Need Help w/ Factorial Simplification

Jun 2014
11
0
earth
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
May 2007
6,932
774
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
Dec 2006
3,267
408
Lexington, MA
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)!}} $

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

 
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