show me how to solve?

Question

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

&

(2n-3)! / (2n-1)!

Answer 1

if you understood the definition of factorial, you wouldn't be asking this,,, so here goes.

3! = 3 x 2 x1 (or 1 x 2 x 3)

6! = 1 x 2 x 3 x...... x 6

Hope you get the idea.

So n! would be
n! = 1 x 2 x 3 x...... n

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

this can be written as
(n+1)! = 1 x 2 x 3 x...... n x (n+1)
= n! (n+1)
= (n-1)! x n x (n+1)


similarly (n-1)! = 1 x 2 x ......x (n-1)

Now if you divide one by the other notice that all terms upto n-1 cancel out. so the only remaining terms would be

nx (n+1)

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


Now go and do the other one on your own

Answer 2

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(n + 1)! / (n - 1)!
= (n + 1) n (n - 1)! / (n - 1)!
= (n + 1) n
= n^2 + n

(2n - 3)! / (2n - 1)!
= (2n - 3)! / (2n - 1) (2n - 2) (2n - 3)!
= 1 / (2n - 1) (2n - 2)
= 1 / (4n^2 - 3n + 2)

Answer 3

(n + 1)! / (n - 1)!
= (n + 1) n (n - 1)! / (n - 1)!
= (n + 1) n
= n^2 + n

(2n - 3)! / (2n - 1)!
= (2n - 3)! / (2n - 1) (2n - 2) (2n - 3)!
= 1 / (2n - 1) (2n - 2)
= 1 / (4n^2 - 6n + 2)

Answer 4

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

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

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

(2n-3)! / (2n-1)! = 1/(2n-1)