Question

(n+1)!(n-1)!= 6(n-1).​

Answer 1

Объяснение:

$(n+1)!*(n-1)!=6*(n-1)\\(n+1)!*n!*(n-1)=6*(n-1)\ |:(n-1)\ (n\neq 1)\\(n+1)!=6\\(n+1)!=1*2*3\\(n+1)!=3!\\n+1=3\\n=2.$

Ответ: n=2.

Answer 2

Let's solve the equation:

(n+1)!(n-1)! = 6(n-1)

First, let's simplify the left side of the equation by expanding the factorials:

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

Now, substitute these into the equation:

(n+1)n(n-1)(n-1)(n-2)... = 6(n-1)

Next, divide both sides by (n-1) to simplify further:

(n+1)n(n-2)... = 6

Now, let's simplify the right side:

6 = 2 * 3

Now, you can see that the equation is:

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

Now, let's analyze the equation. On the left side, you have a product of consecutive integers, starting from (n+1) and decreasing by 1 with each subsequent term. On the right side, you have the product of 2 and 3.

To make the equation true, you can see that n+1 must be equal to 3 because the left side represents the product of three consecutive integers. So:

n + 1 = 3

Subtract 1 from both sides:

n = 2

So, the solution to the equation is n = 2. You can verify this solution by substituting it back into the original equation:

(2+1)! * (2-1)! = 3! * 1! = 6

So, the equation is satisfied when n = 2.