How can I reduce (n+4)!/(n+1)! ?

2008-12-01 8:07 am

(n+4)!
——-
(n+1)!

I cannot figure out the answer and was unable to find any examples of how to solve this problem in my calc book.

回答 (4)

2008-12-01 8:14 am

✔ 最佳答案

(n + 4)!/(n + 1)!
= [(n)(n + 1)(n + 2)(n + 3)(n + 4)]/[(n)(n + 1)]
= (n + 2)(n + 3)(n + 4)/1
= (n + 2)(n + 3)(n + 4)

👍 1 👎 0

2008-12-01 9:22 am

(n + 4) (n + 3) (n + 2) (n + 1) ! / (n + 1) !

(n + 4) (n + 3) (n + 2)

👍 0 👎 0

2008-12-01 8:16 am

(n+4)! = (n+4)(n+3)(n+2) [(n+1)!]
Hence [(n+4)!] / {(n+1)!] = (n+4)(n+3)(n+2) =(nÂ²+7n+12)(n+2)
= nÂ³ +9nÂ² + 26n + 24

AJM

參考: General Ability in Maths

👍 0 👎 0

2008-12-01 8:11 am

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

👍 0 👎 0