# Need Help w/ Factorial Simplification

#### teetar

Hello, I am trying to simplify questions such as:
(n + 1)!/n!
(n + 2)!/n!
((n + 1)! + n! - (n - 1)!)/(n - 1)!

Naturally I don't want answers for them all, however, I do know that the first one equals n + 1, and the second one equals (n + 2)(n + 1), but I don't see how to get to that answer.

If anyone can do out the simplification process simply for me, that would be greatly appreciated, thanks!

#### mathman

Forum Staff
Hello, I am trying to simplify questions such as:
(n + 1)!/n!
(n + 2)!/n!
((n + 1)! + n! - (n - 1)!)/(n - 1)!

Naturally I don't want answers for them all, however, I do know that the first one equals n + 1, and the second one equals (n + 2)(n + 1), but I don't see how to get to that answer.

If anyone can do out the simplification process simply for me, that would be greatly appreciated, thanks!
n! = 1.2.3......n
(n+1)! = 1.2.3......n.(n+1) = n!.(n+1)

#### soroban

Math Team
Hello, teetar!

Do you understand basic factorials?

$\quad \dfrac{5!}{4!} \;=\;\dfrac{5\cdot\cancel{4\cdot3\cdot2\cdot1}}{ \cancel{4 \cdot3\cdot2\cdot1}} \;=\;5$

That is: $\:\dfrac{5!}{4!} \;=\;\dfrac{5\cdot\cancel{4!}}{\cancel{4!}} \;=\;5$

Simplify: $\:\dfrac{(n+1)!}{n!}$

$\dfrac{(n+1)!}{n!} \;=\;\dfrac{(n+1)\,\cancel{n!}}{\cancel{n!}} \;=\;n+1$

$\dfrac{(n + 2)!}{n!}$

$\dfrac{(n+2)!}{n!} \;=\;\dfrac{(n+2)(n+1)\,\cancel{n!}}{\cancel{n!}} \;=\;(n+2)(n+1)$

$\dfrac{(n + 1)! + n! - (n - 1)!}{(n - 1)!}$

$\dfrac{(n+1)! + n! - (n-1)!}{(n-1)!} \;\;=\;\;\dfrac{(n+1)!}{(n-1)!} + \dfrac{n!}{(n-1)!} - \dfrac{(n-1)!}{(n-1)!}$

$\qquad =\;\;\dfrac{(n+1)(n)\cancel{(n-1)!}}{\cancel{(n-1)!}} + \dfrac{n\cancel{(n-1)!}}{\cancel{(n-1)!}} - \dfrac{\cancel{(n-1)!}}{\cancel{(n-1)!}}$

$\qquad =\;\; (n+1)n + n - 1 \;\;=\;\;n^2 + 2n - 1$

• 1 person

#### teetar

Thanks a bunch for your help!