^2+n=56$$Solve the equation $n^2+n-56=0$ and you will find the values $-8$ and $7$.

But since $n \geq 1$ we reject the value $-8$, so we have that $n=7$.

 

[Jameson]

Ok, I had never seen a factorial problem like this, and the answer(n=7) didn't help me much in understand the solution either.

If (n+1)!/(n-1)! = 56 , what's the value of n?

Let's try writing out some terms for the numerator and denominator.

$$\frac{(n+1)!}{(n-1)!}=\frac{(n+1)(n)(n-1)(n-2)...}{(n-1)(n-2)(n-3)...}$$

See anything we can do from here? :)

EDIT: Oops, too late :(

 

[ThomsonKevin]

Yes, Thank you both of you, it makes clear sense now.