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.

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

See anything we can do from here? :)

EDIT: Oops, too late :(

 

[ThomsonKevin]

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

 

[CellCoree]

I would approach this problem by first understanding the concept of factorial and how it works. Factorial is a mathematical operation that multiplies a number by all of the positive integers less than it. For example, 4! (read as "four factorial") is equal to 4 x 3 x 2 x 1 = 24.

Now, looking at the given equation, we can rewrite it as (n+1)(n)(n-1)!/(n-1)! = 56. This simplifies to (n+1)(n) = 56. We can then use algebra to solve for n.

Expanding the brackets, we get n^2 + n = 56. Rearranging the equation, we get n^2 + n - 56 = 0. This is a quadratic equation, which can be solved using the quadratic formula.

The quadratic formula is x = (-b ± √(b^2 - 4ac)) / 2a, where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0. In our case, a = 1, b = 1, and c = -56.

Plugging these values into the formula, we get n = (-1 ± √(1^2 - 4(1)(-56))) / 2(1). This simplifies to n = (-1 ± √(1 + 224)) / 2, which further simplifies to n = (-1 ± 15) / 2.

Therefore, n can have two possible values: (-1 + 15) / 2 = 7 or (-1 - 15) / 2 = -8. However, since factorial is defined only for positive integers, n = -8 is not a valid solution. Therefore, the value of n in this unique factorial problem is 7.

I hope this explanation helps in understanding the solution better. If you have any further questions, please let me know.