The equation (n+1)!/(n-1)! = 30 has to be solved.

The factorial of a number n, n! is equal to the product of all positive integers from 1 till n, n included.

(n+1)! = (n+1)*n*(n-1).... 1

(n-1)! = (n-1)*(n - 2)* ... 1

This gives:

((n+1)*n*(n-1).... 1)/((n-1)*(n - 2)* ... 1) = 30

(n+1)*n = 30

n^2 + n - 30 = 0

n^2 + 6n - 5n - 30 = 0

n(n + 6) - 5(n+6) = 0

(n - 5)(n + 6) = 0

n = 5 and n = -6

But the factorial for negative numbers is not defined, eliminate n = -6

The required solution of the equation is n = 5.

To solve for (n+1)!/(n-1)! = 30.

Solution:

(n+1)! = (n+1)(n)(n-1)(n-2)(n-3)....(3)(2)(1)

Threrefore, (n+1)! = (n+1)(n)*(n-1)!. By dividing both sides , we get :

(n+1)!/(n-1)! = (n+1)(n). Using this on the LHS of the given equation, we get:

(n+1)n = 30. Or

6*5 = 30. So, n = 5 .

(n+1)!/(n-1)! =

(n+1)!= 1x2x3x...(n-1)x(n)x(n+1)

(n-1)!= 1x2x3x...(n-1)

(n+1)!/(n-1)!= 1x2x3x...(n-1)x(n)x(n+1)/1x2x3x...(n-1)

We can simplify the same factors from numerator and denominator: 1x2x3x...(n-1)

Simplifying, it will remain at numerator (n)x(n+1)

(n)x(n+1)= 30

n^2 +n= 30

n^2 +n -30= 0

n1={-1+ [1-4x(-30)]^1/2}/2= (-1 + 121^1/2)/2= (-1 +11)/2

n1=5

n2 = {-1- [1-4x(-30)]^1/2}/2= (-1 - 121^1/2)/2= (-1 -11)/2

n2=-6

Both solution are natural numbers

n= {-6;5}